This course develops the basic structure theory of $C^*$-algebras, the operator-algebraic objects that encode both functional analysis and noncommutative geometry. It begins with Banach algebras and the $C^*$-identity, then moves to commutative $C^*$-algebras, where Gelfand duality shows that the theory recovers ordinary spaces and continuous functions. From there, the course turns to positivity, functional calculus, and states, which provide the analytic and order-theoretic tools used throughout the subject. The later chapters build the representation theory of $C^*$-algebras from these foundations. Positive functionals and the GNS construction show how abstract algebras act concretely on Hilbert spaces. In the overview, $C_0(X)$ denotes continuous complex-valued functions on a locally compact [Hausdorff space](/page/Hausdorff%20Space) $X$ that vanish at infinity, $\mathcal{L}(H)$ denotes bounded operators on a Hilbert space $H$, and $K(H)$ denotes the compact operators on $H$. Ideals, quotients, and approximate units then organize the internal structure of algebras, while multiplier algebras, unitization, tensor products, and nuclearity are deferred until their chapters, where the notation and hypotheses are introduced locally. The course ends with finite-dimensional and AF, or approximately finite-dimensional, $C^*$-algebras and a broader view of noncommutative spaces and further directions. Each topic is chosen to reinforce the central idea that many geometric and algebraic questions can be reformulated in the language of operators. # Introduction This course studies $C^*$-algebras as the meeting point of functional analysis, algebra, topology, and spectral theory. The motivating example is always a norm-closed algebra of bounded operators on a Hilbert space, closed under taking adjoints, but the abstract axioms are strong enough to recover many operator-theoretic phenomena without choosing a particular representation. The course begins with Banach algebra methods, then treats commutative $C^*$-algebras as spaces in disguise, and finally develops the representation theory that turns abstract algebras back into concrete operator algebras. The guiding theme is that a $C^*$-algebra behaves like a noncommutative [algebra of continuous functions](/theorems/197). Commutative $C^*$-algebras are exactly algebras of the form $C_0(X)$ for locally compact Hausdorff spaces $X$, while noncommutative examples such as $\mathcal{L}(H)$, $K(H)$, and matrix algebras retain enough topology, order, and spectral theory to be studied by analogous methods. This chapter fixes the viewpoint, recalls the background objects, and explains the main questions that the later chapters answer. ## Why Operator Algebras Need a Norm Ordinary algebra knows about addition, multiplication, ideals, and homomorphisms, but it does not remember convergence. Operator theory forces convergence into the story: limits of operators matter, spectra depend on the ambient normed algebra, and infinite-dimensional examples cannot be controlled by algebraic manipulation alone. The first problem of the course is therefore to understand what structure remains when an algebra is also a Banach space and multiplication is compatible with the norm. [definition: Banach Algebra] A Banach algebra is a complex Banach space $A$ equipped with an associative bilinear multiplication $A \times A \to A$ such that \begin{align*} \|ab\|_A \le \|a\|_A\|b\|_A \end{align*} for all $a,b \in A$. [/definition] The inequality says that multiplication is continuous, so algebraic constructions interact with limits. This is the minimum analytic framework in which it makes sense to discuss invertibility by perturbation, [power series](/page/Power%20Series) in algebra elements, and spectral theory without explicitly referring to vectors in a Hilbert space. [example: Continuous Functions as a Banach Algebra] Let $X$ be a compact Hausdorff space, and let $C(X)$ be the complex Banach space of continuous functions $f:X\to\mathbb C$ with $\|f\|_\infty=\sup_{x\in X}|f(x)|$. Since $X$ is compact and $|f|:X\to\mathbb R$ is continuous, the image $|f|(X)$ is compact in $\mathbb R$, so the supremum is finite. For $f,g\in C(X)$, define multiplication pointwise by $(fg)(x)=f(x)g(x)$. The product $fg$ is continuous because multiplication $\mathbb C\times\mathbb C\to\mathbb C$ is continuous and $x\mapsto (f(x),g(x))$ is continuous. The constant function $1_X(x)=1$ is continuous, and \begin{align*} (1_Xf)(x)=1_X(x)f(x)=1\cdot f(x)=f(x) \end{align*} for every $x\in X$, with the same calculation giving $f1_X=f$. Thus $1_X$ is the multiplicative identity. It remains to check the norm inequality. For every $x\in X$, \begin{align*} |(fg)(x)|=|f(x)g(x)|=|f(x)|\,|g(x)| \end{align*} and the definitions of the sup norms give $|f(x)|\le \|f\|_\infty$ and $|g(x)|\le \|g\|_\infty$. Hence \begin{align*} |(fg)(x)|\le \|f\|_\infty\|g\|_\infty \end{align*} for every $x\in X$. Taking the supremum over $x\in X$ gives \begin{align*} \|fg\|_\infty\le \|f\|_\infty\|g\|_\infty \end{align*} so pointwise multiplication makes $C(X)$ a unital Banach algebra. This example shows that the topology of $X$ is reflected in an analytic algebra whose norm measures uniform size over the whole space. [/example] Function algebras are commutative, but the examples coming from Hilbert spaces usually are not. To use Hilbert space operators as a model, one must first restrict to operators whose size is controlled by the Hilbert norm; otherwise composition and adjoints need not fit into a Banach-algebra framework. The following definition isolates exactly the bounded operators, where the operator norm, composition, and adjoint coexist in one analytic algebra. [definition: Bounded Operators on a Hilbert Space] Let $H$ be a complex Hilbert space. The algebra $\mathcal{L}(H)$ is the set of bounded linear operators $T:H\to H$, equipped with composition, the operator norm $\|T\|_{\mathcal{L}(H)}$, and the adjoint operation $T\mapsto T^*$ determined by \begin{align*} (Tx,y)_H=(x,T^*y)_H \end{align*} for all $x,y\in H$. [/definition] The adjoint is the extra structure that distinguishes $C^*$-algebras from general Banach algebras. It remembers the Hilbert space geometry and makes positivity possible, which later becomes the route to states and representations. [example: Matrix Algebras as Finite-Dimensional Operator Algebras] Let $H=\mathbb C^n$ with standard basis $e_1,\dots,e_n$ and [inner product](/page/Inner%20Product) $(x,y)=\sum_{j=1}^n x_j\overline{y_j}$. If $T\in\mathcal L(\mathbb C^n)$, write \begin{align*} T e_j=\sum_{i=1}^n a_{ij}e_i \end{align*} and associate to $T$ the matrix $A=(a_{ij})\in M_n(\mathbb C)$. For $x=\sum_{j=1}^n x_j e_j$, \begin{align*} T x=T\left(\sum_{j=1}^n x_j e_j\right)=\sum_{j=1}^n x_j T e_j=\sum_{j=1}^n x_j\sum_{i=1}^n a_{ij}e_i=\sum_{i=1}^n\left(\sum_{j=1}^n a_{ij}x_j\right)e_i \end{align*} so the coordinates of $Tx$ are exactly $Ax$. Composition becomes matrix multiplication. If $S$ corresponds to $B=(b_{ij})$, then \begin{align*} STe_j=S\left(\sum_{k=1}^n a_{kj}e_k\right)=\sum_{k=1}^n a_{kj}S e_k=\sum_{k=1}^n a_{kj}\sum_{i=1}^n b_{ik}e_i=\sum_{i=1}^n\left(\sum_{k=1}^n b_{ik}a_{kj}\right)e_i \end{align*} so the matrix of $ST$ has $(i,j)$-entry $\sum_{k=1}^n b_{ik}a_{kj}$, which is the $(i,j)$-entry of $BA$. The adjoint becomes conjugate transpose. Since \begin{align*} (Te_j,e_i)=\left(\sum_{r=1}^n a_{rj}e_r,e_i\right)=a_{ij} \end{align*} and the defining identity for the adjoint gives \begin{align*} (Te_j,e_i)=(e_j,T^*e_i) \end{align*} the coefficient of $e_j$ in $T^*e_i$ is $\overline{a_{ij}}$. Thus the matrix of $T^*$ has $(j,i)$-entry $\overline{a_{ij}}$, namely $A^*=\overline A^{\,t}$. The operator norm is \begin{align*} \|A\|=\sup_{\|x\|_2=1}\|Ax\|_2 \end{align*} so $M_n(\mathbb C)$ is exactly the finite-dimensional operator algebra $\mathcal L(\mathbb C^n)$ under this identification. For $n\ge 2$ it is noncommutative, for example $E_{12}E_{21}=E_{11}$ while $E_{21}E_{12}=E_{22}$, but [finite dimensionality](/theorems/1534) removes the compactness and closure issues that occur for operators on infinite-dimensional Hilbert spaces. [/example] ## The $C^*$-Identity as the Central Constraint A Banach algebra with an involution still may fail to behave like an algebra of Hilbert space operators. The central constraint of the course is the $C^*$-identity, which ties the norm to the adjoint and multiplication so tightly that the norm is no longer an arbitrary analytic decoration. This identity is the reason the theory has order, functional calculus, and a rigid representation theory. [definition: C Star Algebra] A $C^*$-algebra is a complex Banach algebra $A$ equipped with a map $*:A\to A$, written $a\mapsto a^*$, satisfying \begin{align*} (a+b)^*&=a^*+b^*, & (\lambda a)^*&=\overline{\lambda}a^*, & (ab)^*&=b^*a^*, & (a^*)^*&=a \end{align*} for all $a,b\in A$ and $\lambda\in\mathbb C$, and such that \begin{align*} \|a^*a\|_A=\|a\|_A^2 \end{align*} for all $a\in A$. [/definition] The identity is modeled on the equality $\|T^*T\|_{\mathcal{L}(H)}=\|T\|_{\mathcal{L}(H)}^2$ for bounded operators. The immediate question is whether the concrete operator algebras that motivated the definition really satisfy the abstract axioms, since this is what makes the definition a faithful abstraction rather than a formal analogy. [quotetheorem:8544] [citeproof:8544] This theorem explains why abstract $C^*$-algebras are plausible objects: every norm-closed self-adjoint operator algebra satisfies the axioms. Each hypothesis is doing real work. If the subalgebra is not norm-closed, it need not be complete, so it may fail to be a Banach algebra even when multiplication and adjoints behave well algebraically; the finite-rank operators in an infinite-dimensional Hilbert space are the standard example. If the subalgebra is not closed under adjoints, the operation $T\mapsto T^*$ is not available internally, so the $C^*$-identity is not even a statement inside the algebra. If one changes the norm rather than using the inherited operator norm, the identity $\|T^*T\|=\|T\|^2$ can fail because it is a theorem about the Hilbert-space operator norm, not about an arbitrary equivalent Banach algebra norm. The theorem also does not prove that every abstract $C^*$-algebra has already been realised on a Hilbert space; that converse direction is the representation-theoretic content developed in Chapters 4 and 5 through states and the GNS construction. [example: Compact Operators] Let $K(H)$ denote the compact operators on $H$, meaning those operators $T\in\mathcal L(H)$ for which $T(B_H)$ has compact closure, where $B_H=\{x\in H:\|x\|\le 1\}$. If $T\in K(H)$ and $S\in\mathcal L(H)$, then $TS$ is compact because $S(B_H)$ is bounded, so $S(B_H)\subseteq \|S\|B_H$, and therefore \begin{align*} TS(B_H)\subseteq T(\|S\|B_H)=\|S\|T(B_H), \end{align*} whose closure is compact. Similarly, $ST$ is compact because $T(B_H)$ has compact closure and $S$ is continuous, so $S(\overline{T(B_H)})$ is compact and contains $ST(B_H)$. Thus $K(H)$ is a two-sided ideal in $\mathcal L(H)$. The adjoint of a [compact operator](/page/Compact%20Operator) is compact by *Schauder's theorem on adjoints*, so $K(H)$ is closed under $T\mapsto T^*$. Since $K(H)$ is norm-closed in $\mathcal L(H)$, it is complete in the inherited operator norm. For $T\in K(H)$, we also have $T^*T\in K(H)$, and the inherited operator norm satisfies \begin{align*} \|T^*T\|_{\mathcal L(H)}=\|T\|_{\mathcal L(H)}^2 \end{align*} by the operator-norm $C^*$-identity in $\mathcal L(H)$. Hence $K(H)$ is a $C^*$-algebra. It has no identity when $H$ is infinite-dimensional. If $I_H$ belonged to $K(H)$, then $I_H(B_H)=B_H$ would have compact closure, so the closed unit ball of $H$ would be compact. In an infinite-dimensional Hilbert space this is false: choose an orthonormal sequence $(e_n)$; for $m\ne n$, \begin{align*} \|e_n-e_m\|^2=(e_n-e_m,e_n-e_m)=1+1-0-0=2, \end{align*} so $(e_n)$ has no Cauchy subsequence and therefore no convergent subsequence. Thus $I_H\notin K(H)$, and $K(H)$ is a natural nonunital $C^*$-algebra sitting inside the unital algebra $\mathcal L(H)$. [/example] ## Commutative $C^*$-Algebras as Spaces The first structural question is what the axioms mean in the commutative case. The answer is the bridge from operator theory to topology: a commutative $C^*$-algebra is an algebra of functions on a locally compact Hausdorff space. Thus topology can be recovered from algebra, and noncommutative $C^*$-algebras can be viewed as algebras of functions on hypothetical noncommutative spaces. [definition: Character] Let $A$ be a commutative complex algebra. A character on $A$ is a nonzero algebra homomorphism $\chi:A\to\mathbb C$. [/definition] Characters are the analogue of points. For a function algebra, evaluating at a point gives a character, and the Gelfand theory developed in Chapter 2 shows that, for commutative $C^*$-algebras, these are the right points to use. [example: Evaluation Characters] Let $X$ be a compact Hausdorff space and fix $x\in X$. Define $\operatorname{ev}_x:C(X)\to\mathbb C$ by $\operatorname{ev}_x(f)=f(x)$. For $f,g\in C(X)$ and $\lambda,\mu\in\mathbb C$, \begin{align*} \operatorname{ev}_x(\lambda f+\mu g)=(\lambda f+\mu g)(x)=\lambda f(x)+\mu g(x)=\lambda\operatorname{ev}_x(f)+\mu\operatorname{ev}_x(g) \end{align*} so $\operatorname{ev}_x$ is linear. For multiplication, \begin{align*} \operatorname{ev}_x(fg)=(fg)(x)=f(x)g(x)=\operatorname{ev}_x(f)\operatorname{ev}_x(g) \end{align*} so it is multiplicative. It is not the zero map, because for the constant function $1_X$ we have \begin{align*} \operatorname{ev}_x(1_X)=1_X(x)=1 \end{align*} and therefore $\operatorname{ev}_x$ is a character. The family of evaluation characters separates functions in the following exact sense: if $f,g\in C(X)$ and $f\ne g$, then, since functions are equal precisely when they agree at every point, there exists $y\in X$ such that $f(y)\ne g(y)$. Hence \begin{align*} \operatorname{ev}_y(f)=f(y)\ne g(y)=\operatorname{ev}_y(g) \end{align*} so the evaluations distinguish $f$ from $g$. Thus the algebra $C(X)$ remembers each function through its values against characters, which is the first indication that the topology of $X$ can be recovered from the character space. [/example] Evaluation characters suggest that a space might be reconstructed from the characters of its function algebra. The theorem needed next is the statement that this reconstruction works for every commutative $C^*$-algebra, so the commutative theory is topological in a precise sense. [quotetheorem:2689] [citeproof:2689] This theorem is one of the main reasons $C^*$-algebras belong in analysis rather than pure algebra alone. The algebraic operations, norm, and involution combine to recover an entire [locally compact space](/page/Locally%20Compact%20Space), but the conclusion depends sharply on the hypotheses. Commutativity is essential: $M_n(\mathbb C)$ for $n\ge 2$ is a $C^*$-algebra, but it cannot be isomorphic to any algebra $C_0(X)$ because pointwise multiplication is commutative. The $C^*$-identity is also essential because it forces the Gelfand transform to be isometric and compatible with the involution; for a general commutative Banach algebra, the character space may not recover the normed algebra. Nonunitality is reflected topologically by local compactness rather than compactness: unital algebras correspond to compact spaces, while $C_0(X)$ for noncompact $X$ has no identity. In the noncommutative case, the theorem becomes a guiding analogy rather than a literal reconstruction theorem: there is no underlying point-set space whose ordinary continuous functions are the algebra. ## Positivity, States, and Representations Once $C^*$-algebras are treated as noncommutative function algebras, the next question is how to integrate or measure them. In the commutative case, positive linear functionals on $C_0(X)$ correspond to regular Borel measures by the [Riesz representation theorem](/theorems/218). In the noncommutative case, positive functionals become states, and states generate Hilbert space representations. [definition: Positive Element] Let $A$ be a $C^*$-algebra. An element $a\in A$ is positive if $a=a^*$ and its spectrum satisfies $\sigma(a)\subseteq [0,\infty)$. [/definition] Positivity generalises nonnegative functions and positive operators. Once positivity is available, the next object to define is the analogue of a probability measure: a normalised linear functional that assigns nonnegative numbers to positive elements. [definition: State] Let $A$ be a unital $C^*$-algebra. A state on $A$ is a linear functional $\varphi:A\to\mathbb C$ such that $\varphi(a)\ge 0$ for every positive element $a\in A$ and $\varphi(1_A)=1$. [/definition] A state should be read as a noncommutative probability measure. The representation problem now becomes concrete: given only these expectation values, can we construct a Hilbert space, operators on it, and a vector whose matrix coefficients reproduce the state? [quotetheorem:7110] [citeproof:7110] This result is the course's main mechanism for passing from abstract $C^*$-algebras to concrete operators. The hypotheses are not cosmetic. Positivity is what makes $(a,b)=\varphi(b^*a)$ positive semidefinite; without it the quotient-completion construction would not produce a Hilbert space inner product. The condition $\varphi(1_A)=1$ normalises the cyclic vector, since the class of $1_A$ has squared norm $\varphi(1_A)$; a positive functional with value $0$ on $1_A$ gives the zero construction, and a nonzero positive functional with another value gives the same idea after rescaling. The representation obtained from one state is cyclic, but it need not be faithful: a state can vanish on a nonzero ideal, so its GNS representation may kill nonzero elements. The later faithful representation theorems therefore combine many states, rather than expecting a single state to see the whole algebra. ## Ideals, Approximate Units, and AF Algebras The last part of the course asks how $C^*$-algebras are assembled from smaller pieces. The organizing pattern is local-to-global: first isolate pieces by closed ideals, then handle nonunital pieces with approximate units, and finally study algebras that are exhausted by finite-dimensional subalgebras. This prevents the later theory from becoming only a list of definitions: each construction answers a specific obstruction created by infinite-dimensional examples. Ideals describe quotients and extensions, approximate units replace missing identities, and AF algebras provide a class built from finite-dimensional matrix algebras. These topics connect the structure theory back to explicit examples. A useful failure mode to keep in mind is that algebraic structure alone is too weak: ideals must be closed, approximate identities usually require nets, and finite-dimensional approximations classify only under strong hypotheses. [definition: Closed Two-Sided Ideal] Let $A$ be a $C^*$-algebra. A closed two-sided ideal in $A$ is a closed linear subspace $I\subseteq A$ such that $a x\in I$ and $x a\in I$ for all $a\in A$ and $x\in I$. [/definition] Closed ideals are the correct ideals for $C^*$-algebras because quotients must remain Banach spaces. Algebraic ideals that are not closed can destroy the analytic structure. [example: Ideals in a Function Algebra] Let $X$ be a locally compact Hausdorff space and let $U\subseteq X$ be open. Define \begin{align*} I_U=\{f\in C_0(X): f(x)=0 \text{ for all } x\in X\setminus U\}. \end{align*} We show that $I_U$ is a closed two-sided ideal of $C_0(X)$. First, $I_U$ is a linear subspace. If $f,g\in I_U$ and $\lambda,\mu\in\mathbb C$, then for every $x\in X\setminus U$, \begin{align*} (\lambda f+\mu g)(x)=\lambda f(x)+\mu g(x)=\lambda\cdot 0+\mu\cdot 0=0. \end{align*} Thus $\lambda f+\mu g\in I_U$. Next, $I_U$ is an ideal. Let $f\in I_U$ and $h\in C_0(X)$. For every $x\in X\setminus U$, \begin{align*} (hf)(x)=h(x)f(x)=h(x)\cdot 0=0. \end{align*} Hence $hf\in I_U$. Since multiplication in $C_0(X)$ is pointwise and therefore commutative, $fh=hf\in I_U$ as well. Thus $I_U$ is a two-sided ideal. It remains to check closedness. Suppose $(f_n)$ is a sequence in $I_U$ and $f_n\to f$ in the norm of $C_0(X)$, so $\|f_n-f\|_\infty\to 0$. Fix $x\in X\setminus U$. Since $f_n(x)=0$ for every $n$, \begin{align*} |f(x)|=|f(x)-f_n(x)|\le \|f-f_n\|_\infty. \end{align*} Taking $n\to\infty$ gives $|f(x)|=0$, so $f(x)=0$. Since this holds for every $x\in X\setminus U$, we have $f\in I_U$. Therefore $I_U$ is closed. Thus open subsets of $X$ produce closed two-sided ideals in $C_0(X)$: the functions in $I_U$ are exactly those that vanish outside $U$. Chapter 2 proves the converse commutative ideal theorem, so in the commutative case closed ideals are governed by open subsets of the underlying space. [/example] Nonunital algebras such as $K(H)$ still need elements that behave like a unit for limiting arguments. Approximate units supply this substitute and are indispensable for ideals and representations. [definition: Approximate Unit] Let $A$ be a $C^*$-algebra. An approximate unit for $A$ is a net $(e_i)_{i\in I}$ in $A$ such that \begin{align*} e_i a\to a, \qquad a e_i\to a \end{align*} for every $a\in A$. [/definition] The use of nets rather than sequences reflects the generality of $C^*$-algebras. Approximate units explain how nonunital algebras can still be handled by limiting arguments; the next finite-dimensional approximation idea asks for a stronger kind of limiting structure, where the algebra is built from matrix blocks. [definition: AF Algebra] An AF algebra is a $C^*$-algebra $A$ for which there exists an increasing sequence of finite-dimensional $C^*$-subalgebras \begin{align*} A_1\subset A_2\subset A_3\subset \cdots \subset A \end{align*} such that $\bigcup_{n=1}^\infty A_n$ is dense in $A$. [/definition] AF algebras are built from matrix blocks, so they form a tractable testing ground for the general theory. They also show how infinite-dimensional $C^*$-algebras can retain a strong finite-dimensional approximation property. [example: The Compact Operators as an AF Algebra] Let $H=\ell^2$ with standard [orthonormal basis](/page/Orthonormal%20Basis) $(e_j)_{j\ge 1}$, and let $P_n$ be the [orthogonal projection](/theorems/437) onto $H_n=\operatorname{span}\{e_1,\dots,e_n\}$. Define \begin{align*} A_n=\{T\in K(H):T=P_nTP_n\}. \end{align*} Every $T\in A_n$ is determined by the $n^2$ numbers $a_{ij}=(Te_j,e_i)$ for $1\le i,j\le n$, and for $x=\sum_{j=1}^{\infty}x_je_j$ one has \begin{align*} Tx=\sum_{i=1}^n\left(\sum_{j=1}^n a_{ij}x_j\right)e_i. \end{align*} Thus $A_n$ is exactly the copy of $M_n(\mathbb C)$ acting on $H_n$ and vanishing on $H_n^\perp$. If $S,T\in A_n$, then $ST=P_nSP_nP_nTP_n=P_nSTP_n$, and $T^*=P_nT^*P_n$, so $A_n$ is a finite-dimensional $C^*$-subalgebra of $K(H)$. Since $P_n=P_{n+1}P_n=P_nP_{n+1}$, every operator supported on $H_n$ is also supported on $H_{n+1}$, so \begin{align*} A_1\subset A_2\subset A_3\subset \cdots . \end{align*} We check that $\bigcup_n A_n$ is dense in $K(H)$. Let $T\in K(H)$ and set $T_n=P_nTP_n\in A_n$. Then \begin{align*} T-T_n=T-P_nT+P_nT-P_nTP_n=(I-P_n)T+P_nT(I-P_n). \end{align*} Taking operator norms gives \begin{align*} \|T-T_n\|\le \|(I-P_n)T\|+\|P_nT(I-P_n)\|\le \|(I-P_n)T\|+\|T(I-P_n)\|. \end{align*} Because $T(B_H)$ has compact closure and $P_ny\to y$ for each $y\in H$, compactness gives $\sup_{y\in T(B_H)}\|(I-P_n)y\|\to 0$, hence $\|(I-P_n)T\|\to 0$. Also \begin{align*} \|T(I-P_n)\|=\|(T(I-P_n))^*\|=\|(I-P_n)T^*\|, \end{align*} and $T^*$ is compact, so the same compactness argument gives $\|T(I-P_n)\|\to 0$. Therefore $\|T-T_n\|\to 0$. Thus $K(\ell^2)$ is the norm closure of the increasing union of the finite-dimensional matrix algebras $A_n$, so $K(\ell^2)$ is an AF algebra. The point is that compact operators on $\ell^2$ can be uniformly approximated by operators whose entire action is recorded by a finite matrix. [/example] ## How the Course Fits Together The course has a circular movement: it starts from concrete operators, abstracts the $C^*$-identity, studies the commutative case as topology, and then returns to operators through states and representations. Banach algebra spectral theory supplies the analytic foundation, Gelfand duality supplies the topological interpretation, and the GNS construction supplies the representation-theoretic bridge. The later chapters should be read with three recurring questions in mind. First, which facts are consequences of Banach algebra theory alone, and which require the $C^*$-identity? Second, how does a statement reduce to the familiar case of continuous functions when the algebra is commutative? Third, how can an abstract algebra be represented faithfully on a Hilbert space? These questions organise the proofs and examples throughout the course. With these questions in view, the course now needs a precise setting in which algebraic and analytic structure can be compared on equal footing. Banach algebras provide that setting, and adding an involution together with the C*-identity isolates the class in which the norm is tightly controlled by the algebraic operations. # 1. Banach Algebras and the $C^*$-Identity This first chapter sets up the analytic language used throughout the course. The guiding question is how much of the spectral theory of bounded operators on a Hilbert space depends only on algebra, norm, and completeness. We begin with Banach algebras, add an involution, and then isolate the $C^*$-identity as the condition that makes abstract algebras behave like norm-closed algebras of operators. ## Invertibility in Banach Algebras The first problem is to make sense of inverse elements and spectra without referring to matrices or operators on a fixed space. A Banach algebra packages multiplication and norm estimates so that infinite series may be used algebraically. [definition: Banach Algebra] A complex Banach algebra is a complex Banach space $A$ equipped with an associative bilinear multiplication $A \times A \to A$ such that $\|ab\|_A \le \|a\|_A\|b\|_A$ for all $a,b \in A$. A unital Banach algebra is a Banach algebra $A$ with an element $1_A \in A$ satisfying $1_Aa=a1_A=a$ for all $a\in A$. [/definition] This definition supplies the setting in which algebraic multiplication and analytic convergence can interact. To discuss equations such as $ax=b$ internally, the next issue is which elements admit two-sided inverses inside the same algebra. [definition: Invertible Element] Let $A$ be a unital Banach algebra. An element $a\in A$ is invertible if there exists $b\in A$ such that $ab=ba=1_A$. The set of invertible elements is denoted $A^\times$. [/definition] Invertibility is useful only if it is stable enough for analysis, and the definition above gives no direct construction of an inverse. The motivating problem for the next theorem is whether a small perturbation of the unit can be inverted by summing a geometric series inside the Banach algebra. [quotetheorem:8545] [citeproof:8545] This theorem is the local engine behind spectral theory: invertibility is stable under sufficiently small perturbations. The strict bound $\|a\|_A<1$ is essential for this argument, because it is exactly what makes the geometric series absolutely convergent in the Banach space norm. At the boundary $\|a\|_A=1$, the conclusion can fail: in $A=\mathbb C$, the element $a=1$ has norm $1$, but $1_A-a=0$ is not invertible. The theorem also does not say that every invertible element is obtained from a norm-small perturbation of the identity; rather, it gives a local test that can be moved around by multiplying by an already known inverse. In particular, it implies that $A^\times$ is open in $A$, because if $a$ is invertible and $\|a-b\|_A<\|a^{-1}\|_A^{-1}$, then $b=a(1_A-a^{-1}(a-b))$ is invertible. The same estimate is the operator-theoretic iteration used to solve equations of the form $(I-T)x=y$. [example: Inverting a Small Operator] Let $H$ be a Hilbert space and let $T\in \mathcal{L}(H)$ satisfy $\|T\|_{\mathcal{L}(H)}<1$. Since $\mathcal{L}(H)$ is a unital Banach algebra and $\|T\|_{\mathcal{L}(H)}<1$, the *[Neumann Series Theorem](/theorems/8545)* gives convergence in operator norm of \begin{align*} S=\sum_{n=0}^{\infty}T^n \end{align*} and gives $(I-T)^{-1}=S$. To see the inverse relation on vectors explicitly, let $S_N=\sum_{n=0}^N T^n$. Then \begin{align*} (I-T)S_N=(I-T)(I+T+\cdots+T^N)=I-T^{N+1}. \end{align*} Also \begin{align*} S_N(I-T)=(I+T+\cdots+T^N)(I-T)=I-T^{N+1}. \end{align*} Because $\|T^{N+1}\|_{\mathcal{L}(H)}\le \|T\|_{\mathcal{L}(H)}^{N+1}$ and $\|T\|_{\mathcal{L}(H)}<1$, we have $T^{N+1}\to 0$ in operator norm, so passing to the limit gives $(I-T)S=S(I-T)=I$. For $y\in H$, define \begin{align*} x=\sum_{n=0}^{\infty}T^n y. \end{align*} The vector series converges because \begin{align*} \sum_{n=0}^{\infty}\|T^n y\|_H\le \sum_{n=0}^{\infty}\|T\|_{\mathcal{L}(H)}^n\|y\|_H=\frac{\|y\|_H}{1-\|T\|_{\mathcal{L}(H)}}. \end{align*} Since $x=Sy$, we get \begin{align*} (I-T)x=(I-T)Sy=Iy=y. \end{align*} If $z$ is another solution, then $(I-T)z=y=(I-T)x$, so $(I-T)(z-x)=0$. Applying the inverse $S$ gives \begin{align*} z-x=S(I-T)(z-x)=S0=0, \end{align*} hence $z=x$. Thus the Neumann series is exactly the iterative solution formula for the equation $(I-T)x=y$. [/example] The example shows that Banach algebra invertibility recovers the familiar operator equation solved by iteration. To turn this local criterion into a global invariant of a single element, we ask for which scalar shifts $\lambda1_A-a$ fail to be invertible. [definition: Spectrum] Let $A$ be a unital Banach algebra and let $a\in A$. The spectrum of $a$ in $A$ is \begin{align*} \sigma_A(a)=\{\lambda\in \mathbb C: \lambda 1_A-a \notin A^\times\}. \end{align*} The resolvent set is $\rho_A(a)=\mathbb C\setminus \sigma_A(a)$. [/definition] The spectrum depends on the ambient algebra, because invertibility depends on where the inverse is allowed to live. Once the forbidden scalar shifts have been collected, the next numerical question is how far from the origin they can reach. [definition: Spectral Radius] Let $A$ be a unital Banach algebra and let $a\in A$. The spectral radius of $a$ is \begin{align*} r_A(a)=\sup\{ |\lambda| : \lambda\in \sigma_A(a)\}. \end{align*} [/definition] The definition is meaningful because [the spectrum is nonempty and compact](/theorems/889); nonemptiness is a genuinely complex-analytic consequence of [Liouville's theorem](/theorems/38) applied to the resolvent. The course proves this result early because it guarantees that spectral radius is an actual finite spectral invariant rather than a formal supremum over an uncontrolled set. [quotetheorem:2669] [citeproof:2669] The hypotheses in this theorem are not cosmetic. Complex scalars enter through [Liouville's theorem](/theorems/346); over real Banach algebras the analogous real spectrum may be empty, for example the element $i$ in the algebra $\mathbb C$ viewed as a real Banach algebra has no real scalar $\lambda$ for which $\lambda 1-i$ is noninvertible. The assumption that the algebra is nonzero also avoids the degenerate zero algebra, where the unit and zero coincide and the usual spectral conclusion breaks down. Unitality is needed because the expression $\lambda1_A-a$ lives in the algebra; for nonunital algebras the course will define spectra by first passing to the unitization. The spectral radius initially looks like a spectral definition, but the following formula recovers it from norm growth alone. This will be crucial in Chapter 2, where the $C^*$-identity turns spectral information into norm information for the Gelfand transform. [quotetheorem:2672] [citeproof:2672] The formula says that long-term multiplicative growth contains the same information as the outer edge of the spectrum, but only in the unital complex Banach algebra setting where the preceding spectral theorem applies. It does not identify individual spectral values or eigenvectors; many different spectra may have the same outer radius. It also does not say that the sequence $\|a^n\|_A$ itself behaves like a pure exponential, only that its $n$th-root asymptotics are governed by $r_A(a)$. The completeness and submultiplicativity assumptions are what allow the analytic resolvent argument and the subadditive norm-growth argument to meet. A concrete commutative example shows how the abstract definition reduces to the range of a function. [example: Spectrum in a Function Algebra] Let $X$ be a compact Hausdorff space and let $f\in C(X)$. We show that, in the unital Banach algebra $C(X)$ with pointwise multiplication and unit $1_X$, the spectrum is \begin{align*} \sigma_{C(X)}(f)=f(X). \end{align*} First suppose $\lambda\notin f(X)$. Then $\lambda-f(x)\neq 0$ for every $x\in X$, so the function \begin{align*} g(x)=\frac{1}{\lambda-f(x)} \end{align*} is continuous because $f$ is continuous and $z\mapsto(\lambda-z)^{-1}$ is continuous on $\mathbb C\setminus\{\lambda\}$. For every $x\in X$, \begin{align*} ((\lambda 1_X-f)g)(x)=(\lambda-f(x))\frac{1}{\lambda-f(x)}=1. \end{align*} Similarly, \begin{align*} (g(\lambda 1_X-f))(x)=\frac{1}{\lambda-f(x)}(\lambda-f(x))=1. \end{align*} Thus $g$ is a two-sided inverse for $\lambda1_X-f$, so $\lambda\notin\sigma_{C(X)}(f)$. Conversely, suppose $\lambda=f(x_0)$ for some $x_0\in X$. If $\lambda1_X-f$ had an inverse $h\in C(X)$, then evaluating the identity $(\lambda1_X-f)h=1_X$ at $x_0$ would give \begin{align*} 1=((\lambda1_X-f)h)(x_0)=(\lambda-f(x_0))h(x_0)=(\lambda-\lambda)h(x_0)=0, \end{align*} which is impossible. Hence $\lambda1_X-f$ is not invertible, so $\lambda\in\sigma_{C(X)}(f)$. Therefore the spectral values of $f$ are exactly the values taken by $f$ on $X$. [/example] ## Involutions and the $C^*$-Identity The Banach algebra structure does not remember adjoints of operators. The next question is what algebraic and norm conditions capture the behaviour of $T\mapsto T^*$ on bounded operators on a Hilbert space. [definition: Involutive Banach Algebra] An involutive Banach algebra is a Banach algebra $A$ equipped with a map $a\mapsto a^*$ from $A$ to $A$ such that, for all $a,b\in A$ and $\lambda\in\mathbb C$, \begin{align*} (a+b)^*&=a^*+b^*, \end{align*} \begin{align*} (\lambda a)^*&=\overline{\lambda}a^*, \end{align*} \begin{align*} (ab)^*&=b^*a^*, \end{align*} \begin{align*} (a^*)^*&=a. \end{align*} If $A$ is unital, the involution is required to satisfy $1_A^*=1_A$. [/definition] The reversal in $(ab)^*=b^*a^*$ is what makes the definition match adjoints, but the involution alone does not distinguish elements with special geometric behavior. In operator algebras, some elements behave like observables, some commute with their adjoints, and some preserve norm exactly. The following terminology names these cases so later spectral and functional calculus statements can refer to them without returning to the Hilbert space model each time. [definition: Self-Adjoint, Normal, and Unitary Elements] Let $A$ be a unital involutive Banach algebra. An element $a\in A$ is self-adjoint if $a^*=a$, normal if $a^*a=aa^*$, and unitary if $a^*a=aa^*=1_A$. [/definition] These classes reproduce familiar operator-theoretic distinctions, but a Banach *-algebra can still carry norms that are unrelated to adjoints. The obstruction is that the algebraic involution alone cannot force the norm to behave like the operator norm on Hilbert space. The following definition imposes the decisive compatibility condition: the size of every element is recoverable from the positive-looking product $a^*a$. [definition: $C^*$-Algebra] A C*-algebra is an involutive Banach algebra $A$ such that \begin{align*} \|a^*a\|_A=\|a\|_A^2 \end{align*} for every $a\in A$. [/definition] The identity immediately implies that the involution is isometric, since $\|a^*\|_A^2=\|aa^*\|_A=\|a\|_A^2$. More importantly, it raises a rigidity question: if the same algebraic *-algebra admits two complete norms satisfying the $C^*$-identity, can they differ? [quotetheorem:8546] [citeproof:8546] This theorem explains why C*-algebras are more rigid than general Banach algebras, and the $C^*$-identity is the essential input. Without it, the same involutive algebra may carry inequivalent complete submultiplicative norms; for example, polynomial algebras and group algebras admit many Banach algebra completions that remember different analytic information. The theorem also does not say that every abstract *-algebra has a C*-norm, only that if such a norm exists then it is forced by the *-algebraic spectral structure. The next question is which maps are allowed to compare two involutive algebras. A plain algebra homomorphism preserves sums, products, and scalar multiplication, but it may ignore the operation that distinguishes adjoints from arbitrary algebraic anti-linear maps. Since the $C^*$-identity ties the norm of $a$ to the positive-looking element $a^*a$, any morphism that is meant to preserve C*-algebraic structure must also commute with the involution. [definition: Star Homomorphism] Let $A$ and $B$ be involutive algebras. A *-homomorphism is a complex algebra homomorphism $\varphi:A\to B$ satisfying \begin{align*} \varphi(a^*)=\varphi(a)^* \end{align*} for every $a\in A$. [/definition] A *-homomorphism preserves the algebraic data used in spectra of self-adjoint elements, but continuity is not included in its definition. The motivating problem for the next theorem is whether the $C^*$-identity forces such a map to be bounded without adding continuity as a separate hypothesis. [quotetheorem:8547] [citeproof:8547] The theorem means that *-homomorphisms between C*-algebras are analytic objects even when introduced algebraically. The C*-algebra hypotheses matter: for general Banach *-algebras, algebraic homomorphisms need not be controlled by the given norms, and continuity can be an additional condition rather than a consequence. Contractivity is also weaker than isometry; a non-injective *-homomorphism can collapse a nonzero ideal, so its norm estimate may be strict. This distinction becomes central when quotient C*-algebras and commutative Gelfand duality enter the course. A simple restriction map gives the model example and also previews the contravariant relation between spaces and commutative C*-algebras. [example: Inclusion of Continuous Functions] Let $Y$ be a closed subset of a compact Hausdorff space $X$. Since $Y$ is closed in compact $X$, the space $Y$ is compact, so $C(Y)$ has unit $1_Y$. Define $\rho:C(X)\to C(Y)$ by $\rho(f)=f|_Y$. For $f,g\in C(X)$, $\alpha\in\mathbb C$, and $y\in Y$, we have \begin{align*} \rho(f+g)(y)=(f+g)(y)=f(y)+g(y)=\rho(f)(y)+\rho(g)(y). \end{align*} Also \begin{align*} \rho(\alpha f)(y)=(\alpha f)(y)=\alpha f(y)=\alpha\rho(f)(y). \end{align*} Multiplication is preserved pointwise: \begin{align*} \rho(fg)(y)=(fg)(y)=f(y)g(y)=\rho(f)(y)\rho(g)(y). \end{align*} The involution is preserved because \begin{align*} \rho(f^*)(y)=f^*(y)=\overline{f(y)}=\rho(f)(y)^*=\rho(f)^*(y). \end{align*} Finally, \begin{align*} \rho(1_X)(y)=1_X(y)=1=1_Y(y). \end{align*} Thus $\rho$ is a unital *-homomorphism. By *Automatic Continuity of Star Homomorphisms*, $\rho$ is contractive, so \begin{align*} \|f|_Y\|_{C(Y)}=\|\rho(f)\|_{C(Y)}\le \|f\|_{C(X)}. \end{align*} The same estimate is visible directly from the sup norms: \begin{align*} \|f|_Y\|_{C(Y)}=\sup_{y\in Y}|f(y)|\le \sup_{x\in X}|f(x)|=\|f\|_{C(X)}. \end{align*} So the abstract contractivity theorem exactly matches the elementary fact that restricting the domain cannot increase the supremum norm. [/example] ## Standard Examples and Unitizations The final problem in this chapter is to build a reliable supply of C*-algebras. These examples will reappear as commutative models in Chapter 2, operator models in Chapter 5, ideals and quotients in Chapter 6, and finite-dimensional test cases in Chapter 9. [example: Continuous Functions on a Compact Space] Let $X$ be a compact Hausdorff space. We verify that $C(X)$, with pointwise operations, pointwise multiplication, involution $f^*(x)=\overline{f(x)}$, unit $1_X(x)=1$, and norm \begin{align*} \|f\|_{C(X)}=\sup_{x\in X}|f(x)|, \end{align*} is a unital commutative C*-algebra. For $f,g,h\in C(X)$ and $x\in X$, multiplication is associative because \begin{align*} ((fg)h)(x)=(fg)(x)h(x)=f(x)g(x)h(x)=f(x)(gh)(x)=(f(gh))(x). \end{align*} It is commutative because \begin{align*} (fg)(x)=f(x)g(x)=g(x)f(x)=(gf)(x). \end{align*} The unit acts pointwise: \begin{align*} (1_Xf)(x)=1_X(x)f(x)=1f(x)=f(x). \end{align*} Also \begin{align*} (f1_X)(x)=f(x)1_X(x)=f(x)1=f(x). \end{align*} Since products, sums, scalar multiples, and complex conjugates of continuous functions are continuous, these operations stay inside $C(X)$. The norm is submultiplicative because for every $x\in X$, \begin{align*} |f(x)g(x)|=|f(x)|\,|g(x)|\le \|f\|_{C(X)}\|g\|_{C(X)}. \end{align*} Taking the supremum over $x\in X$ gives \begin{align*} \|fg\|_{C(X)}\le \|f\|_{C(X)}\|g\|_{C(X)}. \end{align*} With the supremum norm, $C(X)$ is complete, so it is a unital commutative Banach algebra. The involution satisfies the required identities pointwise. For example, \begin{align*} (f+g)^*(x)=\overline{f(x)+g(x)}=\overline{f(x)}+\overline{g(x)}=f^*(x)+g^*(x). \end{align*} For $\lambda\in\mathbb C$, \begin{align*} (\lambda f)^*(x)=\overline{\lambda f(x)}=\overline{\lambda}\,\overline{f(x)}=(\overline{\lambda}f^*)(x). \end{align*} For multiplication, \begin{align*} (fg)^*(x)=\overline{f(x)g(x)}=\overline{g(x)}\,\overline{f(x)}=(g^*f^*)(x). \end{align*} Finally, \begin{align*} (f^*)^*(x)=\overline{\overline{f(x)}}=f(x). \end{align*} Thus $C(X)$ is an involutive Banach algebra. It remains to check the C*-identity. For each $x\in X$, \begin{align*} (f^*f)(x)=f^*(x)f(x)=\overline{f(x)}f(x)=|f(x)|^2. \end{align*} Therefore \begin{align*} \|f^*f\|_{C(X)}=\sup_{x\in X}|(f^*f)(x)|=\sup_{x\in X}|f(x)|^2. \end{align*} Since $t\mapsto t^2$ is increasing on $[0,\infty)$, \begin{align*} \sup_{x\in X}|f(x)|^2=\left(\sup_{x\in X}|f(x)|\right)^2=\|f\|_{C(X)}^2. \end{align*} Hence $C(X)$ is a unital commutative C*-algebra, and its C*-norm is exactly the usual supremum norm. [/example] Compactness of $X$ supplies the unit through the constant function $1_X$. To treat locally compact spaces in the same spirit, we need a replacement for bounded continuous functions that still behaves as though the missing point at infinity has value zero. [definition: Functions Vanishing at Infinity] Let $X$ be a locally compact Hausdorff space. The space $C_0(X)$ is the set of continuous functions $f:X\to\mathbb C$ such that, for every $\varepsilon>0$, the set \begin{align*} \{x\in X: |f(x)|\ge \varepsilon\} \end{align*} is compact. [/definition] With pointwise operations, complex conjugation, and the sup norm, $C_0(X)$ is a C*-algebra. It is unital exactly when $X$ is compact, because the constant function $1$ vanishes at infinity precisely in that case. [example: The Algebra $c_0$] Take $X=\mathbb N$ with the [discrete topology](/page/Discrete%20Topology). A function $f:\mathbb N\to\mathbb C$ is the same data as a sequence $a=(a_n)_{n=1}^{\infty}$, where $a_n=f(n)$. Since every subset of a discrete space is open, every function $\mathbb N\to\mathbb C$ is continuous. In the discrete topology, a subset $K\subseteq\mathbb N$ is compact exactly when it is finite. Indeed, if $K$ is compact, the [open cover](/page/Open%20Cover) $\{\{n\}:n\in K\}$ has a finite subcover, so $K$ is finite. Conversely, every finite subset is compact because any open cover has a finite subcover obtained by choosing one [open set](/page/Open%20Set) for each point. Thus $f\in C_0(\mathbb N)$ exactly when, for every $\varepsilon>0$, the set \begin{align*} \{n\in\mathbb N: |a_n|\ge \varepsilon\} \end{align*} is finite. This is equivalent to $a_n\to 0$: if $a_n\to 0$, then for each $\varepsilon>0$ there is $N$ such that $n\ge N$ implies $|a_n|<\varepsilon$, so the displayed set is contained in $\{1,\dots,N-1\}$ and is finite. Conversely, if the displayed set is finite for each $\varepsilon>0$, choose $N$ larger than every element of it; then $n\ge N$ implies $|a_n|<\varepsilon$, which is precisely $a_n\to 0$. Therefore \begin{align*} C_0(\mathbb N)=c_0=\{(a_n)_{n=1}^{\infty}:a_n\to 0\}. \end{align*} The operations are pointwise: \begin{align*} (a+b)_n=a_n+b_n,\quad (\lambda a)_n=\lambda a_n,\quad (ab)_n=a_nb_n,\quad a^*_n=\overline{a_n}. \end{align*} If $a_n\to 0$ and $b_n\to 0$, then $a_n+b_n\to 0$, $\lambda a_n\to 0$, $\overline{a_n}\to 0$, and $a_nb_n\to 0$ because every convergent sequence is bounded and a bounded sequence times a null sequence is null. Hence these operations stay inside $c_0$. The norm is \begin{align*} \|a\|_\infty=\sup_{n\ge 1}|a_n|. \end{align*} For multiplication, \begin{align*} |a_nb_n|\le \|a\|_\infty\|b\|_\infty \end{align*} for every $n$, so taking the supremum gives \begin{align*} \|ab\|_\infty\le \|a\|_\infty\|b\|_\infty. \end{align*} The $C^*$-identity is pointwise: \begin{align*} (a^*a)_n=\overline{a_n}a_n=|a_n|^2. \end{align*} Hence \begin{align*} \|a^*a\|_\infty=\sup_{n\ge 1}|a_n|^2=\left(\sup_{n\ge 1}|a_n|\right)^2=\|a\|_\infty^2. \end{align*} There is no multiplicative unit in $c_0$. If $e$ were a unit, then applying $ea=a$ to the sequence $a$ with $a_k=1$ and $a_n=0$ for $n\neq k$ would give $e_k=1$ for every $k$. Thus $e=(1,1,1,\dots)$, but this sequence does not converge to $0$, so $e\notin c_0$. Thus $c_0$ is the basic commutative nonunital C*-algebra, modeled by the locally [compact space](/page/Compact%20Space) $\mathbb N$. [/example] Operator algebras give the noncommutative prototypes. The basic example is the full algebra of bounded operators on a Hilbert space. [example: Bounded Operators on a Hilbert Space] Let $H$ be a complex Hilbert space, and let $\mathcal{L}(H)$ be the set of bounded linear maps $T:H\to H$. With addition, scalar multiplication, and multiplication given by composition, $\mathcal{L}(H)$ is an algebra: for $S,T\in\mathcal{L}(H)$ and $\xi\in H$, \begin{align*} \|(ST)\xi\|_H=\|S(T\xi)\|_H\le \|S\|_{\mathcal{L}(H)}\|T\xi\|_H\le \|S\|_{\mathcal{L}(H)}\|T\|_{\mathcal{L}(H)}\|\xi\|_H. \end{align*} Taking the supremum over $\|\xi\|_H\le 1$ gives \begin{align*} \|ST\|_{\mathcal{L}(H)}\le \|S\|_{\mathcal{L}(H)}\|T\|_{\mathcal{L}(H)}. \end{align*} The identity operator $I\xi=\xi$ satisfies $IT=TI=T$, so $\mathcal{L}(H)$ is unital. The operator norm is complete. If $(T_n)$ is Cauchy in $\mathcal{L}(H)$, then for each $\xi\in H$ the sequence $(T_n\xi)$ is Cauchy because \begin{align*} \|T_n\xi-T_m\xi\|_H\le \|T_n-T_m\|_{\mathcal{L}(H)}\|\xi\|_H. \end{align*} Define $T\xi=\lim_n T_n\xi$. Linearity follows by passing to limits in $T_n(\alpha\xi+\eta)=\alpha T_n\xi+T_n\eta$. Since a [Cauchy sequence](/page/Cauchy%20Sequence) is bounded, choose $M$ with $\|T_n\|_{\mathcal{L}(H)}\le M$ for all $n$; then \begin{align*} \|T\xi\|_H=\lim_n\|T_n\xi\|_H\le M\|\xi\|_H. \end{align*} Thus $T\in\mathcal{L}(H)$. If $\|T_n-T_m\|_{\mathcal{L}(H)}<\varepsilon$ for $n,m\ge N$, then for $\|\xi\|_H\le 1$, \begin{align*} \|(T_n-T)\xi\|_H=\lim_m\|(T_n-T_m)\xi\|_H\le \varepsilon. \end{align*} Taking the supremum gives $\|T_n-T\|_{\mathcal{L}(H)}\le\varepsilon$, so $\mathcal{L}(H)$ is a Banach algebra. The adjoint operation makes this Banach algebra involutive. For $S,T\in\mathcal{L}(H)$, $\lambda\in\mathbb C$, and $\xi,\eta\in H$, the defining identity of the adjoint gives \begin{align*} ((S+T)\xi,\eta)_H=(S\xi,\eta)_H+(T\xi,\eta)_H=(\xi,S^*\eta)_H+(\xi,T^*\eta)_H=(\xi,(S^*+T^*)\eta)_H. \end{align*} Hence $(S+T)^*=S^*+T^*$. Similarly, \begin{align*} ((\lambda T)\xi,\eta)_H=\lambda(T\xi,\eta)_H=\lambda(\xi,T^*\eta)_H=(\xi,\overline{\lambda}T^*\eta)_H, \end{align*} so $(\lambda T)^*=\overline{\lambda}T^*$. For products, \begin{align*} ((ST)\xi,\eta)_H=(S(T\xi),\eta)_H=(T\xi,S^*\eta)_H=(\xi,T^*S^*\eta)_H, \end{align*} so $(ST)^*=T^*S^*$. Also $(T^*)^*=T$ and $I^*=I$ by the same defining identity. It remains to verify the $C^*$-identity. First, the adjoint has the same operator norm as the original operator. For $\|\eta\|_H\le 1$, \begin{align*} \|T^*\eta\|_H=\sup_{\|\xi\|_H\le 1}|(T^*\eta,\xi)_H|=\sup_{\|\xi\|_H\le 1}|(\eta,T\xi)_H|\le \|T\|_{\mathcal{L}(H)}. \end{align*} Thus $\|T^*\|_{\mathcal{L}(H)}\le \|T\|_{\mathcal{L}(H)}$, and applying this to $T^*$ gives $\|T\|_{\mathcal{L}(H)}\le \|T^*\|_{\mathcal{L}(H)}$. Hence $\|T^*\|_{\mathcal{L}(H)}=\|T\|_{\mathcal{L}(H)}$. The upper bound follows from submultiplicativity: \begin{align*} \|T^*T\|_{\mathcal{L}(H)}\le \|T^*\|_{\mathcal{L}(H)}\|T\|_{\mathcal{L}(H)}=\|T\|_{\mathcal{L}(H)}^2. \end{align*} For the reverse bound, if $\|\xi\|_H\le 1$, then \begin{align*} \|T\xi\|_H^2=(T\xi,T\xi)_H=(T^*T\xi,\xi)_H. \end{align*} By Cauchy-Schwarz and the definition of operator norm, \begin{align*} |(T^*T\xi,\xi)_H|\le \|T^*T\xi\|_H\|\xi\|_H\le \|T^*T\|_{\mathcal{L}(H)}. \end{align*} Therefore $\|T\xi\|_H^2\le \|T^*T\|_{\mathcal{L}(H)}$ for every $\|\xi\|_H\le 1$, and taking the supremum gives \begin{align*} \|T\|_{\mathcal{L}(H)}^2\le \|T^*T\|_{\mathcal{L}(H)}. \end{align*} Combining the two inequalities yields \begin{align*} \|T^*T\|_{\mathcal{L}(H)}=\|T\|_{\mathcal{L}(H)}^2. \end{align*} Thus $\mathcal{L}(H)$ is a unital C*-algebra, and the $C^*$-identity is exactly the Hilbert space adjoint identity expressed in operator norm. [/example] Inside $\mathcal{L}(H)$, compact operators behave like an infinite-dimensional analogue of matrices. They form an ideal, but generally not a unital algebra. [example: Compact Operators] Let $H$ be an infinite-dimensional Hilbert space, and let $\mathcal{K}(H)$ denote the set of compact operators on $H$, meaning bounded operators that send the closed unit ball of $H$ to a relatively compact subset of $H$. First, $\mathcal{K}(H)$ is a two-sided ideal in $\mathcal{L}(H)$. If $K\in\mathcal{K}(H)$ and $S,T\in\mathcal{L}(H)$, then for $\|\xi\|_H\le 1$, \begin{align*} \|T\xi\|_H\le \|T\|_{\mathcal{L}(H)}. \end{align*} Thus \begin{align*} T(B_H)\subseteq \|T\|_{\mathcal{L}(H)}B_H, \end{align*} where $B_H=\{\xi\in H:\|\xi\|_H\le 1\}$. Since $K$ maps bounded sets into relatively compact sets, $K(T(B_H))$ is relatively compact. Applying the continuous operator $S$ to its closure gives another compact set, so \begin{align*} SKT(B_H)=S(K(T(B_H))) \end{align*} is relatively compact. Hence $SKT\in\mathcal{K}(H)$. The set $\mathcal{K}(H)$ is closed in operator norm. Suppose $K_m\in\mathcal{K}(H)$ and $K_m\to T$ in $\mathcal{L}(H)$. Fix $\varepsilon>0$, and choose $m$ such that \begin{align*} \|T-K_m\|_{\mathcal{L}(H)}<\varepsilon/3. \end{align*} Since $K_m(B_H)$ is relatively compact, it is totally bounded, so there are $y_1,\dots,y_r\in H$ such that every $K_m\xi$ with $\|\xi\|_H\le 1$ satisfies \begin{align*} \|K_m\xi-y_j\|_H<\varepsilon/3 \end{align*} for some $j$. Then \begin{align*} \|T\xi-y_j\|_H\le \|T\xi-K_m\xi\|_H+\|K_m\xi-y_j\|_H<\varepsilon/3+\varepsilon/3<\varepsilon. \end{align*} Thus $T(B_H)$ is totally bounded, so $T$ is compact. The compact operators are also closed under adjoints. If $K\in\mathcal{K}(H)$ and $(\eta_n)$ is a bounded sequence in $H$, choose $M$ with $\|\eta_n\|_H\le M$ for all $n$. Since $K(B_H)$ is totally bounded, for each $\varepsilon>0$ choose $\xi_1,\dots,\xi_r\in B_H$ such that every $K\xi$ with $\|\xi\|_H\le 1$ satisfies \begin{align*} \|K\xi-K\xi_j\|_H<\varepsilon/(4M+1) \end{align*} for some $j$. By passing to a subsequence, the finitely many scalar sequences $(\eta_n,K\xi_j)_H$ are all Cauchy. For this subsequence and for sufficiently large $m,n$, \begin{align*} |(\eta_m-\eta_n,K\xi_j)_H|<\varepsilon/2 \end{align*} for every $j$. Given $\|\xi\|_H\le 1$, choose $j$ as above. Then \begin{align*} |((K^*\eta_m-K^*\eta_n),\xi)_H|=|(\eta_m-\eta_n,K\xi)_H|. \end{align*} Also \begin{align*} |(\eta_m-\eta_n,K\xi)_H|\le |(\eta_m-\eta_n,K\xi_j)_H|+|(\eta_m-\eta_n,K\xi-K\xi_j)_H|. \end{align*} Since $\|\eta_m-\eta_n\|_H\le 2M$, the second term is at most \begin{align*} 2M\,\|K\xi-K\xi_j\|_H<\varepsilon/2. \end{align*} Hence \begin{align*} |((K^*\eta_m-K^*\eta_n),\xi)_H|<\varepsilon \end{align*} for every $\|\xi\|_H\le 1$, so $(K^*\eta_n)$ has a Cauchy subsequence. Therefore $K^*$ is compact. Thus $\mathcal{K}(H)$ is a closed two-sided self-adjoint ideal in the C*-algebra $\mathcal{L}(H)$. With the inherited operator norm and adjoint, it satisfies \begin{align*} \|K^*K\|_{\mathcal{L}(H)}=\|K\|_{\mathcal{L}(H)}^2 \end{align*} because this identity already holds in $\mathcal{L}(H)$. Hence $\mathcal{K}(H)$ is a C*-algebra. It is not unital when $H$ is infinite-dimensional. If $I$ were compact, then $I(B_H)=B_H$ would be relatively compact, so the closed unit ball of $H$ would be compact. But an infinite-dimensional Hilbert space has an infinite orthonormal sequence $(e_n)$, and \begin{align*} \|e_n-e_m\|_H^2=\|e_n\|_H^2+\|e_m\|_H^2-2\operatorname{Re}(e_n,e_m)_H=1+1-0=2 \end{align*} for $n\neq m$. Thus $(e_n)$ has no Cauchy subsequence, so $B_H$ is not compact. Therefore $I\notin\mathcal{K}(H)$. The unitization adds exactly one scalar copy of the identity: \begin{align*} \mathcal{K}(H)^+=\{\lambda I+K:\lambda\in\mathbb C,\ K\in\mathcal{K}(H)\}. \end{align*} Multiplication is closed because \begin{align*} (\lambda I+K)(\mu I+L)=\lambda\mu I+(\lambda L+\mu K+KL), \end{align*} and $\lambda L+\mu K+KL\in\mathcal{K}(H)$ by linearity and the ideal property. The adjoint is \begin{align*} (\lambda I+K)^*=\overline{\lambda}I+K^*, \end{align*} and $K^*\in\mathcal{K}(H)$. Thus $\mathcal{K}(H)^+$ is the concrete unital algebra obtained by adjoining the missing identity to $\mathcal{K}(H)$. [/example] The [spectrum of a compact operator](/theorems/220) illustrates why the ambient algebra matters. In the nonunital algebra $\mathcal{K}(H)$ we use the unitization to define spectral data. [example: Spectrum of a Compact Operator in the Unitization] Let $H$ be infinite-dimensional and let $K\in\mathcal{K}(H)$. We identify the unitization $\mathcal{K}(H)^+$ with the concrete subalgebra $\mathbb C I+\mathcal{K}(H)\subseteq\mathcal{L}(H)$, so the unit of $\mathcal{K}(H)^+$ is $I$. We first check that invertibility of $\lambda I-K$ is the same in $\mathcal{K}(H)^+$ and in $\mathcal{L}(H)$. If $\lambda I-K$ has an inverse in $\mathcal{K}(H)^+$, then it also has that inverse in $\mathcal{L}(H)$, because $\mathcal{K}(H)^+\subseteq\mathcal{L}(H)$. Conversely, suppose $\lambda I-K$ is invertible in $\mathcal{L}(H)$. The case $\lambda=0$ cannot occur: if $K$ were invertible in $\mathcal{L}(H)$, then \begin{align*} I=K^{-1}K \end{align*} would be compact because $\mathcal{K}(H)$ is a two-sided ideal in $\mathcal{L}(H)$, contradicting infinite-dimensionality of $H$. Hence $\lambda\neq 0$. Put $R=(\lambda I-K)^{-1}$. Then \begin{align*} R-\lambda^{-1}I=\lambda^{-1}R(\lambda I)-\lambda^{-1}R(\lambda I-K)=\lambda^{-1}RK. \end{align*} Since $K$ is compact and $\mathcal{K}(H)$ is an ideal, $RK\in\mathcal{K}(H)$. Thus \begin{align*} R=\lambda^{-1}I+\lambda^{-1}RK\in \mathbb C I+\mathcal{K}(H)=\mathcal{K}(H)^+. \end{align*} Therefore \begin{align*} \sigma_{\mathcal{K}(H)^+}(K)=\sigma_{\mathcal{L}(H)}(K). \end{align*} For a concrete diagonal example, let $(e_n)$ be an orthonormal basis and suppose \begin{align*} Ke_n=\alpha_ne_n \end{align*} with $\alpha_n\to 0$. If $\lambda=\alpha_m$ for some $m$, then \begin{align*} (\lambda I-K)e_m=(\alpha_m-\alpha_m)e_m=0, \end{align*} so $\lambda I-K$ is not injective and hence not invertible. Also $0\in\sigma(K)$: since $\|e_n\|_H=1$ and \begin{align*} \|Ke_n\|_H=|\alpha_n|\to 0, \end{align*} the operator $K$ is not bounded below, so it cannot be invertible. Now suppose $\lambda\notin \{0\}\cup\{\alpha_n:n\ge 1\}^{\mathrm{cl}}$. Then there is $\delta>0$ such that \begin{align*} |\lambda-\alpha_n|\ge \delta \end{align*} for every $n$. Define $D$ on the basis by \begin{align*} De_n=(\lambda-\alpha_n)^{-1}e_n. \end{align*} The bound $|(\lambda-\alpha_n)^{-1}|\le \delta^{-1}$ makes $D$ a bounded diagonal operator. On each basis vector, \begin{align*} D(\lambda I-K)e_n=D((\lambda-\alpha_n)e_n)=e_n, \end{align*} and \begin{align*} (\lambda I-K)De_n=(\lambda I-K)((\lambda-\alpha_n)^{-1}e_n)=e_n. \end{align*} Thus $D=(\lambda I-K)^{-1}$ in $\mathcal{L}(H)$. Moreover, \begin{align*} (D-\lambda^{-1}I)e_n=\left((\lambda-\alpha_n)^{-1}-\lambda^{-1}\right)e_n=\frac{\alpha_n}{\lambda(\lambda-\alpha_n)}e_n. \end{align*} Since $\alpha_n\to 0$ and $|\lambda-\alpha_n|\ge\delta$, the diagonal entries $\alpha_n/(\lambda(\lambda-\alpha_n))$ tend to $0$, so $D-\lambda^{-1}I$ is compact. Hence $D\in\mathcal{K}(H)^+$. Therefore the spectral values are exactly the diagonal values together with their limit points, including $0$: \begin{align*} \sigma_{\mathcal{K}(H)^+}(K)=\{0\}\cup \{\alpha_n:n\ge 1\}^{\mathrm{cl}}. \end{align*} Repeated diagonal entries do not create repeated spectral points; the spectrum records values, not multiplicities. [/example] Finite-dimensional C*-algebras supply the algebraic test cases. They are built from matrix blocks, and the direct sum norm is forced by the C*-identity. [example: Matrix Algebras and Finite Direct Sums] Represent $M_n(\mathbb C)$ on the Hilbert space $\mathbb C^n$ by matrix multiplication. Then $M_n(\mathbb C)=\mathcal L(\mathbb C^n)$, so with the operator norm, identity matrix $I_n$, and adjoint transpose $a\mapsto a^*$, the matrix algebra is the finite-dimensional case of the C*-algebra $\mathcal L(H)$. Now let \begin{align*} A=\bigoplus_{j=1}^m M_{n_j}(\mathbb C). \end{align*} Elements of $A$ are tuples $a=(a_1,\dots,a_m)$, and the operations are coordinatewise: \begin{align*} a+b=(a_1+b_1,\dots,a_m+b_m),\quad ab=(a_1b_1,\dots,a_mb_m),\quad a^*=(a_1^*,\dots,a_m^*). \end{align*} The unit is \begin{align*} 1_A=(I_{n_1},\dots,I_{n_m}). \end{align*} Indeed, \begin{align*} 1_Aa=(I_{n_1}a_1,\dots,I_{n_m}a_m)=(a_1,\dots,a_m)=a \end{align*} and similarly $a1_A=a$. Define \begin{align*} \|a\|_A=\max_{1\le j\le m}\|a_j\|_{M_{n_j}(\mathbb C)}. \end{align*} For multiplication, \begin{align*} \|ab\|_A=\max_{1\le j\le m}\|a_jb_j\|\le \max_{1\le j\le m}\|a_j\|\|b_j\|\le \left(\max_{1\le j\le m}\|a_j\|\right)\left(\max_{1\le j\le m}\|b_j\|\right)=\|a\|_A\|b\|_A. \end{align*} Since each $M_{n_j}(\mathbb C)$ is finite-dimensional and complete, the finite product norm makes $A$ complete. The involution identities also hold coordinatewise. For example, \begin{align*} (ab)^*=((a_1b_1)^*,\dots,(a_mb_m)^*)=(b_1^*a_1^*,\dots,b_m^*a_m^*)=b^*a^*. \end{align*} The C*-identity follows from the C*-identity in each matrix block: \begin{align*} \|a^*a\|_A=\max_{1\le j\le m}\|a_j^*a_j\|=\max_{1\le j\le m}\|a_j\|^2=\left(\max_{1\le j\le m}\|a_j\|\right)^2=\|a\|_A^2. \end{align*} Thus $A$ is a unital C*-algebra. Finally, let $a=(a_1,\dots,a_m)\in A$. For $\lambda\in\mathbb C$, \begin{align*} \lambda1_A-a=(\lambda I_{n_1}-a_1,\dots,\lambda I_{n_m}-a_m). \end{align*} If $\lambda1_A-a$ has inverse $b=(b_1,\dots,b_m)$ in $A$, then \begin{align*} (\lambda I_{n_j}-a_j)b_j=I_{n_j} \end{align*} and \begin{align*} b_j(\lambda I_{n_j}-a_j)=I_{n_j} \end{align*} for every $j$, so $\lambda I_{n_j}-a_j$ is invertible in each matrix algebra. Conversely, if every $\lambda I_{n_j}-a_j$ is invertible, then \begin{align*} b=((\lambda I_{n_1}-a_1)^{-1},\dots,(\lambda I_{n_m}-a_m)^{-1}) \end{align*} satisfies $(\lambda1_A-a)b=b(\lambda1_A-a)=1_A$. Therefore \begin{align*} \sigma_A(a)=\bigcup_{j=1}^m\sigma_{M_{n_j}(\mathbb C)}(a_j). \end{align*} The direct sum has exactly the spectral values appearing in its matrix blocks; it records finitely many independent matrix systems inside one unital C*-algebra. [/example] The comparison between one matrix algebra and a direct sum already shows a structural theme of the course: direct sums have central projections separating their blocks, while a single full matrix algebra has only scalar central elements. To discuss spectra for nonunital examples with the same formalism as unital examples, we now need a canonical way to add an identity. [definition: Unitization] Let $A$ be a nonunital Banach algebra. Its unitization is the [vector space](/page/Vector%20Space) $A^+=A\oplus \mathbb C$ equipped with the multiplication map $A^+\times A^+\to A^+$ defined by \begin{align*} (a,\lambda)(b,\mu)=(ab+\lambda b+\mu a,\lambda\mu) \end{align*} and unit $(0,1)$. If $A$ is involutive, the unitization involution is the map $A^+\to A^+$ defined by \begin{align*} (a,\lambda)^*=(a^*,\overline{\lambda}). \end{align*} [/definition] For a nonunital C*-algebra, there is a C*-norm on the algebraic unitization compatible with this operation, and it agrees with the concrete norm in standard representations such as $\mathcal{K}(H)^+=\mathcal{K}(H)+\mathbb C I$. Unitization lets the course use spectra and resolvents uniformly, while still keeping track of whether the original algebra had an identity. Once the C*-identity is established, the next step is to see what it forces in the commutative case. There the abstract theory becomes concrete again: spectra and characters identify the algebra with continuous functions on a space, and Gelfand duality explains why this is the right model. # 2. Commutative $C^*$-Algebras and Gelfand Duality This chapter turns the spectral theory of Banach algebras into a representation theorem for commutative C*-algebras. The guiding question is whether an abstract commutative C*-algebra is genuinely more general than an algebra of continuous functions. The answer is no: once the characters are assembled into a [topological space](/page/Topological%20Space), the algebra is recovered as functions on that space. The first chapter introduced spectra, the spectral radius formula, and the C*-identity. Here those tools become geometric: algebra homomorphisms become points, elements become functions, and ideals become open subsets. ## Characters and Maximal Ideal Spaces How can a commutative Banach algebra manufacture a space from its own algebraic structure? The basic idea is to use scalar-valued homomorphisms as points, since evaluation at a point is the model example for algebras of functions. [definition: Character] Let $A$ be a complex Banach algebra. A character on $A$ is a non-zero algebra homomorphism $\tau:A\to\mathbb C$. [/definition] For a function algebra $C(X)$, each point $x\in X$ gives a character $\tau_x(f)=f(x)$. The central problem is whether all characters arise this way after the correct space $X$ has been built. [example: Characters of a Continuous Function Algebra] Let $X$ be compact Hausdorff and let $A=C(X)$. For each $x\in X$, define $\tau_x(f)=f(x)$. If $f,g\in C(X)$ and $\alpha,\beta\in\mathbb C$, then pointwise addition and scalar multiplication give \begin{align*} \tau_x(\alpha f+\beta g)=(\alpha f+\beta g)(x)=\alpha f(x)+\beta g(x)=\alpha\tau_x(f)+\beta\tau_x(g). \end{align*} Pointwise multiplication gives \begin{align*} \tau_x(fg)=(fg)(x)=f(x)g(x)=\tau_x(f)\tau_x(g). \end{align*} Also $\tau_x(1)=1(x)=1$, so $\tau_x$ is non-zero. Hence every point $x\in X$ gives a character on $C(X)$. Conversely, let $\tau:C(X)\to\mathbb C$ be a character and set $M=\ker\tau$. Since $\tau$ is non-zero, choose $g\in C(X)$ with $\tau(g)\ne 0$. Multiplicativity gives \begin{align*} \tau(1)\tau(g)=\tau(1g)=\tau(g). \end{align*} Subtracting $\tau(g)$ from both sides gives \begin{align*} (\tau(1)-1)\tau(g)=0. \end{align*} Since $\tau(g)\ne 0$, we get $\tau(1)=1$. Therefore, for every $\lambda\in\mathbb C$, \begin{align*} \tau(\lambda 1)=\lambda\tau(1)=\lambda. \end{align*} The map $\Phi:C(X)/M\to\mathbb C$ defined by $\Phi(f+M)=\tau(f)$ is well-defined. Indeed, if $f+M=g+M$, then $f-g\in M$, so \begin{align*} 0=\tau(f-g)=\tau(f)-\tau(g). \end{align*} Thus $\tau(f)=\tau(g)$. The map is injective because if $\Phi(f+M)=0$, then $\tau(f)=0$, hence $f\in M$ and $f+M=M$. It is surjective because, for every $\lambda\in\mathbb C$, \begin{align*} \Phi(\lambda 1+M)=\tau(\lambda 1)=\lambda. \end{align*} Thus $C(X)/M\cong\mathbb C$, so $M$ is maximal. We now show that $M$ is the ideal of functions vanishing at one point. Suppose first that no point $x\in X$ has the property that every $f\in M$ vanishes at $x$. Then for each $x\in X$ there is $f_x\in M$ with $f_x(x)\ne 0$. Define \begin{align*} U_x=\{y\in X:f_x(y)\ne 0\}. \end{align*} Since $f_x$ is continuous and $\mathbb C\setminus\{0\}$ is open, $U_x$ is open, and $x\in U_x$. The sets $U_x$ cover $X$, so compactness gives $x_1,\dots,x_n\in X$ such that \begin{align*} X=U_{x_1}\cup\cdots\cup U_{x_n}. \end{align*} Define \begin{align*} h=\sum_{j=1}^n f_{x_j}\overline{f_{x_j}}. \end{align*} For each $j$, the function $\overline{f_{x_j}}$ belongs to $C(X)$. Since $f_{x_j}\in M$ and $M$ is an ideal, $f_{x_j}\overline{f_{x_j}}\in M$. Since $M$ is closed under finite sums, $h\in M$. For any $y\in X$, choose $j$ with $y\in U_{x_j}$. Then $f_{x_j}(y)\ne 0$, so \begin{align*} h(y)=\sum_{k=1}^n f_{x_k}(y)\overline{f_{x_k}(y)}=\sum_{k=1}^n |f_{x_k}(y)|^2\ge |f_{x_j}(y)|^2>0. \end{align*} Thus $h$ has no zeros on $X$. Since $h$ is continuous and never zero, $1/h\in C(X)$, and the ideal property gives \begin{align*} 1=h(1/h)\in M. \end{align*} But $1\in M=\ker\tau$ would imply $\tau(1)=0$, contradicting $\tau(1)=1$. Hence there is some $x\in X$ such that $f(x)=0$ for every $f\in M$. For this point $x$, every $f\in M$ satisfies $\tau_x(f)=f(x)=0$, so $M\subseteq\ker\tau_x$. The ideal $\ker\tau_x$ is proper because $\tau_x(1)=1$. Since $M$ is maximal and \begin{align*} M\subseteq\ker\tau_x\subsetneq C(X), \end{align*} we have $M=\ker\tau_x$. Now let $f\in C(X)$. The function $f-f(x)1$ satisfies \begin{align*} (f-f(x)1)(x)=f(x)-f(x)\cdot 1=0, \end{align*} so $f-f(x)1\in\ker\tau_x=M$. Applying $\tau$ gives \begin{align*} 0=\tau(f-f(x)1)=\tau(f)-\tau(f(x)1)=\tau(f)-f(x)\tau(1)=\tau(f)-f(x). \end{align*} Therefore $\tau(f)=f(x)=\tau_x(f)$ for every $f\in C(X)$, so $\tau=\tau_x$. Distinct points give distinct characters. If $x\ne y$, compact Hausdorff spaces admit continuous functions separating points, so choose $p\in C(X)$ with $p(x)=0$ and $p(y)=1$. Then \begin{align*} \tau_x(p)=p(x)=0. \end{align*} Also \begin{align*} \tau_y(p)=p(y)=1. \end{align*} Hence $\tau_x\ne\tau_y$. Thus the characters of $C(X)$ are exactly the point evaluations, and the algebraic character space recovers the original points of $X$. [/example] The example shows why characters should be treated as points, but it also raises an algebraic question: how can the set of all such points be recognised without already knowing an underlying space? This motivates naming the collection of characters and tying it to maximal ideals, since the kernel of an evaluation map is the ideal of functions vanishing at a point. [definition: Maximal Ideal Space] Let $A$ be a commutative complex Banach algebra. The maximal ideal space of $A$, denoted $\Delta(A)$, is the set of all characters on $A$. [/definition] The definition packages the candidate points of the algebra into a single object, but it is not yet clear that these points are the right algebraic objects to study. The issue is that ideals, not homomorphisms, are often the internal structure visible inside a Banach algebra. The following theorem connects the two viewpoints by showing that characters are exactly the maps whose kernels are maximal ideals, so topology on characters can later translate into ideal theory. [quotetheorem:2676] [citeproof:2676] The unital hypothesis is doing real work here: without an identity, kernels of non-zero homomorphisms need not capture maximal ideals in the same clean quotient-by-scalars form until one passes to the unitization. The theorem also does not say that every proper ideal is a kernel of a character; it says this only for maximal ideals, and later closed ideals in $C_0(X)$ will require a different topological description. Its main role in the chapter is to make the points of the eventual space algebraically visible, by turning scalar quotients into characters. The correspondence reduces maximal ideals to scalar quotients, so the remaining ingredient is a classification of Banach algebras whose non-zero elements are all invertible. This motivates the Gelfand-Mazur theorem: it prevents exotic complex Banach division algebras from appearing in those quotients. [quotetheorem:2670] [citeproof:2670] The complex Banach hypothesis is essential: over $\mathbb R$ there are real Banach division algebras such as $\mathbb C$, and without completeness the spectral argument need not give a non-empty spectrum. The theorem is not a classification of all division rings or all normed division algebras; it is specifically a spectral consequence of complex Banach algebra theory. In this chapter it supplies the scalar quotient step needed to identify maximal ideals with characters. The algebraic set $\Delta(A)$ is now under control, but it is not yet a topological space. To compare it with spaces such as $[0,1]$ or $\mathbb R$, we need a topology whose convergent nets are detected by evaluating fixed algebra elements; this motivates using the weak-* topology from $A^*$. [definition: Weak-Star Topology on the Character Space] Let $A$ be a Banach algebra. The weak-* topology on $\Delta(A)\subset A^*$ is the [subspace topology](/page/Subspace%20Topology) induced by the weak-* topology $\sigma(A^*,A)$. [/definition] In this topology, a net $\tau_i$ converges to $\tau$ exactly when $\tau_i(a)\to\tau(a)$ for every $a\in A$. Pointwise convergence gives a natural topology, but it could still be too large or badly separated to support function-space arguments. The following theorem supplies the missing topological control: in the unital case, the character space is compact Hausdorff, so it can serve as the compact space on which Gelfand transforms will live. [quotetheorem:8548] [citeproof:8548] The unital hypothesis is exactly what forces compactness rather than local compactness: for $C_0(\mathbb R)$ the character space is $\mathbb R$, which is not compact. The theorem does not identify the character space explicitly; it only gives the topological compactness and Hausdorff separation needed for later function theory. This compactness is the point at which Banach-Alaoglu enters Gelfand duality, converting bounded linear functionals into a genuine topological space. Non-unital algebras require a locally compact rather than compact space. Passing to the unitization adds a possible point at infinity, and the original character space appears as the complement of that point. [remark: Non-Unital Character Spaces] For a non-unital commutative Banach algebra $A$, characters are still non-zero homomorphisms $A\to\mathbb C$. The unitization $A^+$ has a compact character space, and characters on $A$ correspond to characters on $A^+$ other than the augmentation character that vanishes on $A$. Thus $\Delta(A)$ is naturally locally compact in the weak-* topology. [/remark] ## The Gelfand Transform for Commutative Banach Algebras Once characters have become points, what does an element of the algebra become? It becomes the function obtained by evaluating that element at every character. [definition: Gelfand Transform] Let $A$ be a commutative complex Banach algebra and let $a\in A$. The Gelfand transform of $a$ is the function $\hat a:\Delta(A)\to\mathbb C$ defined by \begin{align*} \hat a(\tau)=\tau(a). \end{align*} The Gelfand transform of $A$ is the map $\Gamma:A\to C_b(\Delta(A))$, $\Gamma(a)=\hat a$. [/definition] The definition converts algebra elements into scalar functions on the character space, but a representation must preserve structure before it can be useful. This motivates the next theorem, which verifies that the transform respects sums, products, involution when available, and the norm estimate needed for continuity. [quotetheorem:8549] [citeproof:8549] The theorem gives only the formal representation properties; it does not yet prove that the transform remembers all elements or reaches all functions. In a general commutative Banach algebra, non-zero radical elements can have zero Gelfand transform, so injectivity can fail, and uniform algebras show that closed subalgebras of $C(X)$ need not be all of $C(X)$. The C*-identity is the extra structure that will remove both defects, first through the spectral radius formula and then through Stone-Weierstrass. The homomorphism property says that the Gelfand transform is compatible with algebra, but it does not yet say how much spectral information survives. The next theorem motivates the transform as the correct one: the spectrum of an element is recovered as the range of its transformed function. [quotetheorem:2677] [citeproof:2677] Commutativity is essential here because maximal ideals in the algebra generated by one element detect scalar values of that element through characters. The theorem does not assert that the Gelfand transform is onto a function algebra, nor that it is injective; it only identifies the spectrum element by element. Its practical value is that spectral questions become range questions, which is exactly how spectra are computed in the standard examples below. This result is already powerful for classical examples. It says that spectral computations in commutative Banach algebras reduce to ranges of continuous functions on the character space. [example: The Algebra c Zero] Let $c_0$ be the C*-algebra of complex sequences converging to $0$, with pointwise operations and norm $\|(a_k)\|_\infty=\sup_k |a_k|$. For $n\in\mathbb N$, let $e_n$ be the sequence with $1$ in the $n$th coordinate and $0$ elsewhere, and define $\tau_n((a_k))=a_n$. If $a=(a_k)$, $b=(b_k)$, and $\alpha,\beta\in\mathbb C$, then the $n$th coordinate of $\alpha a+\beta b$ is $\alpha a_n+\beta b_n$, so \begin{align*} \tau_n(\alpha a+\beta b)=(\alpha a+\beta b)_n=\alpha a_n+\beta b_n=\alpha\tau_n(a)+\beta\tau_n(b). \end{align*} Since multiplication is pointwise, the $n$th coordinate of $ab$ is $a_nb_n$, hence \begin{align*} \tau_n(ab)=(ab)_n=a_nb_n=\tau_n(a)\tau_n(b). \end{align*} Also \begin{align*} \tau_n(e_n)=(e_n)_n=1, \end{align*} so $\tau_n$ is non-zero. Hence each coordinate evaluation is a character. Conversely, let $\tau:c_0\to\mathbb C$ be a character. Since $e_n^2=e_n$, multiplicativity gives \begin{align*} \tau(e_n)^2=\tau(e_n)\tau(e_n)=\tau(e_n^2)=\tau(e_n). \end{align*} Subtracting $\tau(e_n)$ from both sides gives \begin{align*} \tau(e_n)^2-\tau(e_n)=0. \end{align*} Factoring the left-hand side gives \begin{align*} \tau(e_n)(\tau(e_n)-1)=0. \end{align*} Thus $\tau(e_n)\in\{0,1\}$. If $m\ne n$, then $e_me_n=0$, and therefore \begin{align*} \tau(e_m)\tau(e_n)=\tau(e_me_n)=\tau(0)=0. \end{align*} So at most one of the numbers $\tau(e_n)$ can be equal to $1$. There is at least one such index. Suppose instead that $\tau(e_n)=0$ for every $n$. If $s$ is finitely supported, say $s=\sum_{j=1}^N s_j e_j$, then linearity gives \begin{align*} \tau(s)=\tau\left(\sum_{j=1}^N s_j e_j\right)=\sum_{j=1}^N s_j\tau(e_j)=\sum_{j=1}^N s_j\cdot 0=0. \end{align*} For any $a=(a_k)\in c_0$, let \begin{align*} s^{(N)}=(a_1,\dots,a_N,0,0,\dots). \end{align*} Then $s^{(N)}$ is finitely supported. For $1\le k\le N$, the $k$th coordinate of $a-s^{(N)}$ is $a_k-a_k=0$, while for $k>N$ it is $a_k-0=a_k$. Hence \begin{align*} \|a-s^{(N)}\|_\infty=\sup_{k>N}|a_k|. \end{align*} Since $a_k\to 0$, for every $\varepsilon>0$ there is $N_0$ such that $k>N_0$ implies $|a_k|<\varepsilon$, so for $N\ge N_0$, \begin{align*} \|a-s^{(N)}\|_\infty=\sup_{k>N}|a_k|\le\varepsilon. \end{align*} Thus $s^{(N)}\to a$ in $\|\cdot\|_\infty$, so finitely supported sequences are dense in $c_0$. By *automatic continuity of characters on Banach algebras*, $\tau(s^{(N)})\to\tau(a)$. But each $\tau(s^{(N)})=0$, hence $\tau(a)=0$ for every $a\in c_0$, contradicting that $\tau$ is non-zero. Therefore there is a unique $n$ with $\tau(e_n)=1$. Fix this unique $n$. For any $a=(a_k)\in c_0$, the product $ae_n$ has $a_n$ in the $n$th coordinate and $0$ elsewhere. The sequence $a_ne_n$ has the same coordinates: its $n$th coordinate is $a_n\cdot 1=a_n$, and every other coordinate is $a_n\cdot 0=0$. Therefore \begin{align*} ae_n=a_ne_n. \end{align*} Using $\tau(e_n)=1$, multiplicativity, the identity $ae_n=a_ne_n$, and linearity, \begin{align*} \tau(a)=\tau(a)\cdot 1=\tau(a)\tau(e_n)=\tau(ae_n)=\tau(a_ne_n)=a_n\tau(e_n)=a_n. \end{align*} Thus $\tau(a)=\tau_n(a)$ for every $a\in c_0$, so $\tau=\tau_n$. The characters of $c_0$ are exactly the coordinate evaluations. The weak-* topology on $\Delta(c_0)$ is discrete. For fixed $n$, the function $\tau\mapsto\tau(e_n)$ is weak-* continuous because it is evaluation at the fixed element $e_n\in c_0$. Among characters, \begin{align*} \tau_n(e_n)=1. \end{align*} If $m\ne n$, then the $n$th coordinate of $e_m$ is $0$, so \begin{align*} \tau_m(e_n)=0. \end{align*} Therefore a character $\tau_m$ satisfies $|\tau_m(e_n)-1|<1/2$ exactly when $m=n$, and hence \begin{align*} \{\tau_n\}=\{\tau\in\Delta(c_0):|\tau(e_n)-1|<1/2\}. \end{align*} The set on the right is weak-* open in $\Delta(c_0)$, so each singleton is open. Thus $\Delta(c_0)$ is discrete. Under the identification $\tau_n\leftrightarrow n$, the Gelfand transform sends $a=(a_k)$ to the function \begin{align*} \hat a(n)=\hat a(\tau_n)=\tau_n(a)=a_n. \end{align*} Because $a_k\to 0$, this transformed function vanishes at infinity on the discrete locally compact space $\mathbb N$: for every $\varepsilon>0$, only finitely many $n$ satisfy $|\hat a(n)|=|a_n|\ge\varepsilon$. Conversely, if $f\in C_0(\mathbb N)$ for the discrete topology, then the sequence $(f(n))$ converges to $0$, because for every $\varepsilon>0$ the set $\{n\in\mathbb N:|f(n)|\ge\varepsilon\}$ is compact in a discrete space, hence finite. Therefore the Gelfand transform identifies $c_0$ with $C_0(\mathbb N)$, and the character space is the discrete locally compact space $\mathbb N$. [/example] Finite-dimensional commutative examples show the same idea without topology at infinity. The character space is finite and the algebra decomposes into scalar coordinates. [example: Finite-Dimensional Commutative C-Star Algebras] After choosing the given $*$-isomorphism, identify $A$ with $\mathbb C^n$ with coordinatewise addition, scalar multiplication, multiplication, and involution. For $1\le j\le n$, define $\tau_j(z_1,\dots,z_n)=z_j$. If $z=(z_1,\dots,z_n)$, $w=(w_1,\dots,w_n)$, and $\alpha,\beta\in\mathbb C$, then \begin{align*} \alpha z+\beta w=(\alpha z_1+\beta w_1,\dots,\alpha z_n+\beta w_n). \end{align*} Therefore the $j$th coordinate is $\alpha z_j+\beta w_j$, and \begin{align*} \tau_j(\alpha z+\beta w)=\alpha z_j+\beta w_j=\alpha\tau_j(z)+\beta\tau_j(w). \end{align*} Since multiplication is coordinatewise, \begin{align*} zw=(z_1w_1,\dots,z_nw_n). \end{align*} Hence \begin{align*} \tau_j(zw)=z_jw_j=\tau_j(z)\tau_j(w). \end{align*} The identity element is $(1,\dots,1)$, so \begin{align*} \tau_j(1,\dots,1)=1. \end{align*} Thus $\tau_j$ is a non-zero algebra homomorphism, so each coordinate projection is a character. Conversely, let $\tau:\mathbb C^n\to\mathbb C$ be a character. Since $\tau$ is non-zero, choose $z\in\mathbb C^n$ with $\tau(z)\ne 0$. Multiplicativity gives \begin{align*} \tau(1)\tau(z)=\tau(1z)=\tau(z). \end{align*} Subtracting $\tau(z)$ from both sides gives \begin{align*} (\tau(1)-1)\tau(z)=0. \end{align*} Because $\tau(z)\ne 0$, it follows that $\tau(1)=1$. Let $e_j$ be the vector with $1$ in the $j$th coordinate and $0$ in every other coordinate. Since $e_j^2=e_j$, multiplicativity gives \begin{align*} \tau(e_j)^2=\tau(e_j)\tau(e_j)=\tau(e_j^2)=\tau(e_j). \end{align*} Subtracting $\tau(e_j)$ from both sides gives \begin{align*} \tau(e_j)^2-\tau(e_j)=0. \end{align*} Factoring gives \begin{align*} \tau(e_j)(\tau(e_j)-1)=0. \end{align*} Thus $\tau(e_j)=0$ or $\tau(e_j)=1$. If $i\ne j$, then $e_ie_j=0$, and therefore \begin{align*} \tau(e_i)\tau(e_j)=\tau(e_ie_j)=\tau(0)=0. \end{align*} So two distinct values $\tau(e_i)$ and $\tau(e_j)$ cannot both be $1$. On the other hand, \begin{align*} 1=e_1+\cdots+e_n. \end{align*} Using linearity and $\tau(1)=1$, \begin{align*} 1=\tau(1)=\tau(e_1+\cdots+e_n)=\tau(e_1)+\cdots+\tau(e_n). \end{align*} Each summand is either $0$ or $1$, and at most one summand is $1$, so exactly one index $k$ satisfies $\tau(e_k)=1$. Every $z=(z_1,\dots,z_n)$ has the coordinate decomposition \begin{align*} z=z_1e_1+\cdots+z_ne_n. \end{align*} Applying linearity gives \begin{align*} \tau(z)=\tau(z_1e_1+\cdots+z_ne_n)=z_1\tau(e_1)+\cdots+z_n\tau(e_n). \end{align*} Since $\tau(e_k)=1$ and $\tau(e_j)=0$ for $j\ne k$, the sum reduces to \begin{align*} \tau(z)=z_k. \end{align*} But $\tau_k(z)=z_k$ by definition, so $\tau(z)=\tau_k(z)$ for every $z\in\mathbb C^n$. Hence $\tau=\tau_k$, and the characters are exactly the coordinate maps $\tau_1,\dots,\tau_n$. The weak-* topology is discrete on this finite character space. For each fixed $j$, the map $\tau\mapsto\tau(e_j)$ is weak-* continuous because it is evaluation at the fixed element $e_j\in\mathbb C^n$. It satisfies \begin{align*} \tau_j(e_j)=1. \end{align*} If $m\ne j$, then the $j$th coordinate of $e_m$ is $0$, so \begin{align*} \tau_m(e_j)=0. \end{align*} Therefore a character $\tau_m$ satisfies $|\tau_m(e_j)-1|<1/2$ exactly when $m=j$, and hence \begin{align*} \{\tau_j\}=\{\tau\in\Delta(A):|\tau(e_j)-1|<1/2\}. \end{align*} The set on the right is weak-* open in $\Delta(A)$, so every singleton is open. Thus $\Delta(A)=\{\tau_1,\dots,\tau_n\}$ with the discrete topology. Under the identification $\tau_j\leftrightarrow j$, the Gelfand transform sends $z=(z_1,\dots,z_n)$ to the function \begin{align*} \hat z(j)=\hat z(\tau_j)=\tau_j(z)=z_j. \end{align*} It preserves multiplication coordinate by coordinate: \begin{align*} \widehat{zw}(j)=\tau_j(zw)=z_jw_j=\hat z(j)\hat w(j). \end{align*} It also preserves involution, because $z^*=(\overline{z_1},\dots,\overline{z_n})$, so \begin{align*} \widehat{z^*}(j)=\tau_j(\overline{z_1},\dots,\overline{z_n})=\overline{z_j}=\overline{\hat z(j)}. \end{align*} Thus the Gelfand transform is the usual $*$-isomorphism $\mathbb C^n\cong C(\{1,\dots,n\})$, and the finite-dimensional algebra is recovered as continuous functions on its finite discrete character space. [/example] ## Commutative $C^*$-Algebras as Function Algebras For a general commutative Banach algebra, the Gelfand transform need not be injective or surjective. What extra force does the C*-identity provide? It turns the Gelfand transform into an isometric $*$-isomorphism. [quotetheorem:8550] [citeproof:8550] The C*-hypothesis is again indispensable: a general Banach *-algebra can have characters that do not interact correctly with the involution. The theorem does not say that every algebra homomorphism between C*-algebras is automatically geometric; the dual-space picture becomes functorial only for the appropriate continuous maps between spectra. This result is the final ingredient needed for the Gelfand transform to land in a self-adjoint function algebra, where Stone-Weierstrass can apply. At this point the Gelfand transform has turned a commutative $C^*$-algebra into functions on its character space, but a classification theorem must still rule out missing functions and lost norm information. The key question is whether every commutative $C^*$-algebra is completely recovered from that space, including its $*$-operation and norm. The following [commutative Gelfand-Naimark theorem](/theorems/2689) gives exactly that recovery, with locally compact Hausdorff spaces appearing contravariantly. [quotetheorem:2689] [citeproof:2689] The theorem is strongest when read as a classification statement, but its hypotheses should not be blurred: noncommutative C*-algebras are not generally algebras of scalar-valued functions, and commutative Banach algebras without the C*-identity need not be recovered isometrically from their characters. It also reverses arrows: a proper continuous map $Y\to X$ induces a $*$-homomorphism $C_0(X)\to C_0(Y)$ by composition, so the algebraic direction runs opposite to the topological direction. This contravariance is the precise form of Gelfand duality and is the template for later noncommutative topology, where a general C*-algebra is treated as if it were functions on a virtual space. This theorem turns examples into classification statements. To understand a commutative $C^*$-algebra, find its character space; after that the algebra is $C_0$ of that space. [example: Continuous Functions on an Interval] Let $A=C([0,1])$. For each $x\in[0,1]$, define $\tau_x(f)=f(x)$. If $f,g\in C([0,1])$ and $\alpha,\beta\in\mathbb C$, then pointwise linear operations give \begin{align*} \tau_x(\alpha f+\beta g)=(\alpha f+\beta g)(x)=\alpha f(x)+\beta g(x)=\alpha\tau_x(f)+\beta\tau_x(g). \end{align*} Pointwise multiplication gives \begin{align*} \tau_x(fg)=(fg)(x)=f(x)g(x)=\tau_x(f)\tau_x(g). \end{align*} Also $\tau_x(1)=1(x)=1$, so $\tau_x$ is non-zero. Hence every point $x\in[0,1]$ gives a character on $C([0,1])$. Conversely, let $\tau:C([0,1])\to\mathbb C$ be a character and set $M=\ker\tau$. Since $\tau$ is non-zero, choose $g\in C([0,1])$ with $\tau(g)\ne 0$. Multiplicativity gives \begin{align*} \tau(1)\tau(g)=\tau(1g)=\tau(g). \end{align*} Subtracting $\tau(g)$ from both sides gives \begin{align*} (\tau(1)-1)\tau(g)=0. \end{align*} Since $\tau(g)\ne 0$, it follows that $\tau(1)=1$. Therefore, for every $\lambda\in\mathbb C$, \begin{align*} \tau(\lambda 1)=\lambda\tau(1)=\lambda. \end{align*} The map $\Phi:C([0,1])/M\to\mathbb C$ defined by $\Phi(f+M)=\tau(f)$ is well-defined. Indeed, if $f+M=g+M$, then $f-g\in M$, so \begin{align*} 0=\tau(f-g)=\tau(f)-\tau(g). \end{align*} Thus $\tau(f)=\tau(g)$. The map is injective because if $\Phi(f+M)=0$, then $\tau(f)=0$, hence $f\in M$ and $f+M=M$. It is surjective because, for every $\lambda\in\mathbb C$, \begin{align*} \Phi(\lambda 1+M)=\tau(\lambda 1)=\lambda. \end{align*} Thus $C([0,1])/M\cong\mathbb C$, so $M$ is maximal. We now identify $M$. Suppose first that no point $x\in[0,1]$ has the property that every $f\in M$ vanishes at $x$. Then for each $x\in[0,1]$ there is $f_x\in M$ with $f_x(x)\ne 0$. Define \begin{align*} U_x=\{y\in[0,1]:f_x(y)\ne 0\}. \end{align*} Since $f_x$ is continuous and $\mathbb C\setminus\{0\}$ is open, $U_x$ is open, and $x\in U_x$. The sets $U_x$ cover $[0,1]$, so compactness gives $x_1,\dots,x_n\in[0,1]$ such that \begin{align*} [0,1]=U_{x_1}\cup\cdots\cup U_{x_n}. \end{align*} Define \begin{align*} h=\sum_{j=1}^n f_{x_j}\overline{f_{x_j}}. \end{align*} For each $j$, the function $\overline{f_{x_j}}$ belongs to $C([0,1])$. Since $f_{x_j}\in M$ and $M$ is an ideal, $f_{x_j}\overline{f_{x_j}}\in M$. Since $M$ is closed under finite sums, $h\in M$. For any $y\in[0,1]$, choose $j$ with $y\in U_{x_j}$. Then $f_{x_j}(y)\ne 0$, and \begin{align*} h(y)=\sum_{k=1}^n f_{x_k}(y)\overline{f_{x_k}(y)}=\sum_{k=1}^n |f_{x_k}(y)|^2\ge |f_{x_j}(y)|^2>0. \end{align*} Thus $h$ has no zeros on $[0,1]$. Since $h$ is continuous and never zero, $1/h\in C([0,1])$. The ideal property gives \begin{align*} 1=h(1/h)\in M. \end{align*} But $1\in M=\ker\tau$ would imply $\tau(1)=0$, contradicting $\tau(1)=1$. Hence there is some $x\in[0,1]$ such that $f(x)=0$ for every $f\in M$. For this point $x$, every $f\in M$ satisfies $\tau_x(f)=f(x)=0$, so $M\subseteq\ker\tau_x$. The ideal $\ker\tau_x$ is proper because $\tau_x(1)=1$. Since $M$ is maximal and \begin{align*} M\subseteq\ker\tau_x\subsetneq C([0,1]), \end{align*} we have $M=\ker\tau_x$. Now let $f\in C([0,1])$. The function $f-f(x)1$ satisfies \begin{align*} (f-f(x)1)(x)=f(x)-f(x)\cdot 1=0, \end{align*} so $f-f(x)1\in\ker\tau_x=M$. Applying $\tau$ gives \begin{align*} 0=\tau(f-f(x)1)=\tau(f)-\tau(f(x)1)=\tau(f)-f(x)\tau(1)=\tau(f)-f(x). \end{align*} Therefore $\tau(f)=f(x)=\tau_x(f)$ for every $f\in C([0,1])$, so $\tau=\tau_x$. Distinct points give distinct characters. If $x\ne y$, define $p(t)=t-x$. Then $p\in C([0,1])$ and \begin{align*} p(x)=x-x=0. \end{align*} Also \begin{align*} p(y)=y-x\ne 0. \end{align*} Thus $\tau_x(p)=0$ and $\tau_y(p)\ne 0$, so $\tau_x\ne\tau_y$. The map $x\mapsto\tau_x$ is therefore a bijection from $[0,1]$ onto $\Delta(C([0,1]))$. It is continuous for the weak-* topology: if $x_i\to x$ in $[0,1]$ and $f\in C([0,1])$ is fixed, then continuity of $f$ gives \begin{align*} \tau_{x_i}(f)=f(x_i)\to f(x)=\tau_x(f). \end{align*} The interval $[0,1]$ is compact, and the weak-* topology on $\Delta(C([0,1]))$ is Hausdorff as a subspace of $C([0,1])^*$. Therefore this continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Under this identification, the Gelfand transform satisfies \begin{align*} \hat f(\tau_x)=\tau_x(f)=f(x). \end{align*} So the Gelfand transform sends each [continuous function](/page/Continuous%20Function) to the same function, now regarded as a function on the character space. [/example] The locally compact non-compact case is equally important. Functions vanish at infinity, and the missing point is exactly what distinguishes $C_0(X)$ from $C(X^+)$ on the one-point compactification. [example: Continuous Functions Vanishing at Infinity on the Line] Let $A=C_0(\mathbb R)$. For each $t\in\mathbb R$, define $\tau_t(f)=f(t)$. If $f,g\in C_0(\mathbb R)$ and $\alpha,\beta\in\mathbb C$, then pointwise linear operations give \begin{align*} \tau_t(\alpha f+\beta g)=(\alpha f+\beta g)(t)=\alpha f(t)+\beta g(t)=\alpha\tau_t(f)+\beta\tau_t(g). \end{align*} Pointwise multiplication gives \begin{align*} \tau_t(fg)=(fg)(t)=f(t)g(t)=\tau_t(f)\tau_t(g). \end{align*} To see that $\tau_t$ is non-zero, set $f_t(s)=e^{-(s-t)^2}$. This function is continuous. If $0<\varepsilon<1$, choose $R>|t|+\sqrt{\log(1/\varepsilon)}$. Then $|s|>R$ implies \begin{align*} |s-t|\ge |s|-|t|>R-|t|>\sqrt{\log(1/\varepsilon)}. \end{align*} Squaring gives $(s-t)^2>\log(1/\varepsilon)$, so \begin{align*} |f_t(s)|=e^{-(s-t)^2}<e^{-\log(1/\varepsilon)}=\varepsilon. \end{align*} If $\varepsilon\ge 1$, choose $R>|t|$; then $|s|>R$ implies $s\ne t$, hence $|f_t(s)|<1\le\varepsilon$. Thus $f_t\in C_0(\mathbb R)$, and \begin{align*} \tau_t(f_t)=f_t(t)=e^0=1. \end{align*} So each $\tau_t$ is a character. Conversely, let $\tau:C_0(\mathbb R)\to\mathbb C$ be a character. The unitization $C_0(\mathbb R)^+$ consists of elements $f+\lambda 1$ with multiplication \begin{align*} (f+\lambda 1)(g+\mu 1)=fg+\lambda g+\mu f+\lambda\mu 1. \end{align*} Define \begin{align*} \tau^+(f+\lambda 1)=\tau(f)+\lambda. \end{align*} For linearity, \begin{align*} \tau^+(\alpha(f+\lambda 1)+\beta(g+\mu 1))=\tau(\alpha f+\beta g)+\alpha\lambda+\beta\mu=\alpha\tau(f)+\beta\tau(g)+\alpha\lambda+\beta\mu. \end{align*} The right-hand side equals \begin{align*} \alpha(\tau(f)+\lambda)+\beta(\tau(g)+\mu)=\alpha\tau^+(f+\lambda 1)+\beta\tau^+(g+\mu 1). \end{align*} For multiplicativity, \begin{align*} \tau^+((f+\lambda 1)(g+\mu 1))=\tau(fg+\lambda g+\mu f)+\lambda\mu=\tau(fg)+\lambda\tau(g)+\mu\tau(f)+\lambda\mu. \end{align*} Since $\tau$ is multiplicative on $C_0(\mathbb R)$, \begin{align*} \tau(fg)+\lambda\tau(g)+\mu\tau(f)+\lambda\mu=\tau(f)\tau(g)+\lambda\tau(g)+\mu\tau(f)+\lambda\mu. \end{align*} Expanding the scalar product gives \begin{align*} (\tau(f)+\lambda)(\tau(g)+\mu)=\tau(f)\tau(g)+\tau(f)\mu+\lambda\tau(g)+\lambda\mu. \end{align*} Therefore $\tau^+((f+\lambda 1)(g+\mu 1))=\tau^+(f+\lambda 1)\tau^+(g+\mu 1)$. Also \begin{align*} \tau^+(1)=\tau^+(0+1)=\tau(0)+1=1. \end{align*} Thus $\tau^+$ is a unital character on $C_0(\mathbb R)^+$. Identify $C_0(\mathbb R)^+$ with $C(\mathbb R^+)$, where $\mathbb R^+=\mathbb R\cup\{\infty\}$ is the one-point compactification, by sending $f+\lambda 1$ to the function $F$ satisfying $F(t)=f(t)+\lambda$ for $t\in\mathbb R$ and $F(\infty)=\lambda$. Continuity at $\infty$ follows because $f\in C_0(\mathbb R)$ means $f(t)\to 0$ as $|t|\to\infty$. By the compact continuous-function algebra computation above, the unital character $\tau^+$ is evaluation at some point of $\mathbb R^+$. It cannot be evaluation at $\infty$, because then for every $f\in C_0(\mathbb R)$, \begin{align*} \tau(f)=\tau^+(f+0)=f(\infty)=0, \end{align*} contradicting that $\tau$ is non-zero. Hence there is $t\in\mathbb R$ such that, for every $f\in C_0(\mathbb R)$, \begin{align*} \tau(f)=\tau^+(f+0)=f(t)=\tau_t(f). \end{align*} Therefore the characters of $C_0(\mathbb R)$ are exactly the evaluations $\tau_t$. The map $t\mapsto\tau_t$ is continuous for the weak-* topology: if $t_i\to t$ in $\mathbb R$ and $f\in C_0(\mathbb R)$ is fixed, then continuity of $f$ gives \begin{align*} \tau_{t_i}(f)=f(t_i)\to f(t)=\tau_t(f). \end{align*} Its inverse is also continuous. Suppose $\tau_{t_i}\to\tau_t$ weak-* and let $V$ be an open neighbourhood of $t$. Choose $\delta>0$ with $[t-\delta,t+\delta]\subset V$, and define \begin{align*} h(s)=\max\{0,1-|s-t|/\delta\}. \end{align*} Then $h$ is continuous, $h(t)=1$, and $\operatorname{supp}(h)\subset [t-\delta,t+\delta]\subset V$, so $h\in C_c(\mathbb R)\subset C_0(\mathbb R)$. Weak-* convergence gives \begin{align*} h(t_i)=\tau_{t_i}(h)\to\tau_t(h)=h(t)=1. \end{align*} Therefore eventually $|h(t_i)-1|<1/2$. For those $i$, $h(t_i)>1/2$, so $h(t_i)\ne 0$, hence $t_i\in\operatorname{supp}(h)\subset V$. Thus $t_i\to t$, and $\Delta(C_0(\mathbb R))$ is homeomorphic to $\mathbb R$. Under this identification, the Gelfand transform is \begin{align*} \hat f(\tau_t)=\tau_t(f)=f(t). \end{align*} Thus $C_0(\mathbb R)$ is recovered as the functions vanishing at infinity on the locally compact character space $\mathbb R$, while adjoining a unit adds exactly the point $\infty$ and gives $C(\mathbb R^+)$. [/example] ## Ideals and Open Subsets If commutative C*-algebras are function algebras, what do their closed ideals represent? For $C_0(X)$, an ideal is controlled by where its functions are allowed to be non-zero. [definition: Ideal Supported in an Open Set] Let $X$ be a locally compact Hausdorff space and let $U\subset X$ be open. Define \begin{align*} I_U=\{f\in C_0(X): f(x)=0\text{ for all }x\in X\setminus U\}. \end{align*} [/definition] This definition captures the idea of functions living on $U$ but extended by zero to all of $X$. The obstruction is that an arbitrary closed ideal is described algebraically, while an open subset is described by where functions are allowed to be non-zero. The following theorem proves that, in the commutative $C^*$-setting, these two descriptions coincide: every closed ideal is exactly the collection of functions supported in some open set. [quotetheorem:8551] [citeproof:8551] Closedness of the ideal is essential: non-closed ideals of $C_0(X)$ can be dense in a closed ideal and therefore are not determined by the same open subset. The theorem also classifies only closed ideals in the commutative case; for noncommutative C*-algebras, closed two-sided ideals remain important but no longer literally correspond to open subsets of an ordinary space. The closed complement $X\setminus U$ is the common zero set of the ideal, so the result is the hull-kernel viewpoint in concrete form. Equivalently, the theorem can be read as a hull-kernel correspondence: in the commutative $C^*$-setting, ideal theory is topology. [example: Ideals in $C(I)$ for the Unit Interval] Let $F\subset[0,1]$ be closed and set $U=[0,1]\setminus F$. Since $[0,1]\setminus U=F$, the ideal attached to $U$ is \begin{align*} I_U=\{f\in C([0,1]): f(x)=0\text{ for all }x\in F\}. \end{align*} This is the same as \begin{align*} I_U=\{f\in C([0,1]): f|_F=0\}. \end{align*} For the point $1/2$, take $F=\{1/2\}$ and $U=[0,1]\setminus\{1/2\}$. Then \begin{align*} I_U=\{f\in C([0,1]): f(1/2)=0\}. \end{align*} Define the evaluation map $\varepsilon_{1/2}:C([0,1])\to\mathbb C$ by $\varepsilon_{1/2}(f)=f(1/2)$. If $f,g\in C([0,1])$ and $\alpha,\beta\in\mathbb C$, then \begin{align*} \varepsilon_{1/2}(\alpha f+\beta g)=(\alpha f+\beta g)(1/2)=\alpha f(1/2)+\beta g(1/2)=\alpha\varepsilon_{1/2}(f)+\beta\varepsilon_{1/2}(g). \end{align*} Also, \begin{align*} \varepsilon_{1/2}(fg)=(fg)(1/2)=f(1/2)g(1/2)=\varepsilon_{1/2}(f)\varepsilon_{1/2}(g). \end{align*} Thus $\varepsilon_{1/2}$ is an algebra homomorphism, and its kernel is \begin{align*} \ker\varepsilon_{1/2}=\{f\in C([0,1]):\varepsilon_{1/2}(f)=0\}=\{f\in C([0,1]):f(1/2)=0\}=I_U. \end{align*} The map is onto because, for every $\lambda\in\mathbb C$, the constant function $\lambda 1$ satisfies \begin{align*} \varepsilon_{1/2}(\lambda 1)=(\lambda 1)(1/2)=\lambda. \end{align*} Therefore the quotient by the ideal of functions vanishing at $1/2$ is identified with $\mathbb C$ by \begin{align*} f+I_U\mapsto f(1/2). \end{align*} So closed subsets record common zero conditions, while the complementary open set records where functions in the ideal may be non-zero. [/example] Combining this ideal correspondence with the Gelfand-Naimark theorem gives the first instance of noncommutative topology. A general C*-algebra may not be functions on a space, but its closed two-sided ideals are still treated as analogues of open subsets. Later chapters use this analogy for representations, quotients, approximate units, and AF algebras. The commutative theory shows that spectral data can be read geometrically, but for general C*-algebras one needs more than spectra alone. Positivity and functional calculus supply that extra structure, letting one treat self-adjoint and normal elements as if they were functions of a real or complex variable. # 3. Positive Elements and Functional Calculus This chapter turns the abstract spectral theory from Banach algebras into the order and calculus structure that makes C*-algebras useful. The guiding idea is that a normal element of a C*-algebra behaves like a continuous function on its spectrum, so algebraic questions about $a$ can be translated into pointwise questions about functions on $\sigma(a)$. We use this to define positive elements intrinsically, construct square roots, and explain how absolute values and polar decompositions arise in concrete operator algebras. ## Classes of Elements Distinguished by the Involution The involution in a C*-algebra is more than an algebraic symmetry: it singles out the elements that should be thought of as real-valued, non-negative, unit-modulus, or idempotent functions. The problem in this section is to recognise these classes without already having a representation on a Hilbert space. [definition: Self-Adjoint Element] Let $A$ be a C*-algebra. An element $a \in A$ is self-adjoint if \begin{align*} a^* = a. \end{align*} [/definition] Self-adjoint elements play the role of real-valued functions. In $C(X)$ with pointwise operations and complex conjugation as involution, the condition $f^* = f$ means precisely that $f(x) \in \mathbb R$ for every $x \in X$. [example: Self-Adjoint Matrices] Let $a\in M_n(\mathbb C)$ be self-adjoint, so $a^*=a$; in matrix language this says $a$ is Hermitian. If $\lambda\in\sigma(a)$, choose a nonzero vector $\xi$ with $a\xi=\lambda\xi$. Then \begin{align*} \lambda\|\xi\|^2=\langle a\xi,\xi\rangle \end{align*} and, using $a^*=a$, \begin{align*} \langle a\xi,\xi\rangle=\langle \xi,a\xi\rangle=\langle \xi,\lambda\xi\rangle=\overline{\lambda}\|\xi\|^2. \end{align*} Since $\xi\ne0$, we have $\|\xi\|^2>0$, so $\lambda=\overline{\lambda}$ and hence $\lambda\in\mathbb R$. By the *finite-dimensional spectral theorem*, there is a unitary matrix $u$ and [real numbers](/page/Real%20Numbers) $\lambda_1,\dots,\lambda_n$ such that \begin{align*} a=udu^* \end{align*} where \begin{align*} d=\operatorname{diag}(\lambda_1,\dots,\lambda_n). \end{align*} For a function $f$ on $\sigma(a)$, the corresponding diagonal operation is \begin{align*} f(d)=\operatorname{diag}(f(\lambda_1),\dots,f(\lambda_n)). \end{align*} Thus the finite-dimensional functional calculus is \begin{align*} f(a)=uf(d)u^*. \end{align*} This is the matrix model for the later abstract functional calculus: once $a$ is diagonalised, functions of $a$ are obtained by applying the scalar function to the spectral values. [/example] Once self-adjoint elements have been identified as the real part of the algebra, the next question is which of them should count as non-negative. The spectral answer is the one that survives in every C*-algebra. [definition: Positive Element] Let $A$ be a C*-algebra. An element $a \in A$ is positive, written $a \ge 0$, if $a$ is self-adjoint and \begin{align*} \sigma(a) \subset [0,\infty). \end{align*} The set of positive elements of $A$ is denoted $A_+$. [/definition] In concrete Hilbert space language, positivity means $(a\xi,\xi)_H \ge 0$ for every $\xi \in H$. The spectral definition is designed so that the same condition can be transported to abstract C*-algebras using functional calculus. [example: Positive Functions] Let $X$ be compact Hausdorff and let $A=C(X)$ with pointwise operations and involution $f^*(x)=\overline{f(x)}$. We show that $f$ is positive in the C*-algebraic sense exactly when $f(x)\ge 0$ for every $x\in X$. First suppose $f\ge 0$ in $C(X)$. By definition, $f=f^*$ and $\sigma(f)\subset[0,\infty)$. The equality $f=f^*$ means that for every $x\in X$, \begin{align*} f(x)=f^*(x)=\overline{f(x)}. \end{align*} Hence $f(x)\in\mathbb R$. Also, for each $x\in X$, the value $f(x)$ belongs to $\sigma(f)$: if $f(x)=\lambda$ and $f-\lambda 1$ had an inverse $g\in C(X)$, then evaluating $(f-\lambda 1)g=1$ at $x$ would give \begin{align*} (f(x)-\lambda)g(x)=1, \end{align*} so $0=1$, impossible. Therefore $f(x)\in\sigma(f)\subset[0,\infty)$, so $f(x)\ge0$. Conversely, suppose $f(x)\ge0$ for every $x\in X$. Then $f(x)\in\mathbb R$ for every $x$, so \begin{align*} f^*(x)=\overline{f(x)}=f(x). \end{align*} Thus $f=f^*$. It remains to check that $\sigma(f)\subset[0,\infty)$. If $\lambda\notin f(X)$, then $f(t)-\lambda\ne0$ for every $t\in X$, and the function \begin{align*} g(t)=\frac{1}{f(t)-\lambda} \end{align*} is continuous because $t\mapsto f(t)-\lambda$ is continuous and never zero. Pointwise multiplication gives \begin{align*} ((f-\lambda 1)g)(t)=(f(t)-\lambda)\frac{1}{f(t)-\lambda}=1. \end{align*} Similarly $g(f-\lambda 1)=1$, so $f-\lambda 1$ is invertible. Hence every spectral value of $f$ lies in $f(X)$, and since $f(X)\subset[0,\infty)$, we get $\sigma(f)\subset[0,\infty)$. Thus positivity in $C(X)$ is exactly ordinary pointwise non-negativity. [/example] The previous example shows how positivity is read pointwise in a commutative C*-algebra. To prepare for the same pointwise reading of other algebraic relations, we now name the elements that model reversible symmetries, yes-or-no measurements, and compatibility with the involution. This subsection is unital because unitary elements require a specified identity; projections and normal elements also make sense in non-unital C*-algebras, so the unital assumption here is local to the surrounding discussion rather than part of their intrinsic meaning. [definition: Unitary Projection and Normal Element] Let $A$ be a unital C*-algebra. An element $u \in A$ is unitary if \begin{align*} u^*u = uu^* = 1. \end{align*} An element $p \in A$ is a projection if it is idempotent and self-adjoint: \begin{align*} p^2=p \end{align*} and \begin{align*} p^*=p. \end{align*} An element $a \in A$ is normal if \begin{align*} a^*a = aa^*. \end{align*} [/definition] Unitary elements correspond to functions with values in the unit circle, projections correspond to characteristic functions, and normal elements are the class for which $a$ and $a^*$ generate a commutative C*-algebra. This commutativity is the entrance point for Gelfand duality. [example: Projections as Characteristic Functions] Let $X$ be compact Hausdorff and let $p\in C(X)$ satisfy $p^2=p$ and $p^*=p$. For each $x\in X$, the involution in $C(X)$ gives \begin{align*} p(x)=p^*(x)=\overline{p(x)}. \end{align*} Thus $p(x)\in\mathbb R$. Evaluating $p^2=p$ at $x$ gives \begin{align*} p(x)^2=p(x). \end{align*} Hence \begin{align*} p(x)(p(x)-1)=0. \end{align*} Since $p(x)$ is a scalar, this implies $p(x)=0$ or $p(x)=1$. Therefore $p(X)\subset\{0,1\}$. Let \begin{align*} E=\{x\in X:p(x)=1\}. \end{align*} Then $p(x)=1$ on $E$ and $p(x)=0$ on $X\setminus E$, so $p=\mathbb{1}_E$. Since $p$ is continuous, \begin{align*} E=p^{-1}(\{1\}) \end{align*} is closed, and also \begin{align*} E=p^{-1}(\mathbb C\setminus\{0\}) \end{align*} is open. Thus $E$ is both closed and open. Conversely, if $E\subset X$ is both open and closed, then $\mathbb{1}_E:X\to\mathbb C$ is continuous: the preimage of a set containing $1$ but not $0$ is $E$, the preimage of a set containing $0$ but not $1$ is $X\setminus E$, the preimage of a set containing both is $X$, and the preimage of a set containing neither is $\varnothing$. Hence projections in $C(X)$ are exactly characteristic functions of clopen subsets of $X$. If $X$ is connected, the only clopen subsets are $\varnothing$ and $X$, so the only projections in $C(X)$ are $0$ and $1$. Thus projections in $C(X)$ record the clopen decomposition of the space: connected spaces have only trivial projections, while spaces with many clopen subsets have many nonzero proper projections. [/example] The preceding example shows that algebraic relations impose pointwise restrictions in commutative algebras: the equation $p^2=p=p^*$ forces each value of $p$ to be either $0$ or $1$. Outside the commutative model there are no point evaluations, so it is not obvious how such scalar restrictions should still be detected. The spectral viewpoint supplies the replacement: algebraic identities should force the spectrum of an element to lie in the corresponding scalar solution set, making spectral restrictions the noncommutative analogue of pointwise restrictions. The next obstruction is to make this principle independent of any chosen representation as functions. Before constructing functional calculus, we need a purely C*-algebraic guarantee that the familiar algebraic classes of elements have the spectra suggested by the scalar equations: self-adjoint elements should have real spectrum, positive elements should have nonnegative spectrum, unitaries should have spectrum on the unit circle, and projections should have spectrum contained in the two-point set $\{0,1\}$. [quotetheorem:8552] [citeproof:8552] These spectral restrictions make the coming functional calculus plausible: if an element satisfies an algebraic relation, its spectrum satisfies the corresponding scalar relation. The hypotheses matter. Unitality is used so that spectra are taken inside an algebra with an identity and so that relations such as $u^*u=1$ have their usual meaning; in non-unital algebras the same statements are read in the unitization, with care about the possible extra spectral point $0$. The projection condition also requires self-adjointness: an idempotent matrix can have spectrum contained in $\{0,1\}$ without being an orthogonal projection, so the spectrum alone does not capture the metric content of the involution. Conversely, these spectral restrictions do not classify elements uniquely; many non-conjugate self-adjoint elements have real spectrum, and a spectral subset such as $\{0,1\}$ records only part of the algebraic structure. ## Continuous Functional Calculus for Normal Elements The central question now is whether the function analogy can be made exact. Given a normal element $a$, can we build an isometric copy of $C(\sigma(a))$ inside $A$ in which the coordinate function $z\mapsto z$ is sent to $a$? [definition: C-Star Algebra Generated by a Normal Element] Let $A$ be a unital C*-algebra and let $a\in A$ be normal. The unital C*-subalgebra generated by $a$ is \begin{align*} C^*(a,1)=\bigcap\{B\subset A: B \text{ is a unital C*-subalgebra and } a\in B\}. \end{align*} [/definition] Normality is exactly what makes $C^*(a,1)$ commutative, since it is generated by the commuting elements $a$ and $a^*$. Once this generated algebra is commutative, the previous chapter suggests applying Gelfand duality to identify it with functions on a compact space. The point of the next theorem is that the compact space is not mysterious: it is precisely the spectrum of $a$. [example: Why Normality Is Needed] Let $n\in M_2(\mathbb C)$ be defined by $ne_1=0$ and $ne_2=e_1$. Then $n\ne0$, since $ne_2=e_1$. Also, \begin{align*} n^2e_1=n0=0 \end{align*} and \begin{align*} n^2e_2=ne_1=0, \end{align*} so $n^2=0$. Since $ne_1=0$ and $e_1\ne0$, the operator $n$ is not invertible, hence $0\in\sigma(n)$. If $\lambda\ne0$, then \begin{align*} (\lambda I-n)(\lambda^{-1}I+\lambda^{-2}n)=I+\lambda^{-1}n-\lambda^{-1}n-\lambda^{-2}n^2=I. \end{align*} The same multiplication in the reverse order gives \begin{align*} (\lambda^{-1}I+\lambda^{-2}n)(\lambda I-n)=I+\lambda^{-1}n-\lambda^{-1}n-\lambda^{-2}n^2=I. \end{align*} Thus $\lambda I-n$ is invertible for every $\lambda\ne0$, and therefore \begin{align*} \sigma(n)=\{0\}. \end{align*} The coordinate function on $\sigma(n)=\{0\}$ is $z\mapsto z$, but on the one-point set $\{0\}$ this is the zero function. If a continuous functional calculus on $\sigma(n)$ sent the coordinate function to $n$, it would send the zero function to the nonzero element $n$. That is impossible for any algebra homomorphism, because linearity forces the zero function to map to $0$. The obstruction is visible in the failure of normality. From the defining action of $n$, its adjoint satisfies $n^*e_1=e_2$ and $n^*e_2=0$. Hence \begin{align*} n^*ne_2=n^*e_1=e_2 \end{align*} whereas \begin{align*} nn^*e_2=n0=0. \end{align*} So $n^*n\ne nn^*$, and $n$ is not normal. The spectrum alone is too small to support the continuous functional calculus expected for normal elements. [/example] This failure explains why the theorem below is not stated for arbitrary elements. For non-normal elements, [holomorphic functional calculus](/theorems/2680) still exists, but it is a different and weaker construction; the continuous calculus belongs to the commutative C*-algebra generated by a normal element. [quotetheorem:8416] [citeproof:8416/operator-theory-ii-c-algebras] The Hilbert-space theorem should be read here as the concrete model for the construction, not as the abstract C*-algebra theorem by itself. It shows what continuous functional calculus looks like when a normal element is represented as an operator, and it explains why normality, continuity, and the spectrum are the right ingredients. To use the same notation intrinsically inside an arbitrary unital C*-algebra, we still need the abstract functional-calculus rules for the commutative algebra $C^*(a,1)$. These rules are what make $f(a)$ independent of a chosen representation and record exactly which algebraic, spectral, and norm properties survive. [quotetheorem:8553] [citeproof:8553] These rules are powerful because they allow algebraic identities about $f(a)$ to be checked pointwise on $\sigma(a)$, but this is still a theorem about normal elements and continuous functions. The spectral mapping formula here does not license arbitrary continuous functions of non-normal elements, and it does not say that every spectral relation can be lifted to a projection or discontinuous cutoff inside $A$. The isometry is also part of the conclusion: without the C*-norm and the involution, a homomorphism from functions need not preserve analytic size. With these limits in mind, functional calculus gives a clean intrinsic test for positivity. Earlier we defined positivity by spectral location, while in concrete operator theory positive operators arise as $b^*b$. The next theorem proves that these two ways of recognising positivity coincide in C*-algebras. [quotetheorem:8554] [citeproof:8554] In practice, the theorem lets us move freely between the order notation $a\ge0$, the factorisation $a=b^*b$, and the spectral condition $\sigma(a)\subset[0,\infty)$. The self-adjointness condition is not cosmetic: the nilpotent matrix $n\in M_2(\mathbb C)$ defined by $ne_1=0$ and $ne_2=e_1$ has spectrum $\{0\}\subset[0,\infty)$, but it is not self-adjoint and cannot be positive. Every expression $b^*b$ is automatically self-adjoint, so the theorem says that the spectral condition becomes an order condition only after self-adjointness has been imposed. In non-unital algebras the same criterion is interpreted in the unitization, and the factor $b$ still belongs to the original algebra when $a$ does. The factorisation also should not be read as saying that $b$ is a scalar square root of $a$; the order is noncommutative, and many different $b$ may satisfy $a=b^*b$ even though the positive square root is unique. [example: Spectral Projections for a Normal Matrix] Let $a\in M_n(\mathbb C)$ be normal, and let its distinct eigenvalues be $\lambda_1,\dots,\lambda_m$. By the *finite-dimensional spectral theorem*, choose a unitary matrix $u$ and a diagonal matrix $d=\operatorname{diag}(\mu_1,\dots,\mu_n)$ such that $a=udu^*$, where each $\mu_k$ is one of the $\lambda_j$. For $S\subset\{\lambda_1,\dots,\lambda_m\}$, define $f:\sigma(a)\to\mathbb C$ by $f(\lambda)=1$ if $\lambda\in S$ and $f(\lambda)=0$ otherwise. The finite-dimensional functional calculus gives \begin{align*} p=f(a)=uf(d)u^*. \end{align*} Since \begin{align*} f(d)=\operatorname{diag}(f(\mu_1),\dots,f(\mu_n)), \end{align*} each diagonal entry of $f(d)$ is either $0$ or $1$. Hence \begin{align*} f(d)^2=\operatorname{diag}(f(\mu_1)^2,\dots,f(\mu_n)^2)=\operatorname{diag}(f(\mu_1),\dots,f(\mu_n))=f(d). \end{align*} Therefore \begin{align*} p^2=(uf(d)u^*)(uf(d)u^*)=uf(d)(u^*u)f(d)u^*=uf(d)^2u^*=uf(d)u^*=p. \end{align*} Also, because the diagonal entries of $f(d)$ are real, \begin{align*} p^*=(uf(d)u^*)^*=uf(d)^*u^*=uf(d)u^*=p. \end{align*} Thus $p$ is a projection. To identify its range, write a vector as $\xi=u\eta$, where $\eta=(\eta_1,\dots,\eta_n)$. Then \begin{align*} p\xi=uf(d)\eta. \end{align*} The $k$th coordinate of $f(d)\eta$ is $f(\mu_k)\eta_k$, so coordinates with $\mu_k\notin S$ are killed and coordinates with $\mu_k\in S$ are unchanged. Hence $p$ is exactly the projection onto the direct sum of the eigenspaces whose eigenvalues lie in $S$. Since the spectrum is finite, every function on $\sigma(a)$ is continuous. More concretely, the same $f$ is represented on $\sigma(a)$ by the interpolation polynomial \begin{align*} q(z)=\sum_{\lambda_j\in S}\prod_{i\ne j}\frac{z-\lambda_i}{\lambda_j-\lambda_i}, \end{align*} because substituting $z=\lambda_k$ gives $q(\lambda_k)=1$ when $\lambda_k\in S$ and $q(\lambda_k)=0$ when $\lambda_k\notin S$. Thus these spectral projections are obtained by applying polynomial functions to the matrix $a$. [/example] The finite-dimensional example explains the terminology spectral projection. In infinite-dimensional C*-algebras, characteristic functions of arbitrary spectral subsets are usually not continuous, so continuous functional calculus produces projections only when the relevant subset is clopen in the spectrum. ## Square Roots Absolute Values and Polar Decomposition The last part of the chapter asks how much of Hilbert space operator theory can be recovered inside the C*-algebraic language. Positive square roots are fully internal to every C*-algebra; absolute values are internal as well; polar decompositions require a concrete operator algebra or an enlarged algebra in general. [quotetheorem:8555] [citeproof:8555] This theorem is a workhorse because it turns every element $a$ into a positive element $a^*a$. The positivity hypothesis is essential for uniqueness: in $M_2(\mathbb C)$ the identity matrix has many square roots, including $I$ and $-I$, but only $I$ is positive. Existence can also fail for non-normal elements in subalgebras where an algebraic square root is not present. Functional calculus constructs the distinguished positive square root only for positive elements, where the scalar function $t\mapsto t^{1/2}$ is continuous on the spectrum. Applied to $a^*a$, this square root measures the size of $a$ inside the algebra and gives the C*-algebraic version of modulus. This motivates the next definition. [definition: Absolute Value of an Element] Let $A$ be a C*-algebra and let $a\in A$. The absolute value of $a$ is \begin{align*} |a|=(a^*a)^{1/2}. \end{align*} [/definition] The notation matches the Hilbert space convention: $|a|$ is positive, and $|a|^2=a^*a$. It is not the norm of $a$; it is another element of the algebra. [example: Absolute Value of a Matrix] Let $a\in M_n(\mathbb C)$, and let \begin{align*} a=u\Sigma v^* \end{align*} be a [singular value decomposition](/theorems/3071), where $u$ and $v$ are unitary and \begin{align*} \Sigma=\operatorname{diag}(s_1,\dots,s_n) \end{align*} with $s_i\ge0$. First, \begin{align*} a^*=(u\Sigma v^*)^*=v\Sigma^*u^*=v\Sigma u^* \end{align*} because $\Sigma$ is real diagonal. Hence \begin{align*} a^*a=(v\Sigma u^*)(u\Sigma v^*)=v\Sigma(u^*u)\Sigma v^*=v\Sigma^2v^*. \end{align*} Thus $a^*a$ is unitarily diagonal with diagonal entries $s_1^2,\dots,s_n^2$, all non-negative, so it is positive. Now set \begin{align*} b=v\Sigma v^*. \end{align*} Then \begin{align*} b^*=(v\Sigma v^*)^*=v\Sigma^*v^*=v\Sigma v^*=b, \end{align*} and $b$ is unitarily diagonal with non-negative diagonal entries $s_1,\dots,s_n$, so $b$ is positive. Its square is \begin{align*} b^2=(v\Sigma v^*)(v\Sigma v^*)=v\Sigma(v^*v)\Sigma v^*=v\Sigma^2v^*=a^*a. \end{align*} Therefore, by the defining uniqueness of the positive square root, \begin{align*} |a|=(a^*a)^{1/2}=v\Sigma v^*. \end{align*} Since $|a|$ is unitarily diagonalized by $v$ with diagonal entries $s_1,\dots,s_n$, its eigenvalues are exactly the singular values of $a$, counted with multiplicity. [/example] The previous example shows that $|a|$ contains the singular-value part of a matrix while discarding the directional part. To reconstruct the original operator, we need a phase factor, but non-invertible operators may vanish on a kernel and therefore cannot have a unitary phase. The correct replacement is a partial isometry. [definition: Partial Isometry] Let $H$ be a Hilbert space. A bounded operator $v:H\to H$ is a partial isometry if $v$ is an isometry on $(\ker v)^\perp$ and vanishes on $\ker v$. [/definition] Equivalently, $v^*v$ is the projection onto the initial space $(\ker v)^\perp$, and $vv^*$ is the projection onto the final space $\overline{\operatorname{Range}(v)}$. This formulation is compatible with C*-algebraic calculations and is exactly what is needed to express the missing phase in $a=v|a|$. [quotetheorem:8401] [citeproof:8401] The kernel condition is part of the uniqueness statement, not an extra decoration. If $a$ vanishes on a subspace, the equation $a=v|a|$ does not determine arbitrary values of $v$ away from the initial space unless we require $v$ to vanish on the kernel. The theorem is stated for $B(H)$ because the partial isometry need not lie in an abstract C*-algebra containing $a$; in general it may live only in a multiplier algebra, bidual, or a chosen representation. Thus [polar decomposition](/theorems/3074) is less internal than the absolute value construction, and this distinction becomes important when multiplier algebras and biduals appear later in the course. [example: Polar Decomposition Can Leave the Algebra] Let $H$ be an infinite-dimensional Hilbert space with orthonormal basis $(e_n)_{n=1}^\infty$, and let $A=K(H)$. Define $a:H\to H$ by \begin{align*} ae_n=\frac{1}{n}e_n. \end{align*} For $\xi=\sum_{n=1}^\infty \xi_n e_n$, this gives \begin{align*} a\xi=\sum_{n=1}^\infty \frac{\xi_n}{n}e_n. \end{align*} Since $|\xi_n/n|^2\le |\xi_n|^2$ for every $n$, the series defines a bounded operator and $\|a\xi\|\le \|\xi\|$. The operator is compact because it is the norm limit of finite-rank diagonal truncations. For $N\ge 1$, define $a_Ne_n=n^{-1}e_n$ when $n\le N$, and define $a_Ne_n=0$ when $n>N$. Then $a_N$ has finite-dimensional range. If $\|\xi\|\le1$, then \begin{align*} \|(a-a_N)\xi\|^2=\sum_{n>N}\frac{|\xi_n|^2}{n^2}\le \frac{1}{(N+1)^2}\sum_{n>N}|\xi_n|^2\le \frac{1}{(N+1)^2}. \end{align*} Thus $\|a-a_N\|\le (N+1)^{-1}$, so $a\in K(H)$. The operator is self-adjoint and positive. Indeed, for $\xi=\sum \xi_n e_n$ and $\eta=\sum \eta_n e_n$, \begin{align*} \langle a\xi,\eta\rangle=\sum_{n=1}^\infty \frac{\xi_n\overline{\eta_n}}{n}=\langle \xi,a\eta\rangle. \end{align*} Also, \begin{align*} \langle a\xi,\xi\rangle=\sum_{n=1}^\infty \frac{|\xi_n|^2}{n}\ge0. \end{align*} If $a\xi=0$, then $\xi_n/n=0$ for every $n$, so $\xi_n=0$ for every $n$ and $\xi=0$. Hence $a$ is injective. The range is dense but not closed. It contains every finite linear combination of basis vectors, because \begin{align*} a(ne_n)=e_n. \end{align*} The finite linear combinations of the basis vectors are dense in $H$, so $\overline{\operatorname{Range}(a)}=H$. However, the vector \begin{align*} y=\sum_{n=1}^\infty \frac{1}{n}e_n \end{align*} belongs to $H$, since $\sum_{n=1}^\infty n^{-2}<\infty$. If $y=a\xi$ for some $\xi=\sum \xi_n e_n$, then comparing coefficients gives $\xi_n/n=1/n$, hence $\xi_n=1$ for every $n$. But \begin{align*} \sum_{n=1}^\infty |\xi_n|^2=\sum_{n=1}^\infty 1=\infty, \end{align*} so no such $\xi\in H$ exists. The partial sums of $y$ lie in $\operatorname{Range}(a)$ and converge to $y$, while $y\notin\operatorname{Range}(a)$; therefore the range is not closed. Since $a$ is positive, $a^*=a$ and \begin{align*} |a|=(a^*a)^{1/2}=(a^2)^{1/2}=a. \end{align*} The last equality uses uniqueness of the positive square root, because $a$ is positive and $a^2=a^*a$. Therefore, in $B(H)$, \begin{align*} a=I|a|. \end{align*} The partial isometry in the polar decomposition is $I$, and the kernel condition is satisfied because $\ker I=\{0\}=\ker a$. Finally, $I\notin K(H)$. If $I$ were compact, then the sequence $(Ie_n)=(e_n)$ would have a norm-convergent subsequence, because compact operators send bounded sequences to sequences with convergent subsequences. But for $m\ne n$, \begin{align*} \|e_m-e_n\|^2=\|e_m\|^2+\|e_n\|^2=2, \end{align*} so no subsequence of $(e_n)$ is Cauchy. Hence $I$ is not compact. This shows that the polar partial isometry can live in $B(H)$ while failing to belong to the original C*-algebra $K(H)$. [/example] The compact-operator example marks the boundary of the internal C*-algebraic construction: $|a|$ remains in $K(H)$, while the phase can live outside it. The next example shows the opposite extreme, where invertibility removes the kernel obstruction and the phase is recovered directly from $a$ and $|a|$. [example: Polar Decomposition of an Invertible Operator] Let $a\in B(H)$ be invertible, and write $b=|a|=(a^*a)^{1/2}$. Since $a$ is invertible, $a^*$ is invertible and \begin{align*} (a^*a)^{-1}=a^{-1}(a^*)^{-1}. \end{align*} Indeed, \begin{align*} (a^*a)a^{-1}(a^*)^{-1}=a^*(aa^{-1})(a^*)^{-1}=a^*(a^*)^{-1}=I. \end{align*} The reverse product is also $I$, so $b^2=a^*a$ is invertible. Because $b$ commutes with $b^2$, it also commutes with $(b^2)^{-1}$, and hence \begin{align*} b\bigl(b(b^2)^{-1}\bigr)=b^2(b^2)^{-1}=I \end{align*} and \begin{align*} \bigl(b(b^2)^{-1}\bigr)b=b^2(b^2)^{-1}=I. \end{align*} Thus $|a|=b$ is invertible. Define \begin{align*} v=a|a|^{-1}. \end{align*} Then \begin{align*} v|a|=a|a|^{-1}|a|=a, \end{align*} so this gives the polar factorization $a=v|a|$. Since $|a|$ is positive, it is self-adjoint, and its inverse is also self-adjoint. Therefore \begin{align*} v^*=(a|a|^{-1})^*=|a|^{-1}a^*. \end{align*} Now \begin{align*} v^*v=|a|^{-1}a^*a|a|^{-1}=|a|^{-1}|a|^2|a|^{-1}=I. \end{align*} Also, from $v^*v=I$ we get that $v$ is injective, and since $v=a|a|^{-1}$ is a product of invertible operators, $v$ is invertible. Multiplying $v^*v=I$ on the left by $v$ and on the right by $v^{-1}$ gives \begin{align*} vv^*=I. \end{align*} Thus $v$ is unitary. So when $a$ is invertible, the polar decomposition has no kernel part: the partial isometry is the unitary operator $a|a|^{-1}$, exactly like the phase $z/|z|$ of a nonzero complex number. [/example] Functional calculus therefore supplies the internal positive square root and absolute value constructions, while concrete Hilbert space theory supplies the polar decomposition. The rest of the course uses these tools constantly: positivity defines states, square roots control Cauchy-Schwarz inequalities, and projections encode the first noncommutative shadow of topological subsets. Positivity turns the algebraic calculus of the previous chapter into an order structure that can be tested by scalar-valued maps. States and positive functionals are the natural way to measure elements, and they convert internal positivity into inequalities on complex numbers. # 4. States, Positive Functionals, and Order States are the bridge between the ordered structure of a C*-algebra and the scalar world of complex numbers. In the preceding chapter, positivity was defined internally using spectra and square roots; here we ask how positivity can be detected by linear functionals. The main point is that normalized positive functionals behave like noncommutative probability measures, and their [convex geometry](/page/Convex%20Geometry) supplies the pure states that later generate irreducible representations through the GNS construction. ## Positive Functionals and States The first question is how a linear functional should respect the order on a C*-algebra. Since a positive element is the noncommutative analogue of a nonnegative function, a positive functional should assign nonnegative real numbers to such elements. [definition: Positive Linear Functional] Let $A$ be a C*-algebra. A linear functional $\rho:A\to\mathbb C$ is positive if \begin{align*} \rho(a)\ge 0 \end{align*} for every positive element $a\in A$. [/definition] The definition only mentions positive elements, but a linear functional on a complex algebra must also interact with adjoints and self-adjoint decompositions. Before using positive functionals as order-preserving objects, we need to know that they respect the involution; otherwise the scalar values of self-adjoint elements would not behave like real observables. [quotetheorem:8556] [citeproof:8556] This result says that positive functionals are compatible with the involution, so self-adjoint elements really are sent to real numbers. Positivity is essential here: a general complex-linear functional on a C*-algebra need not satisfy $\rho(a^*)=\overline{\rho(a)}$, for instance on $\mathbb C$ the map $z\mapsto iz$ is linear but not Hermitian. The theorem does not say that every Hermitian functional is positive; positivity is a stronger order condition. This Hermitian behaviour is exactly what makes the next normalization meaningful: for a positive functional on a unital algebra, the scalar $\rho(1)$ is a nonnegative real number, so it can play the role of total mass rather than merely a complex value at the unit. [definition: State] Let $A$ be a C*-algebra. A state on $A$ is a positive linear functional $\rho:A\to\mathbb C$ such that $\|\rho\|=1$. [/definition] The norm condition is the right definition in both unital and non-unital algebras, but in the unital case it should coincide with the more concrete condition of total mass one. Without such a formula, checking that a positive functional is a state would require estimating it on the whole unit ball, even in simple examples such as matrix algebras. Positivity is what makes the shortcut possible: an arbitrary bounded functional can have small or large value at $1$ without its norm being determined by that value. To use states efficiently, especially in examples, we need a formula that computes the norm of a positive functional from its value at the unit. [quotetheorem:8557] [citeproof:8557] The theorem makes states look like normalized positive measures: in a unital algebra, total mass one is exactly the same as norm one for positive functionals. The unit hypothesis matters because the expression $\rho(1)$ is unavailable in a non-unital algebra; for instance, on $C_0((0,1])$ the positive functional $f\mapsto f(1)$ has norm $1$, but there is no constant unit function inside the algebra on which to read off this mass. Positivity also matters: for a general bounded functional on a unital algebra, $\|\rho\|$ need not equal $|\rho(1)|$, and it may even happen that $\rho(1)=0$ while $\rho\ne 0$. The theorem does not classify all states; it only gives a practical normalization test. The basic examples show that this analogy is literal in commutative algebras and genuinely operator-theoretic in noncommutative ones. [example: Vector States on $B(H)$] Here $B(H)$ denotes the C*-algebra of bounded linear operators on the Hilbert space $H$; this is the same operator algebra also often written as $\mathcal L(H)$. Let $H$ be a Hilbert space and let $\xi\in H$ satisfy $\|\xi\|_H=1$. Define $\omega_\xi:B(H)\to\mathbb C$ by \begin{align*} \omega_\xi(T)=(T\xi,\xi)_H. \end{align*} This map is linear because, for $S,T\in B(H)$ and $\alpha,\beta\in\mathbb C$, \begin{align*} \omega_\xi(\alpha S+\beta T)=((\alpha S+\beta T)\xi,\xi)_H=(\alpha S\xi+\beta T\xi,\xi)_H=\alpha(S\xi,\xi)_H+\beta(T\xi,\xi)_H. \end{align*} If $T\ge 0$, then $T=R^*R$ for some $R\in B(H)$, so \begin{align*} \omega_\xi(T)=(R^*R\xi,\xi)_H=(R\xi,R\xi)_H=\|R\xi\|_H^2\ge 0. \end{align*} Thus $\omega_\xi$ is positive. To check the norm, let $\|T\|\le 1$. By the Hilbert-space [Cauchy-Schwarz inequality](/theorems/432), \begin{align*} |\omega_\xi(T)|=|(T\xi,\xi)_H|\le \|T\xi\|_H\|\xi\|_H\le \|T\|\|\xi\|_H^2\le 1. \end{align*} Hence $\|\omega_\xi\|\le 1$. On the other hand, \begin{align*} \omega_\xi(I)=(I\xi,\xi)_H=(\xi,\xi)_H=\|\xi\|_H^2=1, \end{align*} so $\|\omega_\xi\|\ge |\omega_\xi(I)|=1$. Therefore $\|\omega_\xi\|=1$, and $\omega_\xi$ is a state. These states record the expectation of an operator in the vector $\xi$. [/example] Vector states are more than a convenient source of examples: they are the model that the GNS construction will recover from an arbitrary state. Starting with an abstract state, the construction builds a Hilbert space and a representation in which the original state is realised as expectation in a cyclic vector. This is why the example matters structurally rather than only formally: it shows the target form for all states after representation. Matrix algebras also carry a canonical averaged state, whose behaviour contrasts with vector states by spreading mass evenly over an orthonormal basis. [example: Trace State on $M_n$] Let $A=M_n(\mathbb C)$ and define \begin{align*} \tau(a)=\frac{1}{n}\operatorname{Tr}(a). \end{align*} The trace is linear, so for $a,b\in M_n(\mathbb C)$ and $\alpha,\beta\in\mathbb C$, \begin{align*} \tau(\alpha a+\beta b)=\frac{1}{n}\operatorname{Tr}(\alpha a+\beta b)=\alpha\frac{1}{n}\operatorname{Tr}(a)+\beta\frac{1}{n}\operatorname{Tr}(b)=\alpha\tau(a)+\beta\tau(b). \end{align*} If $a\ge 0$, then $a$ is self-adjoint and the finite-dimensional spectral theorem gives a unitary $u$ and numbers $\lambda_1,\ldots,\lambda_n\ge 0$ such that \begin{align*} a=u^*\operatorname{diag}(\lambda_1,\ldots,\lambda_n)u. \end{align*} Using cyclicity of the trace, \begin{align*} \operatorname{Tr}(a)=\operatorname{Tr}(u^*\operatorname{diag}(\lambda_1,\ldots,\lambda_n)u)=\operatorname{Tr}(\operatorname{diag}(\lambda_1,\ldots,\lambda_n)uu^*)=\sum_{j=1}^n\lambda_j. \end{align*} Hence \begin{align*} \tau(a)=\frac{1}{n}\sum_{j=1}^n\lambda_j\ge 0, \end{align*} so $\tau$ is positive. Also, \begin{align*} \tau(I_n)=\frac{1}{n}\operatorname{Tr}(I_n)=\frac{1}{n}\cdot n=1. \end{align*} By the *Norm Formula for Positive Functionals*, a positive functional on a unital C*-algebra with value $1$ at the unit is a state, so $\tau$ is a state. This state is faithful. Indeed, if $a\ge 0$ and $\tau(a)=0$, then with the same spectral decomposition, \begin{align*} 0=\tau(a)=\frac{1}{n}\sum_{j=1}^n\lambda_j. \end{align*} Since every $\lambda_j$ is nonnegative, the equality $\sum_{j=1}^n\lambda_j=0$ forces $\lambda_j=0$ for every $j$. Therefore $\operatorname{diag}(\lambda_1,\ldots,\lambda_n)=0$, and \begin{align*} a=u^*0u=0. \end{align*} So the normalized trace detects nonzero positive matrices. [/example] The commutative example recovers probability theory. It is the reason states are often described as noncommutative probability measures. [example: Probability Measures as States on $C(X)$] Let $X$ be a compact Hausdorff space and let $\mu$ be a Borel probability measure on $X$, so $\mu(X)=1$. Define $\rho_\mu:C(X)\to\mathbb C$ by \begin{align*} \rho_\mu(f)=\int_X f\,d\mu. \end{align*} For $f,g\in C(X)$ and $\alpha,\beta\in\mathbb C$, linearity of the integral gives \begin{align*} \rho_\mu(\alpha f+\beta g)=\int_X(\alpha f+\beta g)\,d\mu=\alpha\int_X f\,d\mu+\beta\int_X g\,d\mu=\alpha\rho_\mu(f)+\beta\rho_\mu(g). \end{align*} Thus $\rho_\mu$ is linear. If $f\ge 0$ in $C(X)$, then $f(x)\ge 0$ for every $x\in X$, so monotonicity of integration gives \begin{align*} \rho_\mu(f)=\int_X f\,d\mu\ge 0. \end{align*} Hence $\rho_\mu$ is positive. It remains to check normalization. Since $1(x)=1$ for every $x\in X$, \begin{align*} \rho_\mu(1)=\int_X 1\,d\mu=\mu(X)=1. \end{align*} Equivalently, the norm can be read directly: if $\|f\|_\infty\le 1$, then $|f(x)|\le 1$ for every $x$, and the triangle inequality for the integral gives \begin{align*} |\rho_\mu(f)|=\left|\int_X f\,d\mu\right|\le \int_X |f|\,d\mu\le \int_X 1\,d\mu=1. \end{align*} Thus $\|\rho_\mu\|\le 1$, while $\|\rho_\mu\|\ge |\rho_\mu(1)|=1$, so $\|\rho_\mu\|=1$. Therefore $\rho_\mu$ is a state on $C(X)$. The *[Riesz representation theorem](/theorems/221)* states conversely that every state on $C(X)$ is integration against a unique Borel probability measure, so in the commutative case states are exactly probability measures. [/example] ## The Cauchy-Schwarz Inequality for Positive Functionals A state behaves like expectation, so the next question is whether it satisfies the same quadratic estimates as integration against a measure. The essential inequality is Cauchy-Schwarz, with $\rho(b^*a)$ playing the role of an inner product. [quotetheorem:8558] [citeproof:8558] This is the positive-functional analogue of the Hilbert-space Cauchy-Schwarz inequality. Positivity is indispensable: an arbitrary Hermitian linear functional need not make $\rho(b^*a)$ into a positive semidefinite form, so Cauchy-Schwarz can fail. For example, on $\mathbb C^2$ the Hermitian functional $\phi(z,w)=z-w$ is not positive; taking $a=(1,0)$ and $b=(0,1)$ gives $|\phi(b^*a)|^2=0$ while $\phi(a^*a)\phi(b^*b)=-1$, contradicting the asserted inequality. The inequality does not say that $\rho(b^*a)$ is an inner product on $A$, because elements with $\rho(a^*a)=0$ may be nonzero. Its importance here is that it gives continuity estimates from positivity alone and prepares the quotient construction used in GNS. [remark: Seminorm from a Positive Functional] For a positive functional $\rho$, the formula \begin{align*} |a|_\rho=\rho(a^*a)^{1/2} \end{align*} defines a seminorm on $A$. The null space $N_\rho=\{a\in A:|a|_\rho=0\}$ will become the subspace quotiented out in the GNS construction. [/remark] The inequality also explains why states interact well with the C*-norm. In particular, a state cannot see more than the operator norm permits. [example: Estimating Matrix Coefficients] Let $\rho$ be a state on a unital C*-algebra $A$, and fix $a\in A$. By the *Norm Formula for Positive Functionals*, $\rho(1)=1$. Applying the *Cauchy-Schwarz Inequality for Positive Functionals* with $b=1$ gives \begin{align*} |\rho(1^*a)|^2\le \rho(a^*a)\rho(1^*1). \end{align*} Since $1^*=1$ and $1^*1=1$, this becomes \begin{align*} |\rho(a)|^2\le \rho(a^*a)\rho(1)=\rho(a^*a). \end{align*} Now $a^*a$ is positive and $\|a^*a\|=\|a\|^2$ by the C*-identity. Hence the spectrum of $a^*a$ lies in $[0,\|a\|^2]$, so $\|a\|^2 1-a^*a\ge 0$. Positivity of $\rho$ gives \begin{align*} 0\le \rho(\|a\|^2 1-a^*a). \end{align*} By linearity and $\rho(1)=1$, \begin{align*} \rho(\|a\|^2 1-a^*a)=\|a\|^2\rho(1)-\rho(a^*a)=\|a\|^2-\rho(a^*a). \end{align*} Therefore $\rho(a^*a)\le \|a\|^2$, and the previous estimate gives \begin{align*} |\rho(a)|^2\le \|a\|^2. \end{align*} Taking nonnegative square roots yields $|\rho(a)|\le \|a\|$. Thus every state is contractive, and the estimate is obtained entirely from positivity and the order structure. [/example] ## The State Space as a Compact Convex Object The final question in this chapter is geometric: what does the collection of all states look like inside the dual Banach space $A^*$? Norm compactness would usually be too much to ask, just as probability measures on an infinite compact space are not compact in total variation norm. The [weak* topology](/page/Weak*%20Topology) is the right substitute because it records pointwise convergence on $A$, matching the way measures converge against test functions. The answer is that the state space is a compact convex set in the weak* topology, and its extreme points are the pure states. [definition: State Space] Let $A$ be a unital C*-algebra. The state space of $A$ is \begin{align*} S(A)=\{\rho\in A^*: \rho\ge 0,\ \rho(1)=1\}. \end{align*} [/definition] Convexity expresses the probabilistic idea of mixing states: if $\rho_1$ and $\rho_2$ are states and $0\le t\le 1$, then $t\rho_1+(1-t)\rho_2$ is again a state. Once mixtures are allowed, the natural next question is which states are indecomposable and therefore represent the noncommutative analogue of points rather than probability distributions. [definition: Pure State] Let $A$ be a unital C*-algebra. A state $\rho\in S(A)$ is pure if whenever \begin{align*} \rho=t\rho_1+(1-t)\rho_2 \end{align*} with $\rho_1,\rho_2\in S(A)$ and $0<t<1$, then $\rho_1=\rho_2=\rho$. [/definition] Pure states are the extreme points of the state space. They are the states that cannot be decomposed as proper statistical mixtures. [example: Pure States on C X] Let $X$ be a compact Hausdorff space and fix $x\in X$. Define $\delta_x:C(X)\to\mathbb C$ by \begin{align*} \delta_x(f)=f(x). \end{align*} For $f,g\in C(X)$ and $\alpha,\beta\in\mathbb C$, \begin{align*} \delta_x(\alpha f+\beta g)=(\alpha f+\beta g)(x)=\alpha f(x)+\beta g(x)=\alpha\delta_x(f)+\beta\delta_x(g), \end{align*} so $\delta_x$ is linear. If $f\ge 0$ in $C(X)$, then $f(y)\ge 0$ for every $y\in X$, and therefore \begin{align*} \delta_x(f)=f(x)\ge 0. \end{align*} Also, \begin{align*} \delta_x(1)=1(x)=1, \end{align*} so $\delta_x$ is a state. Under the *[Riesz representation](/theorems/67) theorem*, $\delta_x$ corresponds to the Dirac probability measure at $x$, denoted by $\mu_x$, where $\mu_x(E)=1$ if $x\in E$ and $\mu_x(E)=0$ if $x\notin E$. To see that $\delta_x$ is pure, suppose \begin{align*} \delta_x=t\rho_1+(1-t)\rho_2 \end{align*} with $\rho_1,\rho_2\in S(C(X))$ and $0<t<1$. Let $\mu_1,\mu_2$ be the probability measures corresponding to $\rho_1,\rho_2$. Evaluating on Borel sets through the representing measures gives \begin{align*} 1=\mu_x(\{x\})=t\mu_1(\{x\})+(1-t)\mu_2(\{x\}). \end{align*} Since $0\le \mu_1(\{x\})\le 1$ and $0\le \mu_2(\{x\})\le 1$, the equality above forces \begin{align*} \mu_1(\{x\})=1 \end{align*} and \begin{align*} \mu_2(\{x\})=1. \end{align*} Thus $\mu_1=\mu_2=\mu_x$, hence $\rho_1=\rho_2=\delta_x$. Conversely, if a probability measure $\mu$ on $X$ is not a Dirac measure, then by regularity there is a Borel set $E\subset X$ with $0<\mu(E)<1$. Put $s=\mu(E)$ and define probability measures \begin{align*} \mu_1(F)=\frac{\mu(F\cap E)}{s} \end{align*} and \begin{align*} \mu_2(F)=\frac{\mu(F\cap (X\setminus E))}{1-s}. \end{align*} For every Borel set $F$, \begin{align*} \mu(F)=\mu(F\cap E)+\mu(F\cap (X\setminus E))=s\mu_1(F)+(1-s)\mu_2(F). \end{align*} The measures $\mu_1$ and $\mu_2$ are different because $\mu_1(E)=1$ while $\mu_2(E)=0$. Therefore the state corresponding to $\mu$ is a nontrivial convex combination and is not pure. Hence the pure states on $C(X)$ are exactly the point evaluations $\delta_x$, so in the commutative case pure states recover the points of the underlying space. [/example] This example is the commutative prototype: pure states are points of the underlying space. In a noncommutative C*-algebra, pure states replace points and lead to irreducible representations. [quotetheorem:8559] [citeproof:8559] Compactness is useful because compact convex sets in locally convex spaces have extreme points, but each hypothesis has a visible role. The weak* topology matters here: Banach-Alaoglu supplies compactness in that topology, while norm compactness would fail for many infinite-dimensional examples such as $C(X)^*$ when $X$ is infinite. Unitality is also part of this formulation, since $S(A)$ is cut out by the affine equation $\rho(1)=1$; for a non-unital algebra the state space is instead normalized by $\|\rho\|=1$, or by passing to the unitization and imposing the corresponding restriction. The theorem does not say that every state is a finite convex combination of pure states; the closure in the weak* topology is essential, just as general probability measures are limits of convex combinations of point masses rather than always finite sums. The next step is therefore not merely to name the extreme points, but to prove that they exist. This requires two inputs at once: the compact convex geometry just established, and nonemptiness of the state space. The nonzero hypothesis rules out the zero algebra with its degenerate unit convention, where the affine condition $\rho(1)=1$ would be incompatible with linearity. Once nonemptiness is secured, Krein-Milman applies in the locally convex weak* topology and turns the abstract geometry of $S(A)$ into the existence and abundance of pure states. [quotetheorem:8560] [citeproof:8560] The theorem has two limitations worth keeping separate. First, the nonzero assumption is used to ensure that a normalized state can exist at all; without it, the scalar subalgebra used in the proof is not available in the same way. Second, Krein-Milman gives the weak* closed convex hull formula, not a finite decomposition theorem and not uniqueness of representing measures. Compactness, convexity, and local convexity are not decorative hypotheses: without compactness, extreme points may be absent, and outside locally convex spaces the usual separation tools behind Krein-Milman are unavailable. The proof of existence also used a hidden extension principle: a state defined on the scalar subalgebra had to be extended to the whole algebra. This is not a harmless formality, because ordinary Hahn-Banach preserves norm but not order in general; a norm-preserving extension of a positive functional must still be checked to be positive. More generally, when studying a C*-subalgebra $B\subset A$, we want states on $B$ to remain available after passing to $A$, otherwise information detected inside $B$ could disappear when the ambient algebra is enlarged. The following theorem supplies this compatibility and is the order-theoretic form of Hahn-Banach needed throughout the representation theory. [quotetheorem:8561] [citeproof:8561] The theorem says that states are plentiful enough to probe subalgebras without losing normalization. The shared-unit hypothesis is doing real work in this formulation: it is what turns the Hahn-Banach equality $\|\widetilde\rho\|=1$ into the concrete normalization $\widetilde\rho(1_A)=\rho(1_B)=1$. If the units are not shared, the same sentence no longer has a single meaning. For example, in $A=\mathbb C^2$ the subalgebra $B=\mathbb C(1,0)$ has unit $(1,0)$ rather than $1_A=(1,1)$; a state on $B$ is normalized at $(1,0)$, while the ambient state condition is normalization at $(1,1)$. This example does not rule out extensions, but it shows why the stated proof and normalization test require the same unit to be visible in both algebras. The theorem also does not claim uniqueness of extensions; even pure states may have several extensions to a larger algebra. In later chapters, this abundance lets us construct representations from chosen states and separate positive elements by evaluating them against states. The [extension theorem](/theorems/59) has a decisive order-theoretic consequence, and it answers a question left open by the internal spectral definition of positivity. Positivity of $a$ certainly forces $\rho(a)\ge 0$ for every state $\rho$, but the converse is not formal: a priori, scalar tests might miss spectral information living inside the C*-algebra. The extension theorem removes this gap by allowing a negative spectral value detected in the commutative algebra $C^*(a,1)$ to be promoted to a state on all of $A$. Thus the next theorem is needed to justify using states not merely as examples of positive functionals, but as a complete family of order tests for self-adjoint elements. [quotetheorem:8562] [citeproof:8562] Thus positivity can be checked externally by all states, not only internally by spectra and square roots. The self-adjointness hypothesis is essential because positivity is an order relation on the self-adjoint part of a C*-algebra; for a general element $a$, the inequalities $\rho(a)\ge 0$ are not even well-typed unless $\rho(a)$ is known to be real. Unitality is also part of this version of the statement because the proof uses the unital commutative subalgebra $C^*(a,1)$ and the state space normalized by $\rho(1)=1$. Non-unital variants exist, but they require either the unitization or the norm-one definition of state. The theorem should also not be read as saying that a single state detects the whole order, or that a fixed small family of states is enough without additional hypotheses. It says that the full state space separates the positive cone from self-adjoint elements outside it. This is the order-theoretic reason states will remain central in the representation theory: once a representation produces enough vector states, positivity questions become scalar inequalities. Once positivity is available through states, the next question is how to turn abstract C*-algebras into operators on Hilbert spaces. The GNS construction provides exactly that bridge, showing that every state comes from a representation and that abstract algebras can be studied concretely. # 5. Representations and the $\mathrm{GNS}$ Construction A C*-algebra is abstractly defined by algebraic operations, a norm, and the C*-identity, but the motivating examples are concrete norm-closed *-subalgebras of $\mathcal{L}(H)$ for Hilbert spaces $H$. This chapter explains why the abstract and concrete viewpoints are equivalent. The bridge is the Gelfand-Naimark-Segal construction: from a state on a C*-algebra it builds a Hilbert space, a *-representation, and a distinguished cyclic vector whose vector state recovers the original state. The chapter begins by isolating what it means for a C*-algebra to act on a Hilbert space. It then constructs the GNS representation from a single state, and finally combines sufficiently many states to obtain faithful representations. The endpoint is the abstract Gelfand-Naimark theorem: every C*-algebra can be realised isometrically as an algebra of bounded operators on a Hilbert space. ## *-Representations and Cyclic Vectors The first problem is to say precisely what structure is preserved when an abstract C*-algebra is viewed as operators on a Hilbert space. Multiplication should become composition, the involution should become the Hilbert-space adjoint, and the norm should interact with bounded operators in a controlled way. This leads to *-representations. [definition: *-Representation] Let $A$ be a C*-algebra and let $H$ be a Hilbert space. A *-representation of $A$ on $H$ is a *-homomorphism \begin{align*} \pi : A \to \mathcal{L}(H). \end{align*} If $A$ is unital, a unital *-representation also satisfies $\pi(1_A)=I_H$. [/definition] The definition is deliberately algebraic, but its analytic content comes from the C*-identity. Since *-homomorphisms between C*-algebras are contractive, every representation is automatically bounded. Contractivity alone is not enough to preserve the algebra: the zero representation $a\mapsto 0$ is contractive but forgets every non-zero element. The next definition isolates the obstruction to this collapse, and the theorem after it explains why removing that obstruction forces norm preservation. [definition: Kernel and Faithfulness of a Representation] Let $\pi:A\to \mathcal{L}(H)$ be a *-representation. The kernel of $\pi$ is \begin{align*} \ker \pi := \{a\in A : \pi(a)=0\}. \end{align*} The representation $\pi$ is faithful if $\ker\pi=\{0\}$. [/definition] Faithfulness is the condition needed if the representation is to lose no information about the algebra. In the unfaithful case, the representation is really seeing the quotient $A/\ker\pi$ rather than all of $A$. The next theorem explains why faithfulness is the correct hypothesis for treating an abstract C*-algebra as a concrete operator algebra without changing its norm. [quotetheorem:8563] [citeproof:8563] This theorem explains why faithful representations are the right concrete models of abstract C*-algebras: after faithfulness is known, the norm is not an extra choice. The hypothesis cannot be weakened to mere contractivity or non-zero representation. For instance, evaluation at a point $x_0$ gives a non-zero representation $C(X)\to\mathbb C$, $f\mapsto f(x_0)$, but any function vanishing at $x_0$ lies in its kernel, so the representation cannot recover the sup norm of such functions. Thus the chapter must next build representations and then combine enough of them to remove all kernels. Cyclic representations give the smallest useful building blocks for that construction. [definition: Cyclic Vector and Cyclic Representation] Let $\pi:A\to \mathcal{L}(H)$ be a *-representation. A vector $\xi\in H$ is cyclic for $\pi$ if \begin{align*} \overline{\{\pi(a)\xi : a\in A\}}=H. \end{align*} The pair $(\pi,H)$ is cyclic if it has a cyclic vector. [/definition] Cyclic representations are the representation-theoretic analogue of generating a module from one element. They are small enough to be constructed from a single functional, yet rich enough to recover many examples. [example: Multiplication Representation of C(X)] Let $X$ be compact Hausdorff, let $\mu$ be a finite regular Borel measure, and put $H=L^2(X,\mu)$. For $f\in C(X)$ define $\pi(f)g=fg$. Since $f$ is continuous on compact $X$, $\|f\|_\infty<\infty$, and for $g\in L^2(X,\mu)$ we have \begin{align*} \|\pi(f)g\|_2^2=\int_X |f(x)g(x)|^2\,d\mu(x)\leq \|f\|_\infty^2\int_X |g(x)|^2\,d\mu(x)=\|f\|_\infty^2\|g\|_2^2. \end{align*} Thus $\pi(f)$ is a bounded operator with $\|\pi(f)\|\leq \|f\|_\infty$. The algebraic identities are pointwise. For $f,h\in C(X)$ and $g\in L^2(X,\mu)$, \begin{align*} \pi(fh)g=(fh)g=f(hg)=\pi(f)(\pi(h)g). \end{align*} Also $\pi(1)g=g$, so $\pi(1)=I_H$. Using the $L^2$ inner product $(u,v)=\int_X u\overline v\,d\mu$, for $g,k\in L^2(X,\mu)$, \begin{align*} (\pi(f)g,k)=\int_X f g\overline{k}\,d\mu=\int_X g\,\overline{\overline f k}\,d\mu=(g,\pi(\overline f)k). \end{align*} Hence $\pi(f)^*=\pi(\overline f)$, so $\pi$ is a unital *-representation. The constant function $1$ belongs to $L^2(X,\mu)$ because $\mu(X)<\infty$, and \begin{align*} \{\pi(f)1:f\in C(X)\}=\{f:f\in C(X)\}. \end{align*} Therefore $1$ is cyclic exactly when $C(X)$ is dense in $L^2(X,\mu)$. To see the density, finite linear combinations of characteristic functions are dense in $L^2(X,\mu)$, and regularity plus [Urysohn's lemma](/theorems/887) approximates each $\mathbf 1_E$ in $L^2$ by continuous functions: choose closed $F\subseteq E\subseteq U$ with $\mu(U\setminus F)$ small, choose $u\in C(X)$ with $u=1$ on $F$, $u=0$ on $X\setminus U$, and $0\leq u\leq 1$, then \begin{align*} \|\mathbf 1_E-u\|_2^2\leq \mu(U\setminus F). \end{align*} Thus the multiplication representation is cyclic with cyclic vector $1$, and it realizes the commutative C*-algebra $C(X)$ as multiplication operators on the measure Hilbert space $L^2(X,\mu)$. [/example] The multiplication example already suggests the GNS pattern. A state behaves like integration against a measure, and the cyclic vector behaves like the constant function $1$. ## Constructing the GNS Representation from a State The next problem is constructive: given only a C*-algebra $A$ and a state $\phi$, how can one manufacture the Hilbert space on which $A$ acts? The guiding formula is that the state should become a vector state, \begin{align*} \phi(a)=(\pi(a)\xi,\xi)_H. \end{align*} To create such a vector, we first turn $\phi$ into an inner product. [definition: GNS Null Space] Let $A$ be a C*-algebra and let $\phi:A\to\mathbb{C}$ be a positive linear functional. Define \begin{align*} N_\phi:=\{a\in A: \phi(a^*a)=0\}. \end{align*} [/definition] The set $N_\phi$ consists of exactly the vectors that must be collapsed to make the formula $(a,b)=\phi(b^*a)$ positive definite. Without positivity this construction can fail immediately: a general linear functional may make $\phi(a^*a)$ negative or zero while still pairing $a$ non-negligibly with other elements, so quotienting by zero-length vectors would not be well-defined. The key point is that positivity gives a Cauchy-Schwarz inequality, so the quotient is compatible with left multiplication. The next theorem supplies the estimate that makes this quotient construction legitimate. [quotetheorem:8558] [citeproof:8558] The inequality just proved shows that zero-length elements pair to zero with every element, so $N_\phi$ can be divided out without ambiguity. Positivity is essential here, not cosmetic: take $A=\mathbb C^2$ with coordinatewise operations and define $\psi(z,w)=z-w$. Then $\psi((0,1)^*(0,1))=-1$, so the would-be squared length of $(0,1)$ is negative. Even worse for the quotient construction, $\psi((1,1)^*(1,1))=0$ while $\psi((1,0)^*(1,1))=1$, so a zero-length vector would not be orthogonal to all vectors. The theorem is therefore exactly the analytic estimate needed before we can define vectors as equivalence classes of algebra elements. This motivates defining the GNS Hilbert space as a completion of the quotient where the state has become an inner product. [definition: GNS Hilbert Space] Let $A$ be a C*-algebra and let $\phi:A\to\mathbb{C}$ be a positive linear functional. The pre-Hilbert space associated to $\phi$ is $A/N_\phi$ with inner product \begin{align*} (a+N_\phi,b+N_\phi)_\phi:=\phi(b^*a). \end{align*} The GNS Hilbert space $H_\phi$ is the Hilbert-space completion of $A/N_\phi$. [/definition] The quotient remembers elements of $A$ as vectors, and left multiplication by $A$ is the natural candidate for the representation. The construction so far gives a Hilbert space, but the next theorem is needed to prove that left multiplication gives bounded adjointable operators and recovers the original state as a vector state. [quotetheorem:7110] [citeproof:7110] The GNS construction packages a state as geometry. It turns positivity into an inner product, multiplication into operators, and the unit into a cyclic vector. The theorem does not say that this representation is faithful: for the point-evaluation state on $C(X)$, every function vanishing at the chosen point acts as zero in the resulting one-dimensional representation. It also does not say that different states lead to equivalent representations; point masses at different points of $X$ can produce different characters. Thus GNS solves the construction problem for one state, and the remaining task is to choose enough states so that no non-zero element is invisible to all of them. [remark: Non-Unital Variant] For a non-unital C*-algebra, the same construction is applied using the quotient of $A$ by $N_\phi$ and the left action of $A$. There need not be a vector represented by $1_A$, but a cyclic vector is obtained by working with the unitisation or by using an approximate unit when the state has norm one. [/remark] The unital case is enough for the main theorem in this chapter, while the non-unital variant is important for $C_0(X)$ and compact operators. We now compute the construction in three guiding examples. [example: Point Mass on C(X)] Let $X$ be compact Hausdorff, fix $x_0\in X$, and define $\phi(f)=f(x_0)$ for $f\in C(X)$. This is a state because $\phi(1)=1$ and, for every $f\in C(X)$, \begin{align*} \phi(f^*f)=\phi(\overline f f)=|f(x_0)|^2\geq 0. \end{align*} Hence the GNS null space is \begin{align*} N_\phi=\{f\in C(X):\phi(f^*f)=0\}=\{f\in C(X):|f(x_0)|^2=0\}=\{f\in C(X):f(x_0)=0\}. \end{align*} Define $T:C(X)/N_\phi\to\mathbb C$ by $T(f+N_\phi)=f(x_0)$. If $f+N_\phi=h+N_\phi$, then $f-h\in N_\phi$, so $(f-h)(x_0)=0$ and $f(x_0)=h(x_0)$; thus $T$ is well-defined. It is onto because $T(c\cdot 1+N_\phi)=c$, and it is one-to-one because $T(f+N_\phi)=0$ means $f(x_0)=0$, hence $f\in N_\phi$ and $f+N_\phi=0+N_\phi$. The inner product is preserved: \begin{align*} (f+N_\phi,h+N_\phi)_\phi=\phi(h^*f)=\phi(\overline h f)=\overline{h(x_0)}f(x_0)=T(f+N_\phi)\overline{T(h+N_\phi)}. \end{align*} So the GNS Hilbert space is naturally $\mathbb C$. Under this identification, the GNS action is scalar multiplication by the value at $x_0$. Indeed, for $f,h\in C(X)$, \begin{align*} T(\pi_\phi(f)(h+N_\phi))=T(fh+N_\phi)=(fh)(x_0)=f(x_0)h(x_0)=f(x_0)T(h+N_\phi). \end{align*} The cyclic vector is $\xi_\phi=1+N_\phi$, and $T(\xi_\phi)=1$. Thus the point-evaluation state produces the one-dimensional representation $f\mapsto f(x_0)$ on $\mathbb C$, with cyclic vector $1$. [/example] Point masses give the smallest commutative examples. At the opposite finite-dimensional extreme, a trace state produces a representation on the algebra itself. [example: Normalized Trace on a Matrix Algebra] Let $A=M_n(\mathbb C)$ and define $\phi(a)=n^{-1}\operatorname{Tr}(a)$. For $a=(a_{ij})\in M_n(\mathbb C)$, the $(j,j)$-entry of $a^*a$ is \begin{align*} (a^*a)_{jj}=\sum_{i=1}^n \overline{a_{ij}}a_{ij}=\sum_{i=1}^n |a_{ij}|^2. \end{align*} Therefore \begin{align*} \phi(a^*a)=\frac{1}{n}\operatorname{Tr}(a^*a)=\frac{1}{n}\sum_{j=1}^n(a^*a)_{jj}=\frac{1}{n}\sum_{j=1}^n\sum_{i=1}^n |a_{ij}|^2. \end{align*} This quantity is non-negative, and it is zero exactly when every entry $a_{ij}$ is zero. Hence \begin{align*} N_\phi=\{a\in M_n(\mathbb C):\phi(a^*a)=0\}=\{0\}. \end{align*} Since the null space is zero, the quotient $A/N_\phi$ is naturally identified with $M_n(\mathbb C)$ itself. Under this identification, the GNS inner product is \begin{align*} (a,b)_\phi=\phi(b^*a)=\frac{1}{n}\operatorname{Tr}(b^*a), \end{align*} which is the Hilbert-Schmidt inner product scaled by $n^{-1}$. The GNS action is left multiplication: \begin{align*} \pi_\phi(a)b=ab. \end{align*} For the identity matrix $I_n$, every $b\in M_n(\mathbb C)$ satisfies \begin{align*} \pi_\phi(b)I_n=bI_n=b. \end{align*} Thus $\{\pi_\phi(b)I_n:b\in M_n(\mathbb C)\}=M_n(\mathbb C)$, so $I_n$ is cyclic. The normalized trace state therefore produces the left regular representation of $M_n(\mathbb C)$ on itself, equipped with the scaled Hilbert-Schmidt geometry. [/example] This matrix example is the finite-dimensional prototype of the regular representation. The same idea appears for group C*-algebras, where multiplication is replaced by convolution. [example: Left Regular Representation of a Discrete Group] Let $G$ be a discrete group and let $\mathbb C[G]$ be the algebra of finite sums $a=\sum_{g\in G} c_g g$, with involution $a^*=\sum_{g\in G}\overline{c_g}g^{-1}$. Define $\phi(a)=c_e$, the coefficient of the identity element $e$. Then $\phi(1)=\phi(e)=1$. If $a=\sum_{g\in G}c_g g$, then \begin{align*} a^*a=\sum_{h\in G}\sum_{g\in G}\overline{c_h}c_g h^{-1}g. \end{align*} The coefficient of $e$ occurs exactly when $h^{-1}g=e$, equivalently $g=h$, so \begin{align*} \phi(a^*a)=\sum_{g\in G}|c_g|^2\geq 0. \end{align*} Thus $\phi$ is positive, and $\phi(a^*a)=0$ forces every coefficient $c_g$ to be zero. Hence the GNS null space is $N_\phi=\{0\}$. The GNS inner product on $\mathbb C[G]$ is therefore \begin{align*} (a,b)_\phi=\phi(b^*a). \end{align*} For group elements $g,h\in G$, \begin{align*} (g,h)_\phi=\phi(h^{-1}g). \end{align*} This equals $1$ when $g=h$ and equals $0$ when $g\neq h$, because $h^{-1}g=e$ exactly when $g=h$. Therefore the group elements are an orthonormal basis in the completion, so the GNS Hilbert space is naturally $\ell^2(G)$, with $g$ corresponding to the basis vector $\delta_g$. Under this identification, left multiplication gives \begin{align*} \pi_\phi(s)\delta_h=\delta_{sh} \end{align*} for $s,h\in G$. For a general element $a=\sum_{s\in G}c_s s$, the operator is \begin{align*} \pi_\phi(a)\delta_h=\sum_{s\in G}c_s\delta_{sh}. \end{align*} Thus the GNS representation is the left regular representation $\lambda$, where $\lambda_s\delta_h=\delta_{sh}$. The reduced group C*-algebra $C_r^*(G)$ is the norm closure of $\lambda(\mathbb C[G])$ inside $\mathcal L(\ell^2(G))$. [/example] These examples show that GNS is not a formal trick: it recovers familiar Hilbert spaces from familiar states. The measure example is the commutative bridge from states to integration, the matrix example is the finite-dimensional bridge from traces to Hilbert-Schmidt geometry, and the group example is the convolution bridge from algebraic multiplication to regular representations. In each case, however, the representation sees only what the chosen state sees. The remaining issue is whether enough states exist to build a representation that sees every element of an arbitrary C*-algebra. ## Faithful Representations from Separating Families of States The final problem is separation. A single GNS representation may have a large kernel, so it may not represent the whole C*-algebra faithfully. For example, one point mass on $C(X)$ detects only the value of a function at one point; a non-zero function supported away from that point is invisible to the corresponding GNS representation. To embed $A$ into bounded operators, we need a family of states large enough to detect every non-zero element. [definition: Separating Family of States] Let $A$ be a C*-algebra. A family $S$ of states on $A$ is separating if, for every non-zero $x\in A$, there exists $\phi\in S$ such that \begin{align*} \phi(x)\neq 0. \end{align*} [/definition] For the representation problem, this point-separation condition is used on the positive element $a^*a$. If $a\neq 0$, then $a^*a\neq 0$, so separation gives some $\phi\in S$ with $\phi(a^*a)\neq 0$; since states are positive, this means $\phi(a^*a)>0$. In the GNS representation associated to $\phi$, the quantity $\phi(a^*a)$ is exactly $\|\pi_\phi(a)\xi_\phi\|_{H_\phi}^2$. The next theorem is needed to turn this state-by-state detection into a single faithful representation. [quotetheorem:8564] [citeproof:8564] The theorem reduces the abstract representation problem to the existence of enough point-separating states. Applied to $a^*a$, that hypothesis is necessary in precisely the expected way: if all chosen states vanish on $a^*a$, then every GNS summand kills the vector $\pi_\phi(a)\xi_\phi$; in the extreme case $S$ consists of one point evaluation on $C(X)$, functions vanishing at that point remain in the kernel. Thus direct sums do not create information that the chosen states never measured. What remains is to prove that the full state space is large enough, and that existence is a Hahn-Banach consequence applied to the positive cone. The next result gives exactly the separation property needed for the direct-sum argument. [quotetheorem:8565] [citeproof:8565] This result supplies the separating family required above: take all states on $A$. The unital hypothesis is part of the statement because states are normalised by $\phi(1_A)=1$ and the proof extends a state from the unital subalgebra $C^*(a,1_A)$. For non-unital C*-algebras, the corresponding separation statement is usually proved by passing to the unitisation or by using positive norm-one functionals on $A$; the representation theorem then has a non-unital version with the same detection mechanism. The condition $a\geq 0$ is not an artificial restriction, because arbitrary elements are detected through the positive element $a^*a$ in every representation. If we tried to separate arbitrary elements directly by demanding $\phi(a)\neq 0$, cancellation could obscure the operator norm; positivity converts detection into a squared norm. The next theorem packages all GNS representations at once, producing a canonical faithful representation that will serve as the abstract algebra's universal concrete model. [quotetheorem:8566] [citeproof:8566] The universal representation is often too large for computation, but it is conceptually decisive. Taking fewer states can fail: on $C(X)$, using only evaluations on a proper closed subset $F\subsetneq X$ kills every function vanishing on $F$, so the resulting direct sum represents $C(F)$ rather than all of $C(X)$. The universal representation avoids this limitation by using the entire state space, and the preceding separation theorem proves that this choice has zero kernel. The final theorem is needed to translate faithfulness into the concrete realisation promised by the C*-algebra axioms themselves. [quotetheorem:8567] [citeproof:8567] This theorem closes the circle opened at the start of the course. Its conclusion depends on faithfulness: a non-faithful representation can still have a norm-closed image, but that image is a concrete model only of the quotient $A/\ker\pi$, not of $A$ itself. The theorem also does not choose a small or canonical computational representation; the universal representation may be very large, and later work often replaces it by more economical faithful representations when available. We began with examples such as $C(X)$, $M_n(\mathbb C)$, $\mathcal{L}(H)$, and $K(H)$; the theorem says that the abstract definition captures exactly the norm-closed *-algebras of operators, up to isometric *-isomorphism. Later chapters use this representation theorem while studying ideals, approximate units, and finite-dimensional approximation, often passing between abstract C*-algebras and their concrete faithful representations without changing the normed *-algebraic content. The representation picture makes ideals and quotients easier to interpret: kernels become the basic examples, and approximate units describe how nonunital pieces still act like unital ones locally. From there, the theory of extensions packages an algebra together with an ideal and the corresponding quotient. # 6. Ideals, Quotients, and Approximate Units Ideals are the mechanism by which a C*-algebra is decomposed into a part we want to ignore and the quotient that remains. In the commutative picture $C_0(X)$, this corresponds to restricting attention away from an open subset or collapsing functions that vanish on a [closed set](/page/Closed%20Set). This chapter develops the noncommutative version: closed two-sided ideals, quotient C*-algebras, approximate identities inside ideals, hereditary subalgebras, and the first examples of extensions. ## Closed Ideals and Quotient $C^*$-Algebras If $A$ is a C*-algebra and $I \subset A$ is a two-sided ideal, the algebraic quotient $A/I$ is available immediately. The analytic question is whether the quotient norm and involution retain the C*-structure, since without closure the quotient may fail to be complete. [definition: Closed Two-Sided Ideal] Let $A$ be a C*-algebra. A closed two-sided ideal in $A$ is a closed linear subspace $I \subset A$ such that $a x \in I$ and $x a \in I$ for all $a \in A$ and $x \in I$. [/definition] The word closed is part of the structure because C*-algebras are Banach algebras. Algebraic ideals still appear, but the quotient construction in this course is built from closed ideals. [example: Compact Operators Inside $\mathcal{L}(H)$] Let $H$ be an infinite-dimensional Hilbert space, and write $B(H)=\mathcal L(H)$. The compact operators are \begin{align*} K(H)=\{T\in B(H):T \text{ sends bounded subsets of }H\text{ to relatively compact subsets of }H\}. \end{align*} We show that $K(H)$ is a closed two-sided ideal in $B(H)$, and that the inclusion $K(H)\subset B(H)$ is proper. First, $K(H)$ is a linear subspace. If $T,R\in K(H)$, $\alpha,\beta\in\mathbb C$, and $B\subset H$ is bounded, then $\overline{T(B)}$ and $\overline{R(B)}$ are compact. The set $\alpha\overline{T(B)}+\beta\overline{R(B)}$ is compact because it is the continuous image of $\overline{T(B)}\times \overline{R(B)}$ under $(u,v)\mapsto \alpha u+\beta v$. Since \begin{align*} (\alpha T+\beta R)(B)\subset \alpha\overline{T(B)}+\beta\overline{R(B)}, \end{align*} the set $(\alpha T+\beta R)(B)$ is relatively compact. Hence $\alpha T+\beta R\in K(H)$. Next, $K(H)$ is two-sided. Let $T\in K(H)$ and $S\in B(H)$. If $B\subset H$ is bounded, then $T(B)$ is relatively compact, so $\overline{T(B)}$ is compact. Since $S$ is continuous, $S(\overline{T(B)})$ is compact, and \begin{align*} ST(B)\subset S(\overline{T(B)}). \end{align*} Thus $ST(B)$ is relatively compact, so $ST\in K(H)$. Also, $S(B)$ is bounded because $S$ is bounded, and then \begin{align*} TS(B)=T(S(B)) \end{align*} is relatively compact because $T$ is compact. Hence $TS\in K(H)$. Now suppose $T_n\in K(H)$ and $\|T_n-T\|\to 0$ in $B(H)$. Let $B_1=\{\xi\in H:\|\xi\|\le 1\}$ and fix $\varepsilon>0$. Choose $n$ such that $\|T-T_n\|<\varepsilon/3$. Since $T_n(B_1)$ is relatively compact, there are vectors $y_1,\dots,y_m\in H$ such that every point of $T_n(B_1)$ is within $\varepsilon/3$ of some $y_j$. For $\xi\in B_1$, choose $j$ with $\|T_n\xi-y_j\|<\varepsilon/3$. Then the triangle inequality gives \begin{align*} \|T\xi-y_j\|\le \|T\xi-T_n\xi\|+\|T_n\xi-y_j\|<\varepsilon. \end{align*} Thus $T(B_1)$ is totally bounded. Since $H$ is complete, the closure of $T(B_1)$ is compact, so $T\in K(H)$. Therefore $K(H)$ is norm closed. Finally, $1_H\notin K(H)$. If $1_H$ were compact, then $1_H(B_1)=B_1$ would have compact closure, so the closed unit ball of $H$ would be compact. But $H$ contains an orthonormal sequence $(\xi_n)$, and for $n\ne m$, \begin{align*} \|\xi_n-\xi_m\|^2=\|\xi_n\|^2+\|\xi_m\|^2-2\operatorname{Re}\langle \xi_n,\xi_m\rangle=1+1-0=2. \end{align*} Hence no subsequence of $(\xi_n)$ is Cauchy, so the closed unit ball is not compact. Therefore the identity operator is not compact, and the ideal $K(H)$ is a proper closed two-sided ideal of $B(H)$. [/example] This example is the model for many extensions: a large operator algebra has a distinguished ideal of compact, or lower-order, operators. The quotient then records the part of an operator that remains after compact perturbations are ignored. [definition: Quotient Norm] Let $A$ be a C*-algebra and let $I \trianglelefteq A$ be a closed two-sided ideal. The quotient norm is the function \begin{align*} \|\cdot\|_{A/I}:A/I\to [0,\infty) \end{align*} defined by \begin{align*} \|a + I\|_{A/I} = \inf_{x \in I} \|a + x\|_A. \end{align*} [/definition] The quotient norm measures the distance from $a$ to the ideal $I$. Closedness is essential here: if $J$ is a nonclosed ideal, then the quotient seminorm vanishes on the larger set $\overline{J}/J$, so the algebraic quotient $A/J$ is not Hausdorff as a normed space. Two-sidedness is also essential, because multiplication on cosets is well-defined only when both left and right products by elements of the ideal remain in the ideal. The next result says that once these two hypotheses are imposed, no additional completion or correction is needed: the quotient is again a C*-algebra. [quotetheorem:8568] [citeproof:8568] The theorem explains why quotients are part of the internal category of C*-algebras: closed two-sided ideals are precisely the subobjects that can be divided out while remaining in the same theory. The closedness hypothesis cannot be weakened to an algebraic ideal, because the quotient norm would only be a seminorm when the ideal has a larger closure. The two-sided hypothesis cannot be weakened to a left or right ideal, because the product $(a+I)(b+I)$ would then depend on the chosen representatives. With the theorem in place, commutative quotients can be computed by identifying the ideal as the kernel of a familiar map, which is the point of the next example. [example: A Point-Evaluation Quotient] Let $A=C([0,1])$ and let $I=\{f\in C([0,1]):f(0)=0\}$. Define \begin{align*} \operatorname{ev}_0:C([0,1])\to \mathbb C,\qquad \operatorname{ev}_0(f)=f(0). \end{align*} For $f,g\in C([0,1])$ and $\alpha,\beta\in\mathbb C$, \begin{align*} \operatorname{ev}_0(\alpha f+\beta g)=(\alpha f+\beta g)(0)=\alpha f(0)+\beta g(0)=\alpha\operatorname{ev}_0(f)+\beta\operatorname{ev}_0(g). \end{align*} Also, \begin{align*} \operatorname{ev}_0(fg)=(fg)(0)=f(0)g(0)=\operatorname{ev}_0(f)\operatorname{ev}_0(g), \end{align*} and \begin{align*} \operatorname{ev}_0(f^*)=f^*(0)=\overline{f(0)}=\overline{\operatorname{ev}_0(f)}. \end{align*} Thus $\operatorname{ev}_0$ is a *-homomorphism. Its kernel is exactly $I$, because \begin{align*} f\in\ker(\operatorname{ev}_0)\Longleftrightarrow \operatorname{ev}_0(f)=0\Longleftrightarrow f(0)=0\Longleftrightarrow f\in I. \end{align*} The induced map $\Phi:C([0,1])/I\to\mathbb C$ is \begin{align*} \Phi(f+I)=f(0). \end{align*} It is well-defined: if $f+I=g+I$, then $f-g\in I$, so $(f-g)(0)=0$, hence $f(0)=g(0)$. It is injective because $\Phi(f+I)=0$ means $f(0)=0$, so $f\in I$ and $f+I=I$. It is surjective because, for every $\lambda\in\mathbb C$, the constant function $c_\lambda(t)=\lambda$ satisfies $\Phi(c_\lambda+I)=\lambda$. Therefore $C([0,1])/I\cong\mathbb C$. Finally, the quotient norm becomes the usual absolute value. For every $g\in I$, \begin{align*} \|f+g\|_\infty\ge |(f+g)(0)|=|f(0)+g(0)|=|f(0)|. \end{align*} Taking the infimum over $g\in I$ gives $\|f+I\|\ge |f(0)|$. For the reverse inequality, let $c(t)=f(0)$ for all $t\in[0,1]$. Then $(c-f)(0)=f(0)-f(0)=0$, so $c-f\in I$, and \begin{align*} \|f+I\|\le \|f+(c-f)\|_\infty=\|c\|_\infty=|f(0)|. \end{align*} Hence $\|f+I\|=|f(0)|$, so the quotient remembers exactly the value of $f$ at the point $0$. [/example] This example is the commutative prototype: quotienting by functions vanishing on a closed subset produces the algebra of functions on that closed subset. In the noncommutative setting, closed subsets are not literal spaces, so we need a representation-theoretic replacement that detects exactly which elements have been collapsed. Kernels of representations provide that replacement and connect quotients to concrete operator algebras. [quotetheorem:8569] [citeproof:8569] The result turns quotient algebras into representation-theoretic objects. Closedness is forced because kernels of continuous representations are closed; a dense proper ideal can never be the kernel of a representation unless the representation is zero on all of $A$. Two-sidedness is forced because kernels must absorb multiplication from both sides. This prepares the next section: once ideals are viewed both algebraically and representation-theoretically, approximate identities explain how a nonunital ideal can still act locally like a unital algebra. ## Approximate Identities and Hereditary Subalgebras A nonunital C*-algebra often has enough approximate units to behave locally as though a unit were present. The question is how to replace $1a=a$ when no identity element exists, especially inside an ideal such as $K(H)$ or $C_0(U)$. [definition: Approximate Identity] Let $A$ be a C*-algebra. An approximate identity for $A$ is a net $(e_\lambda)_{\lambda \in \Lambda}$ in $A$ such that \begin{align*} \|e_\lambda a-a\|_A \to 0, \qquad \|a e_\lambda-a\|_A \to 0 \end{align*} for every $a\in A$. [/definition] This definition only asks for asymptotic left and right multiplication by a unit. In C*-algebras, however, positivity controls order and functional calculus, while contractivity prevents norm estimates from growing during approximation. We therefore isolate the form of approximate identity that can be used safely in ideal arguments. [definition: Contractive Positive Approximate Identity] Let $A$ be a C*-algebra. A contractive positive approximate identity is an approximate identity $(e_\lambda)$ such that $e_\lambda \ge 0$ and $\|e_\lambda\|_A\le 1$ for every $\lambda$. [/definition] For a nonunital C*-algebra, many arguments that would normally multiply by $1$ have no literal identity element available. Simply choosing arbitrary approximating elements is not enough: without positivity, order and functional-calculus estimates can fail, and without contractivity, norm bounds can be lost in the limiting process. The needed replacement is therefore a net that behaves like a unit on each fixed element while remaining positive and norm-controlled throughout the approximation. [quotetheorem:8570] [citeproof:8570] Approximate identities should be read as a convergence device rather than as a substitute element. Positivity is what lets order and functional calculus enter the estimates, and contractivity prevents the approximating multipliers from enlarging norms while passing to limits. Norm convergence to an actual identity occurs only when the algebra is already unital, so the theorem does not manufacture a unit inside a genuinely nonunital algebra. The next examples make this distinction concrete in the two guiding cases: compact operators and functions vanishing at infinity. [example: Finite-Rank Projections Approximating K H] Let $H$ be a separable Hilbert space with orthonormal basis $(\xi_n)_{n\ge 1}$, and define $p_N$ by \begin{align*} p_N\eta=\sum_{k=1}^N \langle \eta,\xi_k\rangle \xi_k. \end{align*} Then $p_N$ has finite-dimensional range, so $p_N\in K(H)$. Also $p_N^*=p_N$ and, for every $\eta\in H$, \begin{align*} \langle p_N\eta,\eta\rangle=\sum_{k=1}^N |\langle \eta,\xi_k\rangle|^2\ge 0. \end{align*} Thus $p_N$ is positive. By [Bessel's inequality](/theorems/540), \begin{align*} \|p_N\eta\|^2=\sum_{k=1}^N |\langle \eta,\xi_k\rangle|^2\le \|\eta\|^2, \end{align*} so $\|p_N\|\le 1$; since $p_N\xi_1=\xi_1$ for $N\ge 1$, we have $\|p_N\|=1$. We first check convergence on finite-rank operators. Let $T\in K(H)$ have finite rank. Choose $M$ such that $\operatorname{ran}(T)\subset \operatorname{span}\{\xi_1,\dots,\xi_M\}$. If $N\ge M$, then $p_NT\eta=T\eta$ for every $\eta\in H$, hence \begin{align*} \|p_NT-T\|=0. \end{align*} For right multiplication, write $T$ on the finite-dimensional subspace supporting its action as a finite sum of rank-one operators, \begin{align*} T\eta=\sum_{j=1}^r \langle \eta,\alpha_j\rangle \beta_j. \end{align*} Since each $\alpha_j$ lies in $H$, we have $\|p_N\alpha_j-\alpha_j\|\to 0$. For $\|\eta\|\le 1$, \begin{align*} \|Tp_N\eta-T\eta\|=\left\|\sum_{j=1}^r \langle \eta,p_N\alpha_j-\alpha_j\rangle \beta_j\right\|\le \sum_{j=1}^r \|p_N\alpha_j-\alpha_j\|\,\|\beta_j\|. \end{align*} The right-hand side tends to $0$, so $\|Tp_N-T\|\to 0$. Now let $T\in K(H)$ be arbitrary. Since compact operators are norm limits of finite-rank operators, choose a finite-rank operator $F$ with $\|T-F\|<\varepsilon/3$. For all $N$ large enough, the finite-rank case gives $\|p_NF-F\|<\varepsilon/3$. Using $\|p_N\|\le 1$, \begin{align*} \|p_NT-T\|\le \|p_N(T-F)\|+\|p_NF-F\|+\|F-T\|<\varepsilon. \end{align*} Similarly, for all $N$ large enough with $\|Fp_N-F\|<\varepsilon/3$, \begin{align*} \|Tp_N-T\|\le \|(T-F)p_N\|+\|Fp_N-F\|+\|F-T\|<\varepsilon. \end{align*} Therefore $(p_N)$ is a contractive positive approximate identity for $K(H)$: it approximates compact operators by first acting exactly on finite-dimensional pieces and then passing to norm limits. [/example] The preceding example shows the geometric meaning of an approximate identity: it captures larger and larger finite-dimensional pieces of $H$. For commutative algebras the same idea becomes exhaustion by compact subsets. [example: Cutoff Functions In $C_0(U)$] Let $X$ be a locally compact Hausdorff space and let $U\subset X$ be open. If $f\in C_0(U)$, extend it by $0$ outside $U$ by setting $\widetilde f(x)=f(x)$ for $x\in U$ and $\widetilde f(x)=0$ for $x\in X\setminus U$. This extension is continuous at points of $U$ because $f$ is continuous. If $x\in X\setminus U$ and $\varepsilon>0$, the set \begin{align*} K_\varepsilon=\{u\in U:|f(u)|\ge \varepsilon\} \end{align*} is compact in $U$, hence compact in $X$, and does not contain $x$. Since $X$ is Hausdorff and locally compact, there is a neighborhood of $x$ disjoint from $K_\varepsilon$, so $|\widetilde f|<\varepsilon$ there. Thus $\widetilde f\in C_0(X)$. Under this embedding, $C_0(U)$ is an ideal of $C_0(X)$. Indeed, if $g\in C_0(X)$ and $f\in C_0(U)$, then $g\widetilde f$ is zero on $X\setminus U$, and on $U$ it is the function $(g|_U)f$. Since $g|_U$ is bounded and continuous and $f$ vanishes at infinity on $U$, the product $(g|_U)f$ lies in $C_0(U)$. Index the compact subsets $K\subset U$ by inclusion. For each compact $K\subset U$, choose $e_K\in C_c(U)$ such that $0\le e_K\le 1$ and $e_K=1$ on $K$, using the locally compact Hausdorff cutoff lemma. Pointwise positivity gives $e_K\ge 0$ in the commutative C*-algebra $C_0(U)$, and \begin{align*} \|e_K\|_\infty=\sup_{x\in U}|e_K(x)|\le 1. \end{align*} So every $e_K$ is a positive contraction. Let $f\in C_0(U)$ and let $\varepsilon>0$. Since $f$ vanishes at infinity, there is a compact set $K_0\subset U$ such that $|f(x)|<\varepsilon$ for all $x\in U\setminus K_0$. If $K\supset K_0$, then $e_K(x)=1$ on $K_0$, so \begin{align*} |e_K(x)f(x)-f(x)|=0 \end{align*} for $x\in K_0$. For $x\in U\setminus K_0$, the bounds $0\le e_K(x)\le 1$ give $|e_K(x)-1|\le 1$, hence \begin{align*} |e_K(x)f(x)-f(x)|=|e_K(x)-1|\,|f(x)|<\varepsilon. \end{align*} Taking the supremum over $U$ gives \begin{align*} \|e_Kf-f\|_\infty<\varepsilon. \end{align*} Because multiplication in $C_0(U)$ is pointwise and commutative, $fe_K=e_Kf$, so also $\|fe_K-f\|_\infty<\varepsilon$. Therefore $(e_K)$ is a contractive positive approximate identity for $C_0(U)$: the cutoff functions become equal to $1$ on larger compact regions, while every function in $C_0(U)$ is uniformly small off some compact region. [/example] This example motivates looking beyond two-sided multiplication and asking which subalgebras are closed under passing to smaller positive elements. Corners such as $pAp$ behave like local pieces without being two-sided. This motivates the following definition of hereditary subalgebras, which treats ideals and corners through the same order-theoretic condition. [definition: Hereditary C Star Subalgebra] Let $A$ be a C*-algebra. A C*-subalgebra $B\subset A$ is hereditary if whenever $0\le a\le b$ with $b\in B_+$ and $a\in A_+$, then $a\in B$. [/definition] The order condition says that $B$ contains every positive element of $A$ dominated by one of its positive elements. This property is stronger than being a subalgebra and is especially useful for corners. [example: Corners As Hereditary Subalgebras] Let $A$ be a unital C*-algebra and let $p\in A$ be a projection. We first check that \begin{align*} pAp=\{pap:a\in A\} \end{align*} is a C*-subalgebra. If $x=pap$ and $y=pbp$, then $\alpha x+\beta y=p(\alpha a+\beta b)p$, so $pAp$ is a linear subspace. Also \begin{align*} xy=(pap)(pbp)=pa p b p=p(apb)p\in pAp. \end{align*} Since $p^*=p$, \begin{align*} x^*=(pap)^*=pa^*p\in pAp. \end{align*} Finally, $pAp$ is closed: if $x_n\in pAp$ and $x_n\to x$ in norm, then $px_np=x_n$ for every $n$, and continuity of multiplication gives $pxp=x$, so $x\in pAp$. Now we prove heredity. Suppose $0\le a\le pbp$ with $b\in A_+$. Let $q=1-p$. Then $q$ is also a projection and \begin{align*} qpbpq=(1-p)pbp(1-p)=(p-p^2)bp(1-p)=0. \end{align*} From $0\le a\le pbp$, multiplying on the left and right by $q$ gives \begin{align*} 0\le qaq\le qpbpq=0. \end{align*} Hence $qaq=0$. Since $a\ge 0$, it has a positive square root $a^{1/2}$, and \begin{align*} (a^{1/2}q)^*(a^{1/2}q)=qa q=0. \end{align*} Therefore $a^{1/2}q=0$, so \begin{align*} aq=a^{1/2}(a^{1/2}q)=0. \end{align*} Taking adjoints gives $qa=0$. Since $1=p+q$, \begin{align*} a=(p+q)a(p+q)=pap+paq+qap+qaq=pap. \end{align*} Thus $a\in pAp$, so the corner $pAp$ is hereditary. This shows that a corner behaves like the part of $A$ supported on the projection $p$: every positive element dominated by a positive element of the corner is forced back into the same corner. [/example] Hereditary subalgebras are linked to ideals because every ideal is hereditary as a C*-subalgebra, but hereditary subalgebras need not be two-sided ideals. The distinction separates local corners from globally invariant pieces. [quotetheorem:8571] [citeproof:8571] This theorem justifies treating ideals as closed regions of a C*-algebra: if a positive element belongs to the ideal, all smaller positive pieces belong to it as well. The closedness hypothesis is used at the final norm-limit step; without it the approximating compressions need not force the limit to remain in the ideal. The two-sided hypothesis is used when passing from $b$ to $c_n^*bc_n$, so a one-sided ideal would not support the same hereditary conclusion. Hereditary subalgebras therefore include all closed ideals but also include corners such as $pAp$, which are usually not ideals unless $p$ is central or otherwise invariant under multiplication by all of $A$. ## Simple $C^*$-Algebras and Extensions Ideals measure the ways a C*-algebra can have internal closed invariant pieces. The extreme case is a simple C*-algebra, where no nonzero proper closed two-sided ideal exists; the opposite perspective packages an algebra with an ideal and quotient into an extension. [definition: Simple C Star Algebra] A C*-algebra $A$ is simple if its only closed two-sided ideals are $\{0\}$ and $A$. [/definition] Simplicity is not the same as being small. Matrix algebras are finite-dimensional simple examples, while many infinite-dimensional C*-algebras arising in dynamics and classification theory are also simple. [example: Matrix Algebras Are Simple] Let $n\ge 1$, and let $I\trianglelefteq M_n(\mathbb C)$ be a nonzero closed two-sided ideal. Choose a nonzero matrix $A=(a_{rs})_{r,s=1}^n\in I$. Since $A\ne 0$, there are indices $r_0,s_0$ such that $a_{r_0s_0}\ne 0$. Write $E_{ij}$ for the standard matrix units. For arbitrary $k,l\in\{1,\dots,n\}$, the two-sided ideal property gives \begin{align*} E_{k r_0} A E_{s_0 l}\in I. \end{align*} We compute its entries. For $1\le i,j\le n$, \begin{align*} (E_{k r_0} A)_{ij}=\sum_{m=1}^n (E_{k r_0})_{im}a_{mj}=\sum_{m=1}^n \delta_{ik}\delta_{r_0m}a_{mj}=\delta_{ik}a_{r_0j}. \end{align*} Therefore \begin{align*} (E_{k r_0} A E_{s_0 l})_{ij}=\sum_{m=1}^n (E_{k r_0}A)_{im}(E_{s_0 l})_{mj}=\sum_{m=1}^n \delta_{ik}a_{r_0m}\delta_{ms_0}\delta_{lj}=\delta_{ik}a_{r_0s_0}\delta_{lj}. \end{align*} Thus \begin{align*} E_{k r_0} A E_{s_0 l}=a_{r_0s_0}E_{kl}. \end{align*} Since $a_{r_0s_0}\ne 0$ and $I$ is a complex linear subspace, it follows that \begin{align*} E_{kl}=a_{r_0s_0}^{-1}E_{k r_0} A E_{s_0 l}\in I. \end{align*} Because $k$ and $l$ were arbitrary, every matrix unit $E_{kl}$ belongs to $I$. Now let $B=(b_{kl})\in M_n(\mathbb C)$. The standard matrix-unit expansion is \begin{align*} B=\sum_{k=1}^n\sum_{l=1}^n b_{kl}E_{kl}. \end{align*} Each $E_{kl}$ lies in $I$, and $I$ is a linear subspace, so $B\in I$. Hence $I=M_n(\mathbb C)$. The only closed two-sided ideals are therefore $\{0\}$ and $M_n(\mathbb C)$, so $M_n(\mathbb C)$ is simple. [/example] This matrix-unit computation motivates the infinite-dimensional analogue for compact operators. Rank-one operators play the role of matrix units, while the density of finite-rank operators supplies the final passage from algebraic manipulation to the norm-closed algebra. The next theorem records that $K(H)$ is the main simple ideal appearing inside a much larger algebra. [quotetheorem:8572] [citeproof:8572] The assumption $H\ne\{0\}$ removes the degenerate zero algebra, whose ideal structure carries no information. Closedness is again essential: the finite-rank operators form a dense proper algebraic ideal in $K(H)$ when $H$ is infinite-dimensional, so algebraic ideals alone would not give simplicity. The theorem also separates internal and external ideal structure: $K(H)$ is simple as an algebra in its own right, yet it is a proper ideal in $B(H)$ when $H$ is infinite-dimensional. This motivates a formal device for studying how a simple algebra can sit inside a larger one while producing a meaningful quotient. That device is a short exact sequence of C*-algebras. [definition: Extension Of C Star Algebras] An extension of C*-algebras is a short exact sequence \begin{align*} 0\longrightarrow I \xrightarrow{\iota} E \xrightarrow{q} A \longrightarrow 0 \end{align*} of *-homomorphisms, where $\iota$ identifies $I$ with a closed two-sided ideal of $E$ and $q$ induces a *-isomorphism $E/I\cong A$. [/definition] Extensions record both an ideal and the quotient by that ideal. They are a major organizing device because many operator algebras are best understood by identifying a known ideal and a known quotient. [example: The Calkin Extension] Let $H$ be an infinite-dimensional Hilbert space, and let $\iota:K(H)\to B(H)$ be the inclusion $\iota(T)=T$. The earlier compact-operator example shows that $K(H)$ is a proper closed two-sided ideal in $B(H)$, so the quotient C*-algebra $B(H)/K(H)$ is defined. Let \begin{align*} q:B(H)\to B(H)/K(H),\qquad q(T)=T+K(H). \end{align*} For $S,T\in B(H)$ and $\alpha,\beta\in\mathbb C$, \begin{align*} q(\alpha S+\beta T)=(\alpha S+\beta T)+K(H)=\alpha(S+K(H))+\beta(T+K(H))=\alpha q(S)+\beta q(T). \end{align*} Also, \begin{align*} q(ST)=ST+K(H)=(S+K(H))(T+K(H))=q(S)q(T), \end{align*} and \begin{align*} q(T^*)=T^*+K(H)=(T+K(H))^*=q(T)^*. \end{align*} Thus $q$ is a *-homomorphism, while $\iota$ is a *-homomorphism because it preserves the same operations by inclusion. Now compute exactness. The map from $0$ into $K(H)$ has image $\{0\}$, and \begin{align*} \ker(\iota)=\{T\in K(H):\iota(T)=0\}=\{T\in K(H):T=0\}=\{0\}. \end{align*} Next, \begin{align*} \ker(q)=\{T\in B(H):T+K(H)=K(H)\}=K(H), \end{align*} and this is exactly $\operatorname{im}(\iota)$. Finally, $q$ is surjective because every coset in $B(H)/K(H)$ has the form $T+K(H)=q(T)$ for some $T\in B(H)$. Therefore \begin{align*} 0\longrightarrow K(H)\xrightarrow{\iota} B(H)\xrightarrow{q} B(H)/K(H)\longrightarrow 0 \end{align*} is a short exact sequence of C*-algebras. The quotient map identifies two operators precisely when their difference is compact: \begin{align*} q(S)=q(T)\Longleftrightarrow S+K(H)=T+K(H)\Longleftrightarrow S-T\in K(H). \end{align*} Thus the Calkin extension records the passage from operators on $H$ to operators modulo compact perturbations. [/example] The Toeplitz algebra is the guiding noncommutative example in which the quotient is commutative. Its extension records how an operator algebra can carry compact-operator error terms while retaining a commutative symbol algebra, a pattern that later becomes central in index theory and $K$-theory. [example: Toeplitz Extension] Let $(e_n)_{n\ge 0}$ be the standard orthonormal basis of $\ell^2(\mathbb N)$, and let $S$ be the unilateral shift, so $Se_n=e_{n+1}$. The Toeplitz algebra is $\mathcal T=C^*(S)\subset B(\ell^2(\mathbb N))$. We first see where the compact ideal comes from. Since $S^*S=1$ and $SS^*$ is the projection onto $\overline{\operatorname{span}}\{e_n:n\ge 1\}$, the operator \begin{align*} P_0=1-SS^* \end{align*} is the rank-one projection onto $\mathbb C e_0$. For $i,j\ge 0$, define \begin{align*} E_{ij}=S^iP_0(S^*)^j. \end{align*} If $n=j$, then $(S^*)^j e_j=e_0$, so $E_{ij}e_j=S^iP_0e_0=S^ie_0=e_i$. If $n<j$, then $(S^*)^j e_n=0$, so $E_{ij}e_n=0$. If $n>j$, then $(S^*)^j e_n=e_{n-j}$ with $n-j\ge 1$, so $P_0e_{n-j}=0$ and again $E_{ij}e_n=0$. Thus $E_{ij}$ is the usual rank-one matrix unit sending $e_j$ to $e_i$. Hence every finite-rank operator lies in $\mathcal T$, and since $K(\ell^2(\mathbb N))$ is the norm closure of the finite-rank operators, we have \begin{align*} K(\ell^2(\mathbb N))\subset \mathcal T. \end{align*} Let $q:\mathcal T\to \mathcal T/K(\ell^2(\mathbb N))$ be the quotient map. In the quotient, the coset $q(S)$ is unitary because \begin{align*} q(S)^*q(S)=q(S^*S)=q(1) \end{align*} and \begin{align*} q(S)q(S)^*=q(SS^*)=q(1-P_0)=q(1), \end{align*} where $q(P_0)=0$ since $P_0$ is compact. The Toeplitz symbol theorem identifies the quotient $\mathcal T/K(\ell^2(\mathbb N))$ with $C(S^1)$ by sending this unitary coset to the coordinate function \begin{align*} u:S^1\to\mathbb C,\qquad u(z)=z. \end{align*} Equivalently, there is a surjective *-homomorphism $\sigma:\mathcal T\to C(S^1)$ such that \begin{align*} \sigma(S)=u \end{align*} and \begin{align*} \ker(\sigma)=K(\ell^2(\mathbb N)). \end{align*} Now exactness is visible. The inclusion $\iota:K(\ell^2(\mathbb N))\to\mathcal T$ is a *-homomorphism, and \begin{align*} \ker(\iota)=\{T\in K(\ell^2(\mathbb N)):T=0\}=\{0\}. \end{align*} The symbol map satisfies \begin{align*} \ker(\sigma)=K(\ell^2(\mathbb N))=\operatorname{im}(\iota), \end{align*} and $\sigma$ is surjective because $C(S^1)=C^*(u)$ and $u=\sigma(S)$ lies in its range. Therefore \begin{align*} 0\longrightarrow K(\ell^2(\mathbb N))\xrightarrow{\iota}\mathcal T\xrightarrow{\sigma}C(S^1)\longrightarrow 0 \end{align*} is a short exact sequence of C*-algebras. The extension says that a Toeplitz operator has a commutative symbol in $C(S^1)$, and the ambiguity in recovering the operator from its symbol is exactly compact-operator error. [/example] The chapter’s main lesson is that closed ideals, quotient C*-algebras, and approximate identities form a single package. Closed ideals are exactly kernels of representations, quotients remain C*-algebras, approximate identities let nonunital ideals be handled by unital methods locally, and extensions keep track of an algebra together with an ideal and the quotient it determines. After understanding how ideals sit inside an algebra, the remaining issue is how to enlarge a nonunital algebra without losing its structure. Multiplier algebras and unitization answer that question by providing the natural ambient algebra in which external multiplication and essential ideals can be handled uniformly. # 7. Multiplier Algebras and Unitization This chapter explains how a nonunital C*-algebra can still behave as if it sits inside a larger unital algebra. The guiding question is not only how to adjoin a formal identity, but also how to find the largest natural algebra of operators that multiply a given algebra from the outside. This leads from unitization to essential ideals, then to double centralizers, multiplier algebras, and the strict topology. ## Adjoining a Unit A nonunital algebra often appears as an ideal in a unital algebra, and many arguments about spectra, functional calculus, and invertibility are most natural in the presence of an identity. The first problem is to adjoin a unit without changing the original multiplication or its C*-norm. [definition: Algebraic Unitization] Let $A$ be a nonunital complex algebra. The algebraic unitization of $A$ is the vector space \begin{align*} A^+ = A \oplus \mathbb C \end{align*} equipped with the binary operation $A^+\times A^+\to A^+$ given by \begin{align*} (a,\lambda)(b,\mu)=(ab+\lambda b+\mu a,\lambda\mu). \end{align*} [/definition] The element $(0,1)$ is the identity of $A^+$, and $A$ is identified with the ideal $A\oplus\{0\}$. The formula is forced by the requirement that $(0,1)$ act as an identity and that $A$ remain an ideal. Without the cross-terms $\lambda b+\mu a$, the new scalar summand would fail to multiply elements of $A$ as a unit does. When $A$ is a *-algebra, the involution is extended by $(a,\lambda)^*=(a^*,\overline{\lambda})$; this is the only extension fixing $A$ and making $(0,1)$ self-adjoint. This motivates the following theorem, which says that the construction is determined not by the chosen formula alone but by how maps out of $A$ extend to unital algebras. [quotetheorem:8573] [citeproof:8573] The scalar-overlap clause is part of the content, not a technical afterthought. For example, embed $C_0(0,1)$ into $C([0,1])$ by extension by zero on the boundary; the image contains no nonzero scalar multiple of $1$, so the unitized map remains faithful. By contrast, the quotient map $C_0(0,1]\to \mathbb C$ given by evaluation at $1$ has image equal to the scalars, and its unitized extension kills nonzero elements of the form $(a,-\pi(a))$. The theorem does not say that every unital extension is faithful, nor that $A^+$ contains all possible multipliers; it only describes the smallest unital algebra forced by a map out of $A$. This limitation points toward multiplier algebras, where external left and right multiplication are retained rather than collapsed to scalars. [example: Adjoining a Unit to C Naught] Let $A=c_0$, with pointwise operations and the sup norm, and write $1=(1,1,1,\ldots)$. In the algebraic unitization, an element $(x,\lambda)\in c_0^+$ acts like the sequence $x+\lambda 1$, whose $n$th coordinate is $x_n+\lambda$. Since $x_n\to0$, we have \begin{align*} \lim_{n\to\infty}(x_n+\lambda)=\lambda. \end{align*} Thus every element of $c_0^+$ gives a convergent sequence. Conversely, if $y=(y_n)$ is a convergent complex sequence and $\lambda=\lim_{n\to\infty}y_n$, then the sequence $x=y-\lambda 1$ satisfies \begin{align*} x_n=y_n-\lambda\to0. \end{align*} Hence $x\in c_0$ and $y=x+\lambda 1$. The decomposition is unique: if $x+\lambda 1=x'+\lambda'1$ with $x,x'\in c_0$, then \begin{align*} x-x'=(\lambda'-\lambda)1. \end{align*} The left side converges to $0$, while the right side is the constant sequence with value $\lambda'-\lambda$, so $\lambda'=\lambda$ and then $x=x'$. Under this identification, multiplication agrees with the unitization formula because the $n$th coordinate of $(x+\lambda 1)(z+\mu 1)$ is \begin{align*} (x_n+\lambda)(z_n+\mu)=x_nz_n+\lambda z_n+\mu x_n+\lambda\mu. \end{align*} Therefore \begin{align*} (x+\lambda 1)(z+\mu 1)=xz+\lambda z+\mu x+\lambda\mu 1, \end{align*} which is exactly the product corresponding to $(xz+\lambda z+\mu x,\lambda\mu)$. Thus $c_0^+$ identifies with $c$, the C*-algebra of all convergent complex sequences, and the adjoined unit is $1=(1,1,1,\ldots)$. The quotient map $c\to c/c_0$ is represented by $y\mapsto \lim_n y_n$: its kernel is $c_0$, and $x+\lambda 1$ is sent to $\lambda$, the scalar component. [/example] This example shows why unitization belongs to noncompact topology. Under Gelfand duality, $c_0=C_0(\mathbb N)$, and adjoining a unit corresponds to passing from the locally compact space $\mathbb N$ to its one-point compactification. ## Essential Ideals The next question is when an ideal is large enough to detect multiplication in the ambient algebra. This matters because multiplier algebras will be characterized by how they act on an ideal from both sides. [definition: Essential Ideal] Let $B$ be a C*-algebra. A closed two-sided ideal $I\trianglelefteq B$ is essential if for every nonzero closed two-sided ideal $J\trianglelefteq B$, \begin{align*} I\cap J\ne \{0\}. \end{align*} [/definition] The definition is ideal-theoretic, but in applications we often test essentiality by multiplying elements. The following criterion converts the condition into an annihilator statement, which is the form needed when $A$ is embedded into a candidate multiplier algebra. [quotetheorem:8574] [citeproof:8574] The closed two-sided ideal hypotheses are the C*-algebra setting in which the annihilator test is stable under the operations used in the proof. Two-sidedness ensures that products with an ambient element remain inside the tested subalgebra, while closedness ensures that the ideal generated by $b^*b$ is again an object covered by the definition of essentiality. If $B=B(H)$ and $I=K(H)$, multiplication by rank-one operators detects every bounded operator. By contrast, if one tests only against a non-ideal linear subspace, such as a single one-dimensional span of a projection, many nonzero operators vanish on that subspace from one side, and the products no longer generate a closed two-sided ideal whose intersection with $I$ can be controlled. The criterion does not classify all ideals of $B$, and it does not say that an essential ideal is large in norm or contains an identity. It says exactly that annihilation on the ideal is impossible for a nonzero ambient element, which is the detection property needed for multiplier embeddings. [example: Compact Operators as an Essential Ideal] Let $H$ be a Hilbert space and let $K(H)\trianglelefteq B(H)$ be the closed ideal of compact operators. For $\xi,\eta\in H$, write the rank-one operator $\xi\otimes\eta$ as \begin{align*} (\xi\otimes\eta)(\zeta)=\langle \zeta,\eta\rangle \xi \end{align*} for $\zeta\in H$. If $T\in B(H)$ satisfies $TK=0$ for every $K\in K(H)$, then $T(\xi\otimes\eta)=0$ for all $\xi,\eta\in H$ because every rank-one operator is compact. Choose a unit vector $\eta$. For any $\xi\in H$, \begin{align*} 0=(T(\xi\otimes\eta))(\eta)=T((\xi\otimes\eta)(\eta))=T(\langle \eta,\eta\rangle \xi)=T\xi. \end{align*} Thus $T\xi=0$ for every $\xi\in H$, so $T=0$. This annihilator computation also matches the ideal-theoretic definition of essentiality. If $J\trianglelefteq B(H)$ is a nonzero closed two-sided ideal, choose $S\in J$ with $S\ne0$. Then there is $\alpha\in H$ such that $\beta=S\alpha\ne0$. The operator $(\beta\otimes\beta)S$ belongs to $J$ because $J$ is a left ideal, and it is compact because $\beta\otimes\beta$ is finite-rank. It is nonzero, since \begin{align*} ((\beta\otimes\beta)S)(\alpha)=(\beta\otimes\beta)(\beta)=\langle \beta,\beta\rangle \beta=\|\beta\|^2\beta\ne0. \end{align*} Hence $J\cap K(H)$ contains the nonzero compact operator $(\beta\otimes\beta)S$, so every nonzero closed two-sided ideal of $B(H)$ meets $K(H)$ nontrivially. Therefore $K(H)$ is an essential ideal in $B(H)$. [/example] The example suggests a general principle: an essential ideal can determine elements of the larger algebra by how they multiply the ideal. The multiplier algebra is built by reversing this observation. ## Double Centralizers A multiplier of a nonunital algebra should act like left and right multiplication by some element of a larger algebra. Recording only one side loses information in noncommutative examples: a bounded left module map on $K(H)$ may determine left composition, but without the compatible right action there is no guarantee that it comes from multiplication by a single ambient operator on both sides. Since the larger element may not be present inside $A$, we record the two actions directly. [definition: Double Centralizer] Let $A$ be a C*-algebra. A double centralizer on $A$ is a pair $(L,R)$ of bounded linear maps $L,R:A\to A$ such that $L(ab)=L(a)b$, $R(ab)=aR(b)$, and $aL(b)=R(a)b$ for all $a,b\in A$. [/definition] The first two identities say that $L$ is a left module map and $R$ is a right module map. The compatibility identity says that the two maps are the left and right shadows of the same multiplier. If the compatibility condition is dropped, independent left and right module maps can be paired even when they cannot arise from a single element in any ambient algebra. This motivates the following definition, which packages all compatible pairs into a single algebraic object. [definition: Multiplier Algebra] Let $A$ be a C*-algebra. The multiplier algebra $M(A)$ is the set of all double centralizers on $A$, with linear operations defined componentwise, multiplication defined by \begin{align*} (L_1,R_1)(L_2,R_2)=(L_1L_2,R_2R_1), \end{align*} and involution defined by \begin{align*} (L,R)^*=(R^*,L^*). \end{align*} [/definition] The reversal in the second component of the product reflects the fact that right multiplication composes in the opposite order. The definition gives the candidate object; the next result verifies that it is a C*-algebra and that the original algebra sits inside it as the ideal whose multiplications generated the construction. [quotetheorem:8575] [citeproof:8575] The essentiality conclusion is the feature that makes the construction useful: if $A$ sits inside $B=A\oplus C$ as the first summand, then the second summand annihilates $A$ and cannot be recovered from left and right multiplication on $A$. The theorem does not identify $M(A)$ with a pre-existing concrete algebra by itself; it constructs the abstract C*-algebra and the essential copy of $A$. The next theorem makes the maximality slogan precise by comparing $M(A)$ with any unital algebra in which $A$ already appears as an essential ideal. [quotetheorem:8576] [citeproof:8576] Essentiality is the hypothesis that prevents invisible ambient summands. In the direct-sum extension $A\trianglelefteq A\oplus C$, every element of $C$ multiplies $A$ to zero on both sides, so the formula $b\mapsto(L_b,R_b)$ cannot be injective. The theorem does not say that every unital extension of $A$ equals $M(A)$, and it does not give surjectivity of $\Phi$ without an additional maximality condition on $B$. It says that every essential unital extension maps faithfully into the multiplier algebra, which is why concrete identifications of $M(A)$ are proved by showing that a familiar extension supplies all multipliers. ## Concrete Multiplier Algebras The double-centralizer definition is intrinsic, but it is not yet apparent how to recognize its elements in algebras students already know. The problem in this section is to identify when an abstract pair of compatible left and right actions is forced to be multiplication by a familiar object. The two model cases are $K(H)$, where multipliers come from bounded operators, and $C_0(X)$, where multipliers come from bounded continuous functions. [quotetheorem:8577] [citeproof:8577] The compactness hypothesis is doing real work: composing a bounded operator with a compact operator is compact, and finite-rank operators give enough tests to reconstruct the bounded operator. Essentiality is the boundary condition that prevents invisible summands. For example, if $K(H)$ is embedded as $K(H)\oplus\{0\}$ inside $K(H)\oplus \mathbb C$, then the element $(0,1)$ multiplies every element of $K(H)\oplus\{0\}$ to zero on both sides, so it cannot be detected by multiplication on the ideal. The theorem therefore depends on the specific essential inclusion $K(H)\trianglelefteq B(H)$, not merely on placing $K(H)$ inside some unital algebra. Infinite-dimensionality also explains why the result is larger than unitization; for infinite-dimensional $H$, the Calkin algebra $B(H)/K(H)$ is much larger than the scalar quotient coming from $K(H)^+$. The theorem does not say that $B(H)=K(H)^+$, nor that every bounded operator is compact. It identifies the external operators that preserve compactness under left and right composition. [example: Rank-One Reconstruction] Fix a unit vector $\eta\in H$ and let $(L,R)$ be a multiplier of $K(H)$. For $\xi\in H$, put \begin{align*} T\xi=L(\xi\otimes\eta)\eta. \end{align*} Let $P_\eta=\eta\otimes\eta$. Since $\|\eta\|=1$, we have $(\xi\otimes\eta)P_\eta=\xi\otimes\eta$. Using the identity $L(ab)=L(a)b$ for double centralizers, \begin{align*} L(\xi\otimes\eta)=L((\xi\otimes\eta)P_\eta)=L(\xi\otimes\eta)P_\eta. \end{align*} Therefore, for every $\zeta\in H$, \begin{align*} L(\xi\otimes\eta)\zeta=L(\xi\otimes\eta)(\langle \zeta,\eta\rangle\eta)=\langle \zeta,\eta\rangle L(\xi\otimes\eta)\eta=\langle \zeta,\eta\rangle T\xi. \end{align*} Thus \begin{align*} L(\xi\otimes\eta)=(T\xi)\otimes\eta. \end{align*} The map $T$ is linear because $L$ is linear. It is bounded, since \begin{align*} \|T\xi\|=\|(T\xi)\otimes\eta\|=\|L(\xi\otimes\eta)\|\le \|L\|\,\|\xi\otimes\eta\|=\|L\|\,\|\xi\|. \end{align*} Now let $\zeta\in H$. Since $\xi\otimes\zeta=(\xi\otimes\eta)(\eta\otimes\zeta)$, the same centralizer identity gives \begin{align*} L(\xi\otimes\zeta)=L((\xi\otimes\eta)(\eta\otimes\zeta))=L(\xi\otimes\eta)(\eta\otimes\zeta)=((T\xi)\otimes\eta)(\eta\otimes\zeta)=(T\xi)\otimes\zeta. \end{align*} Hence $L$ agrees with left composition by $T$ on every rank-one operator. If \begin{align*} F=\sum_{j=1}^n \xi_j\otimes\zeta_j \end{align*} is finite-rank, then by linearity \begin{align*} L(F)=\sum_{j=1}^n L(\xi_j\otimes\zeta_j)=\sum_{j=1}^n (T\xi_j)\otimes\zeta_j=TF. \end{align*} Finite-rank operators are norm dense in $K(H)$, and both $L$ and $F\mapsto TF$ are bounded maps on $K(H)$, so $L(K)=TK$ for every $K\in K(H)$. Thus the left half of the multiplier is already forced by its values on rank-one operators, and the same reconstruction is the concrete reason that multipliers of $K(H)$ come from bounded operators on $H$. [/example] The rank-one example gives the operator-theoretic pattern: local tests on small pieces determine the global multiplier. This motivates the commutative theorem below, where the small pieces are functions supported in compact neighbourhoods and the global multipliers become bounded continuous functions. ## Multipliers of $C_0$ of a Locally Compact Space The rank-one reconstruction has a commutative analogue: instead of probing an operator with finite-rank maps, we probe a multiplier on $C_0(X)$ with functions supported near a point. Write $C_c(X)$ for the continuous complex-valued functions on $X$ with compact support. Local compactness supplies enough functions in $C_c(X)$ to make pointwise reconstruction possible. Without local compactness, the algebra $C_0(X)$ may not have enough local bump functions to recover a multiplier from neighbourhood tests. [quotetheorem:8578] [citeproof:8578] The Hausdorff and local compactness assumptions are what supply the compactly supported tests used in the proof. At a point $x\in X$, local compactness and the Hausdorff property give a neighbourhood $U$ of $x$ with compact closure and a function $f\in C_c(X)$ with $f(x)=1$ and $\operatorname{supp} f\subset U$. If this supply of local bump functions fails, the quotient formula $g(x)=(Lf)(x)/f(x)$ may have no valid denominator at some points, so the pointwise reconstruction of $g$ can break down before continuity is even addressed. For a locally compact noncompact space such as $\mathbb R$, the theorem produces many multipliers, while if one only adjoins a scalar unit then oscillatory bounded functions such as $e^{ix}$ are missed. The theorem does not say that $C_b(X)=C_0(X)^+$, and it does not identify the one-point compactification with the Stone-Cech compactification except in special cases. It separates the compactification attached to unitization from the larger compactification attached to all bounded continuous functions. [example: Bounded Continuous Functions Acting as Multipliers] Let $X=\mathbb R$ and let $g(x)=e^{ix}$. For every $x\in\mathbb R$, \begin{align*} |g(x)|=|e^{ix}|=1. \end{align*} Thus $g$ is bounded, and $g$ is continuous because $x\mapsto e^{ix}$ is continuous, so $g\in C_b(\mathbb R)$. It is not in $C_0(\mathbb R)$: taking $\varepsilon=\frac12$, the inequality $|g(x)|<\frac12$ fails for every $x\in\mathbb R$. Now suppose, toward a contradiction, that $g=f+\lambda$ for some $f\in C_0(\mathbb R)$ and $\lambda\in\mathbb C$. Then \begin{align*} f(x)=e^{ix}-\lambda. \end{align*} Since $f\in C_0(\mathbb R)$, we must have $f(x)\to0$ as $x\to\infty$, so $e^{ix}\to\lambda$ as $x\to\infty$. But along $x_n=2\pi n$ we have \begin{align*} e^{ix_n}=e^{i2\pi n}=1, \end{align*} while along $y_n=(2n+1)\pi$ we have \begin{align*} e^{iy_n}=e^{i(2n+1)\pi}=-1. \end{align*} A function with a limit at infinity has the same limit along every sequence tending to infinity, so this would force $\lambda=1$ and $\lambda=-1$, impossible. Hence $g$ is not supplied by the ordinary unitization $C_0(\mathbb R)^+$. For every $h\in C_0(\mathbb R)$, the product $gh$ is continuous and \begin{align*} |g(x)h(x)|=|g(x)|\,|h(x)|=|h(x)|. \end{align*} Given $\varepsilon>0$, since $h\in C_0(\mathbb R)$ there is $R>0$ such that $|h(x)|<\varepsilon$ whenever $|x|>R$. Therefore $|g(x)h(x)|<\varepsilon$ whenever $|x|>R$, so $gh\in C_0(\mathbb R)$. Thus multiplication by $e^{ix}$ defines a multiplier of $C_0(\mathbb R)$, even though it does not come from adjoining only a scalar unit. [/example] This example is the commutative counterpart of $B(H)$ being larger than $K(H)^+$. The multiplier algebra usually contains the unitization, but it may be much larger. ## The Strict Topology The multiplier algebra is generally too large for its norm topology to reflect approximation by elements of $A$. The final question in this chapter is which topology on $M(A)$ records multiplication against elements of the original algebra. [definition: Strict Topology] Let $A$ be a C*-algebra. The strict topology on $M(A)$ is the locally convex topology generated by the seminorms \begin{align*} p_a:M(A)\to [0,\infty) \end{align*} and \begin{align*} q_a:M(A)\to [0,\infty) \end{align*} defined by \begin{align*} p_a(m)=\|ma\|_A \end{align*} and \begin{align*} q_a(m)=\|am\|_A \end{align*} for $a\in A$ and $m\in M(A)$. [/definition] Strict convergence therefore means convergence after multiplying on either side by every element of $A$. A one-sided version would miss noncommutative behaviour: controlling $ma$ for all $a$ need not control $am$ unless the algebra is commutative or extra symmetry is present. It is weaker than norm convergence on $M(A)$, but it is designed so that approximate identities in $A$ converge to the multiplier identity. This motivates the following theorem, which gives the basic density statement for $A$ inside $M(A)$ in the strict topology. [quotetheorem:8579] [citeproof:8579] Strict topology is the right replacement for norm topology when multipliers are approximated from within the original nonunital algebra. It also explains why multiplier algebras are natural in representation theory, where nondegenerate representations extend from $A$ to $M(A)$ by strict continuity. The two-sided approximate identity hypothesis is essential because the strict topology has both left and right seminorms. A merely left approximate identity would control $\|e_i a-a\|_A$ but could fail to control $\|a e_i-a\|_A$, so it would not give strict convergence. The theorem does not upgrade strict convergence to norm convergence: in $A=K(H)$ with $H$ infinite-dimensional, finite-rank projections increasing strongly to $1_H$ converge strictly in $B(H)$ but have $\|1_H-e_i\|=1$ for every proper projection $e_i$. This is the key reason strict topology, rather than norm topology, is used for extending nondegenerate representations. [remark: Unitization Versus Multiplier Algebra] For a nonunital C*-algebra $A$, there is always a canonical unital inclusion $A^+\to M(A)$ extending the embedding of $A$. Injectivity uses nonunitality: if $a+\lambda 1_{M(A)}=0$ with $\lambda\ne0$, then $-\lambda^{-1}a$ would act as an identity for $A$. The image consists of multipliers of the form $a+\lambda 1_{M(A)}$. This inclusion may be onto, but the examples $A=K(H)$ for infinite-dimensional $H$ and $A=C_0(X)$ with many bounded continuous functions show that $M(A)$ often contains more multipliers than the adjoined scalar unit provides. [/remark] The chapter's main lesson is that unitization solves the formal problem of missing identities, while multiplier algebras solve the structural problem of external multiplication. Essential ideals provide the bridge: if $A$ is essential in a unital C*-algebra, then the ambient algebra is faithfully represented inside $M(A)$, and in the best cases it is the whole multiplier algebra. With unitizations and multipliers in place, tensor products become the next major operation to organize. The goal is to understand how C*-algebras combine, when the completion is unique, and why this behaves so cleanly in the commutative and finite-dimensional cases. # 8. Tensor Products and Nuclearity: First View After Chapter 7's treatment of unitizations and multiplier algebras, this chapter assumes the basic theory of C*-algebras, *-homomorphisms, ideals and unitizations, Hilbert-space representations, and the GNS construction. It also uses the algebraic [tensor product](/page/Tensor%20Product) of complex vector spaces, the Hilbert-space tensor product, finite-dimensional matrix algebras, compact Hausdorff spaces, and the [Stone-Weierstrass theorem](/theorems/886). The main notation from earlier chapters is that $A\odot B$ denotes the algebraic tensor product before completion, while completed C*-tensor products carry a subscript such as $\otimes_{\min}$. Tensor products enter C*-algebra theory as the operation that should model independent systems: if $A$ and $B$ are C*-algebras, then an element of the tensor product should be a finite or completed sum of elementary tensors $a \otimes b$, with multiplication, involution, and norm reflecting the two factors. The new difficulty is that the algebraic tensor product does not carry a canonical C*-norm in general. This chapter develops the minimal tensor product, computes it in the main finite-dimensional and commutative cases, and introduces nuclearity as the condition that no competing C*-tensor norm appears. ## Algebraic Tensor Products and Spatial Representations The first problem is algebraic: before asking for a C*-completion, we need an algebra that records bilinear multiplication from two *-algebras. The algebraic tensor product does this at the vector-space level, but the C*-algebraic point is that the product and involution are forced on elementary tensors. [definition: Algebraic Tensor Product Of Star Algebras] Let $A$ and $B$ be complex *-algebras. Their algebraic tensor product $A \odot B$ is the algebraic vector-space tensor product equipped with multiplication and involution determined by \begin{align*} (a_1 \otimes b_1)(a_2 \otimes b_2) = a_1a_2 \otimes b_1b_2 \end{align*} and \begin{align*} (a \otimes b)^* = a^* \otimes b^* \end{align*} for $a,a_1,a_2 \in A$ and $b,b_1,b_2 \in B$, extended linearly and conjugate-linearly respectively. [/definition] The notation $\odot$ is useful because it reminds us that no norm-completion has yet been taken. The formula says that the two factors commute in separate tensor slots, rather than inside either factor. A concrete algebraic example helps separate the tensor-coordinate bookkeeping from any analytic completion. [example: Tensoring Polynomial $*$-Algebras] Let $A=\mathbb C[z,\bar z]$ and $B=\mathbb C[w,\bar w]$, with involution determined by $z^*=\bar z$, $\bar z^*=z$, $w^*=\bar w$, and $\bar w^*=w$. Define a [linear map](/page/Linear%20Map) \begin{align*} \Phi:A\odot B\to \mathbb C[z,\bar z,w,\bar w] \end{align*} on monomial tensors by \begin{align*} \Phi(z^k\bar z^\ell\otimes w^r\bar w^s)=z^k\bar z^\ell w^r\bar w^s. \end{align*} The monomials $z^k\bar z^\ell$ form a vector-space basis for $A$, the monomials $w^r\bar w^s$ form a vector-space basis for $B$, and the monomials $z^k\bar z^\ell w^r\bar w^s$ form a vector-space basis for $\mathbb C[z,\bar z,w,\bar w]$, so $\Phi$ is a vector-space isomorphism. The product is the tensor product algebra product: \begin{align*} (z\otimes 1)(1\otimes w)=z\cdot 1\otimes 1\cdot w=z\otimes w. \end{align*} Applying $\Phi$ gives \begin{align*} \Phi((z\otimes 1)(1\otimes w))=\Phi(z\otimes w)=zw, \end{align*} which is the product of the two independent coordinate functions in $\mathbb C[z,\bar z,w,\bar w]$. For the involution, use conjugate-linearity and the rule $(a\otimes b)^*=a^*\otimes b^*$. Since \begin{align*} (z^k\bar z^\ell)^*=(\bar z^\ell)^*(z^k)^*=z^\ell\bar z^k \end{align*} and \begin{align*} (w^r\bar w^s)^*=(\bar w^s)^*(w^r)^*=w^s\bar w^r, \end{align*} we get \begin{align*} (z^k\bar z^\ell\otimes w^r\bar w^s)^*=z^\ell\bar z^k\otimes w^s\bar w^r. \end{align*} Thus the algebraic tensor product keeps the $z$-variables and $w$-variables in separate tensor coordinates, while the polynomial algebra $\mathbb C[z,\bar z,w,\bar w]$ records the same bookkeeping as ordinary multiplication of independent variables. [/example] The example shows how the algebraic tensor product keeps the two sources of variables independent. To put a C*-norm on such an algebra, we need to let the two factors act on Hilbert spaces and then make their elementary tensors act on a Hilbert-space tensor product. [definition: Spatial Product Representation] Let $A$ and $B$ be C*-algebras, let $\pi:A\to B(H)$ and $\rho:B\to B(K)$ be *-representations on Hilbert spaces $H$ and $K$. The spatial product representation is the *-representation $\pi \odot \rho:A\odot B \to B(H\otimes K)$ determined by \begin{align*} (\pi\odot\rho)(a\otimes b)(\xi\otimes\eta) = \pi(a)\xi \otimes \rho(b)\eta \end{align*} for $a\in A$, $b\in B$, $\xi\in H$, and $\eta\in K$. [/definition] This construction is the bridge from algebra to norm, but it is only useful if the formula respects multiplication and adjoints. The next result checks that the spatial formula is genuinely a representation of the algebraic tensor product, so its operator norm can be used to measure elements of $A\odot B$. [quotetheorem:8580] [citeproof:8580] The bounded representation hypotheses matter. For example, representing the polynomial *-algebra $\mathbb C[x]$ by the unbounded multiplication operator on $L^2(\mathbb R)$ would not give a map into $B(H)$, so the tensor formula would not define a C*-algebra representation. The theorem also does not say that $\pi\odot\rho$ is faithful, nor that its norm is independent of the representations. Its role is more modest but essential: every pair of concrete C*-representations supplies a seminorm $x\mapsto \| (\pi\odot\rho)(x)\|$ on $A\odot B$, and the minimal tensor norm is obtained by making the faithful spatial choice intrinsic. ## The Minimal Tensor Product The next question is whether the spatial construction gives a norm that depends on how $A$ and $B$ were represented. The answer is that faithful representations give the same C*-norm, so there is a canonical spatial completion. [definition: Minimal Tensor Norm] Let $A$ and $B$ be C*-algebras. Choose faithful *-representations $\pi:A\to B(H)$ and $\rho:B\to B(K)$. The minimal tensor norm associated to these representations is the function $\|\cdot\|_{\min}:A\odot B\to [0,\infty)$ defined by \begin{align*} \|x\|_{\min}:=\|(\pi\odot\rho)(x)\|_{B(H\otimes K)} \end{align*} for $x\in A\odot B$. [/definition] The definition would be unsatisfactory if a different faithful representation changed the value of the norm. The next theorem is therefore the main technical input in the construction: it says the spatial norm is intrinsic to $A$ and $B$, not to the chosen Hilbert spaces. [quotetheorem:8581] [citeproof:8581] Faithfulness cannot be omitted. For example, if $A=\mathbb C^2$ and $\pi$ is evaluation on the first coordinate, then $(0,1)\otimes 1$ has zero image under $\pi\odot\rho$ although it is a nonzero element of $A\odot B$. The theorem also does not say that no other C*-norm exists on $A\odot B$; it only constructs the least spatial norm and proves that this norm is representation-independent. This distinction is exactly what leads to nuclearity: later we ask when the minimal norm is the only possible C*-tensor norm. [definition: Minimal Tensor Product] Let $A$ and $B$ be C*-algebras. Their minimal tensor product $A\otimes_{\min}B$ is the completion of $A\odot B$ with respect to $\|\cdot\|_{\min}$. [/definition] The notation records that this is a completed C*-algebra, not just a vector-space tensor product. The word minimal refers to comparison with other C*-norms: any C*-norm arising from a representation of $A\odot B$ dominates the spatial norm in the standard ordering of C*-tensor norms. A first calculation is obtained by tensoring with matrices. [example: Matrix Amplification] Every element of $M_n(\mathbb C)\odot A$ has a unique form \begin{align*} x=\sum_{i=1}^n\sum_{j=1}^n e_{ij}\otimes a_{ij} \end{align*} because the matrix units $e_{ij}$ form a basis of $M_n(\mathbb C)$. Define \begin{align*} \Psi(x)=(a_{ij})_{i,j}\in M_n(A). \end{align*} For elementary tensors, $\Psi(e_{ij}\otimes a)$ is the matrix with $a$ in the $(i,j)$ entry and $0$ elsewhere. The multiplication agrees with block-matrix multiplication. Since $e_{ij}e_{kl}=\delta_{jk}e_{il}$, \begin{align*} (e_{ij}\otimes a)(e_{kl}\otimes b)=\delta_{jk}e_{il}\otimes ab. \end{align*} Applying $\Psi$ gives the matrix with $\delta_{jk}ab$ in the $(i,l)$ entry, which is exactly the product of the single-entry matrices $\Psi(e_{ij}\otimes a)$ and $\Psi(e_{kl}\otimes b)$. The involution also agrees: \begin{align*} (e_{ij}\otimes a)^*=e_{ji}\otimes a^*. \end{align*} Thus $\Psi((e_{ij}\otimes a)^*)$ has $a^*$ in the $(j,i)$ entry, which is the adjoint of the matrix with $a$ in the $(i,j)$ entry. Now let $\pi:A\to B(H)$ be faithful. Identify $\mathbb C^n\otimes H$ with $H^{\oplus n}$ by sending $u_j\otimes \xi$ to the vector whose $j$th component is $\xi$ and whose other components are $0$. For $x=\sum_{i,j}e_{ij}\otimes a_{ij}$, the spatial representation satisfies \begin{align*} (\operatorname{id}\odot\pi)(x)(u_j\otimes \xi)=\sum_{i=1}^n u_i\otimes \pi(a_{ij})\xi. \end{align*} Under the identification with $H^{\oplus n}$ this is the block operator \begin{align*} (\xi_1,\dots,\xi_n)\mapsto \left(\sum_{j=1}^n\pi(a_{1j})\xi_j,\dots,\sum_{j=1}^n\pi(a_{nj})\xi_j\right). \end{align*} Therefore the minimal norm on $M_n(\mathbb C)\odot A$ is exactly the operator norm of the corresponding block matrix on $H^{\oplus n}$, so the completion is the usual C*-algebra $M_n(A)$. [/example] Matrix amplification is one of the main reasons tensor products are unavoidable in C*-algebra theory. Positivity in $M_n(A)$ will later govern completely positive maps, while tensoring by $M_n(\mathbb C)$ gives a compact way to express all matrix levels simultaneously. ## Finite-Dimensional Computations A reliable test for a tensor product construction is whether it recovers the expected matrix identities. In finite dimensions the algebraic tensor product is already complete once the correct norm is identified, and the minimal tensor product reduces to familiar linear algebra. [quotetheorem:8582] [citeproof:8582] The matrix-algebra hypotheses are doing real work here. If $M_n(\mathbb C)$ is replaced by an infinite-dimensional C*-algebra, the algebraic tensor product need not already be complete, and different C*-norms may compete before completion; the contrast with tensor products involving $B(H)$ is a standard source of non-nuclear phenomena. The theorem also does not identify arbitrary tensor products of finite-dimensional vector spaces with matrix algebras as C*-algebras unless the *-algebra structure and matrix units are respected. Its forward use is that every finite-dimensional C*-algebra is a finite direct sum of matrix blocks, so this computation becomes the basic local model for finite-dimensional nuclearity. [example: Finite-Dimensional Direct Sums] Let \begin{align*} A=\bigoplus_{r=1}^p M_{m_r}(\mathbb C), \qquad B=\bigoplus_{s=1}^q M_{n_s}(\mathbb C). \end{align*} Write $a=(a_1,\dots,a_p)$ and $b=(b_1,\dots,b_q)$. Define a linear map on elementary tensors by \begin{align*} \Theta(a\otimes b)=(a_r\otimes b_s)_{1\le r\le p,\ 1\le s\le q}. \end{align*} For $a,a'\in A$ and $b,b'\in B$, multiplication in the direct sums is componentwise, so \begin{align*} \Theta((a\otimes b)(a'\otimes b'))=\Theta(aa'\otimes bb')=((a_ra'_r)\otimes(b_sb'_s))_{r,s}. \end{align*} On the other hand, multiplication in each block tensor product gives \begin{align*} \Theta(a\otimes b)\Theta(a'\otimes b')=((a_r\otimes b_s)(a'_r\otimes b'_s))_{r,s}=((a_ra'_r)\otimes(b_sb'_s))_{r,s}. \end{align*} Thus $\Theta$ preserves multiplication on elementary tensors, hence on finite sums by bilinearity. Similarly, \begin{align*} \Theta((a\otimes b)^*)=\Theta(a^*\otimes b^*)=(a_r^*\otimes b_s^*)_{r,s}=(a_r\otimes b_s)^*_{r,s}=\Theta(a\otimes b)^*. \end{align*} So $\Theta$ is a *-homomorphism from $A\odot B$ into the finite direct sum of the block tensor products. The map is bijective at the algebraic level. Indeed, the $(r,s)$-block receives $a_r\otimes b_s$ by tensoring the element of $A$ with only $a_r$ in its $r$th coordinate and the element of $B$ with only $b_s$ in its $s$th coordinate. Therefore \begin{align*} A\odot B\cong \bigoplus_{r=1}^p\bigoplus_{s=1}^q \left(M_{m_r}(\mathbb C)\odot M_{n_s}(\mathbb C)\right) \end{align*} as *-algebras. By *Tensor Product Of Matrix Algebras*, each block completes as \begin{align*} M_{m_r}(\mathbb C)\otimes_{\min}M_{n_s}(\mathbb C)\cong M_{m_rn_s}(\mathbb C). \end{align*} Since a finite direct sum of C*-algebras has norm equal to the maximum of the summand norms, completing the algebraic direct sum gives \begin{align*} A\otimes_{\min}B\cong \bigoplus_{r=1}^p\bigoplus_{s=1}^q M_{m_rn_s}(\mathbb C). \end{align*} Thus tensoring finite-dimensional direct sums distributes over the summands, and each pair of matrix blocks contributes one matrix block of size $m_rn_s$. [/example] A second finite-dimensional form appears when only one factor is a matrix algebra and the other factor varies over a compact space. This combines matrix amplification with the commutative picture developed in the next section. [example: Continuous Matrix-Valued Functions] Let $X$ be compact Hausdorff, and let $e_{ij}$ denote the matrix units in $M_n(\mathbb C)$. Define \begin{align*} \Gamma:C(X)\odot M_n(\mathbb C)\to C(X,M_n(\mathbb C)) \end{align*} by \begin{align*} \Gamma(f\otimes T)(x)=f(x)T. \end{align*} For elementary tensors, \begin{align*} \Gamma((f\otimes S)(g\otimes T))(x)=\Gamma(fg\otimes ST)(x)=f(x)g(x)ST. \end{align*} Also, \begin{align*} \Gamma(f\otimes S)(x)\Gamma(g\otimes T)(x)=f(x)Sg(x)T=f(x)g(x)ST, \end{align*} because $f(x)$ and $g(x)$ are scalars. Hence $\Gamma$ preserves multiplication on elementary tensors, and therefore on finite sums by bilinearity. For the involution, \begin{align*} \Gamma((f\otimes T)^*)(x)=\Gamma(\overline f\otimes T^*)(x)=\overline{f(x)}T^*. \end{align*} On the other hand, \begin{align*} \Gamma(f\otimes T)(x)^*=(f(x)T)^*=\overline{f(x)}T^*, \end{align*} so $\Gamma$ is a *-homomorphism. The map is onto already at the algebraic level. If $F\in C(X,M_n(\mathbb C))$ and $F_{ij}(x)$ is its $(i,j)$ entry, then each $F_{ij}\in C(X)$ and \begin{align*} F(x)=\sum_{i=1}^n\sum_{j=1}^n F_{ij}(x)e_{ij}. \end{align*} Thus \begin{align*} F=\Gamma\left(\sum_{i=1}^n\sum_{j=1}^n F_{ij}\otimes e_{ij}\right). \end{align*} It is injective because the matrix units are linearly independent: if $\Gamma(\sum_{i,j} f_{ij}\otimes e_{ij})=0$, then the $(i,j)$ entry gives $f_{ij}(x)=0$ for every $x\in X$, so $f_{ij}=0$ for every $i,j$. Under this identification, an element $\sum_{i,j} f_{ij}\otimes e_{ij}$ becomes the matrix-valued function \begin{align*} x\mapsto (f_{ij}(x))_{i,j}. \end{align*} By *Matrix Amplification*, the minimal norm is the operator norm of this matrix at each point, and the norm on the continuous field is therefore \begin{align*} \left\|\sum_{i,j} f_{ij}\otimes e_{ij}\right\|_{\min}=\sup_{x\in X}\|(f_{ij}(x))_{i,j}\|_{M_n(\mathbb C)}. \end{align*} So $C(X)\otimes_{\min}M_n(\mathbb C)$ is exactly $C(X,M_n(\mathbb C))$ with its usual uniform operator norm. [/example] These examples suggest that the minimal tensor product behaves like the expected product operation whenever the factors have enough finite-dimensional or commutative structure. The next theorem makes the commutative case exact. ## Commutative Tensor Products For commutative C*-algebras, tensor products should match products of spaces. If $C(X)$ represents functions on $X$ and $C(Y)$ represents functions on $Y$, then $f\otimes g$ should represent the function $(x,y)\mapsto f(x)g(y)$ on $X\times Y$. [quotetheorem:8583] [citeproof:8583] Compactness is part of the statement because it keeps the notation unital and identifies the Gelfand transform with $C(X)$ rather than $C_0(X)$. For example, when $X=\mathbb N$ is locally compact but not compact, the corresponding commutative C*-algebra is $C_0(\mathbb N)$, and the product formula must be written with $C_0(X\times Y)$ instead of $C(X\times Y)$. The theorem also does not say that tensor products of noncommutative algebras are functions on an ordinary product space; it is a commutative calculation. Its forward role is to make the geometric meaning of the minimal tensor product explicit before nuclearity abstracts the uniqueness feature away from commutativity. [example: Functions On A Finite Product Space] Let $X=\{1,\dots,m\}$ and $Y=\{1,\dots,n\}$ with the discrete topology. The identification $C(X)\cong \mathbb C^m$ sends $f$ to $(f(1),\dots,f(m))$, and the identification $C(Y)\cong \mathbb C^n$ sends $g$ to $(g(1),\dots,g(n))$. Since $X\times Y$ has the $mn$ points $(i,j)$ with $1\le i\le m$ and $1\le j\le n$, the identification $C(X\times Y)\cong \mathbb C^{mn}$ sends $F$ to the tuple $(F(i,j))_{i,j}$. By *Minimal Tensor Product Of Commutative C Star Algebras*, the map \begin{align*} \Phi:C(X)\otimes_{\min}C(Y)\to C(X\times Y) \end{align*} is determined on elementary tensors by \begin{align*} \Phi(f\otimes g)(i,j)=f(i)g(j). \end{align*} Therefore, under $C(X)\cong \mathbb C^m$ and $C(Y)\cong \mathbb C^n$, \begin{align*} \mathbb C^m\otimes_{\min}\mathbb C^n\cong C(X)\otimes_{\min}C(Y)\cong C(X\times Y)\cong \mathbb C^{mn}. \end{align*} For the coordinate functions, let $\delta_i\in C(X)$ and $\epsilon_j\in C(Y)$ be defined by $\delta_i(k)=1$ if $k=i$ and $0$ otherwise, and $\epsilon_j(\ell)=1$ if $\ell=j$ and $0$ otherwise. Then \begin{align*} \Phi(\delta_i\otimes \epsilon_j)(k,\ell)=\delta_i(k)\epsilon_j(\ell). \end{align*} This value is $1$ exactly when $k=i$ and $\ell=j$, and it is $0$ otherwise. Thus $\delta_i\otimes\epsilon_j$ becomes the coordinate function of the point $(i,j)$ in $X\times Y$, so the tensor product basis is indexed by pairs of coordinates. [/example] This finite example matches the direct-sum computation above, since finite-dimensional commutative C*-algebras are direct sums of copies of $\mathbb C$. The agreement is an important consistency check: matrix blocks and finite spaces are being treated by the same tensor product. ## Nuclearity as Uniqueness of Tensor Norm The minimal tensor product is canonical, but it is not the only possible C*-completion of $A\odot B$ in general. The basic problem behind nuclearity is to identify algebras $A$ for which tensoring with $A$ leaves no choice of C*-norm, regardless of the other factor. [definition: C Star Tensor Norm] Let $A$ and $B$ be C*-algebras. A C*-tensor norm on $A\odot B$ is a function $\|\cdot\|_\alpha:A\odot B\to [0,\infty)$ that is a C*-norm for the algebraic *-algebra structure on $A\odot B$. Its completion $A\otimes_\alpha B$ is required to contain the canonical images of $A$ and $B$ isometrically: in the unital case these are the *-homomorphisms $A\to A\otimes_\alpha B$, $a\mapsto a\otimes 1_B$, and $B\to A\otimes_\alpha B$, $b\mapsto 1_A\otimes b$. In the nonunital case, after passing to unitizations, these are encoded by the nondegenerate multiplier-algebra maps \begin{align*} \iota_A:A\to M(A\otimes_\alpha B),\qquad \iota_A(a)(c\otimes d)=ac\otimes d, \end{align*} and \begin{align*} \iota_B:B\to M(A\otimes_\alpha B),\qquad \iota_B(b)(c\otimes d)=c\otimes bd, \end{align*} initially on elementary tensors and then extended to the completion. [/definition] This definition is deliberately broad because later chapters compare the minimal norm with the maximal norm. The distinction is not cosmetic: for nonabelian free-group C*-algebras, the minimal and maximal tensor norms on suitable algebraic tensor products differ, a phenomenon closely tied to the failure of finite-dimensional approximation properties. For the first view of nuclearity, the essential question is whether this broad definition ever permits more than one answer. [definition: Nuclear C Star Algebra] A C*-algebra $A$ is nuclear if for every C*-algebra $B$, the algebraic tensor product $A\odot B$ admits exactly one C*-tensor norm. [/definition] The definition reduces nuclearity to a uniqueness property for tensor norms, but it gives no examples by itself. Since different tensor norms can genuinely disagree, one must identify classes where every possible completion is forced to coincide. The commutative and finite-dimensional computations already proved suggest two places where no hidden tensor-norm choice should remain: continuous functions on compact spaces and matrix-block algebras. To turn nuclearity from a definition into a usable class, the first task is to verify these basic test cases. The commutative case should behave like functions on a product space, while the finite-dimensional case should behave like a finite matrix block that cannot support competing completions. Establishing these examples supplies the initial stock of nuclear algebras used in later permanence arguments. The next formal result is needed because nuclearity is quantified over every partner algebra $B$, so isolated computations of particular tensor products are not enough on their own. It packages the product-space behavior of commutative algebras and the rigidity of finite matrix blocks into genuine universal uniqueness statements, giving the first reliable sources of nuclear C*-algebras. [quotetheorem:8584] [citeproof:8584] The hypotheses describe classes of algebras, not all C*-algebras. A standard counterexample to unrestricted uniqueness comes from full group C*-algebras of nonabelian free groups, where minimal and maximal tensor norms can differ on algebraic tensor products. The theorem also does not say that every extension, quotient, or infinite limit built from these examples is automatically nuclear without further permanence results. Its importance is that it anchors nuclearity in two familiar worlds: ordinary compact spaces and finite-dimensional matrix mechanics, with later chapters connecting the same condition to approximation by matrix algebras, exactness questions, and operator-space tensor products. [example: Nuclearity Of Matrix-Valued Continuous Functions] Let $X$ be compact Hausdorff. By *Continuous Matrix-Valued Functions*, the map \begin{align*} \Gamma:C(X)\odot M_n(\mathbb C)\to C(X,M_n(\mathbb C)) \end{align*} defined by $\Gamma(f\otimes T)(x)=f(x)T$ extends to a *-isomorphism \begin{align*} C(X)\otimes_{\min}M_n(\mathbb C)\cong C(X,M_n(\mathbb C)). \end{align*} Thus an elementary tensor $f\otimes e_{ij}$ corresponds to the matrix-valued function whose value at $x$ has the scalar $f(x)$ in the $(i,j)$ entry and $0$ in every other entry. We show why this algebra is nuclear. Let $B$ be any C*-algebra. Using the matrix units $e_{ij}$, every element of $C(X,M_n(\mathbb C))\odot B$ can be written uniquely as \begin{align*} \sum_{i=1}^n\sum_{j=1}^n e_{ij}\otimes h_{ij} \end{align*} with $h_{ij}\in C(X)\odot B$, after identifying $C(X,M_n(\mathbb C))$ with $M_n(C(X))$. Multiplication is block multiplication because \begin{align*} (e_{ij}\otimes h)(e_{kl}\otimes h')=\delta_{jk}e_{il}\otimes hh', \end{align*} and the involution is \begin{align*} (e_{ij}\otimes h)^*=e_{ji}\otimes h^*. \end{align*} Since $C(X)$ is nuclear by *Commutative And Finite-Dimensional Algebras Are Nuclear*, the algebraic tensor product $C(X)\odot B$ has exactly one C*-tensor norm. Matrix amplification then fixes the norm on $M_n(C(X)\odot B)$, because an element $(h_{ij})$ must have the usual operator norm as a block matrix in $M_n(C(X)\otimes_{\min}B)$. Hence $C(X,M_n(\mathbb C))\odot B$ has only one C*-tensor norm for every $B$, so $C(X,M_n(\mathbb C))$ is nuclear. Under the same identification, the minimal tensor product with $B$ is \begin{align*} C(X,M_n(\mathbb C))\otimes_{\min}B\cong M_n(C(X)\otimes_{\min}B). \end{align*} Equivalently, its elements are continuous functions on $X$ whose values are $n\times n$ matrices with entries in $B$, with norm \begin{align*} \|(F_{ij})\|=\sup_{x\in X}\|(F_{ij}(x))\|_{M_n(B)}. \end{align*} So matrix-valued continuous functions are nuclear for the same visible reason as the earlier examples: the commutative part contributes the unique tensor norm, and the matrix part only records that norm at finitely many matrix levels. [/example] At this stage nuclearity should be viewed as a promise that tensor products behave as they do in the commutative and finite-dimensional examples. The later theory replaces this promise by intrinsic approximation criteria, but the guiding intuition remains: nuclear C*-algebras are those for which the algebraic tensor product has a unique C*-algebraic completion. Tensor products reveal that finite-dimensional examples already contain much of the structural intuition behind the general theory. In fact, finite-dimensional C*-algebras split into matrix blocks, and AF algebras arise by assembling those blocks through inductive limits. # 9. Finite-Dimensional and $\mathrm{AF}$ $C^*$-Algebras Finite-dimensional $C^*$-algebras are the first place where the abstract axioms become completely concrete. This chapter explains why no hidden finite-dimensional examples exist: every finite-dimensional $C^*$-algebra is built from full matrix algebras by taking finite direct sums. We then use these building blocks to construct larger algebras as inductive limits, leading to approximately finite-dimensional algebras and their Bratteli diagrams. The guiding theme is approximation by finite matrix data. In finite dimension, projections and minimal ideals give a rigid structure theorem; in the AF setting, the same finite-dimensional pieces are assembled along directed systems. The chapter ends with the first contact with dimension groups, the invariant that records how projections pass through the system. ## Recognising Finite-Dimensional $C^*$-Algebras The basic question is: if a $C^*$-algebra has finite vector-space dimension, how much freedom is left in its multiplication and involution? The answer is surprisingly rigid. Finite-dimensional $C^*$-algebras are exactly finite sums of matrix algebras, so the noncommutative part is entirely accounted for by full matrix blocks. Before stating the theorem, we isolate the central algebraic object that separates the blocks. In a direct sum such as $M_2(\mathbb C) \oplus M_3(\mathbb C)$, the two summands are picked out by central projections; these projections are the finite-dimensional replacement for connected components in the commutative case. [definition: Central Projection] Let $A$ be a $C^*$-algebra. A central projection in $A$ is an element $p \in A$ such that $p=p^*=p^2$ and $pa=ap$ for every $a \in A$. [/definition] A central projection $p$ splits the algebra into the ideal $pA$ and the complementary ideal $(1-p)A$ when $A$ is unital. Thus finding enough central projections is the same as decomposing the algebra into independent pieces. [example: Central Projections In A Direct Sum] Let $A=M_m(\mathbb C)\oplus M_n(\mathbb C)$, with coordinatewise operations and identity $1_A=(I_m,I_n)$. For $p=(I_m,0)$, we have \begin{align*} p^*=(I_m^*,0^*)=(I_m,0)=p \end{align*} and \begin{align*} p^2=(I_m,0)(I_m,0)=(I_m^2,0^2)=(I_m,0)=p. \end{align*} Thus $p$ is a projection. Similarly, \begin{align*} q^*=(0^*,I_n^*)=(0,I_n)=q \end{align*} and \begin{align*} q^2=(0,I_n)(0,I_n)=(0^2,I_n^2)=(0,I_n)=q, \end{align*} so $q$ is also a projection. For any $(a,b)\in A$, coordinatewise multiplication gives \begin{align*} p(a,b)=(I_m,0)(a,b)=(a,0) \end{align*} and \begin{align*} (a,b)p=(a,b)(I_m,0)=(a,0), \end{align*} so $p(a,b)=(a,b)p$. Likewise, \begin{align*} q(a,b)=(0,I_n)(a,b)=(0,b) \end{align*} and \begin{align*} (a,b)q=(a,b)(0,I_n)=(0,b), \end{align*} so $q(a,b)=(a,b)q$. Hence both $p$ and $q$ are central projections. Their sum and product are \begin{align*} p+q=(I_m,0)+(0,I_n)=(I_m,I_n)=1_A \end{align*} and \begin{align*} pq=(I_m,0)(0,I_n)=(0,0). \end{align*} Finally, \begin{align*} pA=\{p(a,b):(a,b)\in A\}=\{(a,0):a\in M_m(\mathbb C)\}=M_m(\mathbb C)\oplus 0 \end{align*} and \begin{align*} qA=\{q(a,b):(a,b)\in A\}=\{(0,b):b\in M_n(\mathbb C)\}=0\oplus M_n(\mathbb C). \end{align*} Thus the two central projections separate the direct sum exactly into its two matrix summands. [/example] The example shows how central projections split a known direct sum. The next obstruction is the case where there is no nontrivial central projection at all. Such a summand should be an indivisible finite-dimensional $C^*$-algebra, and the theorem says that the only indivisible building blocks are full matrix algebras. [quotetheorem:8587] [citeproof:8587] The finite-dimensional hypothesis is essential: infinite-dimensional simple $C^*$-algebras such as the compact operators on a separable Hilbert space, after unitisation issues are handled appropriately, are not full matrix algebras. Simplicity is also essential, since $M_m(\mathbb C)\oplus M_n(\mathbb C)$ is finite-dimensional but has nontrivial ideals and nontrivial central projections. The theorem does not classify representations of a finite-dimensional algebra with several summands; it only identifies the indivisible blocks. General finite-dimensional algebras may have several central pieces, so the remaining task is to find the centre, split it into minimal central projections, and apply the simple theorem on each resulting ideal. [quotetheorem:8588] [citeproof:8588] The finite-dimensional hypothesis is doing real work: in infinite dimension the centre need not split into finitely many minimal projections, and even commutative algebras such as $C([0,1])$ are not finite sums of copies of $\mathbb C$. The $C^*$-hypothesis is also stronger than finite-dimensional algebra alone; arbitrary finite-dimensional complex algebras can have nilpotent radicals, while a $C^*$-algebra cannot hide nilpotent ideal structure in this way. The theorem does not choose canonical matrix units inside each block, so it classifies the algebra up to *-isomorphism rather than fixing coordinates. This also explains the commutative case, where every block has size one and the algebra is a function algebra on a finite space. [example: Commutative Finite-Dimensional $C^*$-Algebras] By the finite-dimensional structure theorem, a finite-dimensional $C^*$-algebra has the form $\bigoplus_{j=1}^r M_{n_j}(\mathbb C)$. If $A$ is commutative, then each block must have size $1$: for $n\ge 2$ in $M_n(\mathbb C)$, the standard matrix units satisfy \begin{align*} E_{12}E_{21}=E_{11} \end{align*} and \begin{align*} E_{21}E_{12}=E_{22}, \end{align*} and $E_{11}\neq E_{22}$ because their $(1,1)$ entries are $1$ and $0$, respectively. Thus no block $M_n(\mathbb C)$ with $n\ge 2$ can occur in a commutative finite-dimensional $C^*$-algebra, so $A\cong \mathbb C^r$. Let $X=\{1,\dots,r\}$ with the discrete topology. The map $C(X)\to \mathbb C^r$ given by \begin{align*} f\mapsto (f(1),\dots,f(r)) \end{align*} is a $*$-isomorphism, with pointwise addition, multiplication, and complex conjugation on $C(X)$. A projection $p\in C(X)$ satisfies $p=p^*=p^2$, so for each $x\in X$ we have $\overline{p(x)}=p(x)$ and \begin{align*} p(x)^2=p(x). \end{align*} Hence $p(x)\in\{0,1\}$ for every $x$, and therefore $p=\chi_S$ for the subset $S=\{x\in X:p(x)=1\}$. Since $C(X)$ is commutative, all projections are central. The nonzero minimal central projections are exactly the singleton characteristic functions $\chi_{\{x\}}$: if $S$ has at least two points, then $\chi_{\{x\}}$ and $\chi_{S\setminus\{x\}}$ are nonzero orthogonal projections below $\chi_S$, while a singleton has no nonempty proper subset. [/example] The previous example shows the special case where all blocks are scalar and the centre sees every coordinate. In a genuinely noncommutative block, central projections no longer describe the internal entries of a matrix. To identify and manipulate those entries inside an abstract algebra, we need matrix units. [definition: System Of Matrix Units] Let $A$ be a $C^*$-algebra and let $n \in \mathbb N$. A system of matrix units of size $n$ in $A$ is a family $(e_{ij})_{1\le i,j\le n}$ of elements of $A$ such that $e_{ij}^*=e_{ji}$ and \begin{align*} e_{ij}e_{kl}=\delta_{jk}e_{il} \end{align*} for all $1\le i,j,k,l\le n$. [/definition] The span of a system of matrix units is a copy of $M_n(\mathbb C)$, with $e_{ij}$ playing the role of the standard matrix unit. In finite-dimensional algebras, choosing matrix units in each summand gives an explicit basis adapted to multiplication. [example: Matrix Units In A Block] Inside $M_n(\mathbb C)$, let $E_{ij}$ be the matrix whose $(r,s)$ entry is $1$ when $(r,s)=(i,j)$ and is $0$ otherwise. Its adjoint is obtained by conjugating entries and transposing, so the $(r,s)$ entry of $E_{ij}^*$ is the complex conjugate of the $(s,r)$ entry of $E_{ij}$. Hence this entry is $1$ exactly when $(s,r)=(i,j)$, equivalently when $(r,s)=(j,i)$, and is $0$ otherwise. Therefore $E_{ij}^*=E_{ji}$. For multiplication, the $(r,s)$ entry of $E_{ij}E_{kl}$ is \begin{align*} \sum_{t=1}^n (E_{ij})_{rt}(E_{kl})_{ts}. \end{align*} The factor $(E_{ij})_{rt}$ is nonzero only when $r=i$ and $t=j$, while $(E_{kl})_{ts}$ is nonzero only when $t=k$ and $s=l$. Thus the whole sum is nonzero exactly when $r=i$, $s=l$, and $j=k$. In that case the unique nonzero summand is $1\cdot 1=1$. Therefore \begin{align*} E_{ij}E_{kl}=\delta_{jk}E_{il}. \end{align*} So the standard matrix units form a system of matrix units in $M_n(\mathbb C)$. Now let $(e_{ij})_{1\le i,j\le n}$ be any system of matrix units in a $C^*$-algebra $A$, and define a linear map $\Phi:M_n(\mathbb C)\to A$ by \begin{align*} \Phi\left(\sum_{i,j=1}^n \alpha_{ij}E_{ij}\right)=\sum_{i,j=1}^n \alpha_{ij}e_{ij}. \end{align*} For two matrices $x=\sum_{i,j}\alpha_{ij}E_{ij}$ and $y=\sum_{k,l}\beta_{kl}E_{kl}$, multiplication in $M_n(\mathbb C)$ gives \begin{align*} xy=\sum_{i,j,k,l}\alpha_{ij}\beta_{kl}E_{ij}E_{kl}=\sum_{i,j,l}\alpha_{ij}\beta_{jl}E_{il}. \end{align*} Applying $\Phi$ gives \begin{align*} \Phi(xy)=\sum_{i,j,l}\alpha_{ij}\beta_{jl}e_{il}. \end{align*} On the other hand, using the matrix-unit relation $e_{ij}e_{kl}=\delta_{jk}e_{il}$, \begin{align*} \Phi(x)\Phi(y)=\left(\sum_{i,j}\alpha_{ij}e_{ij}\right)\left(\sum_{k,l}\beta_{kl}e_{kl}\right)=\sum_{i,j,l}\alpha_{ij}\beta_{jl}e_{il}. \end{align*} Thus $\Phi(xy)=\Phi(x)\Phi(y)$. Also, \begin{align*} x^*=\sum_{i,j}\overline{\alpha_{ij}}E_{ji}=\sum_{i,j}\overline{\alpha_{ji}}E_{ij}, \end{align*} so \begin{align*} \Phi(x^*)=\sum_{i,j}\overline{\alpha_{ji}}e_{ij}. \end{align*} Since $e_{ij}^*=e_{ji}$, \begin{align*} \Phi(x)^*=\left(\sum_{i,j}\alpha_{ij}e_{ij}\right)^*=\sum_{i,j}\overline{\alpha_{ij}}e_{ji}=\sum_{i,j}\overline{\alpha_{ji}}e_{ij}. \end{align*} Hence $\Phi(x^*)=\Phi(x)^*$. Therefore $\Phi$ is a $*$-homomorphism, and its range is exactly the linear span of the $e_{ij}$, the $C^*$-subalgebra generated by this system of matrix units. [/example] ## Embeddings Between Finite-Dimensional Blocks After classifying individual finite-dimensional $C^*$-algebras, the next problem is to understand maps between them. AF algebras are assembled from finite-dimensional stages, so the connecting *-homomorphisms must be encoded in a form that survives passage to a limit. A unital *-homomorphism $M_n(\mathbb C)\to M_m(\mathbb C)$ can exist only when $n$ divides $m$, and then it is, up to unitary conjugacy, an amplification $a\mapsto a\otimes I_k$ with $m=nk$. Direct sums allow several such amplifications at once. [definition: Multiplicity Matrix] Let \begin{align*} A=\bigoplus_{j=1}^r M_{n_j}(\mathbb C), \qquad B=\bigoplus_{i=1}^s M_{m_i}(\mathbb C), \end{align*} and let $\varphi:A\to B$ be a unital *-homomorphism. The multiplicity matrix of $\varphi$ is the matrix $R_\varphi=(r_{ij})\in M_{s\times r}(\mathbb N\cup\{0\})$ such that the representation of the $j$-th summand of $A$ on the $i$-th summand of $B$ occurs with multiplicity $r_{ij}$. [/definition] The entries satisfy the dimension equations \begin{align*} m_i=\sum_{j=1}^r r_{ij}n_j, \qquad 1\le i\le s, \end{align*} when $\varphi$ is unital. Thus a finite-dimensional *-homomorphism is recorded by a nonnegative integer matrix, together with unitary conjugacies that do not affect the later invariant. [example: A Diagonal Embedding] Let $\varphi:M_2(\mathbb C)\oplus \mathbb C\to M_5(\mathbb C)$ be defined by \begin{align*} \varphi(a,\lambda)=\operatorname{diag}(a,a,\lambda). \end{align*} Writing $a=(a_{ij})_{1\le i,j\le 2}$, this means \begin{align*} \varphi(a,\lambda)=\operatorname{diag}((a_{ij})_{1\le i,j\le 2},(a_{ij})_{1\le i,j\le 2},\lambda). \end{align*} Thus the first summand $M_2(\mathbb C)$ acts on two separate $2$-dimensional diagonal blocks, while the second summand $\mathbb C=M_1(\mathbb C)$ acts on one $1$-dimensional diagonal block. The target algebra has one summand, so the multiplicity matrix has one row. The source has two summands, $M_2(\mathbb C)$ and $M_1(\mathbb C)$, so it has two columns. The first entry is $2$, because $a$ appears in two copies, and the second entry is $1$, because $\lambda$ appears in one copy: \begin{align*} R_\varphi=\begin{pmatrix}2 & 1\end{pmatrix}. \end{align*} The corresponding dimension equation is \begin{align*} 5=2\cdot 2+1\cdot 1. \end{align*} So this embedding records the $5$-dimensional target representation as two copies of the defining $2$-dimensional representation of $M_2(\mathbb C)$ together with one scalar copy. [/example] Multiplicity matrices are the bridge from operator algebras to ordered abelian groups. They describe how ranks of projections transform under the connecting maps. [remark: Projections And Multiplicity] In $M_n(\mathbb C)$, projections are classified up to Murray-von Neumann equivalence by their rank. In a direct sum $\bigoplus_{j=1}^r M_{n_j}(\mathbb C)$, a projection therefore has a rank vector $(k_1,\dots,k_r)$ with $0\le k_j\le n_j$. Under a connecting map with multiplicity matrix $R$, the rank vector is multiplied by $R$. [/remark] ## Directed Systems And $C^*$-Inductive Limits The next question is how to turn a sequence, or more generally a directed family, of $C^*$-algebras into a single $C^*$-algebra that contains all stages compatibly. Algebraic direct limits are not enough because the norm and completion must be controlled. For example, if a connecting map kills a nonzero element, the algebraic representative still exists but its eventual norm should become zero in the limit; without a $C^*$-seminorm quotient, the construction would remember elements that have disappeared. The $C^*$-inductive limit is the construction that preserves both the algebraic maps and the operator norm. [definition: Directed Set] A directed set is a partially ordered set $(I,\le)$ such that for every $i,j\in I$ there exists $k\in I$ with $i\le k$ and $j\le k$. [/definition] The directed condition says that any two stages can be compared after passing to a later stage. To use this comparison for algebras, the stages must come with maps that agree whenever we pass through intermediate stages. [definition: Directed System Of $C^*$-Algebras] A directed system of $C^*$-algebras consists of $C^*$-algebras $(A_i)_{i\in I}$ over a directed set $(I,\le)$ and *-homomorphisms $\varphi_{ij}:A_i\to A_j$ for $i\le j$ such that $\varphi_{ii}=\operatorname{id}_{A_i}$ and \begin{align*} \varphi_{jk}\circ \varphi_{ij}=\varphi_{ik} \end{align*} whenever $i\le j\le k$. [/definition] The compatibility equation means that the route through intermediate stages does not matter. We now state the $C^*$-algebraic limit by its universal property, which is the right way to express both the dense union and the uniqueness of the resulting object. [definition: $C^*$-Inductive Limit] Let $(A_i,\varphi_{ij})$ be a directed system of $C^*$-algebras. A $C^*$-inductive limit is a $C^*$-algebra $A$ together with *-homomorphisms $\varphi_i:A_i\to A$ such that $\varphi_j\circ\varphi_{ij}=\varphi_i$ for $i\le j$, the union $\bigcup_i \varphi_i(A_i)$ is dense in $A$, and the pair $(A,(\varphi_i))$ has the universal property for compatible families of *-homomorphisms into any $C^*$-algebra. [/definition] The universal property determines the inductive limit uniquely up to canonical *-isomorphism. Existence requires constructing a pre-$C^*$-algebra from compatible equivalence classes and then completing in the induced $C^*$-seminorm. [quotetheorem:8590] [citeproof:8590] The existence theorem needs directedness because addition and multiplication of two representatives require a common later stage. It also needs the seminorm quotient because non-injective connecting maps can force elements to vanish in the limit, so the limit is not always a literal union. When the connecting maps are injective, this obstruction disappears: the canonical maps are isometric on their images, and the theorem has the concrete reading that the limit is obtained by placing each $A_n$ inside $A_{n+1}$ and then completing the union. Most AF examples in this course will be of that form. [example: The CAR Algebra As An Inductive Limit] Identify $M_{2^n}(\mathbb C)$ with the tensor product $M_2(\mathbb C)^{\otimes n}$. The connecting map is \begin{align*} \varphi_n:M_2(\mathbb C)^{\otimes n}\to M_2(\mathbb C)^{\otimes(n+1)}, \qquad \varphi_n(x)=x\otimes I_2. \end{align*} For $x,y\in M_2(\mathbb C)^{\otimes n}$, tensor-product multiplication gives \begin{align*} \varphi_n(xy)=(xy)\otimes I_2=(x\otimes I_2)(y\otimes I_2)=\varphi_n(x)\varphi_n(y). \end{align*} Also, \begin{align*} \varphi_n(x^*)=x^*\otimes I_2=(x\otimes I_2)^*=\varphi_n(x)^*. \end{align*} Since $\varphi_n(1)=I_{2^n}\otimes I_2=I_{2^{n+1}}$, each $\varphi_n$ is a unital $*$-homomorphism. Thus the system is \begin{align*} M_2(\mathbb C)\xrightarrow{a\mapsto a\otimes I_2} M_2(\mathbb C)^{\otimes 2}\xrightarrow{x\mapsto x\otimes I_2} M_2(\mathbb C)^{\otimes 3}\xrightarrow{x\mapsto x\otimes I_2}\cdots . \end{align*} Under these inclusions, an element $a_1\otimes\cdots\otimes a_n$ at stage $n$ is identified at stage $n+1$ with \begin{align*} a_1\otimes\cdots\otimes a_n\otimes I_2. \end{align*} The algebraic union therefore consists of finite tensor words in $2\times2$ matrices, with identity factors appended beyond the last nontrivial tensor factor. Completing this union in the induced $C^*$-norm gives the CAR algebra, also called the UHF algebra of type $2^\infty$. [/example] ## Approximately Finite-Dimensional $C^*$-Algebras The central approximation question is now: which $C^*$-algebras can be recovered from finite-dimensional subalgebras up to arbitrarily small norm error? Exact finite-dimensional containment is too rigid: an infinite-dimensional algebra such as the CAR algebra is not itself finite-dimensional, but every finite calculation in it takes place arbitrarily close to a matrix stage. The definition must therefore approximate finite sets rather than demand that the whole algebra equal one finite-dimensional subalgebra. These are the AF algebras, retaining much of the combinatorial rigidity of finite-dimensional $C^*$-algebras while being infinite-dimensional enough to show new $C^*$-phenomena. [definition: Approximately Finite-Dimensional $C^*$-Algebra] A $C^*$-algebra $A$ is approximately finite-dimensional, or AF, if for every finite set $F\subset A$ and every $\varepsilon>0$ there exists a finite-dimensional $C^*$-subalgebra $B\subset A$ such that for every $a\in F$ there exists $b\in B$ with $\|a-b\|<\varepsilon$. [/definition] This definition is local in norm: every finite portion of $A$ can be approximated inside a finite-dimensional algebra. To build or recognize examples, however, one needs more than isolated finite-dimensional approximations that may have no relation to one another. The useful constructive question is whether, in the separable case, those local approximations can be organized into an increasing sequence of finite-dimensional subalgebras whose union is dense. [quotetheorem:8591] [citeproof:8591] The separability hypothesis is what allows a sequence rather than a general directed family; without it, there may be no countable dense list to drive the construction. The finite-dimensional hypothesis on the stages is also essential, because the theorem is not saying that every separable $C^*$-algebra is an inductive limit of convenient subalgebras, only that finite-dimensional local approximation is equivalent to such a presentation. The perturbation step is the technical point: approximate containment alone does not literally give nested subalgebras until almost matrix units are corrected to genuine matrix units. The first examples split into commutative systems, where finite-dimensional algebras are diagonal, and highly noncommutative systems, where each stage is a full matrix algebra. [example: Diagonal AF Algebra] Let $D_{2^n}\subset M_{2^n}(\mathbb C)$ be the diagonal subalgebra, identified with functions on the $2^n$ binary words $\{0,1\}^n$ by \begin{align*} \operatorname{diag}(\lambda_w)_{w\in\{0,1\}^n}\longleftrightarrow f,\qquad f(w)=\lambda_w. \end{align*} The embedding $D_{2^n}\to D_{2^{n+1}}$ repeats each diagonal entry twice, so under this identification it sends $f\in C(\{0,1\}^n)$ to the function $g\in C(\{0,1\}^{n+1})$ defined by \begin{align*} g(w,\varepsilon)=f(w) \end{align*} for $w\in\{0,1\}^n$ and $\varepsilon\in\{0,1\}$. Thus the connecting map is pullback along the truncation map $\pi_n:\{0,1\}^{n+1}\to\{0,1\}^n$ given by $\pi_n(w,\varepsilon)=w$. Let $X=\{0,1\}^{\mathbb N}$ with the [product topology](/page/Product%20Topology). For each $n$, let $\rho_n:X\to\{0,1\}^n$ be the map taking a sequence to its first $n$ coordinates. The stage $D_{2^n}$ embeds into $C(X)$ by \begin{align*} f\longmapsto f\circ\rho_n. \end{align*} Compatibility holds because $\rho_n=\pi_n\circ\rho_{n+1}$, hence \begin{align*} (f\circ\pi_n)\circ\rho_{n+1}=f\circ(\pi_n\circ\rho_{n+1})=f\circ\rho_n. \end{align*} The union of the images consists exactly of the locally constant functions depending on finitely many coordinates. This union is dense in $C(X)$. Indeed, if $h\in C(X)$ and $\varepsilon>0$, compactness of $X$ gives [uniform continuity](/page/Uniform%20Continuity), so there is $n$ such that two sequences with the same first $n$ coordinates have $|h(x)-h(y)|<\varepsilon$. Choose one point $x_w$ in each cylinder $\rho_n^{-1}(\{w\})$ and define $f(w)=h(x_w)$. Then for every $x\in X$, with $w=\rho_n(x)$, the points $x$ and $x_w$ have the same first $n$ coordinates, so \begin{align*} |h(x)-f(\rho_n(x))|=|h(x)-h(x_w)|<\varepsilon. \end{align*} Hence the inductive limit is $C(X)$. The space $X=\{0,1\}^{\mathbb N}$ is the Cantor space: it is compact as a product of finite discrete compact spaces, and its basic cylinder sets are both open and closed, so it is totally disconnected. Thus this diagonal AF algebra is a commutative algebra obtained by refining finite binary partitions. [/example] The previous example is AF because it refines finite partitions and remains commutative at every stage. The contrasting problem is to build AF algebras that are as matrix-like as possible while still being infinite-dimensional. This leads to systems where every stage is a single full matrix algebra and every connecting map is an amplification. [definition: UHF Algebra] A UHF algebra is a $C^*$-algebra isomorphic to an inductive limit \begin{align*} M_{n_1}(\mathbb C) \longrightarrow M_{n_2}(\mathbb C) \longrightarrow M_{n_3}(\mathbb C) \longrightarrow \cdots \end{align*} where each connecting map is a unital embedding between full matrix algebras and $n_k\mid n_{k+1}$ for all $k$. [/definition] The acronym stands for uniformly hyperfinite. The divisibility condition ensures that the standard embedding $a\mapsto a\otimes I_{n_{k+1}/n_k}$ is available after choosing matrix identifications. [example: A UHF Algebra Of Type $6^\infty$] Let $A_n=M_{6^n}(\mathbb C)$ for $n\ge 1$, and identify $M_{6^{n+1}}(\mathbb C)$ with $M_{6^n}(\mathbb C)\otimes M_6(\mathbb C)$. Define \begin{align*} \varphi_n:A_n\to A_{n+1},\qquad \varphi_n(a)=a\otimes I_6. \end{align*} For $a,b\in A_n$, tensor-product multiplication gives \begin{align*} \varphi_n(ab)=(ab)\otimes I_6=(a\otimes I_6)(b\otimes I_6)=\varphi_n(a)\varphi_n(b). \end{align*} The involution is also preserved, since \begin{align*} \varphi_n(a^*)=a^*\otimes I_6=(a\otimes I_6)^*=\varphi_n(a)^*. \end{align*} Finally, \begin{align*} \varphi_n(I_{6^n})=I_{6^n}\otimes I_6=I_{6^{n+1}}, \end{align*} so each $\varphi_n$ is a unital $*$-homomorphism. Since the stage sizes satisfy \begin{align*} 6^n=(2\cdot 3)^n=2^n3^n, \end{align*} only the primes $2$ and $3$ appear in the matrix sizes. For each $k\in\mathbb N$, the stage $A_k=M_{6^k}(\mathbb C)$ contains both prime exponents $2^k$ and $3^k$ in its size, so the exponents of $2$ and $3$ are unbounded along the system. The associated supernatural number is therefore \begin{align*} 2^\infty 3^\infty=6^\infty. \end{align*} Thus the inductive limit of this system is a UHF algebra of type $6^\infty$. [/example] ## Bratteli Diagrams Once an AF algebra is presented as a sequence of finite-dimensional algebras, the remaining question is how to draw the connecting maps in a stable way. Listing matrix sizes alone loses information: two systems can have the same summands at every level but different multiplicity matrices, and hence different projection growth. Recording the actual *-homomorphisms up to unitary conjugacy is too detailed for the invariant that will survive telescoping. Bratteli diagrams keep exactly the integer multiplicity data by turning those matrices into a graph. [definition: Bratteli Diagram] A Bratteli diagram is a graded directed graph with finite vertex sets $V_n$ for $n\ge 0$, edge sets from $V_n$ to $V_{n+1}$, and a distinguished initial vertex in $V_0$. For an AF system $A_n=\bigoplus_{j=1}^{r_n}M_{d_{n,j}}(\mathbb C)$ with connecting multiplicity matrices $R_n$, the vertices in $V_n$ correspond to the summands of $A_n$, and the number of edges from vertex $j\in V_n$ to vertex $i\in V_{n+1}$ is $(R_n)_{ij}$. [/definition] The diagram does not record unitary conjugacies inside finite-dimensional target algebras. This is intentional: those conjugacies do not change the inductive limit up to the [equivalence relation](/page/Equivalence%20Relation) used for AF classification. [example: Bratteli Diagram For The CAR Algebra] For the CAR system, the $n$-th finite stage is $A_n=M_{2^n}(\mathbb C)$, which has one simple summand. Hence the Bratteli diagram has one vertex at level $n$. The connecting map is \begin{align*} \varphi_n:M_{2^n}(\mathbb C)\to M_{2^{n+1}}(\mathbb C),\qquad \varphi_n(a)=a\otimes I_2. \end{align*} Identify $\mathbb C^{2^{n+1}}$ with $\mathbb C^{2^n}\otimes \mathbb C^2$. If $(e_1,e_2)$ is the standard basis of $\mathbb C^2$, then \begin{align*} (a\otimes I_2)(\xi\otimes e_1)=a\xi\otimes e_1 \end{align*} and \begin{align*} (a\otimes I_2)(\xi\otimes e_2)=a\xi\otimes e_2 \end{align*} for every $\xi\in\mathbb C^{2^n}$. Thus the target representation splits as one copy on $\mathbb C^{2^n}\otimes \mathbb C e_1$ and one copy on $\mathbb C^{2^n}\otimes \mathbb C e_2$. The multiplicity of the unique source summand inside the unique target summand is therefore $2$. Equivalently, the dimension equation is \begin{align*} 2^{n+1}=2\cdot 2^n. \end{align*} So the multiplicity matrix between consecutive levels is the $1\times 1$ matrix \begin{align*} R_n=\begin{pmatrix}2\end{pmatrix}. \end{align*} By the definition of the Bratteli diagram, the single entry $2$ means that there are two edges from the unique vertex at level $n$ to the unique vertex at level $n+1$. Hence the CAR Bratteli diagram is a single vertical chain with double edges between every pair of consecutive levels. [/example] The CAR diagram shows the one-vertex case. Multiple summands produce several vertices per level, and the multiplicity matrix controls the number of edges between every pair of adjacent levels. [example: Bratteli Diagram With Two Summands] Let $A_n=M_{a_n}(\mathbb C)\oplus M_{b_n}(\mathbb C)$ and let the two vertices at level $n$ correspond, in order, to the summands $M_{a_n}(\mathbb C)$ and $M_{b_n}(\mathbb C)$. Suppose the connecting multiplicity matrix has entries $r_{11}=1$, $r_{12}=1$, $r_{21}=1$, and $r_{22}=2$, where rows index target summands and columns index source summands. By the definition of a Bratteli diagram, the number of edges from source vertex $j$ at level $n$ to target vertex $i$ at level $n+1$ is $r_{ij}$. For the first source vertex, $j=1$, the relevant entries are \begin{align*} r_{11}=1 \quad \text{and} \quad r_{21}=1. \end{align*} Thus there is one edge from the first vertex to the first target vertex and one edge from the first vertex to the second target vertex. For the second source vertex, $j=2$, the entries are \begin{align*} r_{12}=1 \quad \text{and} \quad r_{22}=2. \end{align*} Thus there is one edge from the second vertex to the first target vertex and two edges from the second vertex to the second target vertex. The dimension vector at level $n$ is $(a_n,b_n)$. Multiplication by the multiplicity matrix gives the next dimension vector component by component: \begin{align*} a_{n+1}=r_{11}a_n+r_{12}b_n=1\cdot a_n+1\cdot b_n=a_n+b_n. \end{align*} \begin{align*} b_{n+1}=r_{21}a_n+r_{22}b_n=1\cdot a_n+2\cdot b_n=a_n+2b_n. \end{align*} So the diagram records that the first target block receives one copy of each previous block, while the second target block receives one copy of the first previous block and two copies of the second. [/example] Bratteli diagrams are not unique presentations of AF algebras. Telescoping levels, inserting intermediate finite-dimensional algebras, and changing finite-dimensional identifications can produce different diagrams for the same limit. [remark: Telescoping A Bratteli Diagram] Telescoping replaces several consecutive levels by a single connecting level. Algebraically, this replaces multiplicity matrices $R_n,R_{n+1},\dots,R_m$ by their product. The inductive limit is unchanged because the universal property only depends on the compatible maps through the directed system. [/remark] ## Dimension Groups And The First Classification Theorem The final question in this chapter is what invariant records an AF algebra independently of the chosen diagram. A raw Bratteli diagram still depends on choices: levels may be telescoped, finite-dimensional presentations may be refined, and vertices may be relabelled. What survives these changes is the ordered arithmetic of projection ranks. Projections in finite-dimensional $C^*$-algebras give ordered groups of rank vectors, and the connecting maps between them are exactly the multiplicity matrices. Passing to the limit produces the ordered $K_0$ group. Building on the Bratteli-diagram multiplicity matrices introduced above, this is the first point where AF algebras visibly connect operator algebras with ordered abelian groups and dynamics. The same ordered groups arise from dimension theory for totally disconnected dynamical systems, while the operator-algebraic construction packages them through projections. For a finite-dimensional algebra $A=\bigoplus_{j=1}^r M_{n_j}(\mathbb C)$, the group $K_0(A)$ is naturally $\mathbb Z^r$, with positive cone $\mathbb N_0^r$. The order unit of a unital algebra is the class of $1_A$, which corresponds to the dimension vector $(n_1,\dots,n_r)$. [definition: Scaled Ordered $K_0$ Group] For a unital AF algebra $A$, the scaled ordered $K_0$ group is the triple \begin{align*} (K_0(A),K_0(A)^+,[1_A]), \end{align*} where $K_0(A)^+$ is the positive cone generated by projection classes and $[1_A]$ is the order unit represented by the identity projection. [/definition] This invariant is computed from any Bratteli diagram by taking the inductive limit of the ordered groups attached to the levels. At a finite stage, the ordered group is $\mathbb Z^{r_n}$, and the connecting map is multiplication by the multiplicity matrix. [example: $K_0$ Of The CAR Algebra] At stage $n$, identify $K_0(M_{2^n}(\mathbb C))$ with $\mathbb Z$ by sending the class of a rank-one projection to $1$. Then the identity projection $I_{2^n}$ has rank $2^n$, so its class corresponds to $2^n$. The connecting map for the CAR system is $a\mapsto a\otimes I_2$. If $p\in M_{2^n}(\mathbb C)$ is a projection of rank $k$, then $p\otimes I_2$ acts as the identity on $\operatorname{ran}(p)\otimes \mathbb C^2$, whose dimension is $2k$. Hence the induced map on $K_0$ is \begin{align*} \mathbb Z\to \mathbb Z,\qquad k\mapsto 2k. \end{align*} Thus the ordered group is the inductive limit \begin{align*} \mathbb Z\xrightarrow{\times 2}\mathbb Z\xrightarrow{\times 2}\mathbb Z\xrightarrow{\times 2}\cdots . \end{align*} Represent an element of this limit by a pair $(n,k)$, where $k\in\mathbb Z$ lies at stage $n$. Define \begin{align*} \Psi([(n,k)])=\frac{k}{2^n}. \end{align*} This is well-defined because the connecting map sends $(n,k)$ to $(n+1,2k)$, and \begin{align*} \frac{2k}{2^{n+1}}=\frac{k}{2^n}. \end{align*} Every value of $\Psi$ lies in $\mathbb Z[1/2]$, and every dyadic rational $k/2^n$ is obtained from the class of $(n,k)$. Also, \begin{align*} \Psi([(n,k)]+[(n,l)])=\Psi([(n,k+l)])=\frac{k+l}{2^n}=\frac{k}{2^n}+\frac{l}{2^n}, \end{align*} so $\Psi$ is a group isomorphism from the inductive limit onto $\mathbb Z[1/2]$. The positive cone comes from projection classes, so at stage $n$ it consists of integers $k\ge 0$. Under $\Psi$, these become exactly the nonnegative dyadic rationals: \begin{align*} K_0(\mathrm{CAR})^+=\mathbb Z[1/2]\cap [0,\infty). \end{align*} Finally, the identity class at stage $n$ corresponds to $2^n\in\mathbb Z$, and \begin{align*} \Psi([(n,2^n)])=\frac{2^n}{2^n}=1. \end{align*} Therefore the scaled ordered group of the CAR algebra is $\bigl(\mathbb Z[1/2],\mathbb Z[1/2]\cap[0,\infty),1\bigr)$. [/example] The ordered group just computed for the CAR algebra is obtained from one inductive-limit presentation, but AF algebras can have many different Bratteli diagrams and many different finite-dimensional approximating sequences. The classification problem is whether the resulting scaled ordered $K_0$ group remembers enough to recover the algebra itself, or whether it is only a computable shadow of the construction. For AF algebras, the invariant is strong enough to answer that problem completely. [quotetheorem:8593] This theorem is quoted here as the structural endpoint of the finite-dimensional and AF material, not as a result proved in these notes. Its role is to identify what the preceding constructions have been building toward: for unital AF algebras, the scaled ordered $K_0$ group is not merely computable from a Bratteli diagram, but is a complete isomorphism invariant. [remark: Why The Scale Is Needed] The ordered group alone records stable projection data, but the unit records the size of the algebra as a unital object. For example, different corners of the same AF algebra may have related ordered groups but different distinguished order units. The scale prevents a classification statement for unital algebras from forgetting which projection is the identity. [/remark] This completes the finite-dimensional-to-AF passage. Finite-dimensional $C^*$-algebras are sums of matrix blocks; inductive limits assemble those blocks into AF algebras; Bratteli diagrams encode the assembly; and scaled ordered $K_0$ is the invariant that survives changes of presentation. The AF classification picture now points toward the broader theme of the course: C*-algebras as noncommutative spaces. Having seen how matrix blocks, ideals, states, and tensor products encode structure, we are ready to ask how far topological ideas extend beyond the commutative world. # 10. Noncommutative Spaces and Further Directions ## $C^*$-Algebras as Noncommutative Locally Compact Spaces The guiding question is how much topology can be reconstructed from an algebra of functions. For a locally compact Hausdorff space $X$, the algebra $C_0(X)$ remembers $X$ through its characters, its ideals, and its positive elements. The noncommutative slogan is that an arbitrary C*-algebra $A$ should be read as $C_0(X)$ for a generalized space $X$, with the failure of commutativity recording the failure of the underlying space to be classical. [explanation: The Commutative Dictionary] For a locally compact Hausdorff space $X$, points of $X$ correspond to characters of $C_0(X)$ by evaluation: $x\mapsto \operatorname{ev}_x$, where $\operatorname{ev}_x(f)=f(x)$. Open subsets $U\subset X$ correspond to closed ideals $C_0(U)\subset C_0(X)$, where a function in $C_0(U)$ is extended by $0$ outside $U$. Positive functions encode order and measure-theoretic data, while projections in matrix algebras over $C(X)$ encode vector bundles when $X$ is compact. This dictionary is not merely analogy. The commutative Gelfand-Naimark theorem says that every commutative C*-algebra is of the form $C_0(X)$ for a locally compact Hausdorff space $X$, unique up to homeomorphism. Thus the passage from spaces to commutative C*-algebras is reversible. [/explanation] The dictionary above gives a precise reconstruction problem: if the algebra is commutative, recover the underlying space from its multiplicative linear functionals. The theorem below is the point at which the analytic material from the first half of the course becomes topology. It also fixes the model for every later use of the phrase noncommutative space, since the noncommutative theory is obtained by retaining the algebra after the point space ceases to be adequate. [quotetheorem:2689] [citeproof:2689] This theorem is the reason $C_0(X)$ is not only an example but the model for all commutative C*-algebras. The hypotheses are doing real work. Commutativity is essential because characters multiply scalar values; for instance $M_n(\mathbb C)$ has no character when $n\ge 2$, so its character space cannot recover the algebra. The distinction between compact and locally compact spaces is also essential: $C(X)$ is unital exactly when $X$ is compact, while $C_0(X)$ is the right algebra for a noncompact locally compact space such as $\mathbb R$. The theorem also has a sharp limitation. It reconstructs a classical point space only from a commutative C*-algebra; it does not say that an arbitrary noncommutative C*-algebra has enough points, or that its primitive ideal space captures all its geometry. The next definition is therefore a change of language rather than a new axiom: it tells us to treat the algebra as the primary geometric object once the point-space reconstruction fails. [definition: Noncommutative Locally Compact Space] A noncommutative locally compact space is a C*-algebra $A$ regarded as the algebra of continuous functions vanishing at infinity on a generalized locally compact space. [/definition] After this definition, geometric words become algebraic instructions. In the proper-map convention used in this chapter, a map of spaces is represented contravariantly by a *-homomorphism of algebras, and closed ideals play the role of open subspaces. For nonunital algebras this convention is narrower than the most general topological dictionary: arbitrary continuous maps $X\to Y$ correspond to nondegenerate *-homomorphisms $C_0(Y)\to M(C_0(X))$ into the multiplier algebra, while proper maps are exactly the case where the pullback lands in $C_0(X)$ itself. [example: Matrix Algebra as a Finite Noncommutative Space] For $n\ge 2$, $M_n(\mathbb C)$ is noncommutative because, for the matrix units $E_{ij}$, one has \begin{align*} E_{12}E_{21}=E_{11} \end{align*} while \begin{align*} E_{21}E_{12}=E_{22}, \end{align*} and $E_{11}\ne E_{22}$. It is simple: if $I\subset M_n(\mathbb C)$ is a nonzero two-sided ideal and $a=(a_{ij})\in I$ has $a_{ij}\ne 0$, then for any $k,\ell$, \begin{align*} E_{ki}aE_{j\ell}=a_{ij}E_{k\ell}. \end{align*} Since $I$ is a two-sided ideal, $a_{ij}E_{k\ell}\in I$, and since $a_{ij}\ne 0$, also $E_{k\ell}\in I$. Thus every matrix unit belongs to $I$, so every matrix $\sum_{k,\ell} b_{k\ell}E_{k\ell}$ belongs to $I$, and therefore $I=M_n(\mathbb C)$. Thus $M_n(\mathbb C)$ has no nonzero proper closed ideals, so at the level of ideals it behaves like a finite connected object with no proper open-and-closed decomposition. Its pure states still see internal geometry: every unit vector $\xi\in\mathbb C^n$ gives a vector state \begin{align*} \omega_\xi(a)=\langle a\xi,\xi\rangle, \end{align*} and $\omega_{\lambda\xi}=\omega_\xi$ whenever $|\lambda|=1$, so these states are naturally indexed by lines in $\mathbb C^n$. This is the first sign that $M_n(\mathbb C)$ has more geometry than the one-point algebra $\mathbb C$, even though both are simple and unital. [/example] The preceding example shows why noncommutative spaces should not be reduced to their primitive ideal spaces or state spaces alone. Those spaces retain useful shadows, but multiplication in $A$ contains the extra geometry. This also explains why maps must be read through algebra homomorphisms rather than through point functions. [remark: Contravariance] A continuous proper map $f:X\to Y$ induces a *-homomorphism $C_0(Y)\to C_0(X)$ by $g\mapsto g\circ f$. Therefore a *-homomorphism $B\to A$ is read geometrically as a map from the noncommutative space represented by $A$ to the one represented by $B$. [/remark] The next question is how classical vector bundles appear in this algebraic language. The contravariant dictionary handles spaces and maps, but vector bundles require modules over function algebras. For a compact Hausdorff space $X$, the continuous Serre-Swan principle says that complex vector bundles over $X$ are encoded by finitely generated projective modules over $C(X)$. Concretely, sections of a vector bundle form such a module, and every finitely generated projective $C(X)$-module is realized as a direct summand of a trivial bundle, equivalently as the range of a projection in $M_n(C(X))$ for some $n$. This is the classical object whose noncommutative analogue will be a projection over a C*-algebra. The compact Hausdorff hypothesis is not cosmetic. If $X$ is locally compact but noncompact, the algebra becomes $C_0(X)$, which is nonunital, and finitely generated projective $C_0(X)$-modules no longer describe all vector bundles over $X$ in the same direct way. One usually imposes compact-support or vanishing-at-infinity conditions, or passes to the one-point compactification and compares what happens at the added point. This is why the noncompact theory of bundles over $C_0(X)$ is formulated through unitizations, multiplier algebras, and projections satisfying the correct boundary condition at infinity. [example: A Projection-Valued Function Gives a Bundle] Let $X$ be compact and let $p\in M_n(C(X))$ satisfy $p(x)^2=p(x)=p(x)^*$ for every $x\in X$. For each $x$, the matrix $p(x)$ is a projection on $\mathbb C^n$, so its range $E_x=p(x)\mathbb C^n$ is a finite-dimensional subspace. Fix $x_0\in X$. Since $p:X\to M_n(\mathbb C)$ is norm-continuous, there is a neighborhood $U$ of $x_0$ such that $\|p(x)-p(x_0)\|<1$ for all $x\in U$. If $P$ and $Q$ are projections with $\|P-Q\|<1$, then $Q:\operatorname{ran}P\to\operatorname{ran}Q$ is injective: if $\xi\in\operatorname{ran}P$ and $Q\xi=0$, then \begin{align*} \|\xi\|=\|P\xi-Q\xi\|\leq \|P-Q\|\|\xi\|<\|\xi\|. \end{align*} Thus $\xi=0$. Interchanging $P$ and $Q$ gives the reverse inequality of dimensions, so $\dim\operatorname{ran}P=\dim\operatorname{ran}Q$. Applying this to $P=p(x_0)$ and $Q=p(x)$ shows that $\dim E_x$ is constant on $U$. Let $E_0=E_{x_0}$. For $x\in U$, the map \begin{align*} E_0\longrightarrow E_x,\quad \eta\longmapsto p(x)\eta \end{align*} is injective by the same inequality, and the two spaces have the same dimension, so it is an isomorphism. Therefore \begin{align*} U\times E_0\longrightarrow \bigsqcup_{x\in U}E_x,\quad (x,\eta)\longmapsto (x,p(x)\eta) \end{align*} is a local trivialization; its matrix entries are continuous because the entries of $p$ are continuous. These local trivializations show that the ranges $p(x)\mathbb C^n$ assemble into a vector bundle over $X$. The associated $C(X)$-module of sections is exactly $pC(X)^n$. Indeed, if $f\in C(X)^n$, then \begin{align*} (pf)(x)=p(x)f(x)\in p(x)\mathbb C^n=E_x. \end{align*} Conversely, if $s$ is a continuous section of the subbundle, then $s(x)\in E_x$ for every $x$, so $p(x)s(x)=s(x)$ for every $x$, hence $s=ps\in pC(X)^n$. Thus the projection-valued function $p$ encodes both the vector bundle and its module of continuous sections. [/example] This example motivates treating projections over a general C*-algebra as noncommutative vector bundles. The next section develops the equivalence relation on projections that underlies the first K-theory group encountered in the course. ## Projections, Partial Isometries, and $K_0$ as a Shadow of Geometry The problem is to compare vector bundles without using points of the base space. For compact spaces, a vector bundle can be encoded by a projection $p\in M_n(C(X))$, and isomorphism of bundles becomes algebraic equivalence of projections. In a noncommutative C*-algebra, partial isometries supply the correct replacement for bundle isomorphisms. [definition: Projection] Let $A$ be a C*-algebra. A projection in $A$ is an element $p\in A$ such that $p^2=p$ and $p^*=p$. [/definition] Projections are the operator-algebraic analogue of characteristic functions and of vector subbundles. Unlike characteristic functions on a connected compact space, matrix-valued projections can vary continuously and carry nontrivial bundle data. To compare two such projection ranges, we need an algebraic object that identifies an initial projection with a final projection inside the same ambient algebra. [definition: Partial Isometry] Let $A$ be a C*-algebra. A partial isometry in $A$ is an element $v\in A$ such that $v^*v$ and $vv^*$ are projections. [/definition] The projections $v^*v$ and $vv^*$ are called the initial and final projections of $v$. This terminology comes from Hilbert space operators: $v$ is isometric on the range of $v^*v$ and maps it onto the range of $vv^*$. The next problem is to turn this witnessing element into a named equivalence relation on projections, because K-theory will only use projection classes rather than the particular partial isometries chosen between representatives. [definition: Murray-von Neumann Equivalence] Let $A$ be a C*-algebra, and let $p,q\in A$ be projections. The projections $p$ and $q$ are Murray-von Neumann equivalent, written $p\sim q$, if there exists a partial isometry $v\in A$ such that \begin{align*} v^*v&=p, & vv^*&=q. \end{align*} [/definition] This relation is designed to say that two projections determine the same noncommutative vector bundle. Before it can be used to build an invariant, it must behave like equality up to isomorphism rather than as a merely suggestive comparison. The following result provides that formal foundation. [quotetheorem:8594] [citeproof:8594] This result says that a partial isometry is not merely a witness of similarity between two projections; it gives the correct notion of isomorphism for projection ranges. The hypotheses matter: replacing partial isometries by arbitrary elements with the same initial equation would not give a symmetric relation, because the adjoint would no longer necessarily identify the final projection in the required way. The relation also has a limitation: in a fixed algebra $A$, two projections may fail to be equivalent even if they become comparable after adding extra matrix summands. For example, projections arising from vector bundles of different ranks over a compact space cannot be Murray-von Neumann equivalent, but they can still be compared stably after embedding into larger trivial bundles. The equivalence relation is stable under passage to matrices, which is essential because vector bundles may need different ambient trivial bundles before they can be compared. Matrix amplification is therefore not an optional technicality; it is part of the geometry. We first name the amplified algebras so that projections from all matrix sizes can later be combined. [definition: Matrix Amplification] Let $A$ be a C*-algebra. The $n$th matrix amplification of $A$ is the C*-algebra $M_n(A)$ of $n\times n$ matrices with entries in $A$. [/definition] Matrix amplifications allow projections of different sizes to be compared in common larger algebras. For this to give a consistent invariant, two possible ambiguities must be ruled out. First, passing from $A$ to matrices over matrices should agree with passing directly to one larger matrix algebra. Second, if two projections are already Murray-von Neumann equivalent, enlarging the ambient matrix algebra should not destroy that equivalence. These stability facts are what make it possible to form a single projection semigroup independent of bookkeeping choices. The construction therefore needs a compatibility result before the semigroup can be named. Without it, the class of a projection could depend on whether one first regards it in $M_n(A)$, then in $M_k(M_n(A))$, or instead directly in $M_{kn}(A)$, and equivalence could fail to be respected after adding unused matrix coordinates. The following stability theorem removes this bookkeeping ambiguity. [quotetheorem:8596] [citeproof:8596] The theorem justifies forming one large universe of projections, taken over all matrix algebras $M_n(A)$, without remembering the order in which matrix algebras were introduced. The matrix hypothesis cannot simply be omitted: projections in $A$ alone do not contain enough room to add or compare vector bundles of different ranks, just as line bundles and rank-two bundles cannot both live as subbundles of the same fixed trivial line bundle. The stability statement also has a boundary: it preserves Murray-von Neumann equivalence under amplification, but it does not say that inequivalent projections become equivalent after amplification. What it provides is a common arena in which direct sums, stabilization, and later Grothendieck completion are well-defined. Once projections can be moved into common matrix levels, they can be added by direct sum. The construction now needs a named semigroup whose elements are stable projection classes and whose addition is block direct sum. [definition: Projection Semigroup] Let $A$ be a unital C*-algebra. The projection semigroup $V(A)$ is the abelian semigroup of Murray-von Neumann equivalence classes of projections in all matrix algebras $M_n(A)$, with operation \begin{align*} [p]+[q]=[\operatorname{diag}(p,q)]. \end{align*} [/definition] The semigroup $V(A)$ remembers the positive, bundle-like part of the geometry, but semigroups do not allow formal subtraction. Classical K-theory compares vector bundles by allowing differences of bundle classes, and the same idea is needed here. The next definition performs the Grothendieck completion of the projection semigroup. [definition: K Zero Group of a Unital Algebra] Let $A$ be a unital C*-algebra. The group $K_0(A)$ is the Grothendieck group of the projection semigroup $V(A)$. [/definition] The positive cone $K_0(A)^+$ is the image of $V(A)$ inside $K_0(A)$, and the class $[1_A]$ is an order unit when $A$ is unital. These extra ordered features are crucial in AF algebras, where $K_0$ becomes a computable invariant. [example: $K_0$ of a Full Matrix Algebra] Let $A=M_n(\mathbb C)$. A projection in a matrix amplification $M_k(A)$ is a projection $p\in M_k(M_n(\mathbb C))\cong M_{kn}(\mathbb C)$. Since $p=p^*=p^2$, every eigenvalue $\lambda$ of $p$ satisfies $\lambda^2=\lambda$, hence $\lambda\in\{0,1\}$, and the finite-dimensional spectral theorem gives a unitary $u$ such that \begin{align*} upu^*=\operatorname{diag}(1,\dots,1,0,\dots,0). \end{align*} The count of diagonal entries equal to one is $\operatorname{rank}(p)$. We now check that Murray-von Neumann equivalence is exactly equality of rank. If $p\sim q$, choose $v$ with $v^*v=p$ and $vv^*=q$. For $\xi\in\operatorname{ran}p$, we have $p\xi=\xi$, so \begin{align*} \|v\xi\|^2=\langle v^*v\xi,\xi\rangle=\langle p\xi,\xi\rangle=\|\xi\|^2. \end{align*} Also $q(v\xi)=vv^*v\xi=vp\xi=v\xi$, so $v$ maps $\operatorname{ran}p$ isometrically into $\operatorname{ran}q$. Applying the same argument to $v^*$ gives an isometry from $\operatorname{ran}q$ into $\operatorname{ran}p$, hence $\operatorname{rank}(p)=\operatorname{rank}(q)$. Conversely, if $\operatorname{rank}(p)=\operatorname{rank}(q)=r$, choose unitaries $u,w$ such that \begin{align*} upu^*=wqw^*=\operatorname{diag}(1,\dots,1,0,\dots,0). \end{align*} Then $v=w^*d u$, where $d=\operatorname{diag}(1,\dots,1,0,\dots,0)$ with $r$ ones, satisfies \begin{align*} v^*v=u^*d w w^* d u=u^*d^2u=u^*du=p. \end{align*} Similarly, \begin{align*} vv^*=w^*d u u^* d w=w^*d^2w=w^*dw=q. \end{align*} Thus the class of a projection is determined by its rank, and block direct sum adds ranks: \begin{align*} \operatorname{rank}(\operatorname{diag}(p,q))=\operatorname{rank}(p)+\operatorname{rank}(q). \end{align*} Therefore $V(M_n(\mathbb C))\cong\mathbb N\cup\{0\}$, and its Grothendieck group is $\mathbb Z$. Under this rank identification, the identity $1_{M_n(\mathbb C)}$ is the identity matrix on $\mathbb C^n$, so $[1_{M_n(\mathbb C)}]$ corresponds to $n$. [/example] The same computation extends from a single matrix block to a finite direct sum by recording one rank for each summand. This matters because finite-dimensional C*-algebras are exactly the building blocks used in AF algebras. [example: $K_0$ of a Finite-Dimensional Algebra] Let \begin{align*} A=\bigoplus_{j=1}^r M_{n_j}(\mathbb C). \end{align*} For each $k\geq 1$, \begin{align*} M_k(A)\cong \bigoplus_{j=1}^r M_k(M_{n_j}(\mathbb C))\cong \bigoplus_{j=1}^r M_{kn_j}(\mathbb C). \end{align*} Thus a projection $p\in M_k(A)$ is a tuple $p=(p_1,\dots,p_r)$, where each $p_j\in M_{kn_j}(\mathbb C)$ satisfies $p_j^2=p_j=p_j^*$. Define \begin{align*} \rho([p])=(\operatorname{rank}(p_1),\dots,\operatorname{rank}(p_r)). \end{align*} This is well-defined on Murray-von Neumann classes. Indeed, if $p\sim q$ in $M_k(A)$, then there is $v=(v_1,\dots,v_r)$ with $v^*v=p$ and $vv^*=q$, so for each $j$, \begin{align*} v_j^*v_j=p_j \quad\text{and}\quad v_jv_j^*=q_j. \end{align*} The finite-dimensional calculation for full matrix algebras gives $\operatorname{rank}(p_j)=\operatorname{rank}(q_j)$ for every $j$, hence $\rho([p])=\rho([q])$. Conversely, suppose two projections $p=(p_1,\dots,p_r)$ and $q=(q_1,\dots,q_r)$ have the same rank vector. For each $j$, equality of ranks in the full matrix algebra gives a partial isometry $v_j$ such that \begin{align*} v_j^*v_j=p_j \quad\text{and}\quad v_jv_j^*=q_j. \end{align*} Then $v=(v_1,\dots,v_r)\in M_k(A)$ satisfies \begin{align*} v^*v=(v_1^*v_1,\dots,v_r^*v_r)=(p_1,\dots,p_r)=p. \end{align*} Also, \begin{align*} vv^*=(v_1v_1^*,\dots,v_rv_r^*)=(q_1,\dots,q_r)=q. \end{align*} Thus $p\sim q$, so the Murray-von Neumann class is exactly the rank vector. Block direct sum adds these vectors componentwise, because \begin{align*} \operatorname{rank}(\operatorname{diag}(p_j,q_j))=\operatorname{rank}(p_j)+\operatorname{rank}(q_j) \end{align*} for each matrix block. Therefore \begin{align*} V(A)\cong (\mathbb N\cup\{0\})^r. \end{align*} Taking the Grothendieck group replaces each copy of $\mathbb N\cup\{0\}$ by $\mathbb Z$, so \begin{align*} K_0(A)\cong \mathbb Z^r. \end{align*} Under this identification, the identity of $A$ is $(1_{n_1},\dots,1_{n_r})$, and $\operatorname{rank}(1_{n_j})=n_j$, so the order unit is $(n_1,\dots,n_r)$. [/example] These examples explain why ordered $K_0$ is visible in AF algebras. An AF algebra is built as an inductive limit of finite-dimensional C*-algebras, and the induced maps on $K_0$ are integer matrices recording how matrix blocks embed into later matrix blocks. A practical computation follows this pattern. For a finite-dimensional algebra $\bigoplus_{j=1}^r M_{n_j}(\mathbb C)$, compute $V(A)$ by recording the rank vector of a projection on each summand, so $V(A)\cong(\mathbb N\cup\{0\})^r$ and $K_0(A)\cong\mathbb Z^r$. For an AF inductive system, replace each finite stage by its ordered group $\mathbb Z^{r_s}$ and replace each connecting *-homomorphism by the nonnegative integer matrix whose $(i,j)$ entry records how many copies of the $j$th source block land in the $i$th target block. The ordered $K_0$ group is then the algebraic direct limit of these ordered groups and connecting matrices, with the order unit tracked by the images of the finite-stage identity projections. [example: First Glimpse of $K_0$ for an AF Limit] Consider the inductive system \begin{align*} \mathbb Z \xrightarrow{\times 2} \mathbb Z \xrightarrow{\times 2} \mathbb Z \xrightarrow{\times 2} \cdots . \end{align*} Write an element represented by $m\in\mathbb Z$ at the $s$th stage as $(m,s)$. In the algebraic direct limit, $(m,s)$ and $(m',t)$ represent the same element exactly when they agree after passing to some later stage. If $u\ge s,t$, this condition is \begin{align*} 2^{u-s}m=2^{u-t}m' . \end{align*} Define \begin{align*} \Phi([(m,s)])=\frac{m}{2^{s-1}}\in\mathbb Q . \end{align*} This is well-defined: if $2^{u-s}m=2^{u-t}m'$, then multiplying the desired equality by $2^{u-1}$ gives \begin{align*} 2^{u-1}\frac{m}{2^{s-1}}=2^{u-s}m=2^{u-t}m'=2^{u-1}\frac{m'}{2^{t-1}} . \end{align*} Hence $m/2^{s-1}=m'/2^{t-1}$. The map $\Phi$ is additive because representatives add stagewise: \begin{align*} \Phi([(m,s)]+[(m',s)])=\Phi([(m+m',s)])=\frac{m+m'}{2^{s-1}}=\frac{m}{2^{s-1}}+\frac{m'}{2^{s-1}} . \end{align*} Its image is exactly \begin{align*} \mathbb Z[1/2]=\left\{\frac{m}{2^r}:m\in\mathbb Z,\ r\ge 0\right\}, \end{align*} because $\Phi([(m,r+1)])=m/2^r$, and every value of $\Phi$ has this form. Thus the algebraic direct limit is identified with $\mathbb Z[1/2]\subset\mathbb Q$. The positive cone is obtained by starting with the usual positive cone $\mathbb N\cup\{0\}$ at each finite stage. Under $\Phi$, a class represented by $m\ge 0$ at stage $s$ maps to \begin{align*} \frac{m}{2^{s-1}}\ge 0, \end{align*} and every nonnegative dyadic rational $m/2^r$ is the image of the positive representative $(m,r+1)$. Therefore the positive cone is \begin{align*} \mathbb Z[1/2]\cap[0,\infty). \end{align*} When the connecting maps are unital in the chosen normalization, the distinguished order unit is the common direct-limit class of the finite-stage units. In the standard CAR presentation this ordered group is therefore $\bigl(\mathbb Z[1/2],\mathbb Z[1/2]\cap[0,\infty),[1]\bigr)$, showing how repeated doubling at finite stages becomes division by powers of $2$ in the limit. [/example] This calculation is small but representative: a complicated infinite-dimensional C*-algebra can have a computable ordered group extracted from finite-dimensional approximations. ## Boundaries of the Course The final question is what lies beyond the C*-algebraic framework developed here. Three directions appear naturally from the material already covered: von Neumann completions, modular theory, and deeper K-theory. Each begins from familiar objects in the course, but each requires new analytic or homological tools. [explanation: Von Neumann Completions] Given a representation $\pi:A\to B(H)$, the norm closure of $\pi(A)$ is again a C*-algebra, but the strong operator and weak operator closures usually give a larger algebra. A von Neumann algebra is a *-subalgebra $M\subset B(H)$ closed in the weak operator topology, or equivalently satisfying $M=M''$ by the bicommutant theorem. This changes the geometry: measure-theoretic phenomena become built into the algebra, projections become abundant, and duality with preduals replaces the locally compact topology of $C_0(X)$. Chapters 5 and 7 have already touched this boundary through representations and multiplier algebras. Passing from $A$ to $\pi(A)''$ is a completion process depending on the representation, and it is central in the study of factors, traces, and measurable noncommutative spaces. [/explanation] The next boundary appears when states are no longer tracial. The GNS construction still represents the algebra, but the cyclic vector need not behave symmetrically with respect to left and right multiplication. [explanation: Modular Theory] Tomita-Takesaki modular theory studies a von Neumann algebra $M$ together with a faithful normal state or weight. It produces a modular automorphism group that measures the failure of the state to be tracial. In finite-dimensional matrix algebras this failure is already visible: a state of the form $\phi(a)=\operatorname{Tr}(ha)$ with $h>0$ is tracial exactly when $h$ is a scalar multiple of the identity. The full theory requires unbounded operators arising from the closure of $a\Omega\mapsto a^*\Omega$ in the GNS Hilbert space. That analytic layer belongs to the full von Neumann algebra theory, but the motivating point is close to this course: states are not only linear functionals; they generate representations with internal dynamics. [/explanation] K-theory is the third boundary. We defined $K_0$ as a group completion of projection classes, but this is only the first part of a larger invariant. [explanation: Deeper K-Theory] For general C*-algebras, $K_0$ must be defined with care for nonunital algebras using unitization and the kernel of the map induced by $A^+\to\mathbb C$. The companion group $K_1(A)$ is built from homotopy classes of unitaries in matrix algebras over the unitization. Bott periodicity then organizes all higher K-groups into a two-periodic theory. This deeper theory connects operator algebras with topology, index theory, and noncommutative geometry. In this course, the role of $K_0$ is to show how projections turn analytic C*-algebra data into computable ordered algebraic invariants. [/explanation] The course therefore ends with a change of perspective rather than a terminal result. C*-algebras began as norm-closed algebras of operators; through the Gelfand theorem they became locally compact spaces in contravariant algebraic form; through projections and $K_0$ they began to carry vector-bundle-like geometry. The later theories keep these ideas but add new completions, new equivalence relations, and new invariants suited to the infinite-dimensional noncommutative setting. ## Beyond and Connections The analytic background for the course is the geometry of [Banach Space](/page/Banach%20Space) and [Hilbert Space](/page/Hilbert%20Space). Banach-space completeness is what makes spectra, quotients, and infinite series stable, while Hilbert-space geometry supplies adjoints, bounded operators, compact operators, and the representation-theoretic model for abstract $C^*$-algebras. The spectral material continues naturally in [Operator Theory I: Spectral Theory](/page/Operator%20Theory%20I%3A%20Spectral%20Theory). That viewpoint explains why resolvents, spectral mapping, compact operators, and functional calculus are central tools before the additional $*$-algebraic structure is imposed. The commutative side connects back to [Topology](/page/Topology): Gelfand duality turns locally compact Hausdorff spaces into commutative $C^*$-algebras and reverses continuous maps into algebra homomorphisms. The state-space and von Neumann algebra directions also touch [Measure Space](/page/Measure%20Space), where scalar integration becomes the commutative model for positive linear functionals and noncommutative integration.