This course develops the structural theory and classification of finite-dimensional Lie algebras over the complex numbers, with a focus on the semisimple case. After the introductory chapter, it studies how the Killing form detects semisimplicity and how Cartan subalgebras organize the internal geometry of a Lie algebra. From there, the course turns to the root space decomposition, which breaks a semisimple Lie algebra into a Cartan subalgebra together with weight spaces that encode its essential structure. The central theme is that semisimple Lie algebras can be understood through their roots. The course first builds the abstract theory of root systems, then introduces bases, positive roots, and Weyl chambers to impose combinatorial structure on the root data. Weyl groups, Cartan matrices, and Dynkin diagrams then provide the symmetry and encoding needed to pass from linear-algebraic information to a finite classification problem. The later chapters complete that classification. They show how finite root systems are classified, how a Lie algebra can be reconstructed from its root data, and why the resulting Lie algebra is determined uniquely up to isomorphism. In this way, the course moves from intrinsic structural tools to a complete and explicit description of semisimple Lie algebras, culminating in the classification theorem. # Introduction This course develops the classification of finite-dimensional semisimple Lie algebras over the complex numbers. The main problem is to turn a noncommutative algebra into intrinsic data that survives change of basis: first the Killing form and Cartan subalgebras, then root spaces, root systems, Weyl groups, Cartan matrices, and Dynkin diagrams. The path is deliberately circular. We begin with a semisimple Lie algebra and extract a finite root system from its adjoint action. We then classify the possible root systems by their diagrams, and finally reconstruct the Lie algebra from the resulting Cartan matrix by generators and relations. The point of the course is that this circle closes: no hidden continuous parameter remains once the finite-type Dynkin diagram is known. The notes assume the usual first-course material on ideals, solvable and nilpotent Lie algebras, Engel's theorem, Lie's theorem, the Killing form, and Cartan's criteria. Those facts are used as tools rather than reproved as a separate preliminary chapter. Chapter 1 begins the classification argument by isolating semisimplicity and explaining why the Killing form is the right bilinear invariant for the rest of the course. Throughout these notes the standing field is $\mathbb C$. When formulas use the letter $k$, read $k=\mathbb C$ unless a local example explicitly says otherwise. This keeps the quoted theorem cards, which are mostly stated in the classical complex form, aligned with the surrounding course text. # 1. Semisimple Lie Algebras and the Killing Form This opening chapter fixes the structural language used throughout the classification course. The central question is how much of a finite-dimensional Lie algebra is forced by the absence of solvable ideals, and why the Killing form is the correct bilinear form for detecting that condition. We work over $k=\mathbb C$, and all Lie algebras in this chapter are finite-dimensional over $k$ unless stated otherwise. ## Structural Recap: Radicals, Semisimplicity, and the Adjoint Action What is the obstruction to treating a Lie algebra as a direct sum of indecomposable non-abelian pieces? The first obstruction is the solvable radical: it collects all solvable ideal behaviour into a single canonical ideal. Once that ideal vanishes, the Lie algebra becomes rigid enough for the Killing form and the adjoint action to control its structure. [definition: Solvable Radical] Let $L$ be a Lie algebra. The solvable radical of $L$, denoted $\operatorname{rad} L$, is the largest solvable ideal of $L$. [/definition] The existence of $\operatorname{rad} L$ follows because the sum of two solvable ideals is solvable. Thus the sum of all solvable ideals is again a solvable ideal, and it contains every other solvable ideal. With the radical available as a canonical obstruction, the next step is to name the case where that obstruction has vanished. This is the structural setting in which the Killing form will become nondegenerate. [definition: Semisimple Lie Algebra] A Lie algebra $L$ is semisimple if $\operatorname{rad} L = 0$. [/definition] This definition rules out abelian ideals, since every abelian ideal is solvable. It also rules out nonzero centres, because $Z(L)$ is an abelian ideal. Semisimplicity removes solvable pieces, but it does not require the ideal structure to be indivisible. To describe the building blocks of semisimple algebras, we next isolate the algebras with no proper nontrivial ideals at all. [definition: Simple Lie Algebra] A Lie algebra $L$ is simple if $L$ is non-abelian and its only ideals are $0$ and $L$. [/definition] Simple Lie algebras are the irreducible objects for the ideal structure, but semisimple Lie algebras need not be simple. The point of this chapter is that semisimple Lie algebras are still built from simple ideals in a controlled direct-sum way. [definition: Adjoint Representation] Let $L$ be a Lie algebra. The adjoint representation is the Lie algebra homomorphism \begin{align*} \operatorname{ad}: L &\longrightarrow \mathfrak{gl}(L), & x &\longmapsto \operatorname{ad}_x, \end{align*} where $\operatorname{ad}_x(y)=[x,y]$ for $y \in L$. [/definition] The kernel of the adjoint representation is $Z(L)$. Thus, for semisimple $L$, the adjoint representation is faithful. This is why many structural questions about $L$ can be converted into questions about the linear operators $\operatorname{ad}_x$. [example: The Adjoint Action of $\mathfrak{sl}_2$] Let $L=\mathfrak{sl}_2(k)$ with basis \begin{align*} e=\begin{pmatrix}0&1\\0&0\end{pmatrix},\quad f=\begin{pmatrix}0&0\\1&0\end{pmatrix},\quad h=\begin{pmatrix}1&0\\0&-1\end{pmatrix}. \end{align*} Using $[a,b]=ab-ba$, the first bracket is \begin{align*} [h,e] &=\begin{pmatrix}1&0\\0&-1\end{pmatrix} \begin{pmatrix}0&1\\0&0\end{pmatrix} - \begin{pmatrix}0&1\\0&0\end{pmatrix} \begin{pmatrix}1&0\\0&-1\end{pmatrix} \\ &=\begin{pmatrix}0&1\\0&0\end{pmatrix} - \begin{pmatrix}0&-1\\0&0\end{pmatrix} =\begin{pmatrix}0&2\\0&0\end{pmatrix} =2e. \end{align*} Similarly, \begin{align*} [h,f] &=\begin{pmatrix}1&0\\0&-1\end{pmatrix} \begin{pmatrix}0&0\\1&0\end{pmatrix} - \begin{pmatrix}0&0\\1&0\end{pmatrix} \begin{pmatrix}1&0\\0&-1\end{pmatrix} \\ &=\begin{pmatrix}0&0\\-1&0\end{pmatrix} - \begin{pmatrix}0&0\\1&0\end{pmatrix} =\begin{pmatrix}0&0\\-2&0\end{pmatrix} =-2f, \end{align*} and \begin{align*} [e,f] &=\begin{pmatrix}0&1\\0&0\end{pmatrix} \begin{pmatrix}0&0\\1&0\end{pmatrix} - \begin{pmatrix}0&0\\1&0\end{pmatrix} \begin{pmatrix}0&1\\0&0\end{pmatrix} \\ &=\begin{pmatrix}1&0\\0&0\end{pmatrix} - \begin{pmatrix}0&0\\0&1\end{pmatrix} =\begin{pmatrix}1&0\\0&-1\end{pmatrix} =h. \end{align*} Therefore \begin{align*} \operatorname{ad}_h(e)&=[h,e]=2e,& \operatorname{ad}_h(f)&=[h,f]=-2f,& \operatorname{ad}_h(h)&=[h,h]=0, \end{align*} so $\operatorname{ad}_h$ is diagonal in the ordered basis $(e,f,h)$, with eigenvalues $2,-2,0$. The remaining adjoint maps move among the same basis vectors: \begin{align*} \operatorname{ad}_e(e)&=[e,e]=0,& \operatorname{ad}_e(f)&=[e,f]=h,& \operatorname{ad}_e(h)&=[e,h]=-[h,e]=-2e,\\ \operatorname{ad}_f(e)&=[f,e]=-[e,f]=-h,& \operatorname{ad}_f(f)&=[f,f]=0,& \operatorname{ad}_f(h)&=[f,h]=-[h,f]=2f. \end{align*} Thus the adjoint action already separates $L$ into eigenspaces for $\operatorname{ad}_h$, and $\operatorname{ad}_e,\operatorname{ad}_f$ move between those eigenspaces; this is the prototype for the root-space decompositions introduced later. [/example] The adjoint action therefore provides both a representation and a supply of concrete linear operators whose traces can be computed. The structural question is whether those traces see the radical of the Lie algebra, rather than merely producing an invariant bilinear form. Cartan's criterion answers this: nondegeneracy of the trace form is exactly what rules out nonzero solvable ideals in the finite-dimensional characteristic-zero setting. [quotetheorem:3813] [citeproof:3813] The criterion is the bridge between ideal-theoretic semisimplicity and linear algebra. Finite dimensionality is essential because the proof uses traces of endomorphisms and dimension arguments for radicals; in infinite-dimensional Lie algebras the Killing form may not even be defined in this trace form. Characteristic $0$ is also part of the mechanism behind Cartan's criterion, since trace arguments and nilpotence criteria behave differently in small positive characteristic. Algebraic closure is not needed for this particular criterion in its most general form, but it is part of the standing course setting because the later root-space theory diagonalises adjoint operators over $k$. A basic limitation is that a degenerate Killing form does not merely mean that the Lie algebra is hard to decompose: the one-dimensional abelian Lie algebra has identically zero Killing form and is solvable rather than semisimple. From now on, nondegeneracy of the Killing form will often replace direct arguments about solvable radicals. ## The Killing Form and Orthogonal Ideals How can a bilinear form remember the Lie bracket? The Killing form is not just a trace pairing; its invariance says that Lie brackets can be moved from one argument to the other. This compatibility makes orthogonal complements into ideals when the subspace being complemented is an ideal. [definition: Killing Form] Let $L$ be a finite-dimensional Lie algebra. The Killing form of $L$ is the bilinear map \begin{align*} \kappa_L: L\times L &\longrightarrow k, & (x,y)&\longmapsto \operatorname{tr}(\operatorname{ad}_x\operatorname{ad}_y). \end{align*} [/definition] The subscript is often omitted when the Lie algebra is fixed. The form is symmetric because trace satisfies $\operatorname{tr}(AB)=\operatorname{tr}(BA)$ for endomorphisms of a finite-dimensional vector space. [quotetheorem:3808] [citeproof:3808] Invariance is the algebraic analogue of integration by parts: a bracket can be transferred across the form at the cost of changing which two elements are bracketed. The hypothesis needed here is only finite dimensionality, so that the trace expression defining $\kappa_L$ is available; semisimplicity is not required for invariance. What invariance does not say is that the form is nondegenerate: on an abelian Lie algebra every adjoint map is zero, so the Killing form is invariant but identically zero. The theorem becomes structurally powerful only when paired with nondegeneracy, because then orthogonality can be converted into a direct-sum statement. This is the mechanism behind orthogonal decompositions. [definition: Orthogonal Complement for the Killing Form] Let $L$ be a finite-dimensional Lie algebra and let $U\subset L$ be a subspace. The Killing-orthogonal complement of $U$ is \begin{align*} U^\perp=\{x\in L: \kappa_L(x,u)=0 \text{ for all } u\in U\}. \end{align*} [/definition] When the Killing form is nondegenerate, the dimensions satisfy $\dim U+\dim U^\perp=\dim L$. If $U$ is an ideal in a semisimple Lie algebra, invariance upgrades this linear complement into an ideal-theoretic complement. [quotetheorem:3809] [citeproof:3809] This theorem supplies the ideal-theoretic part of the orthogonal decomposition argument. Invariance makes $I^\perp$ stable under brackets with the whole algebra; when this is paired with nondegeneracy of the Killing form in the semisimple case, orthogonality can then be used to build direct-sum decompositions. The theorem itself does not assert the full splitting statement on its own, nor does it classify the ideal $I$; it explains why the Killing form is compatible with the ideal structure. [example: The Killing Form on $\mathfrak{sl}_n$] Let $L=\mathfrak{sl}_n(k)$ for $n\ge 2$, and write $M_n=M_n(k)$. For $x\in L$, the adjoint map on $M_n$ is \begin{align*} \operatorname{ad}_x(A)=xA-Ax. \end{align*} Let $L_x(A)=xA$ and $R_x(A)=Ax$. Then $\operatorname{ad}_x=L_x-R_x$, so \begin{align*} \operatorname{ad}_x\operatorname{ad}_y &=(L_x-R_x)(L_y-R_y)\\ &=L_xL_y-L_xR_y-R_xL_y+R_xR_y\\ &=L_{xy}-L_xR_y-R_xL_y+R_{yx}. \end{align*} Using the matrix-unit basis $E_{ij}$ of $M_n$, one has \begin{align*} L_{xy}(E_{ij})=(xy)E_{ij} =\sum_p (xy)_{pi}E_{pj}, \end{align*} so the coefficient of $E_{ij}$ in $L_{xy}(E_{ij})$ is $(xy)_{ii}$. Summing over all $i,j$ gives \begin{align*} \operatorname{Tr}_{M_n}(L_{xy}) =\sum_{i,j}(xy)_{ii} =n\operatorname{tr}(xy). \end{align*} Similarly, \begin{align*} R_{yx}(E_{ij})=E_{ij}(yx) =\sum_q (yx)_{jq}E_{iq}, \end{align*} so \begin{align*} \operatorname{Tr}_{M_n}(R_{yx}) =\sum_{i,j}(yx)_{jj} =n\operatorname{tr}(yx) =n\operatorname{tr}(xy). \end{align*} For the mixed term, \begin{align*} L_xR_y(E_{ij}) =xE_{ij}y =\sum_{p,q}x_{pi}y_{jq}E_{pq}, \end{align*} so the coefficient of $E_{ij}$ is $x_{ii}y_{jj}$, and hence \begin{align*} \operatorname{Tr}_{M_n}(L_xR_y) =\sum_{i,j}x_{ii}y_{jj} =\left(\sum_i x_{ii}\right)\left(\sum_j y_{jj}\right) =\operatorname{tr}(x)\operatorname{tr}(y). \end{align*} The same calculation gives \begin{align*} \operatorname{Tr}_{M_n}(R_xL_y) =\operatorname{tr}(x)\operatorname{tr}(y). \end{align*} Therefore \begin{align*} \operatorname{Tr}_{M_n}(\operatorname{ad}_x\operatorname{ad}_y) &=n\operatorname{tr}(xy)-\operatorname{tr}(x)\operatorname{tr}(y) -\operatorname{tr}(x)\operatorname{tr}(y)+n\operatorname{tr}(xy)\\ &=2n\operatorname{tr}(xy)-2\operatorname{tr}(x)\operatorname{tr}(y). \end{align*} Since $x,y\in \mathfrak{sl}_n(k)$, $\operatorname{tr}(x)=\operatorname{tr}(y)=0$, so \begin{align*} \operatorname{Tr}_{M_n}(\operatorname{ad}_x\operatorname{ad}_y) =2n\operatorname{tr}(xy). \end{align*} Also $M_n=kI\oplus \mathfrak{sl}_n(k)$, and $\operatorname{ad}_x\operatorname{ad}_y$ is zero on $kI$. Thus its trace on $M_n$ equals its trace on $\mathfrak{sl}_n(k)$, and \begin{align*} \kappa_L(x,y)=2n\operatorname{tr}(xy). \end{align*} It remains to see that this form is nondegenerate. Let $0\ne x=(x_{ij})\in \mathfrak{sl}_n(k)$. If some off-diagonal entry $x_{ij}$ with $i\ne j$ is nonzero, take $y=E_{ji}$. Then $y\in\mathfrak{sl}_n(k)$ and \begin{align*} \operatorname{tr}(xy)=x_{ij}\ne 0. \end{align*} If all off-diagonal entries of $x$ vanish, then $x=\operatorname{diag}(\lambda_1,\dots,\lambda_n)$ with $\sum_i\lambda_i=0$. Since $x\ne 0$, two diagonal entries differ, say $\lambda_a\ne \lambda_b$. Taking $y=E_{aa}-E_{bb}$ gives $y\in\mathfrak{sl}_n(k)$ and \begin{align*} \operatorname{tr}(xy)=\lambda_a-\lambda_b\ne 0. \end{align*} Thus the trace form is nondegenerate on $\mathfrak{sl}_n(k)$, and since $2n\ne 0$ in characteristic $0$, the Killing form is nondegenerate as well. By the *Cartan Semisimplicity Criterion*, $\mathfrak{sl}_n(k)$ is semisimple. In fact, for $n\ge 2$ it is simple, so this example is the basic simple Lie algebra of type $A_{n-1}$. [/example] ## Decomposition into Simple Ideals If every ideal in a semisimple Lie algebra splits off, what happens after splitting repeatedly? Finite dimension forces the process to stop, and the terminal summands must be simple. This gives the structural form used throughout the classification: semisimple Lie algebras are finite direct sums of simple ideals. [quotetheorem:3821] [citeproof:3821/lie-algebras-ii-structure-and-classification] The theorem tells us that classification of semisimple Lie algebras reduces to classification of simple Lie algebras. The hypothesis excludes solvable examples such as a nonzero abelian Lie algebra, which has many one-dimensional ideals but is not a direct sum of simple non-abelian ideals. It also excludes reductive but non-semisimple algebras such as $\mathfrak{gl}_n(k)$, whose scalar centre is an abelian ideal and must be separated from the semisimple part $\mathfrak{sl}_n(k)$. In practice, the theorem gives a method: compute or use the Killing form to test semisimplicity, find a nonzero ideal, split it by its Killing-orthogonal complement, and repeat until the summands are simple. Chapters 2 through 8 classify the simple summands by choosing Cartan subalgebras, extracting root systems, and reading off Dynkin diagrams. [example: A Semisimple Algebra That Is Not Simple] Let $S=\mathfrak{sl}_2(k)$, so $L=S\oplus S$ has bracket \begin{align*} [(a,b),(c,d)]=([a,c],[b,d]). \end{align*} The subspaces $S\oplus 0$ and $0\oplus S$ are ideals: for $(a,b)\in L$ and $u,v\in S$, \begin{align*} [(a,b),(u,0)]&=([a,u],[b,0])=([a,u],0)\in S\oplus 0,\\ [(a,b),(0,v)]&=([a,0],[b,v])=(0,[b,v])\in 0\oplus S. \end{align*} They are nonzero proper ideals, so $L$ is not simple. Their mutual bracket is zero because \begin{align*} [(u,0),(0,v)]=([u,0],[0,v])=(0,0). \end{align*} We compute the Killing form on $L$. For $x=(a,b)\in L$, \begin{align*} \operatorname{ad}_x(u,v) &=[(a,b),(u,v)]\\ &=([a,u],[b,v])\\ &=(\operatorname{ad}_a^S(u),\operatorname{ad}_b^S(v)). \end{align*} Thus $\operatorname{ad}_{(a,b)}^L$ is block diagonal with blocks $\operatorname{ad}_a^S$ and $\operatorname{ad}_b^S$. Hence, for $x=(a,b)$ and $y=(c,d)$, \begin{align*} \operatorname{ad}_x^L\operatorname{ad}_y^L(u,v) &=\operatorname{ad}_x^L([c,u],[d,v])\\ &=([a,[c,u]],[b,[d,v]])\\ &=(\operatorname{ad}_a^S\operatorname{ad}_c^S(u),\operatorname{ad}_b^S\operatorname{ad}_d^S(v)). \end{align*} The trace of a block diagonal operator is the sum of the traces of its diagonal blocks, so \begin{align*} \kappa_L((a,b),(c,d)) &=\operatorname{tr}_L(\operatorname{ad}_{(a,b)}^L\operatorname{ad}_{(c,d)}^L)\\ &=\operatorname{tr}_S(\operatorname{ad}_a^S\operatorname{ad}_c^S)+\operatorname{tr}_S(\operatorname{ad}_b^S\operatorname{ad}_d^S)\\ &=\kappa_S(a,c)+\kappa_S(b,d). \end{align*} The preceding computation for $\mathfrak{sl}_n(k)$ gives $\kappa_S(p,q)=4\operatorname{tr}(pq)$ for $S=\mathfrak{sl}_2(k)$, and this form is nondegenerate. If $(a,b)\ne (0,0)$, then either $a\ne 0$ or $b\ne 0$. If $a\ne 0$, choose $c\in S$ with $\kappa_S(a,c)\ne 0$ and take $(c,0)$; then \begin{align*} \kappa_L((a,b),(c,0))=\kappa_S(a,c)+\kappa_S(b,0)=\kappa_S(a,c)\ne 0. \end{align*} If $a=0$ and $b\ne 0$, choose $d\in S$ with $\kappa_S(b,d)\ne 0$ and take $(0,d)$; then \begin{align*} \kappa_L((0,b),(0,d))=\kappa_S(0,0)+\kappa_S(b,d)=\kappa_S(b,d)\ne 0. \end{align*} Therefore $\kappa_L$ is nondegenerate, so by the Cartan Semisimplicity Criterion, $L$ is semisimple. Thus $L=\mathfrak{sl}_2(k)\oplus\mathfrak{sl}_2(k)$ is semisimple but not simple: semisimplicity allows a direct sum of simple ideals, while simplicity forbids any nonzero proper ideal. [/example] ## Complete Reducibility for the Adjoint Action Which form of complete reducibility is needed once semisimplicity has been detected? The decomposition of $L$ into simple ideals gives complete reducibility for the adjoint action, but the representation-theoretic theorem is stronger: every finite-dimensional representation of a semisimple Lie algebra splits into irreducible subrepresentations. We record that full result here because it explains why semisimple Lie algebras behave like diagonalizable objects far beyond the adjoint representation. [definition: Adjoint Submodule] Let $L$ be a Lie algebra. An adjoint submodule of $L$ is a subspace $U\subset L$ such that $[x,u]\in U$ for all $x\in L$ and all $u\in U$. [/definition] An adjoint submodule of $L$ is the same thing as an ideal of $L$. Thus the preceding decomposition theorem is a complete reducibility statement for the adjoint representation. [quotetheorem:3755] [citeproof:3755] This statement is the full complete reducibility theorem for finite-dimensional representations of a semisimple Lie algebra. The adjoint representation is one important case: its subrepresentations are ideals, so the theorem is compatible with the earlier decomposition of $L$ into simple ideals. Semisimplicity is essential; for the two-dimensional non-abelian Lie algebra with basis $x,y$ and $[x,y]=y$, the ideal $ky$ is an adjoint submodule, but it has no adjoint-stable complement. Finite dimensionality is also part of the theorem, since the conclusion is a finite direct-sum decomposition into irreducible modules. [remark: Why the Killing Form Is Central] The Killing form detects semisimplicity, converts ideals into orthogonal direct summands, and identifies the adjoint action as a completely reducible internal representation. These three roles explain why the form appears before Cartan subalgebras and roots in the course. Once a Cartan subalgebra is chosen, the same adjoint action will be diagonalised into root spaces. [/remark] Once the Killing form has isolated the semisimple case, the next task is to choose a Cartan subalgebra and use it as a coordinate system for the algebra itself. That choice lets us pass from an intrinsic noncommutative object to a setting where the adjoint action can be broken into manageable pieces. # 2. Cartan Subalgebras Cartan subalgebras provide the first internal coordinate system for a semisimple Lie algebra. The guiding question is how to find a subalgebra that behaves like the diagonal matrices inside $\mathfrak{sl}_n$, so that the ambient algebra can later be decomposed into simultaneous eigenspaces. This chapter begins with the intrinsic definition, then explains existence and conjugacy over the complex numbers, and ends with the construction from regular elements. ## Finding the Right Replacement for Diagonal Matrices The diagonal traceless matrices in $\mathfrak{sl}_n$ are small, commutative, and detect the off-diagonal matrix units by their commutators. For an abstract Lie algebra there may be no preferred matrix realisation, so the first problem is to describe the same object without mentioning diagonal entries. [definition: Normalizer of a Subalgebra] Let $L$ be a Lie algebra over a field $k$, and let $H \le L$ be a Lie subalgebra. The normalizer of $H$ in $L$ is \begin{align*} N_L(H) = \{x \in L : [x,H] \subset H\}. \end{align*} [/definition] The condition $N_L(H)=H$ says that $H$ is maximal with respect to being preserved by the adjoint action of its own ambient normalizer. It is stronger than being maximal as a subalgebra, and it is tailored to the adjoint representation. This condition is needed because nilpotence alone is far too weak: in $\mathfrak{sl}_2(k)$, the one-dimensional subalgebra $kE_{12}$ is nilpotent, but it is normalized by the upper triangular subalgebra, so it cannot play the role of a full diagonal direction. A maximal diagonal subalgebra in matrices is useful because it is large enough to detect weights but small enough not to contain irrelevant normalized directions. In an abstract Lie algebra, the obstruction is that nilpotent subalgebras can sit inside larger normalizing subalgebras, so the correct intrinsic substitute must require both nilpotence and self-normalization. [definition: Cartan Subalgebra] Let $L$ be a finite-dimensional Lie algebra over a field $k$. A Cartan subalgebra of $L$ is a Lie subalgebra $H \le L$ such that $H$ is nilpotent and \begin{align*} N_L(H)=H. \end{align*} [/definition] Nilpotence replaces commutativity in the general definition because Cartan subalgebras in arbitrary Lie algebras need not be abelian. In semisimple Lie algebras over $\mathbb C$, a later theorem will force them to be abelian, recovering the diagonal picture. [example: Diagonal Cartan Subalgebra of Sl N] Let $L=\mathfrak{sl}_n(k)$, where $k$ is algebraically closed and $\operatorname{char} k=0$, and let \begin{align*} H=\{\operatorname{diag}(a_1,\dots,a_n):a_1+\cdots+a_n=0\}. \end{align*} If $h=\operatorname{diag}(a_1,\dots,a_n)$ and $h'=\operatorname{diag}(b_1,\dots,b_n)$ lie in $H$, then \begin{align*} hh'&=\operatorname{diag}(a_1b_1,\dots,a_nb_n),\\ h'h&=\operatorname{diag}(b_1a_1,\dots,b_na_n), \end{align*} so $[h,h']=hh'-h'h=0$. Thus $H$ is abelian, and hence nilpotent. For $i\ne j$, compute the commutator with the matrix unit $E_{ij}$. Since $hE_{ij}$ has only one possibly nonzero entry, namely \begin{align*} (hE_{ij})_{ij}=h_{ii}(E_{ij})_{ij}=a_i, \end{align*} we have $hE_{ij}=a_iE_{ij}$. Similarly, \begin{align*} (E_{ij}h)_{ij}=(E_{ij})_{ij}h_{jj}=a_j, \end{align*} so $E_{ij}h=a_jE_{ij}$. Therefore \begin{align*} [h,E_{ij}] &=hE_{ij}-E_{ij}h\\ &=a_iE_{ij}-a_jE_{ij}\\ &=(a_i-a_j)E_{ij}. \end{align*} It remains to prove that $H$ is self-normalizing. Let $x=(x_{pq})\in \mathfrak{sl}_n(k)$ and suppose $x\in N_L(H)$. Then for every $h=\operatorname{diag}(a_1,\dots,a_n)\in H$, the commutator $[x,h]$ lies in $H$, so it is diagonal. For $p\ne q$, the $(p,q)$-entry of $xh$ is \begin{align*} (xh)_{pq}=x_{pq}h_{qq}=x_{pq}a_q, \end{align*} while the $(p,q)$-entry of $hx$ is \begin{align*} (hx)_{pq}=h_{pp}x_{pq}=a_px_{pq}. \end{align*} Hence \begin{align*} ([x,h])_{pq} &=(xh-hx)_{pq}\\ &=x_{pq}a_q-a_px_{pq}\\ &=(a_q-a_p)x_{pq}. \end{align*} Because $[x,h]$ is diagonal, this entry is $0$ for every $h\in H$. Fix $p\ne q$. Choose $h\in H$ with $a_p=1$, $a_q=-1$, and all other diagonal entries equal to $0$ if $n>2$; when $n=2$, this is still traceless. Then \begin{align*} a_q-a_p=-1-1=-2. \end{align*} Since $\operatorname{char} k=0$, $-2\ne 0$, so \begin{align*} 0=(a_q-a_p)x_{pq}=-2x_{pq} \end{align*} implies $x_{pq}=0$. Thus every off-diagonal entry of $x$ is zero, so $x$ is diagonal. Since $x\in\mathfrak{sl}_n(k)$, its trace is zero, and therefore $x\in H$. We have shown $N_L(H)\subset H$. The reverse inclusion holds because $H$ is abelian: if $h_0\in H$, then $[h_0,H]=0\subset H$, so $h_0\in N_L(H)$. Therefore \begin{align*} N_L(H)=H. \end{align*} Thus the traceless diagonal matrices form a nilpotent self-normalizing subalgebra of $\mathfrak{sl}_n(k)$, so they are a Cartan subalgebra. [/example] This example is the model for the root space decomposition in Chapter 3. The scalars $a_i-a_j$ are the first appearance of roots, although the formal language of roots will be introduced only after Cartan subalgebras have been fixed. [example: Block-Compatible Cartan Subalgebras in Classical Lie Algebras] For each of the classical Lie algebras, use the paired basis $e_1,\dots,e_n,f_1,\dots,f_n$, and in the odd orthogonal case add one vector $u$. Let \begin{align*} h(a)=\operatorname{diag}(a_1,\dots,a_n,-a_1,\dots,-a_n) \end{align*} in the even-dimensional paired basis, and in $\mathfrak{so}_{2n+1}(k)$ use \begin{align*} h(a)=\operatorname{diag}(a_1,\dots,a_n,0,-a_1,\dots,-a_n). \end{align*} The standard Cartan candidate is \begin{align*} H=\{h(a):a_1,\dots,a_n\in k\}. \end{align*} In $\mathfrak{sp}_{2n}(k)$ take the symplectic matrix \begin{align*} J=\begin{pmatrix} 0&I_n\\ -I_n&0 \end{pmatrix}. \end{align*} Writing $h=h(a)$, the transpose is $h^T=h$, and \begin{align*} hJ &= \begin{pmatrix} \operatorname{diag}(a_1,\dots,a_n)&0\\ 0&-\operatorname{diag}(a_1,\dots,a_n) \end{pmatrix} \begin{pmatrix} 0&I_n\\ -I_n&0 \end{pmatrix}\\ &= \begin{pmatrix} 0&\operatorname{diag}(a_1,\dots,a_n)\\ \operatorname{diag}(a_1,\dots,a_n)&0 \end{pmatrix}, \end{align*} while \begin{align*} Jh &= \begin{pmatrix} 0&I_n\\ -I_n&0 \end{pmatrix} \begin{pmatrix} \operatorname{diag}(a_1,\dots,a_n)&0\\ 0&-\operatorname{diag}(a_1,\dots,a_n) \end{pmatrix}\\ &= \begin{pmatrix} 0&-\operatorname{diag}(a_1,\dots,a_n)\\ -\operatorname{diag}(a_1,\dots,a_n)&0 \end{pmatrix}. \end{align*} Thus $h^TJ+Jh=hJ+Jh=0$, so $h\in\mathfrak{sp}_{2n}(k)$. The same calculation with the symmetric paired form \begin{align*} S=\begin{pmatrix} 0&I_n\\ I_n&0 \end{pmatrix} \end{align*} gives $h^TS+Sh=0$, and in the odd orthogonal case the extra middle row and column contribute only zeros. Hence these diagonal paired matrices lie in the corresponding orthogonal Lie algebras as well. If $h(a),h(b)\in H$, then both are diagonal, so their product is \begin{align*} h(a)h(b) = \operatorname{diag}(a_1b_1,\dots,a_nb_n,a_1b_1,\dots,a_nb_n) \end{align*} in the even paired cases, with an additional middle $0$ in the odd orthogonal case. The same product is obtained in the opposite order, so \begin{align*} [h(a),h(b)]=h(a)h(b)-h(b)h(a)=0. \end{align*} Therefore $H$ is abelian, hence nilpotent. Now let $x=(x_{rs})$ lie in the ambient classical Lie algebra and suppose $x\in N_L(H)$. Label the basis vectors by weights \begin{align*} \lambda(e_i)=\varepsilon_i,\qquad \lambda(f_i)=-\varepsilon_i, \end{align*} and in $\mathfrak{so}_{2n+1}(k)$ also $\lambda(u)=0$, where $\varepsilon_i(h(a))=a_i$. For any basis vectors $v_r,v_s$, \begin{align*} (xh(a))_{rs}=x_{rs}\lambda(v_s)(h(a)), \qquad (h(a)x)_{rs}=\lambda(v_r)(h(a))x_{rs}, \end{align*} so \begin{align*} ([x,h(a)])_{rs} &=(xh(a)-h(a)x)_{rs}\\ &=\bigl(\lambda(v_s)(h(a))-\lambda(v_r)(h(a))\bigr)x_{rs}. \end{align*} Since $[x,h(a)]\in H$, it is diagonal for every $a$, so the left side is $0$ whenever $r\ne s$. If $r\ne s$, then the weights $\lambda(v_r)$ and $\lambda(v_s)$ are distinct. Choose $a$ so that \begin{align*} \lambda(v_s)(h(a))-\lambda(v_r)(h(a))\ne 0. \end{align*} For example, if $v_r=e_i$ and $v_s=f_i$, take $a_i=1$; then the difference is $-1-1=-2$, which is nonzero because $\operatorname{char}k=0$. Therefore \begin{align*} 0=\bigl(\lambda(v_s)(h(a))-\lambda(v_r)(h(a))\bigr)x_{rs} \end{align*} implies $x_{rs}=0$. Thus every off-diagonal entry of $x$ is zero. It remains only to use the defining form condition. A diagonal matrix \begin{align*} x=\operatorname{diag}(c_1,\dots,c_n,d_1,\dots,d_n) \end{align*} satisfies $x^TJ+Jx=0$ or $x^TS+Sx=0$ exactly when \begin{align*} c_i+d_i=0 \end{align*} for every $i$. In the odd orthogonal case, \begin{align*} x=\operatorname{diag}(c_1,\dots,c_n,c_0,d_1,\dots,d_n) \end{align*} also has the middle condition $2c_0=0$, hence $c_0=0$. Therefore every diagonal element of the classical Lie algebra has the paired form $h(c_1,\dots,c_n)$, so $x\in H$. We have proved $N_L(H)\subset H$. Since $H$ is abelian, every element of $H$ normalizes $H$, so $H\subset N_L(H)$. Hence $N_L(H)=H$, and these block-compatible diagonal subalgebras are Cartan subalgebras. The calculation shows that they play exactly the same role as traceless diagonal matrices in $\mathfrak{sl}_n(k)$, with the signs forced by the bilinear or symplectic form. [/example] The orthogonal and symplectic examples show that Cartan subalgebras are not an artefact of trace-zero matrices. They are the part of the algebra that can be diagonalized simultaneously in the adjoint representation, subject to the form preserved by the classical Lie algebra. ## Existence and Conjugacy Over Algebraically Closed Fields The definition is useful only if Cartan subalgebras exist and are unique up to the symmetries of the Lie algebra. The next problem is therefore structural: over the field used for classification, can every finite-dimensional Lie algebra be supplied with such a subalgebra, and how many choices are there? [quotetheorem:4680] [citeproof:4680] Elements whose centralizer has the minimum possible dimension are called regular, and they will be studied more directly in the final section of this chapter. The infinite-field hypothesis is part of the usable range of the existence statement: outside it, the expected supply of regular elements over the ground field can fail. The theorem asserts existence of some Cartan subalgebra, not uniqueness, abelianness, or compatibility with a preferred splitting over non-algebraically closed fields. We need a separate conjugacy result because existence alone still leaves a possible hidden choice in the classification. If two Cartan subalgebras could lead to unrelated root systems, the Dynkin diagram would not be an invariant of the Lie algebra. [quotetheorem:4681] [citeproof:4681] For semisimple Lie algebras this means that later constructions do not depend on a hidden choice. A root system is built after selecting $H$, but a different Cartan subalgebra gives an isomorphic root system after conjugation. The hypotheses are essential to this clean statement: over $\mathbb R$, split and compact Cartan subalgebras of the same complex Lie algebra need not be conjugate by the real adjoint group, as in real forms of $\mathfrak{sl}_2$. In positive characteristic, exponentials of inner derivations may not define the expected algebraic group action, and regular-element arguments can fail without extra restrictions. [remark: Field Hypotheses] Algebraic closedness and characteristic zero are not cosmetic assumptions. Over non-algebraically closed fields, split and nonsplit Cartan subalgebras can behave differently; over fields of positive characteristic, the usual exponential and regular-element arguments require substantial modification. [/remark] The classification course works over the complex numbers precisely so that Cartan subalgebras behave uniformly. This is the setting in which root spaces, Dynkin diagrams, and Serre presentations have their cleanest form. ## Cartan Subalgebras Inside Semisimple Lie Algebras The general definition permits nilpotent nonabelian Cartan subalgebras, but semisimple Lie algebras are more rigid. The main question for classification is whether a Cartan subalgebra in a semisimple Lie algebra really acts like a diagonal subalgebra. [quotetheorem:4682] [citeproof:4682] This theorem is the point at which the definition by nilpotence turns into the diagonal intuition. In the semisimple setting, a Cartan subalgebra is a commutative subalgebra whose adjoint action can be used to decompose the whole Lie algebra. Outside this setting the conclusion can fail: Cartan subalgebras of solvable or general nonsemisimple Lie algebras may be nilpotent but nonabelian, and over non-algebraically closed fields the available diagonalisation may be only after scalar extension. Abelianness alone also does not identify a Cartan subalgebra; it must still be self-normalizing, as a one-dimensional diagonal subalgebra inside a larger diagonal algebra in $\mathfrak{sl}_n(k)$ shows. Granting that a Cartan subalgebra $H$ diagonalises the whole algebra, the immediate question is how large the zero-weight space of this action turns out to be. If some element outside $H$ commuted with all of $H$, that element would lie in the zero-weight space and $H$ could no longer be recovered from its own adjoint action; the decomposition would then carry an inflated central part. We therefore need to know that $H$ is exactly its own centraliser. [quotetheorem:4683] [citeproof:4683] The centralizer statement will be used constantly in the root decomposition. It says that the zero-weight space for the adjoint action of $H$ contains no extra part beyond $H$ itself. The abelianness and self-normalizing hypotheses both matter: if $A$ is merely abelian, its centralizer may be much larger than $A$, while if a nilpotent subalgebra is not self-normalizing, elements outside it can centralize or normalize it. The theorem also does not say that every self-centralizing subalgebra is Cartan; nilpotence is still part of the Cartan condition. [example: Centralizer in Sl N] Let \begin{align*} H=\{\operatorname{diag}(a_1,\dots,a_n):a_1+\cdots+a_n=0\}\le \mathfrak{sl}_n(k), \end{align*} and suppose $x=(x_{pq})\in C_{\mathfrak{sl}_n}(H)$. We show that $x\in H$. For any $h=\operatorname{diag}(a_1,\dots,a_n)\in H$, the equation $[h,x]=0$ means $hx-xh=0$. The $(p,q)$-entry of $hx$ is \begin{align*} (hx)_{pq} &=\sum_{r=1}^n h_{pr}x_{rq}\\ &=h_{pp}x_{pq}\\ &=a_px_{pq}, \end{align*} because $h_{pr}=0$ unless $r=p$. Similarly, the $(p,q)$-entry of $xh$ is \begin{align*} (xh)_{pq} &=\sum_{r=1}^n x_{pr}h_{rq}\\ &=x_{pq}h_{qq}\\ &=a_qx_{pq}, \end{align*} because $h_{rq}=0$ unless $r=q$. Hence \begin{align*} 0=([h,x])_{pq} &=(hx-xh)_{pq}\\ &=a_px_{pq}-a_qx_{pq}\\ &=(a_p-a_q)x_{pq}. \end{align*} Fix $p\ne q$. Choose $h\in H$ by setting $a_p=1$, $a_q=-1$, and all other $a_r=0$. Then $a_1+\cdots+a_n=0$, so $h\in H$, and \begin{align*} a_p-a_q=1-(-1)=2. \end{align*} Since $\operatorname{char} k=0$, $2\ne 0$, so \begin{align*} 0=(a_p-a_q)x_{pq}=2x_{pq} \end{align*} implies $x_{pq}=0$. Thus every off-diagonal entry of $x$ is zero, so \begin{align*} x=\operatorname{diag}(c_1,\dots,c_n). \end{align*} Because $x\in\mathfrak{sl}_n(k)$, its trace is zero: \begin{align*} c_1+\cdots+c_n=0. \end{align*} Therefore $x\in H$, and we have proved $C_{\mathfrak{sl}_n}(H)\subset H$. Conversely, if $h_0,h\in H$, then both are diagonal, so \begin{align*} h_0h=hh_0. \end{align*} Hence \begin{align*} [h_0,h]=h_0h-hh_0=0 \end{align*} for every $h\in H$, which means $h_0\in C_{\mathfrak{sl}_n}(H)$. Thus $H\subset C_{\mathfrak{sl}_n}(H)$, and consequently \begin{align*} C_{\mathfrak{sl}_n}(H)=H. \end{align*} This direct calculation shows that no off-diagonal matrix unit can commute with all traceless diagonal matrices. [/example] The example also explains why roots are nonzero on off-diagonal directions. Every matrix unit $E_{ij}$ is moved by some element of $H$, so it cannot lie in the centralizer. ## Regular Elements and Generalized Zero Eigenspaces The existence proof suggests a more systematic question: can Cartan subalgebras be manufactured from single elements of the Lie algebra? Regular elements answer this question and provide a practical way to locate Cartan subalgebras in examples. [definition: Generalized Zero Eigenspace of an Adjoint Operator] Let $L$ be a finite-dimensional Lie algebra over a field $k$, and let $x\in L$. Define \begin{align*} \operatorname{ad} x &: L \to L, & y &\mapsto [x,y]. \end{align*} The generalized zero eigenspace of $\operatorname{ad} x$ is \begin{align*} L_0(x)=\{y\in L : (\operatorname{ad} x)^m(y)=0 \text{ for some } m\ge 1\}. \end{align*} [/definition] This space is the nilpotent part of $L$ for the operator $\operatorname{ad} x$. It is stable under the bracket, so it is a Lie subalgebra; this follows from applying powers of $\operatorname{ad} x$ to $[y,z]$ and using the derivation rule. The size of $L_0(x)$ measures how many directions fail to move invertibly under $\operatorname{ad}x$. If this zero part is larger than necessary, it contains accidental degeneracies rather than the intrinsic diagonal core we want; the useful elements are therefore those for which this obstruction is minimized. [definition: Regular Element] Let $L$ be a finite-dimensional Lie algebra over a field $k$. An element $x\in L$ is regular if $\dim L_0(x)$ is minimal among the dimensions $\dim L_0(y)$ for $y\in L$. [/definition] Regularity is a minimal-kernel condition for the family of adjoint operators. In matrix examples it corresponds to having as few repeated eigenvalue relations as the Lie algebra permits. The remaining question is why this minimization condition finds the right subalgebras rather than merely small linear spaces. A priori, $L_0(x)$ is defined from one operator and need not have the self-normalizing nilpotent structure required of a Cartan subalgebra. The theorem below is needed precisely to rule out that mismatch. [quotetheorem:4684] [citeproof:4684] This theorem is the bridge between linear algebra and structure theory. It lets us replace the search for Cartan subalgebras by the study of adjoint operators with minimal generalized zero eigenspace. The assumption that $k$ is infinite again supports the generic perturbation argument: regularity excludes elements whose adjoint action has avoidable extra generalized-zero directions, but it does not imply that $\operatorname{ad} x$ is diagonalizable or that $x$ itself is semisimple. In $\mathfrak{sl}_n(k)$, repeated diagonal entries are exactly the visible obstruction, since they create extra off-diagonal vectors in $L_0(x)$. [example: Regular Diagonal Elements in Sl N] In $\mathfrak{sl}_n(k)$, take \begin{align*} x=\operatorname{diag}(a_1,\dots,a_n), \qquad a_1+\cdots+a_n=0, \end{align*} with $a_i\ne a_j$ whenever $i\ne j$. Let \begin{align*} H=\{\operatorname{diag}(b_1,\dots,b_n): b_1+\cdots+b_n=0\}. \end{align*} If $h=\operatorname{diag}(b_1,\dots,b_n)\in H$, then $xh=hx$ because both matrices are diagonal, so \begin{align*} (\operatorname{ad}x)(h)=[x,h]=0. \end{align*} Thus $H\subset L_0(x)$. For $i\ne j$, the matrix $xE_{ij}$ has only one possibly nonzero entry, and \begin{align*} (xE_{ij})_{ij}=x_{ii}(E_{ij})_{ij}=a_i. \end{align*} Hence $xE_{ij}=a_iE_{ij}$. Similarly, \begin{align*} (E_{ij}x)_{ij}=(E_{ij})_{ij}x_{jj}=a_j, \end{align*} so $E_{ij}x=a_jE_{ij}$. Therefore \begin{align*} (\operatorname{ad}x)(E_{ij}) &=[x,E_{ij}]\\ &=xE_{ij}-E_{ij}x\\ &=a_iE_{ij}-a_jE_{ij}\\ &=(a_i-a_j)E_{ij}. \end{align*} Since $a_i-a_j\ne 0$, repeated application gives \begin{align*} (\operatorname{ad}x)^m(E_{ij})=(a_i-a_j)^mE_{ij}\ne 0 \end{align*} for every $m\ge 1$. Every element of $\mathfrak{sl}_n(k)$ can be written uniquely as \begin{align*} z=h+\sum_{i\ne j} c_{ij}E_{ij}, \end{align*} with $h\in H$. Applying $(\operatorname{ad}x)^m$ gives \begin{align*} (\operatorname{ad}x)^m(z) &=(\operatorname{ad}x)^m(h)+\sum_{i\ne j}c_{ij}(\operatorname{ad}x)^m(E_{ij})\\ &=0+\sum_{i\ne j}c_{ij}(a_i-a_j)^mE_{ij}. \end{align*} The matrices $E_{ij}$ are linearly independent, and each $(a_i-a_j)^m$ is nonzero, so $(\operatorname{ad}x)^m(z)=0$ holds exactly when every $c_{ij}=0$. Hence \begin{align*} L_0(x)=H. \end{align*} Thus \begin{align*} \dim L_0(x)=\dim H=n-1. \end{align*} For any $y\in\mathfrak{sl}_n(k)$, the space $L_0(y)$ contains $\ker(\operatorname{ad}y)=C_{\mathfrak{sl}_n}(y)$, and by the standard *matrix centralizer dimension bound* one has \begin{align*} \dim C_{\mathfrak{sl}_n}(y)\ge n-1. \end{align*} Therefore no element of $\mathfrak{sl}_n(k)$ has generalized zero eigenspace of dimension smaller than $n-1$, so $x$ is regular. This shows that regular diagonal elements with distinct diagonal entries recover exactly the traceless diagonal Cartan subalgebra. [/example] If the diagonal entries are repeated, then additional off-diagonal matrix units commute with $x$, increasing the generalized zero eigenspace. Regular elements are precisely the ones avoiding these extra centralizing directions in the adjoint representation. [remark: Rank] For a finite-dimensional semisimple Lie algebra $L$ over the complex numbers, the common dimension of its Cartan subalgebras is called the rank of $L$. Equivalently, it is the minimal value of $\dim L_0(x)$ over all $x\in L$. [/remark] The rank is the number of independent diagonal directions available for the later root decomposition. For $\mathfrak{sl}_n(k)$ the rank is $n-1$, matching the dimension of the traceless diagonal matrices. ## What Cartan Subalgebras Prepare The output of this chapter is a fixed abelian subalgebra $H\le L$ satisfying $C_L(H)=H$. The next chapter studies the simultaneous eigenspace decomposition of $L$ under the commuting family $\{\operatorname{ad} h : h\in H\}$. For semisimple $L$, the expected form is \begin{align*} L = H \oplus \bigoplus_{\alpha\in \Phi} L_\alpha, \end{align*} where each nonzero functional $\alpha\in H^*$ records how $H$ acts on the corresponding root space. The fact that the zero part is exactly $H$ is the centralizer theorem above, and the fact that all choices of $H$ are conjugate ensures that the resulting root system is an invariant of $L$ rather than an artefact of presentation. A Cartan subalgebra gives the stage, but the real structure appears when the adjoint action is decomposed into root spaces. The next chapter develops that decomposition systematically, showing how the algebra splits into eigenspaces indexed by roots and why this data is independent of the particular Cartan subalgebra chosen. # 3. Root Space Decomposition This chapter begins the passage from an abstract semisimple Lie algebra to combinatorial data. The preceding chapter produced Cartan subalgebras as the subalgebras that control the adjoint action. We now diagonalise that action, decompose the Lie algebra into simultaneous eigenspaces, and study the arithmetic constraints forced by copies of $\mathfrak{sl}_2$ inside $\mathfrak{g}$. The hypotheses in this chapter are not cosmetic: algebraic closedness supplies eigenvalues, characteristic zero gives the usual representation theory of $\mathfrak{sl}_2$, and semisimplicity removes central directions that would otherwise hide from the root data. ## Weights of the Adjoint Action The first question is how a Cartan subalgebra can organise the whole Lie algebra. If $\mathfrak{h}$ is abelian and acts on $\mathfrak{g}$ by the adjoint representation, then each $h \in \mathfrak{h}$ gives a linear map $\operatorname{ad}(h):\mathfrak{g}\to\mathfrak{g}$. Without diagonalisation, this action would only give generalised eigenspaces or Jordan blocks, and the later arithmetic of roots would not be visible. The useful case is when these maps can be diagonalised simultaneously, so that each vector in $\mathfrak{g}$ has a definite weight under every $h$. [definition: Weight Space For The Adjoint Action] Let $\mathfrak{g}$ be a finite-dimensional Lie algebra over $k=\mathbb C$, and let $\mathfrak{h}\le \mathfrak{g}$ be a subalgebra. For $\alpha\in\mathfrak{h}^*$, define \begin{align*} \mathfrak{g}_\alpha=\{x\in\mathfrak{g}: [h,x]=\alpha(h)x\text{ for all }h\in\mathfrak{h}\}. \end{align*} [/definition] The case $\alpha=0$ is the simultaneous zero-eigenspace. When $\mathfrak{h}$ is a Cartan subalgebra in the semisimple setting, this zero space is exactly $\mathfrak{h}$, and the non-zero weights are the roots. The zero weight records the Cartan subalgebra itself, so including it among the directions would blur the distinction between the diagonal part and the off-diagonal structure. The classification needs the nonzero weights that actually occur in $\mathfrak{g}$, because only those weights contribute root spaces whose brackets and angles carry new information. [definition: Root] Let $\mathfrak{g}$ be semisimple and let $\mathfrak{h}$ be a Cartan subalgebra. A root of $\mathfrak{g}$ relative to $\mathfrak{h}$ is a non-zero functional $\alpha\in\mathfrak{h}^*$ such that $\mathfrak{g}_\alpha\ne 0$. The set of roots is denoted $\Phi\subset\mathfrak{h}^*$. [/definition] The definition only names the non-zero eigenspaces; it does not yet say that these spaces account for the whole Lie algebra. The main obstruction is the possible presence of nilpotent or generalized-eigenvector behavior under the adjoint action of $\mathfrak{h}$. The next structural result rules this out in the semisimple setting and identifies the zero-weight part with the Cartan subalgebra itself. [quotetheorem:4677] [citeproof:4677] This theorem changes the study of $\mathfrak{g}$ into the study of the finite set $\Phi$ and the way the corresponding subspaces bracket with each other. The assumption that $\mathfrak{g}$ is semisimple is used twice: it makes the Cartan subalgebra self-centralising and prevents a non-zero centre from appearing as invisible zero-weight material. Algebraic closedness and characteristic zero are what allow the toral adjoint action to split into genuine eigenspaces rather than Jordan components. The statement also explains why the Cartan subalgebra is not an extra summand unrelated to the roots: it is the zero-weight part of the same decomposition, and the next section studies how the Lie bracket interacts with that grading by roots. [example: Roots Of Sl Three] Let $\mathfrak{g}=\mathfrak{sl}_3(k)$ and let $\mathfrak{h}$ be the subalgebra of diagonal trace-zero matrices. For \begin{align*} h=\operatorname{diag}(a_1,a_2,a_3)\in\mathfrak{h}, \qquad a_1+a_2+a_3=0, \end{align*} write $\varepsilon_i(h)=a_i$. If $E_{ij}$ is the matrix unit with $i\ne j$, then multiplication by a diagonal matrix scales rows on the left and columns on the right: \begin{align*} hE_{ij}=a_iE_{ij}, \qquad E_{ij}h=a_jE_{ij}. \end{align*} Therefore \begin{align*} [h,E_{ij}] &=hE_{ij}-E_{ij}h\\ &=a_iE_{ij}-a_jE_{ij}\\ &=(a_i-a_j)E_{ij}\\ &=\bigl(\varepsilon_i(h)-\varepsilon_j(h)\bigr)E_{ij}. \end{align*} Since this equality holds for every $h\in\mathfrak{h}$, the one-dimensional subspace $kE_{ij}$ is contained in the weight space $\mathfrak{g}_{\varepsilon_i-\varepsilon_j}$. Conversely, every element of $\mathfrak{sl}_3(k)$ has a unique expansion \begin{align*} x= \begin{pmatrix} d_1&0&0\\ 0&d_2&0\\ 0&0&d_3 \end{pmatrix} +\sum_{i\ne j} c_{ij}E_{ij}, \qquad d_1+d_2+d_3=0. \end{align*} The diagonal summand lies in $\mathfrak{h}$, and the six off-diagonal summands lie in the six weight spaces computed above. Hence the non-zero roots are \begin{align*} \Phi=\{\varepsilon_i-\varepsilon_j:1\le i,j\le 3,\ i\ne j\}, \end{align*} with \begin{align*} \mathfrak{g}_{\varepsilon_i-\varepsilon_j}=kE_{ij} \qquad (i\ne j). \end{align*} Thus the root-space decomposition of $\mathfrak{sl}_3(k)$ is precisely the splitting of a traceless matrix into its diagonal trace-zero part and its six off-diagonal matrix-unit directions. [/example] The trace-zero condition is not a harmless notational detail. It forces the diagonal functionals $\varepsilon_i$ to satisfy one linear relation after restriction to $\mathfrak{h}$, which is why the root system for $\mathfrak{sl}_n(k)$ has rank $n-1$ rather than $n$. [example: Matrix Unit Roots In Sl N] Let $\mathfrak{g}=\mathfrak{sl}_n(k)$, and let $\mathfrak{h}$ be the subalgebra of diagonal trace-zero matrices. For \begin{align*} h=\operatorname{diag}(a_1,\ldots,a_n)\in\mathfrak{h}, \qquad a_1+\cdots+a_n=0, \end{align*} define $\varepsilon_i(h)=a_i$. If $E_{ij}$ is the matrix unit with a $1$ in the $(i,j)$-entry and $0$ elsewhere, then for $i\ne j$ the $(p,q)$-entry of $hE_{ij}$ is \begin{align*} (hE_{ij})_{pq} &=\sum_{\ell=1}^n h_{p\ell}(E_{ij})_{\ell q} \\ &=h_{pi}(E_{ij})_{iq} \\ &=a_i\delta_{pi}\delta_{jq}, \end{align*} so $hE_{ij}=a_iE_{ij}$. Similarly, \begin{align*} (E_{ij}h)_{pq} &=\sum_{\ell=1}^n (E_{ij})_{p\ell}h_{\ell q} \\ &=(E_{ij})_{pj}h_{jq} \\ &=a_j\delta_{pi}\delta_{jq}, \end{align*} so $E_{ij}h=a_jE_{ij}$. Therefore \begin{align*} [h,E_{ij}] &=hE_{ij}-E_{ij}h \\ &=a_iE_{ij}-a_jE_{ij} \\ &=(a_i-a_j)E_{ij} \\ &=\bigl(\varepsilon_i(h)-\varepsilon_j(h)\bigr)E_{ij}. \end{align*} Since this holds for every $h\in\mathfrak{h}$, the line $kE_{ij}$ lies in the weight space $\mathfrak{g}_{\varepsilon_i-\varepsilon_j}$. Conversely, every $x\in\mathfrak{sl}_n(k)$ has a unique decomposition \begin{align*} x= \operatorname{diag}(d_1,\ldots,d_n)+\sum_{i\ne j}c_{ij}E_{ij}, \qquad d_1+\cdots+d_n=0. \end{align*} The diagonal summand is in $\mathfrak{h}$, and the off-diagonal summand $c_{ij}E_{ij}$ is in $kE_{ij}\subseteq\mathfrak{g}_{\varepsilon_i-\varepsilon_j}$. Hence the non-zero roots are exactly \begin{align*} \Phi=\{\varepsilon_i-\varepsilon_j:1\le i,j\le n,\ i\ne j\}, \end{align*} and for each $i\ne j$, \begin{align*} \mathfrak{g}_{\varepsilon_i-\varepsilon_j}=kE_{ij}. \end{align*} The trace-zero condition also gives, for every $h=\operatorname{diag}(a_1,\ldots,a_n)\in\mathfrak{h}$, \begin{align*} (\varepsilon_1+\cdots+\varepsilon_n)(h) &=\varepsilon_1(h)+\cdots+\varepsilon_n(h) \\ &=a_1+\cdots+a_n \\ &=0. \end{align*} Thus $\varepsilon_1+\cdots+\varepsilon_n=0$ as a functional on $\mathfrak{h}$, so the roots live in the $(n-1)$-dimensional space $\mathfrak{h}^*$. [/example] ## Bracket Rules Between Root Spaces Once the decomposition exists, the next problem is to determine how the Lie bracket moves between the summands. The Jacobi identity says that bracketing a vector of weight $\alpha$ with a vector of weight $\beta$ produces either weight $\alpha+\beta$ or zero. [quotetheorem:4686] [citeproof:4686] There are several immediate consequences. If $\alpha+\beta$ is not a root and is not zero, then $[\mathfrak{g}_\alpha,\mathfrak{g}_\beta]=0$, so many possible brackets are ruled out by weight arithmetic alone. If $\beta=-\alpha$, the bracket lands in $\mathfrak{h}$, and this exceptional case cannot be read as another non-zero root space. This is the source of the embedded $\mathfrak{sl}_2$ attached to $\alpha$, and it is also the first sign that the root data contains both additive and reflective structure. To convert these bracket rules into geometry we need a pairing on the root spaces, and the Killing form is the only invariant one available. Before using it we must know which root spaces it actually connects: if it paired spaces whose weights failed to cancel, its restriction to the decomposition would be intractable and the identification of $\mathfrak{h}$ with $\mathfrak{h}^*$ would break down. The question is thus precisely when $\kappa(\mathfrak{g}_\alpha,\mathfrak{g}_\beta)$ is allowed to be nonzero. [quotetheorem:4687] [citeproof:4687] The orthogonality theorem is the linear-algebra mechanism behind the later geometry of roots. The condition $\alpha+\beta\ne0$ is essential: opposite root spaces are precisely the ones that can pair nontrivially, and their pairing will produce Cartan elements. Nondegeneracy on $\mathfrak{h}$ would fail in the presence of central directions, so this is another point where semisimplicity is doing real work. This lets us identify $\mathfrak{h}$ with $\mathfrak{h}^*$ through the Killing form. For each $\alpha\in\mathfrak{h}^*$, there is a unique element $t_\alpha\in\mathfrak{h}$ satisfying \begin{align*} \kappa(t_\alpha,h)=\alpha(h)\quad\text{for all }h\in\mathfrak{h}. \end{align*} This element is the bridge between a root as a functional and the Cartan element appearing in the corresponding $\mathfrak{sl}_2$ triple. We now know that opposite root spaces are the only ones that pair and that $t_\alpha$ represents the functional $\alpha$ inside $\mathfrak{h}$. To build the $\mathfrak{sl}_2$ triple attached to $\alpha$ we need more than the fact that $[x,y]$ lands in $\mathfrak{h}$ for $x\in\mathfrak{g}_\alpha$ and $y\in\mathfrak{g}_{-\alpha}$: we need the exact element it produces. The obstruction is that membership in $\mathfrak{h}$ alone leaves a whole subspace of possibilities, and only one of them is proportional to $t_\alpha$. [quotetheorem:4688] [citeproof:4688] This theorem explains why opposite roots are different from arbitrary pairs whose sum is zero in the ambient vector space: their bracket is not just constrained to lie in $\mathfrak{h}$, but is determined by the Killing-form pairing. The formula also shows why the element $t_\alpha$ is the correct representative of the functional $\alpha$ inside $\mathfrak{h}$. The next step is to normalise this element so that the three brackets match the standard $\mathfrak{sl}_2(k)$ relations. [example: Brackets In Sl Three] In $\mathfrak{sl}_3(k)$, let $E_{ij}$ denote the matrix unit. The product of two matrix units is determined entrywise by \begin{align*} (E_{ij}E_{kl})_{pq} &=\sum_{r=1}^3 (E_{ij})_{pr}(E_{kl})_{rq} \\ &=(E_{ij})_{pj}(E_{kl})_{jq} \\ &=\delta_{pi}\delta_{jk}\delta_{ql}, \end{align*} so \begin{align*} E_{ij}E_{kl}=\delta_{jk}E_{il}. \end{align*} Similarly, \begin{align*} E_{kl}E_{ij}=\delta_{li}E_{kj}, \end{align*} and therefore \begin{align*} [E_{ij},E_{kl}] &=E_{ij}E_{kl}-E_{kl}E_{ij} \\ &=\delta_{jk}E_{il}-\delta_{li}E_{kj}. \end{align*} For the first bracket, take $E_{12}\in\mathfrak{g}_{\varepsilon_1-\varepsilon_2}$ and $E_{23}\in\mathfrak{g}_{\varepsilon_2-\varepsilon_3}$. Substituting $i=1,j=2,k=2,l=3$ gives \begin{align*} [E_{12},E_{23}] &=\delta_{22}E_{13}-\delta_{31}E_{22} \\ &=E_{13}-0 \\ &=E_{13}. \end{align*} The resulting matrix unit lies in $\mathfrak{g}_{\varepsilon_1-\varepsilon_3}$, and its weight is \begin{align*} (\varepsilon_1-\varepsilon_2)+(\varepsilon_2-\varepsilon_3) &=\varepsilon_1-\varepsilon_3. \end{align*} By contrast, substituting $i=1,j=2,k=1,l=3$ gives \begin{align*} [E_{12},E_{13}] &=\delta_{21}E_{13}-\delta_{31}E_{12} \\ &=0-0 \\ &=0. \end{align*} Here the sum of the two weights is \begin{align*} (\varepsilon_1-\varepsilon_2)+(\varepsilon_1-\varepsilon_3) &=2\varepsilon_1-\varepsilon_2-\varepsilon_3, \end{align*} which is not of the form $\varepsilon_p-\varepsilon_q$ with $p\ne q$. Thus the non-zero bracket occurs exactly when the sum of the two displayed roots is again a root, while the second bracket vanishes when the corresponding sum is not a root. [/example] ## Embedded Sl Two Subalgebras The next problem is to understand a single root in isolation. The pair of opposite root spaces $\mathfrak{g}_\alpha$ and $\mathfrak{g}_{-\alpha}$, together with their bracket in $\mathfrak{h}$, generate a copy of $\mathfrak{sl}_2(k)$. This copy imposes the representation-theoretic arithmetic that later becomes the root system axioms. [definition: Coroot Element] Let $\mathfrak{g}$ be semisimple with Cartan subalgebra $\mathfrak{h}$ and Killing form $\kappa$. For $\alpha\in\Phi$, let $t_\alpha\in\mathfrak{h}$ be the unique element satisfying \begin{align*} \kappa(t_\alpha,h)=\alpha(h) \end{align*} for all $h\in\mathfrak{h}$. The coroot element associated to $\alpha$ is \begin{align*} h_\alpha=\frac{2t_\alpha}{\kappa(t_\alpha,t_\alpha)}. \end{align*} [/definition] This definition is chosen so that $\alpha(h_\alpha)=2$, matching the standard diagonal element in $\mathfrak{sl}_2(k)$. The point of this normalization is not only cosmetic: it asks whether each root direction in an arbitrary semisimple Lie algebra carries a genuine copy of the rank-one algebra $\mathfrak{sl}_2(k)$. If so, the local geometry around one root can be studied by the familiar representation theory of $\mathfrak{sl}_2(k)$, and the later integrality restrictions become possible. [quotetheorem:4689] [citeproof:4689] The root subalgebra theorem is the point where the course imports the representation theory of $\mathfrak{sl}_2(k)$ into the structure theory of $\mathfrak{g}$. The normalisation $\alpha(h_\alpha)=2$ is essential: without it the weights below would be rescaled and would not be the Cartan integers used in root systems. The theorem depends on the nondegenerate pairing between opposite root spaces, so it is special to the semisimple setting rather than a feature of arbitrary Lie algebras with a diagonal action. The subalgebra generated by a root is small, but it acts on the whole Lie algebra by the adjoint representation. Since finite-dimensional $\mathfrak{sl}_2$-modules have integral weights and string-shaped weight diagrams, every root line through $\alpha$ inherits integrality restrictions. [quotetheorem:4690] [citeproof:4690] This is one of the decisive simplifications of semisimple Lie theory. The result would be false for a general simultaneous eigenspace decomposition: multiplicities can occur unless the Lie bracket, Killing form, and rank-one subalgebras impose the semisimple constraints used in the proof. The root decomposition is therefore not merely a decomposition into many eigenspaces; each non-zero eigenspace is a single root vector direction. This fact makes it possible to encode bracket information by constants attached to pairs of roots rather than by higher-dimensional linear maps. ## Root Strings And Integrality The final problem of the chapter is to measure how far one may move from a root $\beta$ by repeatedly adding or subtracting another root $\alpha$. The resulting chain is called a root string, and the embedded $\mathfrak{sl}_2$ for $\alpha$ determines both its shape and its length. [definition: Root String] Let $\alpha,\beta\in\Phi$ with $\alpha\ne0$. The $\alpha$-string through $\beta$ is the set of roots of the form \begin{align*} \beta+n\alpha, \qquad n\in\mathbb{Z}. \end{align*} [/definition] Since $\Phi$ is finite, every root string is finite. We write it as \begin{align*} \beta-r\alpha, \ldots, \beta, \ldots, \beta+q\alpha, \end{align*} where $r,q\ge0$ are maximal integers for which all displayed terms are roots. The point of recording both endpoints is that the geometry of two roots should be recoverable from this one-dimensional string. The obstruction is that an inner product such as $2(\beta,\alpha)/(\alpha,\alpha)$ looks analytic, while the string is only combinatorial; the next structural result explains why these two kinds of data agree. [quotetheorem:4691] [citeproof:4691] The integrality statement is the first appearance of the Cartan integers. Its exceptional cases are not defects in the theory: they correspond to the root paired with itself and its negative, where the value is already forced to be $2$ or $-2$. The theorem is powerful because all other pairs of roots are constrained by the finite-dimensional representation theory of a single embedded $\mathfrak{sl}_2(k)$. After identifying $\mathfrak{h}^*$ with a Euclidean space using the Killing form, these integers become \begin{align*} \frac{2(\beta,\alpha)}{(\alpha,\alpha)}. \end{align*} They control possible angles between roots and will be the input for Dynkin diagrams later in the course. [illustration:a2-alpha-string] [example: Root Strings In Sl Three] In $\mathfrak{sl}_3(k)$, take \begin{align*} \alpha=\varepsilon_1-\varepsilon_2, \qquad \beta=\varepsilon_2-\varepsilon_3. \end{align*} The positive direction in the $\alpha$-string begins with \begin{align*} \beta+\alpha &=(\varepsilon_2-\varepsilon_3)+(\varepsilon_1-\varepsilon_2) \\ &=\varepsilon_1-\varepsilon_3, \end{align*} which is one of the roots $\varepsilon_i-\varepsilon_j$ with $i\ne j$. The next term is \begin{align*} \beta+2\alpha &=(\varepsilon_2-\varepsilon_3)+2(\varepsilon_1-\varepsilon_2) \\ &=\varepsilon_2-\varepsilon_3+2\varepsilon_1-2\varepsilon_2 \\ &=2\varepsilon_1-\varepsilon_2-\varepsilon_3. \end{align*} This is not equal to $\varepsilon_i-\varepsilon_j$ for any $i\ne j$: the coefficients of a root $\varepsilon_i-\varepsilon_j$ are one $1$, one $-1$, and one $0$, while the coefficients here are $2,-1,-1$. Hence $\beta+2\alpha$ is not a root. In the negative direction, \begin{align*} \beta-\alpha &=(\varepsilon_2-\varepsilon_3)-(\varepsilon_1-\varepsilon_2) \\ &=\varepsilon_2-\varepsilon_3-\varepsilon_1+\varepsilon_2 \\ &=2\varepsilon_2-\varepsilon_1-\varepsilon_3. \end{align*} Again this is not of the form $\varepsilon_i-\varepsilon_j$, since its coefficients are $-1,2,-1$ rather than one $1$, one $-1$, and one $0$. Therefore the $\alpha$-string through $\beta$ is \begin{align*} \beta,\ \beta+\alpha, \end{align*} so $r=0$ and $q=1$. The root-string formula gives \begin{align*} \beta(h_\alpha) &=r-q \\ &=0-1 \\ &=-1. \end{align*} Thus the two adjacent simple roots in type $A_2$ have Cartan integer $-1$. [/example] The example shows how a short calculation in matrices becomes a geometric restriction on the possible configuration of roots. In later chapters the same integer will be read directly from a Cartan matrix, without returning to the ambient Lie algebra. [remark: Why Root Strings Matter] Root strings are the local arithmetic shadows of the Lie bracket. The bracket rule says when a single sum of roots can occur; the string theorem says how many repeated sums may occur and packages that information as an integer. These integers later determine reflections, Weyl groups, Cartan matrices, and Dynkin diagrams. [/remark] The root space decomposition turns Lie brackets into combinatorial constraints, but those constraints still live inside the algebra. The next step is to abstract away the ambient Lie algebra and isolate the finite geometric pattern carried by the roots themselves. # 4. Abstract Root Systems This chapter turns the root decomposition of a semisimple Lie algebra into a finite piece of Euclidean geometry. The guiding question is: which finite configurations of vectors can occur as roots, once the Lie algebra has supplied reflections, strings, and integrality? We separate the abstract geometry from its Lie-theoretic origin, because the classification of semisimple Lie algebras will later be reduced to the classification of these root systems. ## Reflections and Cartan Integers The first problem is to decide what geometric structure remains if we forget the Lie algebra but keep the behaviour of its roots under the $\mathfrak{sl}_2$-subalgebras attached to individual roots. The answer is a finite spanning set of non-zero vectors in a Euclidean space, stable under special orthogonal reflections and subject to a discreteness condition. Let $E$ be a finite-dimensional real vector space with positive definite inner product $(\cdot,\cdot)$. For a non-zero vector $\alpha \in E$, the reflection in the hyperplane perpendicular to $\alpha$ is the map \begin{align*} s_\alpha &: E \to E, & s_\alpha(x) &= x - \frac{2(x,\alpha)}{(\alpha,\alpha)}\alpha. \end{align*} It fixes the hyperplane $\alpha^\perp$ pointwise and sends $\alpha$ to $-\alpha$. [definition: Root System] A root system in $E$ is a finite subset $\Phi \subset E \setminus \{0\}$ such that: 1. $\Phi$ spans $E$; 2. if $\alpha \in \Phi$, then $\Phi \cap \mathbb R\alpha = \{\alpha,-\alpha\}$; 3. if $\alpha \in \Phi$, then $s_\alpha(\Phi)=\Phi$; 4. if $\alpha,\beta \in \Phi$, then $\frac{2(\beta,\alpha)}{(\alpha,\alpha)} \in \mathbb Z$. [/definition] The second condition is the reducedness condition: along a line through the origin, the system contains only the two opposite roots. The fourth condition is the crystallographic condition; it is not natural for arbitrary finite reflection arrangements, but it is forced by root strings in Lie algebras. The reflection formula now presents a notational problem: the same normalized version of a root appears every time a vector is reflected. Without naming this normalization, every calculation would have to carry the denominator $(\alpha,\alpha)$, and the distinction between angle data and length data would be harder to see. The coroot records the root as the linear functional that measures displacement in the $\alpha$-direction. [definition: Coroot] For $\alpha \in E \setminus \{0\}$, the coroot of $\alpha$ is \begin{align*} \alpha^\vee = \frac{2\alpha}{(\alpha,\alpha)}. \end{align*} [/definition] With this notation $s_\alpha(x)=x-(x,\alpha^\vee)\alpha$. For roots $\alpha,\beta$, the integer $(\beta,\alpha^\vee)$ records how far $\beta$ moves in the $\alpha$-direction under reflection. To compare different pairs of roots systematically, these integers need their own name. They are the entries that survive when the root system is converted into a Cartan matrix or Dynkin diagram. They are not symmetric in the two roots, because the denominator remembers the length of the root used for reflection; naming them makes that ordered dependence explicit. [definition: Cartan Integer] For roots $\alpha,\beta \in \Phi$, the Cartan integer is the function \begin{align*} \Phi \times \Phi &\to \mathbb Z, & (\beta,\alpha) &\mapsto \langle \beta,\alpha^\vee\rangle = (\beta,\alpha^\vee)=\frac{2(\beta,\alpha)}{(\alpha,\alpha)}. \end{align*} [/definition] The order of the two roots matters. Changing from $\langle \beta,\alpha^\vee\rangle$ to $\langle \alpha,\beta^\vee\rangle$ changes the denominator and therefore detects relative root lengths. This is the first sign that root systems remember more than the unlabelled angles between lines: they also remember which roots are long and which are short. The next example shows how the same reflection formula simultaneously computes a Cartan integer and checks stability of the set of roots. [example: Reflection in A2] Take $E=\{(x_1,x_2,x_3)\in \mathbb R^3:x_1+x_2+x_3=0\}$ with the Euclidean inner product, and let \begin{align*} \Phi=\{e_i-e_j:1\le i\ne j\le 3\}. \end{align*} For $\alpha=e_1-e_2=(1,-1,0)$ and $\beta=e_2-e_3=(0,1,-1)$, we compute the two inner products appearing in the Cartan integer: \begin{align*} (\alpha,\alpha) &=(1,-1,0)\cdot(1,-1,0)\\ &=1^2+(-1)^2+0^2\\ &=2, \\ (\beta,\alpha) &=(0,1,-1)\cdot(1,-1,0)\\ &=0\cdot 1+1\cdot(-1)+(-1)\cdot 0\\ &=-1. \end{align*} Therefore \begin{align*} \langle \beta,\alpha^\vee\rangle &=\frac{2(\beta,\alpha)}{(\alpha,\alpha)}\\ &=\frac{2(-1)}{2}\\ &=-1. \end{align*} The reflection formula gives \begin{align*} s_\alpha(\beta) &=\beta-\langle \beta,\alpha^\vee\rangle\alpha\\ &=(e_2-e_3)-(-1)(e_1-e_2)\\ &=e_2-e_3+e_1-e_2\\ &=e_1-e_3. \end{align*} Since $e_1-e_3\in\Phi$, this particular reflection sends the root $\beta$ to another root. Geometrically, $s_\alpha$ interchanges the first two coordinate directions in the plane $x_1+x_2+x_3=0$, and this calculation displays one edge of the hexagonal $A_2$ root configuration. [/example] [illustration:a2-weyl-reflection] The integrality condition has strong metric consequences because the two Cartan integers multiply to a number depending only on the angle between roots. At this point the classification problem needs a finite menu of possible rank-two interactions, not a continuum of possible angles. The obstruction is that ordinary reflection arrangements can have many other angles; crystallographic root systems are more rigid because both ordered Cartan integers must be integral. [quotetheorem:4692] [citeproof:4692] Thus only a small list of angles can occur: $90^\circ$, $120^\circ$ or $60^\circ$, $135^\circ$ or $45^\circ$, and $150^\circ$ or $30^\circ$. The hypotheses are doing real work here: without integrality, a finite reflection arrangement could contain adjacent lines at an angle such as $\pi/5$, producing the dihedral system $I_2(5)$ rather than a crystallographic root system. The non-proportional assumption excludes the case $\beta=\pm\alpha$, where the product would be $4$ and would not describe an angle between two distinct root lines. The sharper information is ordered, because the signs and sizes of the two Cartan integers distinguish which root is longer, and this is what the next example illustrates in rank two. [example: Long and Short Roots in B2] In $E=\mathbb R^2$, let \begin{align*} \Phi=\{\pm e_1,\pm e_2,\pm e_1\pm e_2\}. \end{align*} We compare the ordered Cartan integers for the short root $\beta=e_1$ and the long root $\alpha=e_1+e_2$. Their squared lengths are \begin{align*} (\beta,\beta) &=e_1\cdot e_1\\ &=1, \\ (\alpha,\alpha) &=(e_1+e_2)\cdot(e_1+e_2)\\ &=e_1\cdot e_1+2e_1\cdot e_2+e_2\cdot e_2\\ &=1+0+1\\ &=2. \end{align*} The inner product between them is \begin{align*} (\beta,\alpha) &=e_1\cdot(e_1+e_2)\\ &=e_1\cdot e_1+e_1\cdot e_2\\ &=1+0\\ &=1. \end{align*} Thus the angle $\theta$ between $e_1$ and $e_1+e_2$ satisfies \begin{align*} \cos\theta &=\frac{(\beta,\alpha)}{\sqrt{(\beta,\beta)}\sqrt{(\alpha,\alpha)}}\\ &=\frac{1}{\sqrt{1}\sqrt{2}}\\ &=\frac{1}{\sqrt2}, \end{align*} so $\theta=45^\circ$. Now compute the two Cartan integers in their two possible orders: \begin{align*} \langle \beta,\alpha^\vee\rangle &=\frac{2(\beta,\alpha)}{(\alpha,\alpha)}\\ &=\frac{2\cdot 1}{2}\\ &=1, \\ \langle \alpha,\beta^\vee\rangle &=\frac{2(\alpha,\beta)}{(\beta,\beta)}\\ &=\frac{2\cdot 1}{1}\\ &=2. \end{align*} Their product is therefore \begin{align*} \langle \beta,\alpha^\vee\rangle\langle \alpha,\beta^\vee\rangle &=1\cdot 2\\ &=2. \end{align*} The unequal values come from the unequal denominators $(\alpha,\alpha)=2$ and $(\beta,\beta)=1$: reflection through the long root subtracts one copy of $\alpha$, while reflection through the short root subtracts two copies of $\beta$. [/example] ## Root Strings from Lie Algebras The next question is why the crystallographic condition belongs in the definition at all. As Chapter 3 showed, integrality is not an extra geometric preference: it comes from finite-dimensional $\mathfrak{sl}_2$ representation theory applied to the string of root spaces in the direction of a fixed root. Suppose $\mathfrak g$ is a finite-dimensional semisimple Lie algebra over the complex numbers, $\mathfrak h$ is a Cartan subalgebra, and \begin{align*} \mathfrak g = \mathfrak h \oplus \bigoplus_{\alpha\in \Phi}\mathfrak g_\alpha \end{align*} is its root space decomposition. For each root $\alpha$, the elements $e_\alpha\in \mathfrak g_\alpha$, $f_\alpha\in \mathfrak g_{-\alpha}$, and $h_\alpha=[e_\alpha,f_\alpha]$ span a copy of $\mathfrak{sl}_2$ after suitable scaling. The adjoint action of this copy on $\mathfrak g$ organises roots into finite strings. [definition: Root String] Let $\Phi$ be a set of roots and let $\alpha,\beta\in\Phi$ with $\alpha$ fixed. The $\alpha$-string through $\beta$ is the set of roots of the form \begin{align*} \beta+n\alpha,\qquad n\in\mathbb Z. \end{align*} [/definition] In a finite root system, every such string is finite. The missing link is quantitative: finiteness alone says the string stops, but it does not say how the two stopping points remember the inner product ratio $2(\beta,\alpha)/(\alpha,\alpha)$. For Lie algebra roots, the embedded $\mathfrak{sl}_2$ representation supplies exactly that link between endpoints and Cartan integers, so the next result is needed to turn a visible string into a numerical invariant. [quotetheorem:4693] [citeproof:4693] The theorem simultaneously explains integrality and gives a computational method. To find $\langle \beta,\alpha^\vee\rangle$, count how many steps the string extends backward and forward in the $\alpha$ direction. The condition $\beta\ne\pm\alpha$ avoids the special three-dimensional adjoint string, where the root spaces $\mathfrak g_{-\alpha}$ and $\mathfrak g_\alpha$ are linked through the Cartan subalgebra rather than through a string of non-zero roots alone. If the Lie algebra were not semisimple, the required $\mathfrak{sl}_2$ triples and finite weight-string structure could fail. This result is the mechanism that turns representation theory into the integrality axiom of the abstract definition. [example: A2 Root String] In the $A_2$ system \begin{align*} \Phi=\{e_i-e_j:1\le i\ne j\le 3\}, \end{align*} take $\alpha=e_1-e_2$ and $\beta=e_2-e_3$. The first forward step in the $\alpha$-direction is \begin{align*} \beta+\alpha &=(e_2-e_3)+(e_1-e_2)\\ &=e_1-e_3, \end{align*} which belongs to $\Phi$. The next forward step is \begin{align*} \beta+2\alpha &=(e_2-e_3)+2(e_1-e_2)\\ &=2e_1-e_2-e_3. \end{align*} This vector is not of the form $e_i-e_j$, since its coordinates are $2,-1,-1$ rather than one $1$, one $-1$, and one $0$. In the backward direction, \begin{align*} \beta-\alpha &=(e_2-e_3)-(e_1-e_2)\\ &=2e_2-e_1-e_3, \end{align*} which is also not of the form $e_i-e_j$. Hence the $\alpha$-string through $\beta$ is exactly \begin{align*} \beta,\qquad \beta+\alpha, \end{align*} so $p=0$ and $q=1$. By the *Root String Theorem*, \begin{align*} \langle \beta,\alpha^\vee\rangle &=p-q\\ &=0-1\\ &=-1. \end{align*} This matches the inner-product formula, since \begin{align*} (\beta,\alpha) &=(0,1,-1)\cdot(1,-1,0)\\ &=0\cdot 1+1\cdot(-1)+(-1)\cdot 0\\ &=-1, \\ (\alpha,\alpha) &=(1,-1,0)\cdot(1,-1,0)\\ &=1^2+(-1)^2+0^2\\ &=2, \end{align*} and therefore \begin{align*} \langle \beta,\alpha^\vee\rangle &=\frac{2(\beta,\alpha)}{(\alpha,\alpha)}\\ &=\frac{2(-1)}{2}\\ &=-1. \end{align*} The string computation recovers the same Cartan integer by counting endpoints rather than by measuring inner products. [/example] The example also shows why reflection invariance should be expected rather than imposed arbitrarily. The reflected root appears at the opposite end of the same finite string, suggesting that root strings contain a hidden Euclidean symmetry. The theorem below turns that string symmetry into the geometric reflection formula and explains why reflections are forced by the Lie algebra, not added by hand. [quotetheorem:4694] [citeproof:4694] This is the precise bridge from semisimple Lie algebras to abstract root systems. The exclusion of $\beta=\pm\alpha$ from the string argument is harmless because those two roots are handled directly by the reflection formula. The theorem depends on the finiteness of the string; without the finite-dimensional $\mathfrak{sl}_2$ representation theory behind it, a formal set of weights need not be symmetric under reflection. The Killing form supplies the Euclidean inner product on the real span of the roots, $\mathfrak{sl}_2$ theory supplies the integers, and the string symmetry supplies the reflections, leaving reducedness as the remaining axiom to justify later in the chapter. ## Rank Two Root Systems The classification begins locally: what can two non-proportional roots generate? Since the angle and relative lengths are restricted by the Cartan integer product, every rank two subsystem must be one of a short list. These rank two pictures are the building blocks encoded later by edges in Dynkin diagrams. [definition: Rank of a Root System] The rank of a root system $\Phi\subset E$ is $\dim E$. [/definition] For rank two, the whole system lives in a Euclidean plane. Reducedness prevents multiple roots on the same line, while reflection invariance forces every reflected vector to appear. The obstruction to a short list is the apparent continuum of possible angles between two lines; the Cartan integer product turns that continuum into a small set of allowed crystallographic angles. [quotetheorem:4695] [citeproof:4695] The theorem is best remembered through the pictures: square cross, hexagon, square with diagonals, and twelve-root hexagonal pattern with alternating lengths. Reducedness is essential here: if both $\alpha$ and $2\alpha$ were allowed on the same line, non-reduced rank two systems such as $BC_2$ would enter the list. Crystallography is also essential, since finite dihedral reflection systems exist with other angles but do not have integral Cartan integers. The algebraic distinction among the last three is the ratio of squared lengths: $1$, $2$, and $3$, and these ratios become the edge multiplicities in Dynkin diagrams. [example: The Orthogonal System A1 x A1] Let $E=\mathbb R^2$ with its standard orthonormal basis $e_1=(1,0)$ and $e_2=(0,1)$, and set \begin{align*} \Phi=\{\pm e_1,\pm e_2\}. \end{align*} The two positive directions are perpendicular because \begin{align*} (e_1,e_2) &=(1,0)\cdot(0,1)\\ &=1\cdot 0+0\cdot 1\\ &=0. \end{align*} Their squared lengths are \begin{align*} (e_1,e_1)&=1, & (e_2,e_2)&=1, \end{align*} so the mixed Cartan integers are \begin{align*} \langle e_2,e_1^\vee\rangle &=\frac{2(e_2,e_1)}{(e_1,e_1)}\\ &=\frac{2\cdot 0}{1}\\ &=0, \\ \langle e_1,e_2^\vee\rangle &=\frac{2(e_1,e_2)}{(e_2,e_2)}\\ &=\frac{2\cdot 0}{1}\\ &=0. \end{align*} The reflection through the hyperplane perpendicular to $e_1$ sends \begin{align*} s_{e_1}(e_1) &=e_1-\frac{2(e_1,e_1)}{(e_1,e_1)}e_1\\ &=e_1-2e_1\\ &=-e_1, \\ s_{e_1}(e_2) &=e_2-\frac{2(e_2,e_1)}{(e_1,e_1)}e_1\\ &=e_2-\frac{2\cdot 0}{1}e_1\\ &=e_2. \end{align*} Similarly, $s_{e_2}(e_2)=-e_2$ and $s_{e_2}(e_1)=e_1$. Hence reflections in the two root directions preserve exactly the four roots $\pm e_1,\pm e_2$. No vector of the form $e_1+e_2=(1,1)$ or $e_1-e_2=(1,-1)$ appears in $\Phi$, so the two rank one systems do not interact; this is the product $A_1\times A_1$. [/example] The next rank two system is what happens when the two simple directions are no longer orthogonal but all roots still have the same length. It is the smallest example where reflections generate a genuinely connected configuration. [example: The Hexagonal System A2] The $A_2$ system is realised in the plane \begin{align*} E=\{(x_1,x_2,x_3)\in\mathbb R^3:x_1+x_2+x_3=0\} \end{align*} by the six roots \begin{align*} \Phi=\{e_i-e_j:1\le i\ne j\le 3\}. \end{align*} Take \begin{align*} \alpha=e_1-e_2=(1,-1,0), \qquad \beta=e_2-e_3=(0,1,-1). \end{align*} Their squared lengths are equal: \begin{align*} (\alpha,\alpha) &=(1,-1,0)\cdot(1,-1,0)\\ &=1^2+(-1)^2+0^2\\ &=2, \\ (\beta,\beta) &=(0,1,-1)\cdot(0,1,-1)\\ &=0^2+1^2+(-1)^2\\ &=2. \end{align*} Their inner product is \begin{align*} (\beta,\alpha) &=(0,1,-1)\cdot(1,-1,0)\\ &=0\cdot 1+1\cdot(-1)+(-1)\cdot 0\\ &=-1. \end{align*} Therefore the two ordered Cartan integers are \begin{align*} \langle \beta,\alpha^\vee\rangle &=\frac{2(\beta,\alpha)}{(\alpha,\alpha)}\\ &=\frac{2(-1)}{2}\\ &=-1, \\ \langle \alpha,\beta^\vee\rangle &=\frac{2(\alpha,\beta)}{(\beta,\beta)}\\ &=\frac{2(-1)}{2}\\ &=-1. \end{align*} The equality of these two values comes from the equality $(\alpha,\alpha)=(\beta,\beta)=2$; there is no long-short distinction. The remaining positive root obtained from these two simple roots is \begin{align*} \alpha+\beta &=(e_1-e_2)+(e_2-e_3)\\ &=e_1-e_3. \end{align*} Thus the positive roots are \begin{align*} \alpha,\qquad \beta,\qquad \alpha+\beta, \end{align*} and including their negatives gives the six roots of $A_2$. The angle between the two simple roots satisfies \begin{align*} \cos\theta &=\frac{(\alpha,\beta)}{\sqrt{(\alpha,\alpha)}\sqrt{(\beta,\beta)}}\\ &=\frac{-1}{\sqrt2\sqrt2}\\ &=-\frac12, \end{align*} so $\theta=120^\circ$ between the chosen simple roots, equivalently $60^\circ$ between adjacent root lines in the hexagonal picture. This is the first rank two case where the simple directions interact, but all roots still have the same length. [/example] Allowing two root lengths gives a new possibility. The first such case is $B_2$, where the Cartan integers record not only the angle but also the long-short asymmetry. [example: The Square-Octagon System B2] The $B_2$ system is \begin{align*} \Phi=\{\pm e_1,\pm e_2,\pm e_1\pm e_2\}\subset\mathbb R^2, \end{align*} where $e_1=(1,0)$ and $e_2=(0,1)$ are orthonormal. The roots $\pm e_1,\pm e_2$ have squared length $1$, while \begin{align*} (e_1+e_2,e_1+e_2) &=e_1\cdot e_1+2e_1\cdot e_2+e_2\cdot e_2\\ &=1+0+1\\ &=2, \\ (e_1-e_2,e_1-e_2) &=e_1\cdot e_1-2e_1\cdot e_2+e_2\cdot e_2\\ &=1-0+1\\ &=2. \end{align*} Thus $\pm e_1,\pm e_2$ are the four short roots, and $\pm e_1\pm e_2$ are the four long roots. Choose the long root $\alpha=e_1-e_2$ and the short root $\beta=e_2$. Their squared lengths and inner product are \begin{align*} (\alpha,\alpha) &=(e_1-e_2)\cdot(e_1-e_2)\\ &=1-0+1\\ &=2, \\ (\beta,\beta) &=e_2\cdot e_2\\ &=1, \\ (\beta,\alpha) &=e_2\cdot(e_1-e_2)\\ &=e_2\cdot e_1-e_2\cdot e_2\\ &=0-1\\ &=-1. \end{align*} Therefore the two ordered Cartan integers are \begin{align*} \langle \beta,\alpha^\vee\rangle &=\frac{2(\beta,\alpha)}{(\alpha,\alpha)}\\ &=\frac{2(-1)}{2}\\ &=-1, \\ \langle \alpha,\beta^\vee\rangle &=\frac{2(\alpha,\beta)}{(\beta,\beta)}\\ &=\frac{2(-1)}{1}\\ &=-2. \end{align*} Their product is \begin{align*} \langle \beta,\alpha^\vee\rangle\langle \alpha,\beta^\vee\rangle &=(-1)(-2)\\ &=2. \end{align*} The angle $\theta$ between $\alpha$ and $\beta$ satisfies \begin{align*} \cos\theta &=\frac{(\alpha,\beta)}{\sqrt{(\alpha,\alpha)}\sqrt{(\beta,\beta)}}\\ &=\frac{-1}{\sqrt2\sqrt1}\\ &=-\frac{1}{\sqrt2}, \end{align*} so $\theta=135^\circ$. The $\beta$-string through the long root $\alpha$ is visible by adding multiples of $\beta=e_2$: \begin{align*} \alpha-\beta &=(e_1-e_2)-e_2\\ &=e_1-2e_2, \\ \alpha &=e_1-e_2, \\ \alpha+\beta &=(e_1-e_2)+e_2\\ &=e_1, \\ \alpha+2\beta &=(e_1-e_2)+2e_2\\ &=e_1+e_2, \\ \alpha+3\beta &=(e_1-e_2)+3e_2\\ &=e_1+2e_2. \end{align*} Among these, $\alpha$, $\alpha+\beta=e_1$, and $\alpha+2\beta=e_1+e_2$ belong to $\Phi$, while $\alpha-\beta=e_1-2e_2$ and $\alpha+3\beta=e_1+2e_2$ do not. Thus the string has two forward steps and no backward step. This is the same long-short asymmetry recorded by the unequal Cartan integers $-1$ and $-2$. [/example] The last crystallographic rank two case pushes the same phenomenon to its limit. The length ratio increases from $2$ to $3$, and the corresponding root string is as long as the product formula permits. [example: The Twelve-Root System G2] Normalize the short root so that $(\alpha,\alpha)=a>0$. Since $\beta$ is long and the squared length ratio is $3$, we have $(\beta,\beta)=3a$. The angle between $\alpha$ and $\beta$ is $150^\circ$, so \begin{align*} (\alpha,\beta) &=\sqrt{(\alpha,\alpha)}\sqrt{(\beta,\beta)}\cos(150^\circ)\\ &=\sqrt a\sqrt{3a}\left(-\frac{\sqrt3}{2}\right)\\ &=-\frac{3a}{2}. \end{align*} Thus the ordered Cartan integers are \begin{align*} \langle \beta,\alpha^\vee\rangle &=\frac{2(\beta,\alpha)}{(\alpha,\alpha)}\\ &=\frac{2\left(-\frac{3a}{2}\right)}{a}\\ &=-3, \\ \langle \alpha,\beta^\vee\rangle &=\frac{2(\alpha,\beta)}{(\beta,\beta)}\\ &=\frac{2\left(-\frac{3a}{2}\right)}{3a}\\ &=-1. \end{align*} For any vector $m\alpha+n\beta$, its squared length is \begin{align*} (m\alpha+n\beta,m\alpha+n\beta) &=m^2(\alpha,\alpha)+2mn(\alpha,\beta)+n^2(\beta,\beta)\\ &=m^2a+2mn\left(-\frac{3a}{2}\right)+3n^2a\\ &=a(m^2-3mn+3n^2). \end{align*} Applying this to the six positive roots gives \begin{align*} (\alpha,\alpha)&=a, & (\beta,\beta)&=3a, \\ (\alpha+\beta,\alpha+\beta) &=a(1-3+3)\\ &=a, \\ (2\alpha+\beta,2\alpha+\beta) &=a(4-6+3)\\ &=a, \\ (3\alpha+\beta,3\alpha+\beta) &=a(9-9+3)\\ &=3a, \\ (3\alpha+2\beta,3\alpha+2\beta) &=a(9-18+12)\\ &=3a. \end{align*} So among the positive roots, \begin{align*} \alpha,\quad \alpha+\beta,\quad 2\alpha+\beta \end{align*} are short, while \begin{align*} \beta,\quad 3\alpha+\beta,\quad 3\alpha+2\beta \end{align*} are long. Including their negatives gives six short roots and six long roots. The $\alpha$-string through the long root $\beta$ is \begin{align*} \beta,\qquad \beta+\alpha,\qquad \beta+2\alpha,\qquad \beta+3\alpha. \end{align*} Indeed, these are exactly \begin{align*} \beta,\qquad \alpha+\beta,\qquad 2\alpha+\beta,\qquad 3\alpha+\beta, \end{align*} which appear in the listed positive roots. The next forward vector is \begin{align*} \beta+4\alpha=4\alpha+\beta, \end{align*} which is not in the list, and the first backward vector is \begin{align*} \beta-\alpha=-\alpha+\beta, \end{align*} which is not in the list. Hence the string has $p=0$ and $q=3$, giving \begin{align*} p-q&=0-3\\ &=-3, \end{align*} the same value as $\langle \beta,\alpha^\vee\rangle$. This is the longest possible root string in a reduced crystallographic rank two system: a long root moves three steps in a short-root direction. [/example] [illustration:rank-two-lengths-comparison] ## Reducedness and Crystallography The final question in this chapter is which parts of the definition are geometric choices and which are forced by the Lie algebra. Reflection invariance and integrality come from the root string theorem, while reducedness comes from the fact that roots occur in $\mathfrak{sl}_2$ strings with tightly controlled multiples. [quotetheorem:4696] [citeproof:4696] Together with the previous section, this explains why the abstract definition was chosen. The theorem rules out the most immediate non-reduced failure, namely having both $\alpha$ and $2\alpha$ on the same root line as in the finite non-reduced systems of type $BC_n$. Its proof uses both the string formula and finiteness; outside semisimple Lie theory, finite reflection arrangements can be deliberately enlarged by such double roots. It is exactly the package needed to capture the finite geometry of roots of a semisimple Lie algebra. Reflection invariance, integrality, and reducedness have now each been established separately. The closing question of the chapter is whether they combine into a single self-contained object that no longer refers to the Lie algebra at all. The risk is that the axioms might constrain the roots only one at a time, leaving the collected root data short of the full abstract definition, so we need a statement that verifies every axiom simultaneously. In the theorem below, the "standard real form of the root lattice" means the real vector space spanned by the roots, obtained from the integer lattice generated by $\Phi$ after extending scalars to $\mathbb R$. The Killing form identifies roots and coroot directions and induces the positive definite inner product used on this real span. [quotetheorem:4697] [citeproof:4697] This theorem identifies the exact point where the Lie algebra course hands the problem to finite geometry. Semisimplicity is essential: it supplies the Killing form, the Cartan decomposition, and the $\mathfrak{sl}_2$ strings used throughout the proof. The result does not classify root systems by itself; it only proves that every semisimple Lie algebra produces one. The next stage is to choose simple roots and encode these finite configurations by Cartan matrices and Dynkin diagrams. The rest of the course now becomes combinatorial. We will choose bases of simple roots, encode their Cartan integers in Cartan matrices and Dynkin diagrams, and then prove that these diagrams classify the possible semisimple Lie algebras. Abstract root systems distill the geometry hidden in the Lie algebra into a finite combinatorial object with strong integrality and reflection properties. From there, the natural question is how to choose a positive half of the system so that the roots can be ordered and built up inductively. # 5. Bases, Positive Roots, and Weyl Chambers Chapter 4 showed that Lie algebra roots form abstract reduced crystallographic root systems; this chapter turns such a root system into a combinatorial object that can be ordered, drawn, and used for induction. The central problem is to choose a consistent notion of positivity among roots so that every root is built from a small distinguished set. These distinguished roots, called simple roots, are the vertices from which Dynkin diagrams and classification data will later be read. ## Choosing Positive Roots from a Hyperplane A finite root system has no preferred direction: if $\alpha$ is a root then $-\alpha$ is also a root. To do induction or choose generators, we need to cut the ambient Euclidean space into two halves without placing any root on the cutting wall. [definition: Regular Hyperplane] Let $E$ be a finite-dimensional real inner product space and let $\Phi \subset E$ be a root system. A hyperplane $H \subset E$ is regular for $\Phi$ if $0 \in H$ and $\alpha \notin H$ for every $\alpha \in \Phi$. [/definition] A regular hyperplane separates the roots into two open half-spaces. Since $\Phi$ is finite, regular hyperplanes exist: choose a vector $v \in E$ such that the inner product of $\alpha$ and $v$ is nonzero for every $\alpha \in \Phi$, and set $H=v^\perp$. The separation alone is not yet the algebraic object used for induction: choosing one side must also be stable under adding roots when their sum remains a root. This compatibility is what makes a choice of positive roots useful rather than merely a sign convention. [definition: Positive System] Let $\Phi \subset E$ be a root system. A positive system is a subset $\Phi^+ \subset \Phi$ such that for every $\alpha \in \Phi$, exactly one of $\alpha$ and $-\alpha$ lies in $\Phi^+$, and if $\alpha,\beta \in \Phi^+$ with $\alpha+\beta \in \Phi$, then $\alpha+\beta \in \Phi^+$. [/definition] The positive roots are the roots lying on one chosen side of a regular hyperplane. The closure condition says that positivity is compatible with the partial addition operation inside the root system. [example: Positive Roots in Type A Two] In the $A_2$ root system the six roots are \begin{align*} \Phi=\{\alpha,\beta,\alpha+\beta,-\alpha,-\beta,-(\alpha+\beta)\}. \end{align*} Choose a regular line through the origin whose positive side contains $\alpha$, $\beta$, and $\alpha+\beta$, and set \begin{align*} \Phi^+=\{\alpha,\beta,\alpha+\beta\}. \end{align*} Then exactly one root from each opposite pair lies in $\Phi^+$: \begin{align*} \{\alpha,-\alpha\}\cap \Phi^+&=\{\alpha\},\\ \{\beta,-\beta\}\cap \Phi^+&=\{\beta\},\\ \{\alpha+\beta,-(\alpha+\beta)\}\cap \Phi^+&=\{\alpha+\beta\}. \end{align*} It remains to check compatibility with sums. Among sums of two elements of $\Phi^+$, we have \begin{align*} \alpha+\beta&\in \Phi^+,\\ \alpha+(\alpha+\beta)&=2\alpha+\beta \notin \Phi,\\ \beta+(\alpha+\beta)&=\alpha+2\beta \notin \Phi,\\ \alpha+\alpha&=2\alpha \notin \Phi,\\ \beta+\beta&=2\beta \notin \Phi,\\ (\alpha+\beta)+(\alpha+\beta)&=2\alpha+2\beta \notin \Phi. \end{align*} Thus the only sum of two positive roots which is again a root is $\alpha+\beta$, and it is already in $\Phi^+$, so $\Phi^+$ is a positive system. The opposite choice is \begin{align*} -\Phi^+=\{-\alpha,-\beta,-(\alpha+\beta)\}, \end{align*} and the same check gives $(-\alpha)+(-\beta)=-(\alpha+\beta)\in -\Phi^+$. By contrast, the choice $\{\alpha,\beta,-(\alpha+\beta)\}$ is not a positive system, because $\alpha,\beta$ are chosen positive and $\alpha+\beta\in\Phi$, but $\alpha+\beta$ is not chosen positive. This is the extra closure condition beyond choosing one root from each pair $\{\gamma,-\gamma\}$. [/example] The example also shows a geometric obstruction: the signs of roots change only when the chosen vector crosses a root hyperplane. Thus the useful choices of positivity are not individual vectors but the regions cut out by all the reflecting hyperplanes at once. [definition: Weyl Chamber] Let $\Phi \subset E$ be a root system. The Weyl chambers are the connected components of \begin{align*} E \setminus \bigcup_{\alpha \in \Phi} \alpha^\perp . \end{align*} [/definition] A point $v$ in a Weyl chamber determines a positive system by declaring $\alpha$ positive when the inner product of $\alpha$ and $v$ is positive. Moving $v$ inside the same chamber does not change any sign, while crossing a wall reverses the sign of at least one root. The remaining question is whether this geometric construction captures all compatible positivity choices, or only some examples produced by hyperplanes. The theorem identifies chambers with positive systems, so the algebraic and geometric languages can be used interchangeably. [quotetheorem:4698] [citeproof:4698] This result turns a geometric choice, a chamber, into an algebraic choice, a positive system. The regularity hypothesis is essential: if $v$ lies on a root hyperplane, the corresponding root is neither positive nor negative by the sign rule. The theorem also does not say that there is a canonical positive system; it says that every chamber gives one and every compatible positivity convention arises this way. For $A_2$, the six chambers around the origin give the six possible compatible choices of positive roots. We next ask which positive roots are fundamental and which are assembled from smaller positive roots. ## Simple Roots and Indecomposable Positivity Once a positive system is fixed, the next problem is to find a minimal list of positive roots from which all others are generated. The right notion of minimality is not linear independence alone, because the coefficients must be non-negative integers. [definition: Simple Root] Let $\Phi^+$ be a positive system in a root system $\Phi$. A root $\alpha \in \Phi^+$ is simple if it cannot be written as $\alpha=\beta+\gamma$ with $\beta,\gamma \in \Phi^+$. [/definition] Indecomposable positive roots are defined relative to a chosen positive system, but later arguments need a formulation that describes the whole root system at once. The issue is whether a small subset can generate every root with coefficients of one sign, so that positive and negative roots are separated by their coordinates. [definition: Base of a Root System] Let $\Phi \subset E$ be a root system. A base, or set of simple roots, is a subset $\Delta \subset \Phi$ such that every root $\gamma \in \Phi$ can be written in the form \begin{align*} \gamma = \sum_{\alpha \in \Delta} n_\alpha \alpha \end{align*} where all $n_\alpha \in \mathbb Z$ are either all non-negative or all non-positive. [/definition] The simple roots attached to a positive system are intended to form a base, but this is not automatic from the definition of indecomposable roots. One must know that decomposition actually stops, that all positive roots are generated by the indecomposable ones, and that no linear dependence remains among them. [quotetheorem:4699] [citeproof:4699] The key hidden estimate in the proof is the sign condition for distinct simple roots. If the inner product of two distinct simple roots $\alpha$ and $\beta$ were positive, then $\alpha-\beta$ would be a root by the root-string criterion, contradicting simplicity after choosing the sign compatible with $\Phi^+$. The closure condition in the definition of positive systems is needed here: without it, an arbitrary choice of one root from each pair need not generate a sensible cone of positive roots. The theorem does not classify all possible bases at once; it constructs the base attached to one chosen positive system. This construction is the point at which the later Dynkin diagram becomes possible, because the diagram records the pairwise angles and lengths among these simple roots. [quotetheorem:4700] [citeproof:4700] This uniqueness is what makes the language of coefficients meaningful. The independence of the simple roots is essential; in a merely spanning set, the same vector could have many different integral expressions. The theorem does not say that the coefficients are invariant under changing the base, only that they are well-defined after the base has been fixed. We can therefore speak about the coefficient of a simple root in a root without specifying a choice of expression, and this prepares the height function used for induction. [definition: Height of a Root] Let $\Delta$ be a base and let $\gamma \in \Phi$ have expansion $\gamma=\sum_{\alpha \in \Delta} n_\alpha\alpha$. The height of $\gamma$ is \begin{align*} \operatorname{ht}(\gamma) := \sum_{\alpha \in \Delta} n_\alpha . \end{align*} [/definition] For positive roots, height is a positive integer. Induction on height is the standard replacement for induction on degree in this finite combinatorial setting. ## Reflections and the Behaviour of Positive Roots The Weyl group acts on the set of all roots, but after choosing $\Phi^+$ most Weyl group elements do not preserve positivity. Reflections in simple roots have a controlled failure: they flip the simple root itself and permute all other positive roots. [definition: Root Reflection] Let $\alpha \in \Phi$. The reflection through $\alpha$ is the linear map $s_\alpha:E \to E$ defined by \begin{align*} s_\alpha(x) = x - \frac{2(x,\alpha)}{(\alpha,\alpha)}\alpha . \end{align*} [/definition] The root-system axioms say that $s_\alpha(\Phi)=\Phi$. For simple $\alpha$, this reflection is adjacent to the chosen chamber: it crosses exactly the wall perpendicular to $\alpha$. The point to control is how much positivity is lost in that crossing. If a simple reflection turned many unrelated positive roots negative, it would be too coarse for building Weyl group combinatorics one wall at a time. [quotetheorem:4701] [citeproof:4701] This theorem is the combinatorial engine behind reduced decompositions in the Weyl group. The assumption that $\alpha$ is simple is essential: reflection in a non-simple positive root can send several positive roots to negative roots. The result does not say that $s_\alpha$ preserves the whole positive system, since it deliberately sends $\alpha$ to $-\alpha$. It explains why simple reflections correspond to the walls of a chamber: crossing one wall changes the sign of exactly one simple root at the first step, which is the local move used later in Weyl group combinatorics. [example: Simple Roots for Type A Three] Realise $A_3$ in the subspace of $\mathbb R^4$ where $x_1+x_2+x_3+x_4=0$, with roots $e_i-e_j$ for $i \ne j$. Choose the positive roots $e_i-e_j$ for $i<j$, and set \begin{align*} \alpha_1=e_1-e_2,\qquad \alpha_2=e_2-e_3,\qquad \alpha_3=e_3-e_4. \end{align*} The six positive roots are exactly \begin{align*} e_1-e_2&=\alpha_1,\\ e_2-e_3&=\alpha_2,\\ e_3-e_4&=\alpha_3,\\ e_1-e_3&=(e_1-e_2)+(e_2-e_3)=\alpha_1+\alpha_2,\\ e_2-e_4&=(e_2-e_3)+(e_3-e_4)=\alpha_2+\alpha_3,\\ e_1-e_4&=(e_1-e_2)+(e_2-e_3)+(e_3-e_4)=\alpha_1+\alpha_2+\alpha_3. \end{align*} Thus every positive root is a consecutive sum of the three displayed roots. The roots $\alpha_1,\alpha_2,\alpha_3$ are simple. Indeed, every positive root in the displayed list has a non-negative expansion in $\alpha_1,\alpha_2,\alpha_3$, and the coefficient sums are \begin{align*} \operatorname{ht}(\alpha_1)&=1,& \operatorname{ht}(\alpha_2)&=1,& \operatorname{ht}(\alpha_3)&=1,\\ \operatorname{ht}(\alpha_1+\alpha_2)&=2,& \operatorname{ht}(\alpha_2+\alpha_3)&=2,& \operatorname{ht}(\alpha_1+\alpha_2+\alpha_3)&=3. \end{align*} A sum of two positive roots has coefficient sum at least $2$, so none of $\alpha_1,\alpha_2,\alpha_3$ can be written as a sum of two positive roots. The remaining positive roots are not simple because \begin{align*} \alpha_1+\alpha_2&=(e_1-e_2)+(e_2-e_3),\\ \alpha_2+\alpha_3&=(e_2-e_3)+(e_3-e_4),\\ \alpha_1+\alpha_2+\alpha_3&=\alpha_1+(\alpha_2+\alpha_3). \end{align*} Hence the simple roots are precisely $\alpha_1,\alpha_2,\alpha_3$. With the standard inner product on $\mathbb R^4$, \begin{align*} (\alpha_1,\alpha_2)&=(e_1-e_2,e_2-e_3)=-1,\\ (\alpha_2,\alpha_3)&=(e_2-e_3,e_3-e_4)=-1,\\ (\alpha_1,\alpha_3)&=(e_1-e_2,e_3-e_4)=0. \end{align*} Also $(\alpha_i,\alpha_i)=2$ for each $i$, so adjacent simple roots meet and the two outer simple roots are orthogonal. This is the chain shape of the $A_3$ Dynkin diagram. [/example] Type $A_3$ shows the simply laced pattern, where adjacency is read directly from nonzero inner products. Type $B_3$ introduces two root lengths, so the same simple-root calculation must also keep track of which endpoint of the double edge is short. [example: Simple Roots for Type B Three] Realise $B_3$ in $\mathbb R^3$ with the standard inner product, with roots $\pm e_i$ and $\pm e_i\pm e_j$ for $i<j$. Choose the positive roots $e_i$ and $e_i\pm e_j$ for $i<j$, and set \begin{align*} \alpha_1=e_1-e_2, \qquad \alpha_2=e_2-e_3, \qquad \alpha_3=e_3. \end{align*} We compute every positive root in these coordinates: \begin{align*} e_1-e_2&=\alpha_1,\\ e_2-e_3&=\alpha_2,\\ e_3&=\alpha_3,\\ e_2&=(e_2-e_3)+e_3=\alpha_2+\alpha_3,\\ e_1-e_3&=(e_1-e_2)+(e_2-e_3)=\alpha_1+\alpha_2,\\ e_2+e_3&=(e_2-e_3)+2e_3=\alpha_2+2\alpha_3,\\ e_1&=(e_1-e_2)+(e_2-e_3)+e_3=\alpha_1+\alpha_2+\alpha_3,\\ e_1+e_3&=(e_1-e_2)+(e_2-e_3)+2e_3=\alpha_1+\alpha_2+2\alpha_3,\\ e_1+e_2&=(e_1-e_2)+2(e_2-e_3)+2e_3=\alpha_1+2\alpha_2+2\alpha_3. \end{align*} Thus every positive root is a non-negative integer combination of $\alpha_1,\alpha_2,\alpha_3$. The roots $\alpha_1,\alpha_2,\alpha_3$ are simple because each has height $1$, while a sum of two positive roots has height at least $2$. The remaining positive roots are not simple, as shown by the displayed decompositions \begin{align*} e_2&=\alpha_2+\alpha_3,\\ e_1-e_3&=\alpha_1+\alpha_2,\\ e_2+e_3&=\alpha_2+2\alpha_3=\alpha_3+(\alpha_2+\alpha_3),\\ e_1&=\alpha_1+(\alpha_2+\alpha_3),\\ e_1+e_3&=(\alpha_1+\alpha_2)+2\alpha_3,\\ e_1+e_2&=\alpha_1+2\alpha_2+2\alpha_3. \end{align*} Hence the simple roots are precisely $\alpha_1,\alpha_2,\alpha_3$. Their squared lengths are \begin{align*} (\alpha_1,\alpha_1)&=(e_1-e_2,e_1-e_2)=1+1=2,\\ (\alpha_2,\alpha_2)&=(e_2-e_3,e_2-e_3)=1+1=2,\\ (\alpha_3,\alpha_3)&=(e_3,e_3)=1. \end{align*} Also \begin{align*} (\alpha_1,\alpha_2)&=(e_1-e_2,e_2-e_3)=-1,\\ (\alpha_2,\alpha_3)&=(e_2-e_3,e_3)=-1,\\ (\alpha_1,\alpha_3)&=(e_1-e_2,e_3)=0. \end{align*} Thus the last simple root is short, while $\alpha_1$ and $\alpha_2$ are long; this length difference is the feature recorded by the double edge in the $B_3$ Dynkin diagram. [/example] ## Highest Roots and Height Induction After a base is chosen, the partial ordering on roots has a largest element in each irreducible finite type. The highest root packages useful numerical data and provides a convenient testing ground for height induction. [definition: Root Order] Let $\Delta$ be a base of $\Phi$. For roots $\mu,\nu \in \Phi$, write $\mu \le \nu$ if \begin{align*} \nu-\mu = \sum_{\alpha \in \Delta} n_\alpha\alpha \end{align*} with all $n_\alpha \in \mathbb Z_{\ge 0}$. [/definition] The root order may have several maximal elements in a general partially ordered set, so the useful question is whether irreducibility forces a single dominant endpoint among the positive roots. That endpoint is the root against which height arguments and extended diagrams are measured. [definition: Highest Root] Let $\Phi$ be an irreducible root system with base $\Delta$ and positive system $\Phi^+$. The highest root is the unique root $\theta \in \Phi^+$ such that $\gamma \le \theta$ for every $\gamma \in \Phi^+$. [/definition] The definition asserts a unique root with a strong domination property, so it requires justification from the structure of irreducible root systems. The obstruction is that ordinary maximality in height need not by itself compare every positive root; the theorem supplies the existence and uniqueness needed for the term to be meaningful. [quotetheorem:4702] [citeproof:4702] In computations, the highest root is found by listing positive roots in simple-root coordinates and selecting the maximal element. Irreducibility is essential: for a reducible root system such as $A_1 \times A_1$, the two simple roots are incomparable maximal positive roots, so there is no single highest root. The theorem does not provide a uniform closed formula for $\theta$ in every type; the coefficients still depend on the Dynkin diagram. In proofs, the highest root often appears as the endpoint of an induction on height and as the extra vertex in the extended Dynkin diagram. [example: Highest Root in Type A Two] For $A_2$ with simple roots $\alpha_1,\alpha_2$, the positive roots are \begin{align*} \alpha_1, \qquad \alpha_2, \qquad \alpha_1+\alpha_2. \end{align*} Their heights are \begin{align*} \operatorname{ht}(\alpha_1)&=1,\\ \operatorname{ht}(\alpha_2)&=1,\\ \operatorname{ht}(\alpha_1+\alpha_2)&=1+1=2. \end{align*} Thus the only positive root of maximal height is \begin{align*} \theta=\alpha_1+\alpha_2. \end{align*} We also check the root order explicitly. For the two simple roots, \begin{align*} \theta-\alpha_1 &=(\alpha_1+\alpha_2)-\alpha_1\\ &=\alpha_2, \end{align*} and $\alpha_2$ is a non-negative integer combination of simple roots. Similarly, \begin{align*} \theta-\alpha_2 &=(\alpha_1+\alpha_2)-\alpha_2\\ &=\alpha_1, \end{align*} and $\alpha_1$ is a non-negative integer combination of simple roots. Finally, \begin{align*} \theta-\theta=0=0\alpha_1+0\alpha_2, \end{align*} so $\theta$ also dominates itself. Hence every positive root is less than or equal to $\theta$, and $\alpha_1+\alpha_2$ is the highest root in type $A_2$. [/example] The simply-laced rank-two case has all roots of the same length, so the highest root has coefficients balanced across the two simple directions. Non-simply-laced examples show why coefficients greater than $1$ appear. [example: Highest Root in Type B Two] For $B_2$ choose simple roots \begin{align*} \alpha_1=e_1-e_2, \qquad \alpha_2=e_2. \end{align*} Then \begin{align*} (\alpha_1,\alpha_1) &=(e_1-e_2,e_1-e_2)\\ &=(e_1,e_1)-2(e_1,e_2)+(e_2,e_2)\\ &=1-0+1\\ &=2, \end{align*} while \begin{align*} (\alpha_2,\alpha_2)=(e_2,e_2)=1. \end{align*} Thus $\alpha_1$ is long and $\alpha_2$ is short. With the usual positive roots for $B_2$, namely $e_1-e_2$, $e_2$, $e_1$, and $e_1+e_2$, their simple-root expansions are \begin{align*} e_1-e_2&=\alpha_1,\\ e_2&=\alpha_2,\\ e_1&=(e_1-e_2)+e_2=\alpha_1+\alpha_2,\\ e_1+e_2&=(e_1-e_2)+2e_2=\alpha_1+2\alpha_2. \end{align*} Their heights are therefore \begin{align*} \operatorname{ht}(\alpha_1)&=1,\\ \operatorname{ht}(\alpha_2)&=1,\\ \operatorname{ht}(\alpha_1+\alpha_2)&=1+1=2,\\ \operatorname{ht}(\alpha_1+2\alpha_2)&=1+2=3. \end{align*} Set $\theta=\alpha_1+2\alpha_2=e_1+e_2$. We check that every positive root is below $\theta$ in the root order: \begin{align*} \theta-\alpha_1 &=(\alpha_1+2\alpha_2)-\alpha_1\\ &=2\alpha_2,\\ \theta-\alpha_2 &=(\alpha_1+2\alpha_2)-\alpha_2\\ &=\alpha_1+\alpha_2,\\ \theta-(\alpha_1+\alpha_2) &=(\alpha_1+2\alpha_2)-(\alpha_1+\alpha_2)\\ &=\alpha_2,\\ \theta-\theta &=0\\ &=0\alpha_1+0\alpha_2. \end{align*} Each difference is a non-negative integer combination of $\alpha_1$ and $\alpha_2$, so $\theta=\alpha_1+2\alpha_2$ is the highest root. This example shows concretely that the short simple root can occur with coefficient greater than $1$ in the highest root. [/example] The $B_2$ calculation is the first sign that root length affects the height data. The exceptional rank-two system pushes this asymmetry further and is the useful test case for checking conventions about long and short roots. [example: Highest Root in Type G Two] For $G_2$, let $\alpha_1$ be the short simple root and $\alpha_2$ the long simple root. With this choice of base, the positive roots are \begin{align*} \alpha_1, \qquad \alpha_2, \qquad \alpha_1+\alpha_2, \qquad 2\alpha_1+\alpha_2, \qquad 3\alpha_1+\alpha_2, \qquad 3\alpha_1+2\alpha_2. \end{align*} Their heights are obtained by adding the simple-root coefficients: \begin{align*} \operatorname{ht}(\alpha_1)&=1,\\ \operatorname{ht}(\alpha_2)&=1,\\ \operatorname{ht}(\alpha_1+\alpha_2)&=1+1=2,\\ \operatorname{ht}(2\alpha_1+\alpha_2)&=2+1=3,\\ \operatorname{ht}(3\alpha_1+\alpha_2)&=3+1=4,\\ \operatorname{ht}(3\alpha_1+2\alpha_2)&=3+2=5. \end{align*} Thus the only positive root of maximal height in this list is \begin{align*} \theta=3\alpha_1+2\alpha_2. \end{align*} We now check the root order explicitly. Subtracting each positive root from $\theta$ gives \begin{align*} \theta-\alpha_1 &=(3\alpha_1+2\alpha_2)-\alpha_1\\ &=2\alpha_1+2\alpha_2,\\ \theta-\alpha_2 &=(3\alpha_1+2\alpha_2)-\alpha_2\\ &=3\alpha_1+\alpha_2,\\ \theta-(\alpha_1+\alpha_2) &=(3\alpha_1+2\alpha_2)-(\alpha_1+\alpha_2)\\ &=2\alpha_1+\alpha_2,\\ \theta-(2\alpha_1+\alpha_2) &=(3\alpha_1+2\alpha_2)-(2\alpha_1+\alpha_2)\\ &=\alpha_1+\alpha_2,\\ \theta-(3\alpha_1+\alpha_2) &=(3\alpha_1+2\alpha_2)-(3\alpha_1+\alpha_2)\\ &=\alpha_2,\\ \theta-\theta &=0\\ &=0\alpha_1+0\alpha_2. \end{align*} Each displayed difference is a non-negative integer combination of $\alpha_1$ and $\alpha_2$, so every positive root is below $\theta$ in the root order. Hence the highest root is $\theta=3\alpha_1+2\alpha_2$, and its coefficients show that the short simple direction contributes more heavily than the long simple direction in type $G_2$. [/example] Highest roots are useful only if the partial order interacts well with subtraction by simple roots. The key induction problem is this: given a positive root that is not already simple, can one step downward while staying inside the root system? Without such a step, arguments about all positive roots would not reduce to the simple roots and the height function would be bookkeeping rather than a proof tool. [quotetheorem:4703] [citeproof:4703] This height-reduction statement is the practical induction lemma used throughout the classification proof. The hypothesis that $\gamma$ is not simple is essential, because subtracting a simple root from itself gives $0$, which is not a root. The theorem does not say that every simple root appearing in the expansion of $\gamma$ can be subtracted; some subtractions land outside the root system. It lets arguments about all positive roots begin at simple roots and climb by adding one simple root at a time, or descend from the highest root by subtracting one carefully chosen simple root at a time. A choice of positive roots and simple roots gives a way to navigate the root system one reflection chamber at a time. With that order in place, the reflections generated by the roots become the next central object, because they encode the symmetry of the entire configuration. # 6. Weyl Groups Weyl groups provide the finite reflection groups that remember the geometry of the root system. In Chapter 5, roots divided the real span of the root system into hyperplanes and open regions; here those regions become the Weyl chambers, and the reflections in root hyperplanes become the main symmetry group. The guiding questions are: how much of the root system is generated by reflections, how does this group move chambers, and how can it be presented by simple generators and Coxeter relations? ## Root Reflections and Weyl Chambers A root system comes with many hyperplanes, and the first problem is to understand how the corresponding reflections organise the complement of those hyperplanes. The striking answer is that the reflection group acts on the chambers with no redundancy: every chamber is reached from a fixed chamber in exactly one way. Let $E$ be a finite-dimensional real inner product space, and let $\Phi\subset E$ be a reduced crystallographic root system spanning $E$, equipped with the inner product induced from the Killing form construction of the previous chapters. For a root $\alpha \in \Phi$, write $\alpha^\vee = 2\alpha/(\alpha,\alpha)$ under the chosen identification of $E$ and $E^*$. [definition: Root Reflection] For $\alpha \in \Phi$, the root reflection $s_\alpha: E \to E$ is the linear map \begin{align*} s_\alpha(\lambda) = \lambda - \langle \lambda, \alpha^\vee \rangle \alpha. \end{align*} [/definition] This reflection fixes the hyperplane $H_\alpha = \{\lambda \in E : \langle \lambda,\alpha^\vee\rangle = 0\}$ and sends $\alpha$ to $-\alpha$. We need the group generated by all these root symmetries, because the individual reflections only describe one wall at a time while the classification uses their combined action on the whole root system. [definition: Weyl Group] The Weyl group of $\Phi$ is the subgroup \begin{align*} W = \langle s_\alpha : \alpha \in \Phi \rangle \le GL(E). \end{align*} [/definition] The Weyl group is finite because it acts faithfully on the finite set $\Phi$. We need the chambers cut out by the reflecting hyperplanes, since these regions are the geometric objects on which the group action becomes visible. [definition: Weyl Chamber] A Weyl chamber is a connected component of \begin{align*} E \setminus \bigcup_{\alpha \in \Phi} H_\alpha. \end{align*} [/definition] If a choice of positive roots $\Phi^+$ has simple roots $\Delta = \{\alpha_1,\dots,\alpha_r\}$, the associated dominant chamber is \begin{align*} C = \{\lambda \in E : \langle \lambda, \alpha_i^\vee\rangle > 0 \text{ for } i=1,\dots,r\}. \end{align*} Its closure is cut out by the weak inequalities $\langle \lambda, \alpha_i^\vee\rangle \ge 0$. We need the first structural theorem to say that these chambers are not merely regions in a picture. The theorem below explains that the Weyl group reaches every chamber from the dominant one, and does so without ambiguity. [quotetheorem:4704] [citeproof:4704] The hypotheses are doing real work here. The chamber is defined using the full hyperplane arrangement of a finite root system, and the reflections are required to preserve that arrangement; an arbitrary collection of hyperplanes with chosen reflections need not act transitively on chambers, and a subgroup generated by only some root reflections may leave several chambers unreachable. The theorem also says less than it might first appear: it labels chambers by group elements, but it does not yet give a small generating set or a practical way to compute the label. This theorem is the geometric reason that the Weyl group is the right symmetry group: it does not merely permute chambers, it labels them. A reduced expression for an element will later be read as a minimal gallery from the dominant chamber to its image. [illustration:rank-two-weyl-chambers] [example: Type A Weyl Chambers] For type $A_{n-1}$, work in \begin{align*} E=\{(x_1,\dots,x_n)\in \mathbb R^n:x_1+\cdots+x_n=0\}, \end{align*} with roots $\alpha_{ij}=e_i-e_j$ for $i\ne j$. Since $(\alpha_{ij},\alpha_{ij})=2$, we have $\alpha_{ij}^\vee=\alpha_{ij}$, and for $x=(x_1,\dots,x_n)$, \begin{align*} \langle x,\alpha_{ij}^\vee\rangle &=(x,e_i-e_j)\\ &=x_i-x_j. \end{align*} Thus the root hyperplane attached to $\alpha_{ij}$ is \begin{align*} H_{\alpha_{ij}} &=\{x\in E:\langle x,\alpha_{ij}^\vee\rangle=0\}\\ &=\{x\in E:x_i-x_j=0\}\\ &=\{x\in E:x_i=x_j\}. \end{align*} Therefore the complement of the root hyperplanes consists exactly of the points of $E$ whose coordinates are pairwise distinct. On each connected component, every sign $x_i-x_j$ is fixed, so the coordinates occur in one strict order; conversely, each strict ordering such as \begin{align*} x_{\sigma(1)}>x_{\sigma(2)}>\cdots>x_{\sigma(n)} \end{align*} defines an open convex region in $E$, hence a chamber. The reflection in the root $\alpha_{ij}$ is \begin{align*} s_{\alpha_{ij}}(x) &=x-\langle x,\alpha_{ij}^\vee\rangle\alpha_{ij}\\ &=x-(x_i-x_j)(e_i-e_j). \end{align*} Its coordinates are \begin{align*} (s_{\alpha_{ij}}(x))_i&=x_i-(x_i-x_j)=x_j,\\ (s_{\alpha_{ij}}(x))_j&=x_j+(x_i-x_j)=x_i,\\ (s_{\alpha_{ij}}(x))_k&=x_k \qquad (k\ne i,j). \end{align*} So $s_{\alpha_{ij}}$ is exactly the transposition swapping the $i$th and $j$th coordinates. Since transpositions generate $S_n$, the Weyl group contains every coordinate permutation; since every root reflection is a coordinate transposition, it contains nothing beyond these permutations. Hence $W\cong S_n$, and the dominant chamber \begin{align*} x_1>x_2>\cdots>x_n \end{align*} is carried to all chambers by permuting the coordinate positions. [/example] ## Simple Reflections, Length, and Inversions Once a chamber is chosen, the next problem is to replace the large generating set $\{s_\alpha : \alpha \in \Phi\}$ by the smaller set of reflections in the walls of that chamber. This is what makes the Weyl group computable: every element is a word in the simple reflections, and the length of the shortest such word has a direct root-theoretic meaning. [definition: Simple Reflection] Let $\Delta=\{\alpha_1,\dots,\alpha_r\}$ be the set of simple roots associated to a positive system $\Phi^+$. The simple reflections are the maps $s_i:E\to E$ defined by \begin{align*} s_i = s_{\alpha_i}, \qquad i=1,\dots,r. \end{align*} [/definition] The simple reflection $s_i$ crosses the wall of the dominant chamber defined by $\langle \lambda,\alpha_i^\vee\rangle=0$. We need to know that these wall-crossings already generate every chamber movement, otherwise the small set of simple reflections would miss part of the Weyl group. [quotetheorem:4705] [citeproof:4705] The chosen base is essential in this statement. Without fixing a positive system, there is no distinguished small set of wall reflections, and the generating set by all roots is too large to measure words efficiently. The theorem says that the walls of one chamber already contain all the reflection data, but it does not say that expressions in the simple reflections are unique; the next definitions measure the failure of uniqueness by looking for shortest words. [definition: Length In A Weyl Group] For $w \in W$, the length $\ell(w)$ is the smallest integer $m \ge 0$ such that \begin{align*} w=s_{i_1}\cdots s_{i_m} \end{align*} for simple reflections $s_{i_1},\dots,s_{i_m}$. [/definition] A word of length $\ell(w)$ is called a reduced expression for $w$. We need a root-theoretic way to compute this length without checking all possible words, and the relevant data are the positive roots whose signs are reversed by $w$. [definition: Inversion Set] For $w \in W$, the inversion set of $w$ is \begin{align*} N(w)=\{\alpha \in \Phi^+ : w\alpha \in -\Phi^+\}. \end{align*} [/definition] The inversion set records which positive root hyperplanes have been crossed on the way from $C$ to $wC$. In this finite Weyl-group setting, we regard $(W,S)$ as the Coxeter system whose simple generators are the simple reflections $S=\{s_i\}$, and we use the same real root system $\Phi=\Phi^+\sqcup(-\Phi^+)$ attached to the chosen base. We need the theorem below because it turns that geometric crossing data into the numerical length function used in computations with reduced expressions. [quotetheorem:4706] [citeproof:4706] The formula depends on the positive system and the corresponding simple generators. Changing the dominant chamber changes which roots are counted as positive and changes the inversion set, although the resulting length function is conjugate to the old one under the corresponding relabelling of chambers. The theorem does not make every word with $|N(w)|$ letters reduced; rather, it gives a numerical test for reducedness and a recursive rule for whether appending a chosen simple reflection makes a word longer or shorter. This is the computational engine behind Bruhat order, reduced galleries, and later highest-weight calculations. [example: Inversions In Type A] In type $A_{n-1}$, identify $W$ with $S_n$ as in the coordinate-permutation model, and choose the positive roots \begin{align*} \Phi^+ = \{e_i-e_j : 1\le i<j\le n\}. \end{align*} For $\sigma\in S_n$ and $i<j$, the action on the root $e_i-e_j$ is \begin{align*} \sigma(e_i-e_j) &=\sigma(e_i)-\sigma(e_j)\\ &=e_{\sigma(i)}-e_{\sigma(j)}. \end{align*} A root $e_a-e_b$ is positive exactly when $a<b$, and it is negative exactly when $a>b$, since \begin{align*} e_a-e_b = -(e_b-e_a) \end{align*} and $e_b-e_a\in \Phi^+$ precisely when $b<a$. Therefore, for $i<j$, \begin{align*} \sigma(e_i-e_j)\in -\Phi^+ &\Longleftrightarrow e_{\sigma(i)}-e_{\sigma(j)}\in -\Phi^+\\ &\Longleftrightarrow \sigma(i)>\sigma(j). \end{align*} Hence the inversion set is \begin{align*} N(\sigma) &=\{e_i-e_j\in \Phi^+:\sigma(e_i-e_j)\in -\Phi^+\}\\ &=\{e_i-e_j: i<j \text{ and } \sigma(i)>\sigma(j)\}. \end{align*} Thus $N(\sigma)$ records exactly the usual inversions of the permutation $\sigma$, and by *Length Equals Number Of Inversions*, \begin{align*} \ell(\sigma)=|N(\sigma)|=\#\{(i,j):i<j \text{ and } \sigma(i)>\sigma(j)\}. \end{align*} So the Weyl group length in type $A$ is the ordinary inversion number of a permutation. [/example] This example is the model for the general theory: a root inversion is the same kind of data as a pair of entries appearing in the wrong order. The difference in other types is that the allowable inversions also remember signs and root lengths. [remark: Reduced Expressions Are Not Unique] Reduced expressions need not be unique. In type $A_2$, if $s_1$ and $s_2$ are the two simple reflections, then the longest element has reduced expressions $s_1s_2s_1$ and $s_2s_1s_2$. This non-uniqueness is controlled exactly by the Coxeter relations discussed below. [/remark] ## Coxeter Relations From Root Angles The final problem is to determine the relations among the simple reflections. Since each $s_i$ is an involution, the remaining data must say how pairs of wall reflections interact; the answer is encoded by the angle between the corresponding simple roots. [definition: Coxeter Matrix Of A Root System] Let $\Delta=\{\alpha_1,\dots,\alpha_r\}$ be a base of $\Phi$. The Coxeter matrix $(m_{ij})$ is defined by \begin{align*} m_{ii}=1, \qquad m_{ij}=\operatorname{ord}(s_is_j) \quad \text{for } i\ne j. \end{align*} [/definition] The possible values of $m_{ij}$ are restricted by the root system axioms. We need a presentation theorem showing that these pairwise orders are the only relations, because otherwise the angle data would not be enough to recover the Weyl group. [quotetheorem:4707] [citeproof:4707] The crystallographic root-system hypotheses are again restrictive. The presentation is built from the simple reflections attached to one base and from the finite orders forced by the rank-two root subsystems; if the reflections did not preserve a root system, the same angle data would not necessarily give the Weyl group of a Lie algebra. For example, a regular pentagon gives a finite dihedral Coxeter group with $m=5$, but it is not crystallographic and hence does not arise as a Weyl group of a semisimple complex Lie algebra. This theorem is the bridge from Lie theory to Coxeter theory. Dynkin diagrams store the same pairwise angle data, so the classification of root systems can be approached by classifying the possible diagrams. [example: Rank Two Dihedral Weyl Groups] In rank two the simple roots span a plane, and the two simple reflections $s_1,s_2$ act as ordinary reflections across the two walls of the dominant chamber. Let the angle between the simple roots be \begin{align*} \theta=\pi-\frac{\pi}{m}. \end{align*} Since each reflecting hyperplane is perpendicular to its simple root, the angle between the two reflecting lines is \begin{align*} \phi=\pi-\theta=\frac{\pi}{m}. \end{align*} Choose coordinates so that the reflecting line for $s_1$ is the $x$-axis and the reflecting line for $s_2$ makes angle $\phi$ with it. Then \begin{align*} s_1 &= \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}, \end{align*} and reflection across the line of angle $\phi$ is \begin{align*} s_2 &= \begin{pmatrix} \cos \phi&-\sin \phi\\ \sin \phi&\cos \phi \end{pmatrix} \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} \begin{pmatrix} \cos \phi&\sin \phi\\ -\sin \phi&\cos \phi \end{pmatrix} \\ &= \begin{pmatrix} \cos \phi&-\sin \phi\\ \sin \phi&\cos \phi \end{pmatrix} \begin{pmatrix} \cos \phi&\sin \phi\\ \sin \phi&-\cos \phi \end{pmatrix} \\ &= \begin{pmatrix} \cos^2\phi-\sin^2\phi&2\sin\phi\cos\phi\\ 2\sin\phi\cos\phi&\sin^2\phi-\cos^2\phi \end{pmatrix} \\ &= \begin{pmatrix} \cos 2\phi&\sin 2\phi\\ \sin 2\phi&-\cos 2\phi \end{pmatrix}. \end{align*} Therefore \begin{align*} s_1s_2 &= \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} \begin{pmatrix} \cos 2\phi&\sin 2\phi\\ \sin 2\phi&-\cos 2\phi \end{pmatrix} \\ &= \begin{pmatrix} \cos 2\phi&\sin 2\phi\\ -\sin 2\phi&\cos 2\phi \end{pmatrix}, \end{align*} which is rotation through angle $-2\phi=-2\pi/m$. Hence \begin{align*} (s_1s_2)^m \end{align*} is rotation through angle $-2\pi$, so $(s_1s_2)^m=e$, and no smaller positive power is the identity because $0<2k\pi/m<2\pi$ for $0<k<m$. Writing $r=s_1s_2$, we have $r^m=e$ and $s_2=s_1r$. Thus every word in $s_1$ and $s_2$ can be rewritten as either $r^k$ or $s_1r^k$ with $0\le k<m$. The $m$ elements $r^k$ are distinct rotations, and the $m$ elements $s_1r^k$ are distinct reflections; moreover rotations have determinant $1$ while reflections have determinant $-1$, so the two lists do not overlap. Hence the Weyl group has exactly \begin{align*} m+m=2m \end{align*} elements, the dihedral group of order $2m$. The crystallographic rank-two possibilities are $m=2,3,4,6$, corresponding respectively to $A_1\times A_1$, $A_2$, $B_2$, and $G_2$. These are precisely the cases where the chamber angle $\pi/m$ is compatible with the integral Cartan data of a crystallographic root system. [/example] [illustration:rank-two-dihedral-products] The rank-two examples explain every Coxeter relation locally, but higher-rank Weyl groups combine these local relations across many pairs of simple roots. Type $B$ gives a first example where the group is still concrete but no longer just an ordinary permutation group. [example: Type B Signed Permutations] For type $B_n$, take \begin{align*} \Phi=\{\pm e_i : 1\le i\le n\}\cup\{\pm e_i\pm e_j : i\ne j\} \end{align*} in $\mathbb R^n$ with the standard inner product. If $\alpha=e_i-e_j$, then $(\alpha,\alpha)=2$, so $\alpha^\vee=\alpha$, and for $x=(x_1,\dots,x_n)$, \begin{align*} s_{e_i-e_j}(x) &=x-\langle x,e_i-e_j\rangle(e_i-e_j)\\ &=x-(x_i-x_j)(e_i-e_j). \end{align*} Thus \begin{align*} (s_{e_i-e_j}(x))_i&=x_i-(x_i-x_j)=x_j,\\ (s_{e_i-e_j}(x))_j&=x_j+(x_i-x_j)=x_i,\\ (s_{e_i-e_j}(x))_k&=x_k \qquad (k\ne i,j), \end{align*} so $s_{e_i-e_j}$ swaps the $i$th and $j$th coordinates. If $\alpha=e_i$, then $(\alpha,\alpha)=1$, so $\alpha^\vee=2e_i$, and \begin{align*} s_{e_i}(x) &=x-\langle x,2e_i\rangle e_i\\ &=x-2x_i e_i. \end{align*} Hence \begin{align*} (s_{e_i}(x))_i&=x_i-2x_i=-x_i,\\ (s_{e_i}(x))_k&=x_k \qquad (k\ne i), \end{align*} so $s_{e_i}$ changes exactly the sign of the $i$th coordinate. The remaining long roots give the same kind of signed coordinate operation. For $\alpha=e_i+e_j$, again $(\alpha,\alpha)=2$ and $\alpha^\vee=\alpha$, so \begin{align*} s_{e_i+e_j}(x) &=x-\langle x,e_i+e_j\rangle(e_i+e_j)\\ &=x-(x_i+x_j)(e_i+e_j), \end{align*} and therefore \begin{align*} (s_{e_i+e_j}(x))_i&=x_i-(x_i+x_j)=-x_j,\\ (s_{e_i+e_j}(x))_j&=x_j-(x_i+x_j)=-x_i,\\ (s_{e_i+e_j}(x))_k&=x_k \qquad (k\ne i,j). \end{align*} Also $s_{-\alpha}=s_\alpha$, since $(-\alpha)^\vee=-\alpha^\vee$ and \begin{align*} s_{-\alpha}(x) &=x-\langle x,-\alpha^\vee\rangle(-\alpha)\\ &=x-\langle x,\alpha^\vee\rangle\alpha\\ &=s_\alpha(x). \end{align*} Thus every root reflection is a signed coordinate permutation, so $W$ is contained in the group of maps \begin{align*} e_i\mapsto \varepsilon_i e_{\sigma(i)}, \qquad \varepsilon_i\in\{\pm1\},\ \sigma\in S_n. \end{align*} Conversely, the reflections $s_{e_i-e_j}$ contain all coordinate transpositions, and transpositions generate $S_n$; the reflections $s_{e_i}$ independently change each coordinate sign. Hence these reflections generate every signed permutation. Therefore \begin{align*} W\cong (\mathbb Z/2\mathbb Z)^n\rtimes S_n. \end{align*} There are $2^n$ independent choices of signs and $n!$ choices of permutation, so \begin{align*} |W|=2^n n!. \end{align*} Type $B_n$ therefore enlarges the type $A$ permutation group by allowing independent sign changes of the coordinate axes. [/example] This example also indicates why Weyl groups matter in representation theory: for a fixed Cartan subalgebra, the Weyl group acts on weights and identifies the different chambers in which highest-weight data may be compared. The combinatorics of signs, permutations, and reduced words later controls character formulae and the ordering of weights. [remark: Weyl Groups As Finite Coxeter Groups] Every Weyl group is a finite Coxeter group, but not every finite Coxeter group is a Weyl group. The extra restriction is crystallographic: the Cartan integers $\langle \alpha_i,\alpha_j^\vee\rangle$ must be integral. This excludes, for instance, the non-crystallographic dihedral groups with $m$ other than $2,3,4,6$ and the exceptional finite Coxeter types $H_3$ and $H_4$. [/remark] The chapter conclusion is that a semisimple Lie algebra supplies a root system, the root system supplies a Weyl group, and the Weyl group supplies a Coxeter presentation controlled by root angles. The next step in the course is to package the same information into Cartan matrices and Dynkin diagrams, where the classification becomes a finite combinatorial problem. Weyl groups capture the symmetry that has been present since the first root decompositions, but now in a purely finite reflection-theoretic form. The remaining structure can be compressed further into Cartan matrices and Dynkin diagrams, where angles, lengths, and reflections become a small amount of combinatorial data. # 7. Cartan Matrices and Dynkin Diagrams The previous chapters reduced the classification problem for semisimple Lie algebras to the combinatorics of root systems. This chapter explains how a choice of simple roots is compressed into a Cartan matrix and then into a Dynkin diagram. The point is that the diagram remembers exactly the angle and relative-length data needed for classification, while being much easier to inspect than the full root system. ## From Simple Roots to Cartan Matrices What numerical data is forced by a simple root basis, and how much of the root system can it remember? Let $\Phi \subset E$ be a reduced crystallographic root system in a Euclidean space $E$, and let $\Delta = \{\alpha_1, \dots, \alpha_n\}$ be a basis of simple roots. The reflections in the hyperplanes orthogonal to the simple roots generate the Weyl group, so the interaction among the $\alpha_i$ should record the local geometry of the chamber. For roots $\alpha,\beta \in \Phi$, set \begin{align*} \langle \beta, \alpha^\vee \rangle = \frac{2(\beta,\alpha)}{(\alpha,\alpha)}. \end{align*} This integer is the amount by which the reflection $s_\alpha$ changes $\beta$ in the $\alpha$-direction: \begin{align*} s_\alpha(\beta)=\beta-\langle \beta,\alpha^\vee\rangle\alpha. \end{align*} [definition: Cartan Matrix of a Simple Root Basis] Let $\Phi \subset E$ be a reduced crystallographic root system with simple root basis $\Delta=\{\alpha_1,\dots,\alpha_n\}$. The Cartan matrix associated to $\Delta$ is the integer matrix $A=(a_{ij})\in \mathbb Z^{n\times n}$ defined by \begin{align*} a_{ij}=\langle \alpha_j,\alpha_i^\vee\rangle=\frac{2(\alpha_j,\alpha_i)}{(\alpha_i,\alpha_i)}. \end{align*} [/definition] The row index records the coroot being tested against, while the column index records the root being tested. This convention is important because, when two roots have different lengths, the two off-diagonal entries need not agree. [quotetheorem:4708] [citeproof:4708] The signs in this theorem are not cosmetic. If an off-diagonal entry were positive, the corresponding two simple roots would meet at an acute angle, and the chosen chamber would fail to be a chamber cut out by positive roots. If $a_{ij}=0$ but $a_{ji}\ne 0$, the matrix would claim that one root is orthogonal to the other only in one direction, which is impossible for a symmetric Euclidean inner product. The symmetrizability condition records the hidden metric data; without it, an integer matrix can look locally like Cartan data but fail to come from any Euclidean root system. This theorem motivates the abstract notion of a generalized Cartan matrix, the same object that will be used in Chapter 9 to write Serre generators and relations. The matrices arising from finite root systems have extra positivity properties, but the axioms below are the part visible before imposing finiteness. The obstruction is that these local axioms do not prevent infinite behaviour: affine and indefinite generalized Cartan matrices satisfy the same sign conditions but do not correspond to finite-dimensional semisimple Lie algebras. [definition: Generalized Cartan Matrix] A generalized Cartan matrix is a matrix $A=(a_{ij})\in \mathbb Z^{n\times n}$ such that: 1. $a_{ii}=2$ for all $i$. 2. $a_{ij}\le 0$ for all $i\ne j$. 3. $a_{ij}=0$ iff $a_{ji}=0$. [/definition] A Cartan matrix coming from a finite-dimensional semisimple Lie algebra is a generalized Cartan matrix, but not every generalized Cartan matrix is of finite type. The finite-type condition is the condition that the symmetrized form is positive definite. [quotetheorem:4709] [citeproof:4709] Positive definiteness is the essential finiteness hypothesis. If the symmetrized form is only positive semidefinite, the affine diagrams appear and the Weyl group acts by Euclidean motions with infinitely many roots. If the form is indefinite, the reflection representation still exists, but the orbit of a simple root is typically infinite and no finite root-system classification follows. Indecomposability is also needed: a block diagonal matrix corresponds to a product of smaller root systems rather than a single connected type. ## Dynkin Diagrams as Encoded Geometry How can the matrix be drawn so that the relevant geometry is visible at a glance? The diagonal entries are always $2$, and the off-diagonal entries are constrained by rank-two root systems. Thus the information to draw is concentrated in each unordered pair of simple roots. A plain graph with only adjacency would lose the distinction between $A_2$, $B_2$, and $G_2$, while a matrix alone hides the same information behind labels and row order. Dynkin diagrams solve both problems by displaying the rank-two angle and length data directly. [illustration:dynkin-angle-length-codes] [definition: Dynkin Diagram] Let $A=(a_{ij})$ be a finite-type Cartan matrix. Its Dynkin diagram is the graph with vertices $1,\dots,n$, one for each simple root, where the vertices $i$ and $j$ are joined according to the product $a_{ij}a_{ji}$: 1. no edge if $a_{ij}a_{ji}=0$; 2. one edge if $a_{ij}a_{ji}=1$; 3. two edges if $a_{ij}a_{ji}=2$; 4. three edges if $a_{ij}a_{ji}=3$. When there is more than one edge, an arrow points toward the shorter simple root. [/definition] The product $a_{ij}a_{ji}$ determines the angle, while the ratio of $a_{ij}$ and $a_{ji}$ determines the length ratio. Indeed, \begin{align*} a_{ij}a_{ji} =4\frac{(\alpha_i,\alpha_j)^2}{(\alpha_i,\alpha_i)(\alpha_j,\alpha_j)} =4\cos^2\theta_{ij}, \end{align*} where $\theta_{ij}$ is the angle between $\alpha_i$ and $\alpha_j$. Since distinct simple roots meet at angles in $\{\pi/2,2\pi/3,3\pi/4,5\pi/6\}$, the possible products are $0,1,2,3$. [example: Rank Two Diagrams] Let $\Delta=\{\alpha_1,\alpha_2\}$ and write $p=a_{12}a_{21}$. For distinct simple roots, \begin{align*} p &=\frac{2(\alpha_2,\alpha_1)}{(\alpha_1,\alpha_1)} \frac{2(\alpha_1,\alpha_2)}{(\alpha_2,\alpha_2)}\\ &=4\frac{(\alpha_1,\alpha_2)^2}{(\alpha_1,\alpha_1)(\alpha_2,\alpha_2)}\\ &=4\cos^2\theta, \end{align*} where $\theta$ is the angle between $\alpha_1$ and $\alpha_2$. The finite rank-two possibilities are therefore \begin{align*} \theta=\frac{\pi}{2},\frac{2\pi}{3},\frac{3\pi}{4},\frac{5\pi}{6}, \end{align*} and hence \begin{align*} 4\cos^2\frac{\pi}{2}=0,\qquad 4\cos^2\frac{2\pi}{3}=1,\qquad 4\cos^2\frac{3\pi}{4}=2,\qquad 4\cos^2\frac{5\pi}{6}=3. \end{align*} These give respectively the diagrams $A_1\times A_1$, $A_2$, $B_2=C_2$, and $G_2$. The corresponding rank-two Cartan matrices, up to interchanging the two simple roots, are \begin{align*} A_1\times A_1:\quad \begin{pmatrix} 2&0\\ 0&2 \end{pmatrix},\qquad A_2:\quad \begin{pmatrix} 2&-1\\ -1&2 \end{pmatrix}, \end{align*} \begin{align*} B_2=C_2:\quad \begin{pmatrix} 2&-2\\ -1&2 \end{pmatrix},\qquad G_2:\quad \begin{pmatrix} 2&-3\\ -1&2 \end{pmatrix}. \end{align*} When $a_{12}$ and $a_{21}$ are both nonzero, \begin{align*} \frac{a_{12}}{a_{21}} &= \frac{2(\alpha_2,\alpha_1)/(\alpha_1,\alpha_1)} {2(\alpha_1,\alpha_2)/(\alpha_2,\alpha_2)}\\ &=\frac{(\alpha_2,\alpha_2)}{(\alpha_1,\alpha_1)}. \end{align*} Thus, if $\alpha_1$ is shorter than $\alpha_2$, then $(\alpha_2,\alpha_2)>(\alpha_1,\alpha_1)$ and $|a_{12}|>|a_{21}|$. The arrow in a double or triple edge therefore points toward the shorter root exactly by pointing toward the vertex whose row contains the off-diagonal entry of larger absolute value. [/example] The examples show how much information is hidden in a diagram, but the classification needs a precise recovery statement. A drawing should determine the Cartan matrix up to relabelling; otherwise two different finite root systems could have the same diagram and the diagram would lose algebraic data. The next theorem confirms that the edge multiplicities and arrow directions retain exactly the finite-type Cartan information needed for classification. [quotetheorem:4710] [citeproof:4710] This is why Dynkin diagrams are a classification tool rather than a drawing convention. They contain the same finite-type Cartan data, with the irrelevant choice of ordering removed. The finite-type hypothesis matters here: outside finite type, diagrams with the same local edge conventions can represent affine or indefinite generalized Cartan matrices, and the diagram alone must be interpreted together with the type condition. The arrow convention is also essential; reversing the arrow interchanges $B_n$ and $C_n$ and changes the Cartan matrix even though the underlying unlabelled graph is the same. This observation lets the course move from matrices to a finite list of diagrams without losing the Lie-theoretic data. ## Connected Diagrams and Irreducible Root Systems When does a diagram represent one simple object rather than a product of independent ones? The answer is that connectedness of the diagram corresponds to irreducibility of the root system, and later to simplicity of the associated Lie algebra. Without a decomposability test, a diagram such as two disconnected $A_1$ vertices could be mistaken for a new rank-two simple type, even though it is only the product of two rank-one systems. Connectedness is therefore the diagrammatic way of separating classification of simple pieces from bookkeeping of direct sums. [definition: Decomposable Root System] A root system $\Phi\subset E$ is decomposable if there are nonempty root systems $\Phi_1\subset E_1$ and $\Phi_2\subset E_2$ such that $E=E_1\oplus E_2$ is an orthogonal direct sum and \begin{align*} \Phi=\Phi_1\sqcup \Phi_2. \end{align*} It is irreducible if it is not decomposable. [/definition] The definition is geometric, while the diagram is combinatorial, so one still has to prove that they detect the same decomposition. The potential problem is that two groups of roots might look connected in the diagram even though their spans are orthogonal, or conversely that a disconnected diagram might conceal an indecomposable configuration. The theorem below makes connected components the correct units for classification. [quotetheorem:4711] [citeproof:4711] The reduced and finite hypotheses keep the argument inside the usual root-system classification: nonreduced systems require extra care because multiples of roots can sit on the same line, and infinite root systems may have subtler connectedness phenomena. The theorem also explains why disconnected diagrams are not listed among the simple finite types. They are still valid semisimple data, but they are assembled from connected components. For semisimple Lie algebras, this statement becomes the diagrammatic form of decomposition into simple ideals. Orthogonal components of the root system correspond to commuting simple summands of the Lie algebra. Classification, however, is carried out with Cartan matrices rather than with the abstract orthogonal decomposition, so the geometric splitting has to be visible in the integer data itself. A disconnected Dynkin diagram has no edges between different components, which means the corresponding off-block Cartan integers vanish. After reordering the simple roots by connected component, the Cartan matrix becomes block diagonal. This block diagonal form is the matrix shadow of the later Lie algebra direct sum. At this stage we use only the combinatorial conclusion: disconnected finite diagrams are not new irreducible types, because each connected component can be classified separately. ## The Classical Families What diagrams occur in the infinite families, and how do they reflect the standard matrix Lie algebras? The four classical series account for most finite-dimensional simple Lie algebras and provide the models against which the exceptional diagrams are compared. [example: Type A] For $\mathfrak{sl}_{n+1}$, realize the roots in the hyperplane $\sum_{i=1}^{n+1}x_i=0$ as \begin{align*} \Phi=\{e_i-e_j:1\le i\ne j\le n+1\}, \end{align*} with simple roots \begin{align*} \alpha_i=e_i-e_{i+1}\qquad (1\le i\le n). \end{align*} Using $(e_i,e_j)=\delta_{ij}$, each simple root has length \begin{align*} (\alpha_i,\alpha_i) &=(e_i-e_{i+1},e_i-e_{i+1})\\ &=(e_i,e_i)-2(e_i,e_{i+1})+(e_{i+1},e_{i+1})\\ &=1-0+1\\ &=2. \end{align*} For adjacent simple roots, \begin{align*} (\alpha_i,\alpha_{i+1}) &=(e_i-e_{i+1},e_{i+1}-e_{i+2})\\ &=(e_i,e_{i+1})-(e_i,e_{i+2})-(e_{i+1},e_{i+1})+(e_{i+1},e_{i+2})\\ &=0-0-1+0\\ &=-1, \end{align*} and similarly $(\alpha_{i+1},\alpha_i)=-1$. If $|i-j|>1$, then the four basis vectors appearing in $e_i-e_{i+1}$ and $e_j-e_{j+1}$ are distinct, so \begin{align*} (\alpha_i,\alpha_j)=0. \end{align*} Therefore the Cartan entries are \begin{align*} a_{ii}=2,\qquad a_{i,i+1}=a_{i+1,i}=\frac{2(-1)}{2}=-1,\qquad a_{ij}=0\quad\text{if }|i-j|>1. \end{align*} Thus the Dynkin diagram has one edge exactly between consecutive vertices and no multiple edges: \begin{align*} 1-2-\cdots-n. \end{align*} All roots $e_i-e_j$ have squared length \begin{align*} (e_i-e_j,e_i-e_j)=1-0-0+1=2, \end{align*} so all roots have the same length and no arrow appears. [/example] The type $A$ model establishes the single-bond chain. The type $B$ model keeps the chain shape but changes the final bond, making root length visible in the diagram. [example: Type B] For $B_n$, use the standard root system in $\mathbb R^n$ with orthonormal basis $e_1,\dots,e_n$: \begin{align*} \Phi=\{\pm e_i\pm e_j:1\le i<j\le n\}\cup\{\pm e_i:1\le i\le n\}, \end{align*} and choose simple roots \begin{align*} \alpha_i=e_i-e_{i+1}\quad(1\le i\le n-1),\qquad \alpha_n=e_n. \end{align*} For $1\le i\le n-1$, \begin{align*} (\alpha_i,\alpha_i) &=(e_i-e_{i+1},e_i-e_{i+1})\\ &=(e_i,e_i)-2(e_i,e_{i+1})+(e_{i+1},e_{i+1})\\ &=1-0+1\\ &=2, \end{align*} while \begin{align*} (\alpha_n,\alpha_n)=(e_n,e_n)=1. \end{align*} Thus the last simple root is shorter than the preceding ones. For adjacent roots away from the end, $1\le i\le n-2$, \begin{align*} (\alpha_i,\alpha_{i+1}) &=(e_i-e_{i+1},e_{i+1}-e_{i+2})\\ &=(e_i,e_{i+1})-(e_i,e_{i+2})-(e_{i+1},e_{i+1})+(e_{i+1},e_{i+2})\\ &=0-0-1+0\\ &=-1. \end{align*} Since both roots have squared length $2$, the corresponding Cartan entries are \begin{align*} a_{i,i+1}=\frac{2(\alpha_{i+1},\alpha_i)}{(\alpha_i,\alpha_i)} =\frac{2(-1)}{2}=-1,\qquad a_{i+1,i}=\frac{2(\alpha_i,\alpha_{i+1})}{(\alpha_{i+1},\alpha_{i+1})} =\frac{2(-1)}{2}=-1. \end{align*} If $|i-j|>1$ and $\{i,j\}\ne\{n-1,n\}$, then the basis vectors appearing in $\alpha_i$ and $\alpha_j$ are disjoint, so $(\alpha_i,\alpha_j)=0$ and hence $a_{ij}=a_{ji}=0$. At the final edge, \begin{align*} (\alpha_{n-1},\alpha_n) &=(e_{n-1}-e_n,e_n)\\ &=(e_{n-1},e_n)-(e_n,e_n)\\ &=0-1\\ &=-1. \end{align*} Therefore \begin{align*} a_{n-1,n} &=\frac{2(\alpha_n,\alpha_{n-1})}{(\alpha_{n-1},\alpha_{n-1})} =\frac{2(-1)}{2} =-1,\\ a_{n,n-1} &=\frac{2(\alpha_{n-1},\alpha_n)}{(\alpha_n,\alpha_n)} =\frac{2(-1)}{1} =-2. \end{align*} Thus \begin{align*} a_{n-1,n}a_{n,n-1}=(-1)(-2)=2, \end{align*} so the last two vertices are joined by a double edge. The larger absolute off-diagonal entry lies in row $n$, and $\alpha_n$ has squared length $1$, so the arrow points toward the shorter last root: \begin{align*} 1-2-\cdots-(n-1)\Rightarrow n. \end{align*} The roots $\pm e_i\pm e_j$ have squared length \begin{align*} (e_i\pm e_j,e_i\pm e_j)=1\pm 0\pm 0+1=2, \end{align*} whereas the roots $\pm e_i$ have squared length $1$. Thus the double edge at the end records exactly the length ratio between the long roots $e_i\pm e_j$ and the short roots $e_i$ in the root system of $\mathfrak{so}_{2n+1}$. [/example] Type $B$ shows one orientation of the double bond. Type $C$ uses the same signed-coordinate geometry with the long and short roles reversed at the end of the chain. [example: Type C] For $C_n$ with $n\ge 3$, realize the roots in $\mathbb R^n$ with orthonormal basis $e_1,\dots,e_n$ as \begin{align*} \Phi=\{\pm e_i\pm e_j:1\le i<j\le n\}\cup\{\pm 2e_i:1\le i\le n\}, \end{align*} and choose simple roots \begin{align*} \alpha_i=e_i-e_{i+1}\quad(1\le i\le n-1),\qquad \alpha_n=2e_n. \end{align*} For $1\le i\le n-1$, \begin{align*} (\alpha_i,\alpha_i) &=(e_i-e_{i+1},e_i-e_{i+1})\\ &=(e_i,e_i)-2(e_i,e_{i+1})+(e_{i+1},e_{i+1})\\ &=1-0+1\\ &=2, \end{align*} while \begin{align*} (\alpha_n,\alpha_n)=(2e_n,2e_n)=4(e_n,e_n)=4. \end{align*} Thus $\alpha_n$ is longer than $\alpha_{n-1}$. For adjacent roots away from the end, $1\le i\le n-2$, \begin{align*} (\alpha_i,\alpha_{i+1}) &=(e_i-e_{i+1},e_{i+1}-e_{i+2})\\ &=(e_i,e_{i+1})-(e_i,e_{i+2})-(e_{i+1},e_{i+1})+(e_{i+1},e_{i+2})\\ &=0-0-1+0\\ &=-1. \end{align*} Since both roots have squared length $2$, \begin{align*} a_{i,i+1}=\frac{2(-1)}{2}=-1,\qquad a_{i+1,i}=\frac{2(-1)}{2}=-1. \end{align*} If $|i-j|>1$ and $\{i,j\}\ne\{n-1,n\}$, the basis vectors appearing in $\alpha_i$ and $\alpha_j$ are disjoint, so \begin{align*} (\alpha_i,\alpha_j)=0,\qquad a_{ij}=a_{ji}=0. \end{align*} At the final edge, \begin{align*} (\alpha_{n-1},\alpha_n) &=(e_{n-1}-e_n,2e_n)\\ &=2(e_{n-1},e_n)-2(e_n,e_n)\\ &=0-2\\ &=-2. \end{align*} Therefore \begin{align*} a_{n-1,n} &=\frac{2(\alpha_n,\alpha_{n-1})}{(\alpha_{n-1},\alpha_{n-1})} =\frac{2(-2)}{2} =-2,\\ a_{n,n-1} &=\frac{2(\alpha_{n-1},\alpha_n)}{(\alpha_n,\alpha_n)} =\frac{2(-2)}{4} =-1. \end{align*} Hence \begin{align*} a_{n-1,n}a_{n,n-1}=(-2)(-1)=2, \end{align*} so the last two vertices are joined by a double edge. The larger absolute off-diagonal entry lies in row $n-1$, and $\alpha_{n-1}$ has squared length $2<4=(\alpha_n,\alpha_n)$, so the arrow points toward the penultimate vertex: \begin{align*} 1-2-\cdots-(n-1)\Leftarrow n. \end{align*} The roots $\pm e_i\pm e_j$ have squared length \begin{align*} (e_i\pm e_j,e_i\pm e_j)=1\pm 0\pm 0+1=2, \end{align*} whereas the roots $\pm 2e_i$ have squared length \begin{align*} (\pm 2e_i,\pm 2e_i)=4(e_i,e_i)=4. \end{align*} Thus $C_n$ has the same underlying chain as $B_n$, but the double-edge arrow is reversed because the long and short roots have exchanged roles; this is the diagram attached to $\mathfrak{sp}_{2n}$. [/example] After the two non-simply-laced chains, type $D$ returns to equal root lengths but changes the topology of the diagram: the chain now forks near the end. [example: Type D] For $n\ge 4$, realize the type $D_n$ root system in $\mathbb R^n$ with orthonormal basis $e_1,\dots,e_n$ as \begin{align*} \Phi=\{\pm e_i\pm e_j:1\le i<j\le n\}, \end{align*} and choose simple roots \begin{align*} \alpha_i=e_i-e_{i+1}\quad(1\le i\le n-1),\qquad \alpha_n=e_{n-1}+e_n. \end{align*} For $1\le i\le n-1$, \begin{align*} (\alpha_i,\alpha_i) &=(e_i-e_{i+1},e_i-e_{i+1})\\ &=(e_i,e_i)-2(e_i,e_{i+1})+(e_{i+1},e_{i+1})\\ &=1-0+1\\ &=2, \end{align*} and for the final simple root, \begin{align*} (\alpha_n,\alpha_n) &=(e_{n-1}+e_n,e_{n-1}+e_n)\\ &=(e_{n-1},e_{n-1})+2(e_{n-1},e_n)+(e_n,e_n)\\ &=1+0+1\\ &=2. \end{align*} Thus all simple roots have the same squared length. For adjacent roots along the chain, $1\le i\le n-2$, \begin{align*} (\alpha_i,\alpha_{i+1}) &=(e_i-e_{i+1},e_{i+1}-e_{i+2})\\ &=(e_i,e_{i+1})-(e_i,e_{i+2})-(e_{i+1},e_{i+1})+(e_{i+1},e_{i+2})\\ &=0-0-1+0\\ &=-1. \end{align*} Since both roots have squared length $2$, this gives \begin{align*} a_{i,i+1}=\frac{2(-1)}{2}=-1,\qquad a_{i+1,i}=\frac{2(-1)}{2}=-1. \end{align*} The extra terminal root is joined to $\alpha_{n-2}$ because \begin{align*} (\alpha_{n-2},\alpha_n) &=(e_{n-2}-e_{n-1},e_{n-1}+e_n)\\ &=(e_{n-2},e_{n-1})+(e_{n-2},e_n)-(e_{n-1},e_{n-1})-(e_{n-1},e_n)\\ &=0+0-1-0\\ &=-1, \end{align*} so \begin{align*} a_{n-2,n}=\frac{2(-1)}{2}=-1,\qquad a_{n,n-2}=\frac{2(-1)}{2}=-1. \end{align*} The two terminal roots are not joined, because \begin{align*} (\alpha_{n-1},\alpha_n) &=(e_{n-1}-e_n,e_{n-1}+e_n)\\ &=(e_{n-1},e_{n-1})+(e_{n-1},e_n)-(e_n,e_{n-1})-(e_n,e_n)\\ &=1+0-0-1\\ &=0. \end{align*} All other non-adjacent pairs involve disjoint basis vectors, so their inner products are $0$ and the corresponding Cartan entries vanish. Therefore the Dynkin diagram has single edges \begin{align*} 1-2-\cdots-(n-2), \end{align*} together with the two additional single edges from $n-2$ to $n-1$ and from $n-2$ to $n$. Since every root $\pm e_i\pm e_j$ has squared length \begin{align*} (\pm e_i\pm e_j,\pm e_i\pm e_j)=1\pm 0\pm 0+1=2, \end{align*} there are no arrows or multiple edges. The branch records the fact that the two terminal simple roots are $e_{n-1}-e_n$ and $e_{n-1}+e_n$, both attached to $e_{n-2}-e_{n-1}$ and orthogonal to each other; this is the diagram attached to $\mathfrak{so}_{2n}$. [/example] These families illustrate three recurring themes: chains encode adjacent simple roots, arrows encode unequal lengths, and branching is a genuine new combinatorial feature rather than a length effect. ## Exceptional Diagrams What remains after the infinite families are removed? The finite-type positivity condition allows five additional connected diagrams. They do not extend to infinite families, but they are indispensable in the final classification. [example: Exceptional Types] We record the five exceptional connected finite diagrams by listing their vertices and edges. The simply-laced exceptional diagrams have only single edges. For $E_6$, take vertices $1,\dots,6$ and single edges \begin{align*} 1-3,\quad 2-3,\quad 3-4,\quad 4-5,\quad 5-6. \end{align*} Thus $E_6$ is a chain of five vertices with one extra vertex attached at the second vertex of the chain. For $E_7$, take vertices $1,\dots,7$ and single edges \begin{align*} 1-3,\quad 2-3,\quad 3-4,\quad 4-5,\quad 5-6,\quad 6-7. \end{align*} This is obtained from the $E_6$ shape by extending the long chain by one vertex. For $E_8$, take vertices $1,\dots,8$ and single edges \begin{align*} 1-3,\quad 2-3,\quad 3-4,\quad 4-5,\quad 5-6,\quad 6-7,\quad 7-8. \end{align*} This is obtained from the $E_7$ shape by extending the same chain by one more vertex. Since every edge in $E_6$, $E_7$, and $E_8$ is single, all adjacent off-diagonal Cartan entries are $-1$ in both directions, and the diagrams have no arrows. The remaining exceptional diagrams are not simply-laced. The diagram $F_4$ has vertices $1,2,3,4$ with a chain \begin{align*} 1-2\Rightarrow 3-4, \end{align*} where the middle edge is double. Equivalently, the adjacent products are \begin{align*} a_{12}a_{21}=1,\qquad a_{23}a_{32}=2,\qquad a_{34}a_{43}=1, \end{align*} and all non-adjacent products are $0$. The diagram $G_2$ has two vertices joined by a triple edge: \begin{align*} 1\Rrightarrow 2, \end{align*} so its two off-diagonal entries have product \begin{align*} a_{12}a_{21}=3. \end{align*} Thus the exceptional finite diagrams consist of three simply-laced branching diagrams, $E_6,E_7,E_8$, together with the double-edge chain $F_4$ and the triple-edge rank-two diagram $G_2$. [/example] The preceding examples account for the exceptional possibilities, but the classification needs an exclusion theorem as well as a list. Positive definiteness, connectedness, and the crystallographic edge restrictions must rule out every other connected diagram, including longer branching patterns and forbidden multiple-edge chains. The following theorem is the point where those numerical constraints become a complete finite list. [quotetheorem:4713] [citeproof:4713] The theorem is the matrix-side list of possible connected finite-type diagrams. Its hypotheses are doing real work: dropping connectedness allows arbitrary disjoint unions of the listed diagrams, and dropping positive definiteness admits affine extensions such as $\widetilde A_n$. The classification also has consequences beyond naming Lie algebras: the diagram controls the Weyl group, the weight lattice, representation highest weights, and later the possible parabolic and flag varieties. Cartan matrices and Dynkin diagrams reduce the classification problem to a finite list of candidates satisfying rigid numerical constraints. The next chapter rederives this list from finite root systems themselves, proving the exclusion arguments that explain why no other crystallographic root geometries occur. # 8. Classification of Finite Root Systems This chapter proves the root-system side of the finite diagram list previewed in Chapter 7. Chapters 4 through 7 encoded root geometry in root systems, Cartan matrices, and Dynkin diagrams; here we supply the exclusion arguments showing which finite diagrams actually occur. The central question is: once the root-string restrictions have been translated into a Dynkin diagram, how much freedom remains? The answer is strikingly rigid. Positive definiteness rules out cycles and excessive branching, while crystallographic integrality restricts multiple bonds to a small list. The outcome matches the candidate list from Cartan matrices: the four infinite classical families and five exceptional diagrams. ## Dynkin Diagrams as Positive Definite Cartan Data The classification begins with a reduction from geometry to a finite labelled graph. We want to know when a diagram can be the diagram of a finite reduced crystallographic root system, so the first problem is to recover the relevant quadratic form from the Cartan matrix. [definition: Cartan Matrix of a Root System] Let $\Phi$ be a finite reduced crystallographic root system with simple roots $\Delta = \{\alpha_1, \dots, \alpha_\ell\}$. The Cartan matrix of $\Phi$ relative to $\Delta$ is the integer matrix $A = (a_{ij})_{1 \le i,j \le \ell}$ defined by \begin{align*} a_{ij} = \langle \alpha_j, \alpha_i^\vee \rangle = \frac{2(\alpha_i,\alpha_j)}{(\alpha_i,\alpha_i)}. \end{align*} [/definition] The convention is that the coroot index appears in the row. Thus $a_{ii}=2$, while for $i \ne j$ the entries are non-positive integers. We need a diagrammatic encoding because the products $a_{ij}a_{ji}$ record exactly the angles and squared-length ratios that survive rescaling. [definition: Dynkin Diagram] Let $A=(a_{ij})$ be the Cartan matrix of a finite reduced crystallographic root system with simple roots $\alpha_1,\dots,\alpha_\ell$. Its Dynkin diagram has one vertex for each simple root. Distinct vertices $i$ and $j$ are joined by $a_{ij}a_{ji}$ bonds, and if the roots have unequal lengths, the bonds are oriented toward the shorter root. [/definition] The diagram suppresses the scale of the inner product but remembers exactly the Cartan integers. In finite type the possible products are $0,1,2,3$, corresponding respectively to no bond, a single bond, a double bond, and a triple bond. [example: Rank Two Diagrams] Let $\alpha$ and $\beta$ be two simple roots, and let $\theta$ be the angle between them. From the definition of coroots, \begin{align*} \langle \alpha,\beta^\vee\rangle\langle \beta,\alpha^\vee\rangle &=\frac{2(\alpha,\beta)}{(\beta,\beta)}\cdot \frac{2(\beta,\alpha)}{(\alpha,\alpha)}\\ &=\frac{4(\alpha,\beta)^2}{(\alpha,\alpha)(\beta,\beta)}\\ &=4\left(\frac{(\alpha,\beta)}{\sqrt{(\alpha,\alpha)}\sqrt{(\beta,\beta)}}\right)^2\\ &=4\cos^2\theta. \end{align*} The two roots are linearly independent, so $|\cos\theta|<1$, and hence \begin{align*} 0\le \langle \alpha,\beta^\vee\rangle\langle \beta,\alpha^\vee\rangle<4. \end{align*} Because the root system is crystallographic, both Cartan integers are integral; because $\alpha$ and $\beta$ are distinct simple roots, they are non-positive. Thus their product is one of \begin{align*} 0,\ 1,\ 2,\ 3. \end{align*} These four cases correspond respectively to $A_1\times A_1$, $A_2$, $B_2=C_2$, and $G_2$. When the product is $2$ or $3$, the two roots have unequal lengths: indeed, if $(\alpha,\beta)\ne 0$, then \begin{align*} \frac{\langle \alpha,\beta^\vee\rangle}{\langle \beta,\alpha^\vee\rangle} =\frac{2(\alpha,\beta)/(\beta,\beta)}{2(\alpha,\beta)/(\alpha,\alpha)} =\frac{(\alpha,\alpha)}{(\beta,\beta)}. \end{align*} Thus the larger absolute Cartan entry occurs on the side of the shorter coroot, and the Dynkin arrow is drawn toward the shorter root. [/example] To classify finite diagrams, it is useful to forget the arrows and start from a Euclidean quadratic form. We need a positivity criterion because the Gram matrix $G=((\alpha_i,\alpha_j))$ is symmetric positive definite, while the Cartan matrix is obtained from $G$ by rescaling rows. [quotetheorem:4714] [citeproof:4714] This theorem is the bridge between root systems and graph theory, but its hypotheses matter. Finiteness gives a genuine Euclidean root system with finitely many simple roots, so the Gram matrix is a finite positive definite matrix; without finite type one meets affine Cartan matrices whose symmetrised forms are only positive semidefinite. Reducedness avoids the ambiguity created by having both $\alpha$ and $2\alpha$ as roots, while crystallographicity is what makes the Cartan entries integral and hence representable by a Dynkin diagram with the usual bond multiplicities. The theorem also does not classify Cartan matrices by itself: it says that diagrams coming from finite root systems pass a positive-definiteness test, not that every labelled graph with a positive definite symmetrisation has already been realised by roots. From this point the problem becomes: which connected labelled graphs support such a positive definite form and satisfy the local rank two restrictions? ## Excluding Cycles and Excessive Branching The next question is how positive definiteness constrains the shape of the underlying graph. The easiest forbidden configurations are those whose vertices support a vector on which the quadratic form is non-positive. [quotetheorem:4715] [citeproof:4715] Thus a connected finite Dynkin diagram is a tree, but the theorem should not be read as saying that every tree is allowed. Connectedness is used only to speak about one irreducible component; a reducible finite root system has a disconnected diagram whose components are trees. Finite type is also essential: affine diagrams such as $\widetilde{A}_\ell$ contain cycles or affine branching patterns precisely because their forms are positive semidefinite rather than positive definite. The next restrictions distinguish the few positive definite trees from the many trees that still fail to come from finite root systems. The contrast between a path and a cycle is already visible in small rank. [example: Why A4 Is Allowed but a Four-Cycle Is Not] For the path $A_4$, take $\alpha_i=e_i-e_{i+1}$ for $1\le i\le 4$ in the hyperplane $x_1+\cdots+x_5=0\subset \mathbb R^5$. If $c_1,\dots,c_4\in \mathbb R$, then \begin{align*} \sum_{i=1}^4 c_i\alpha_i &=c_1(e_1-e_2)+c_2(e_2-e_3)+c_3(e_3-e_4)+c_4(e_4-e_5)\\ &=c_1e_1+(c_2-c_1)e_2+(c_3-c_2)e_3+(c_4-c_3)e_4-c_4e_5. \end{align*} Therefore \begin{align*} \left|\sum_{i=1}^4 c_i\alpha_i\right|^2 =c_1^2+(c_2-c_1)^2+(c_3-c_2)^2+(c_4-c_3)^2+c_4^2. \end{align*} This is $0$ only when $c_1=0$, $c_2=c_1$, $c_3=c_2$, $c_4=c_3$, and $c_4=0$, hence only when all $c_i=0$. Thus the Gram matrix of these four simple roots is positive definite on their span. Now compare a four-cycle with the same single-bond normalization. Let its four proposed simple roots be $\beta_1,\beta_2,\beta_3,\beta_4$, indexed cyclically, with \begin{align*} (\beta_i,\beta_i)=2,\qquad (\beta_i,\beta_{i+1})=-1,\qquad (\beta_1,\beta_3)=(\beta_2,\beta_4)=0. \end{align*} For $v=\beta_1+\beta_2+\beta_3+\beta_4$, bilinearity gives \begin{align*} (v,v) &=\sum_{i=1}^4(\beta_i,\beta_i)+2\sum_{1\le i<j\le 4}(\beta_i,\beta_j)\\ &=(2+2+2+2)+2\bigl((\beta_1,\beta_2)+(\beta_2,\beta_3)+(\beta_3,\beta_4)+(\beta_4,\beta_1)+(\beta_1,\beta_3)+(\beta_2,\beta_4)\bigr)\\ &=8+2((-1)+(-1)+(-1)+(-1)+0+0)\\ &=0. \end{align*} A nonzero vector supported on simple roots with square $0$ is incompatible with positive definiteness, so the path $A_4$ can occur in finite type while the four-cycle cannot. [/example] Positive definiteness also controls branching. We need a name for the vertices where branching occurs, because the classification reduces to measuring how long the arms can be from such a vertex. [definition: Branch Point] A branch point of a graph is a vertex incident with at least three edges. [/definition] For finite Dynkin diagrams, branch points appear only in the simply laced exceptional family $D_\ell$ and $E_6,E_7,E_8$. The following estimate is the key numerical restriction. [quotetheorem:4716] [citeproof:4716] This inequality is the reason the simply laced list stops, and each hypothesis is doing work. The assumption of simply lacedness lets every arm be measured by the same path quadratic form; once multiple bonds are present, the length ratios have to be tracked separately and the statement is no longer the right numerical test. The assumption that there is exactly one branch point isolates a three-arm subdiagram, while two branch points would contain a larger forbidden tree whose quadratic form is already non-positive on a vector supported between the branch points and along three escaping arms. After ordering $p \le q \le r$, the inequality leaves the $D_\ell$ family with $p=q=2$, and the exceptional triples $(2,3,3)$, $(2,3,4)$, and $(2,3,5)$. Equality marks the affine boundary, so the estimate also points forward to the affine Dynkin diagrams that appear when positive definiteness is weakened to positive semidefiniteness. [example: Why E6 E7 and E8 Stop at E8] For a simply laced diagram with one branch point, write the three branch-arm lengths as $(p,q,r)$, where each arm length includes the branch point. The exceptional triples are \begin{align*} (2,3,3),\qquad (2,3,4),\qquad (2,3,5). \end{align*} For these triples the branch length sums are \begin{align*} \frac{1}{2}+\frac{1}{3}+\frac{1}{3} &=\frac{3}{6}+\frac{2}{6}+\frac{2}{6} =\frac{7}{6}>1,\\ \frac{1}{2}+\frac{1}{3}+\frac{1}{4} &=\frac{6}{12}+\frac{4}{12}+\frac{3}{12} =\frac{13}{12}>1,\\ \frac{1}{2}+\frac{1}{3}+\frac{1}{5} &=\frac{15}{30}+\frac{10}{30}+\frac{6}{30} =\frac{31}{30}>1. \end{align*} Thus these triples pass the positivity test from the *Branch Length Inequality*, and they give the finite diagrams $E_6$, $E_7$, and $E_8$, respectively. Extending the longest arm once more changes $(2,3,5)$ into $(2,3,6)$. Then \begin{align*} \frac{1}{2}+\frac{1}{3}+\frac{1}{6} &=\frac{3}{6}+\frac{2}{6}+\frac{1}{6} =\frac{6}{6} =1. \end{align*} The strict inequality required for positive definiteness has become equality, so the extended diagram lies on the affine boundary rather than in finite type. [/example] ## Multiple Bonds and Root Lengths After cycles and branching are excluded, the remaining problem is to understand how double and triple bonds may appear. Multiple bonds are local rank two phenomena, but they also interact with the global tree structure through root lengths. [quotetheorem:4717] [citeproof:4717] The finite Dynkin assumptions are doing real work here: a triple edge with any extra neighbour, two double edges in a chain of length three, and a double edge feeding into a branch all fail the positive-definite determinant test. Irreducibility is used to keep the diagram connected; otherwise the statement applies component by component. The rank hypotheses prevent the rank two coincidences $B_2=C_2$ and $G_2$ from being confused with higher-rank families. The theorem classifies where multiple bonds can sit only after positive definiteness and the tree condition are already in force. [example: The Diagram F4] Take the following four roots in the displayed set: \begin{align*} \alpha_1=e_2-e_3,\qquad \alpha_2=e_3-e_4,\qquad \alpha_3=e_4,\qquad \alpha_4=\frac12(e_1-e_2-e_3-e_4). \end{align*} Their squared lengths are \begin{align*} (\alpha_1,\alpha_1)&=(e_2-e_3,e_2-e_3)=1+1=2,\\ (\alpha_2,\alpha_2)&=(e_3-e_4,e_3-e_4)=1+1=2,\\ (\alpha_3,\alpha_3)&=(e_4,e_4)=1,\\ (\alpha_4,\alpha_4)&=\frac14\bigl((e_1,e_1)+(e_2,e_2)+(e_3,e_3)+(e_4,e_4)\bigr)=1. \end{align*} Thus $\alpha_1,\alpha_2$ are long and $\alpha_3,\alpha_4$ are short. The adjacent inner products are \begin{align*} (\alpha_1,\alpha_2) &=(e_2-e_3,e_3-e_4) =-(e_3,e_3) =-1,\\ (\alpha_2,\alpha_3) &=(e_3-e_4,e_4) =-(e_4,e_4) =-1,\\ (\alpha_3,\alpha_4) &=\left(e_4,\frac12(e_1-e_2-e_3-e_4)\right) =-\frac12. \end{align*} The non-adjacent inner products vanish: \begin{align*} (\alpha_1,\alpha_3)&=(e_2-e_3,e_4)=0,\\ (\alpha_1,\alpha_4)&=\left(e_2-e_3,\frac12(e_1-e_2-e_3-e_4)\right) =-\frac12+\frac12=0,\\ (\alpha_2,\alpha_4)&=\left(e_3-e_4,\frac12(e_1-e_2-e_3-e_4)\right) =-\frac12+\frac12=0. \end{align*} Using $a_{ij}=2(\alpha_i,\alpha_j)/(\alpha_i,\alpha_i)$, the Cartan entries on adjacent vertices are \begin{align*} a_{12}=a_{21}=-1,\qquad a_{23}=-1,\quad a_{32}=-2,\qquad a_{34}=a_{43}=-1. \end{align*} Hence the products $a_{ij}a_{ji}$ along the chain are \begin{align*} a_{12}a_{21}=1,\qquad a_{23}a_{32}=2,\qquad a_{34}a_{43}=1, \end{align*} and all non-adjacent products are $0$. Therefore the Dynkin diagram is a four-vertex chain with one double bond in the middle, joining the long root $\alpha_2$ to the short root $\alpha_3$; this is the diagram $F_4$. [/example] ## The Simply Laced Classification It remains to assemble the restrictions in the equal-length case. The problem is now purely graph-theoretic: classify connected trees with no multiple bonds whose Cartan quadratic form is positive definite. [quotetheorem:4718] [citeproof:4718] This theorem explains the name ADE: it is the finite, connected, simply laced classification. Each adjective is necessary. If connectedness is dropped, disjoint unions such as $A_2\sqcup A_3$ occur and correspond to reducible root systems rather than new irreducible types. If finite type is replaced by affine type, the extra diagrams $\widetilde{A}_\ell$, $\widetilde{D}_\ell$, and $\widetilde{E}_6,\widetilde{E}_7,\widetilde{E}_8$ appear because the quadratic form acquires a one-dimensional kernel. If simply lacedness is dropped, the non-simply laced families $B_\ell,C_\ell,F_4,G_2$ enter. The result is therefore not a classification of all Coxeter diagrams, but of the positive definite crystallographic diagrams with equal root lengths, and it is exactly the input later used for simply laced semisimple Lie algebras. [example: Classical Simply Laced Coordinates] In $\mathbb R^{\ell+1}$, every vector $e_i-e_j$ satisfies \begin{align*} (e_i-e_j)_1+\cdots+(e_i-e_j)_{\ell+1}=1-1=0, \end{align*} so \begin{align*} A_\ell=\{e_i-e_j:1\le i\ne j\le \ell+1\} \end{align*} lies in the hyperplane $\sum_i x_i=0$. Choose simple roots \begin{align*} \alpha_i=e_i-e_{i+1}\qquad 1\le i\le \ell. \end{align*} Then \begin{align*} (\alpha_i,\alpha_i) &=(e_i-e_{i+1},e_i-e_{i+1}) =(e_i,e_i)+(e_{i+1},e_{i+1}) =2, \end{align*} and, for adjacent indices, \begin{align*} (\alpha_i,\alpha_{i+1}) &=(e_i-e_{i+1},e_{i+1}-e_{i+2})\\ &=(e_i,e_{i+1})-(e_i,e_{i+2})-(e_{i+1},e_{i+1})+(e_{i+1},e_{i+2})\\ &=0-0-1+0=-1. \end{align*} If $|i-j|>1$, the four basis vectors appearing in $\alpha_i$ and $\alpha_j$ are distinct, so \begin{align*} (\alpha_i,\alpha_j)=0. \end{align*} Since all simple roots have squared length $2$, the Cartan entry is \begin{align*} a_{ij}=\frac{2(\alpha_i,\alpha_j)}{(\alpha_i,\alpha_i)}=(\alpha_i,\alpha_j). \end{align*} Thus consecutive vertices are joined by one bond and non-consecutive vertices are not joined, giving the path diagram $A_\ell$. In $\mathbb R^\ell$, take \begin{align*} D_\ell=\{\pm e_i\pm e_j:1\le i<j\le \ell\} \end{align*} and choose \begin{align*} \beta_i=e_i-e_{i+1}\quad 1\le i\le \ell-1,\qquad \beta_\ell=e_{\ell-1}+e_\ell. \end{align*} Each $\beta_i$ belongs to the displayed set, and each has squared length $2$: \begin{align*} (\beta_i,\beta_i)=2\quad 1\le i\le \ell. \end{align*} For $1\le i\le \ell-2$, \begin{align*} (\beta_i,\beta_{i+1}) &=(e_i-e_{i+1},e_{i+1}-e_{i+2}) =0-0-1+0=-1. \end{align*} The extra end vertex satisfies \begin{align*} (\beta_{\ell-2},\beta_\ell) &=(e_{\ell-2}-e_{\ell-1},e_{\ell-1}+e_\ell) =0+0-1-0=-1, \end{align*} while \begin{align*} (\beta_{\ell-1},\beta_\ell) &=(e_{\ell-1}-e_\ell,e_{\ell-1}+e_\ell) =1+0-0-1=0. \end{align*} All other non-neighboring pairs use disjoint basis directions, or have the same cancellation as the last displayed line, so their inner product is $0$. Therefore the Cartan matrix has entry $-1$ exactly along the chain \begin{align*} \beta_1-\beta_2-\cdots-\beta_{\ell-2} \end{align*} with two terminal vertices $\beta_{\ell-1}$ and $\beta_\ell$ attached to $\beta_{\ell-2}$. This is the forked simply laced diagram $D_\ell$. [/example] ## The Full Finite Type List The final question is whether the preceding restrictions have missed any non-simply laced diagrams. The answer is no: the rank two possibilities, the tree condition, and the multiple-bond restrictions give the full finite list. [quotetheorem:4719] [citeproof:4719] The classification is a theorem about irreducible finite reduced crystallographic root systems, and changing any of those words changes the answer. If irreducibility is dropped, one obtains orthogonal direct sums and hence disconnected unions of the listed diagrams. If reducedness is dropped, non-reduced systems such as $BC_\ell$ appear, where both $\alpha$ and $2\alpha$ may occur. If crystallographicity is dropped, the non-crystallographic finite Coxeter types $H_3$, $H_4$, and $I_2(m)$ for $m\notin\{2,3,4,6\}$ are no longer excluded by integrality. If finite type is weakened to affine type, the positive definite condition becomes positive semidefinite and the affine Dynkin diagrams enter. Thus the theorem gives exactly the finite crystallographic input needed for semisimple Lie algebras; in the next stage of the course, Serre's theorem shows that these diagrams are not merely necessary but present the corresponding semisimple Lie algebras. [example: Classical Non-Simply Laced Coordinates] In $\mathbb R^\ell$ with orthonormal basis $e_1,\dots,e_\ell$, consider \begin{align*} B_\ell&=\{\pm e_i:1\le i\le \ell\}\cup\{\pm e_i\pm e_j:1\le i<j\le \ell\},\\ C_\ell&=\{\pm 2e_i:1\le i\le \ell\}\cup\{\pm e_i\pm e_j:1\le i<j\le \ell\}. \end{align*} For $B_\ell$, choose simple roots \begin{align*} \alpha_i=e_i-e_{i+1}\quad 1\le i\le \ell-1,\qquad \alpha_\ell=e_\ell. \end{align*} Their squared lengths are \begin{align*} (\alpha_i,\alpha_i)&=(e_i-e_{i+1},e_i-e_{i+1})=1+1=2\quad 1\le i\le \ell-1,\\ (\alpha_\ell,\alpha_\ell)&=(e_\ell,e_\ell)=1. \end{align*} For $1\le i\le \ell-2$, \begin{align*} (\alpha_i,\alpha_{i+1}) &=(e_i-e_{i+1},e_{i+1}-e_{i+2})\\ &=0-0-1+0=-1, \end{align*} while the terminal inner product is \begin{align*} (\alpha_{\ell-1},\alpha_\ell) &=(e_{\ell-1}-e_\ell,e_\ell) =0-1=-1. \end{align*} If $|i-j|>1$, the basis vectors appearing in $\alpha_i$ and $\alpha_j$ are disjoint, except for the already listed terminal case, so $(\alpha_i,\alpha_j)=0$. Hence, using $a_{ij}=2(\alpha_i,\alpha_j)/(\alpha_i,\alpha_i)$, \begin{align*} a_{i,i+1}=a_{i+1,i}=-1\quad 1\le i\le \ell-2, \end{align*} and at the terminal edge \begin{align*} a_{\ell-1,\ell} =\frac{2(\alpha_{\ell-1},\alpha_\ell)}{(\alpha_{\ell-1},\alpha_{\ell-1})} =\frac{2(-1)}{2}=-1,\qquad a_{\ell,\ell-1} =\frac{2(\alpha_\ell,\alpha_{\ell-1})}{(\alpha_\ell,\alpha_\ell)} =\frac{2(-1)}{1}=-2. \end{align*} Thus the diagram is a chain with one double bond at the end, oriented toward the short root $\alpha_\ell$. For $C_\ell$, choose instead \begin{align*} \gamma_i=e_i-e_{i+1}\quad 1\le i\le \ell-1,\qquad \gamma_\ell=2e_\ell. \end{align*} Then \begin{align*} (\gamma_i,\gamma_i)=2\quad 1\le i\le \ell-1,\qquad (\gamma_\ell,\gamma_\ell)=(2e_\ell,2e_\ell)=4. \end{align*} The non-terminal adjacent inner products are the same as before: \begin{align*} (\gamma_i,\gamma_{i+1})=-1\quad 1\le i\le \ell-2. \end{align*} At the terminal edge, \begin{align*} (\gamma_{\ell-1},\gamma_\ell) &=(e_{\ell-1}-e_\ell,2e_\ell) =0-2=-2. \end{align*} Therefore \begin{align*} a_{\ell-1,\ell} =\frac{2(-2)}{2}=-2,\qquad a_{\ell,\ell-1} =\frac{2(-2)}{4}=-1. \end{align*} Again the terminal product is $a_{\ell-1,\ell}a_{\ell,\ell-1}=2$, but now the shorter root is $\gamma_{\ell-1}$, so the double bond has the opposite long-short convention from $B_\ell$. The reflections are the same in both families: roots of the form $e_i-e_j$ interchange coordinates $i$ and $j$, roots of the form $e_i+e_j$ interchange them with signs, and the reflecting hyperplane for $e_i$ is the same as the reflecting hyperplane for $2e_i$, namely $x_i=0$. Thus both systems have the signed permutation Weyl group. In rank two, the diagrams have two vertices joined by one double bond; reversing which endpoint is called long changes $B_2$ into $C_2$, so $B_2=C_2$ as Dynkin diagrams. [/example] The example also points to a boundary of the idea. The following remark, Reducible Root Systems, records that interpretation before the construction is used again. [remark: Reducible Root Systems] A finite reduced crystallographic root system is reducible exactly when its Dynkin diagram is disconnected. Each connected component is one of the irreducible diagrams in the classification theorem, and the whole root system is the orthogonal direct sum of the corresponding irreducible subsystems. [/remark] The classification of finite root systems finishes the forward direction: every semisimple Lie algebra produces one of the allowed Dynkin types. The final step is to reverse the construction and show that the diagrammatic data is enough to rebuild the Lie algebra itself. # 9. Reconstruction from Roots This chapter closes the classification loop. Chapters 3 through 8 extracted a root system, Cartan matrix, and Dynkin diagram from a semisimple Lie algebra; now we reverse the process and rebuild the Lie algebra from that combinatorial data. The guiding question is how much algebraic structure is forced by a Cartan matrix, and the answer is that the Cartan matrix determines generators, relations, and hence the semisimple Lie algebra of the corresponding finite type. The chapter moves from internal normal forms, through generators and relations, to the final classification statement. ## Chevalley Bases and Integral Constants The first reconstruction problem is internal: suppose a semisimple Lie algebra $\mathfrak g$ is already known through its root decomposition. Can we choose root vectors so that all brackets are controlled by integers rather than arbitrary scalars? Such a choice is the bridge between the geometry of root strings and the abstract presentations used later in the chapter. If the root vectors are chosen independently, then replacing $e_\alpha$ by $c_\alpha e_\alpha$ changes the constants by ratios such as $c_\alpha c_\beta/c_{\alpha+\beta}$, so the same Lie algebra can appear to have unrelated bracket constants. The Chevalley basis theorem says that these scaling freedoms can be coordinated globally. [definition: Chevalley Basis] Let $\mathfrak g$ be a finite-dimensional semisimple Lie algebra over $k=\mathbb C$, let $\mathfrak h \subset \mathfrak g$ be a Cartan subalgebra, and let $\Phi \subset \mathfrak h^*$ be its root system. A Chevalley basis of $\mathfrak g$ is a basis consisting of elements \begin{align*} \{h_\alpha : \alpha \in \Delta\} \cup \{e_\alpha : \alpha \in \Phi\}, \end{align*} where $\Delta$ is a fixed base of simple roots, $e_\alpha \in \mathfrak g_\alpha$, and the brackets satisfy \begin{align*} [h_\alpha,h_\beta] &= 0,\\ [h_\alpha,e_\beta] &= \langle \beta,\alpha^\vee\rangle e_\beta,\\ [e_\alpha,e_{-\alpha}] &= h_\alpha \end{align*} for simple $\alpha$, with all remaining structure constants integral. [/definition] The notation $f_\alpha$ is often used for $e_{-\alpha}$ when $\alpha$ is simple. We need the existence theorem because choosing the root vectors independently does not automatically make all the rank-one pieces interact through integral root-string constants. [quotetheorem:4720] [citeproof:4720] The hypotheses are doing real work here. Semisimplicity gives a root-space decomposition with one-dimensional root spaces and embedded $\mathfrak{sl}_2$-subalgebras; without it, nilpotent radicals can introduce brackets not governed by a finite root system. Characteristic zero is also essential for the standard $\mathfrak{sl}_2$ representation theory used in the root-string argument; in small positive characteristic the same normalisation can fail or produce extra degeneracies. The theorem is not yet a classification theorem: it gives integral control inside an already known semisimple Lie algebra, and the next step is to turn that control into an abstract presentation. [example: Chevalley Basis in Sl Three] Let $\mathfrak g=\mathfrak{sl}_3(k)$ and let $\mathfrak h$ be the diagonal trace-zero subalgebra. For the simple roots $\alpha_1=\varepsilon_1-\varepsilon_2$ and $\alpha_2=\varepsilon_2-\varepsilon_3$, choose \begin{align*} e_{\alpha_1}=E_{12},\qquad e_{\alpha_2}=E_{23},\qquad e_{\alpha_1+\alpha_2}=E_{13}, \end{align*} and choose the opposite root vectors \begin{align*} e_{-\alpha_1}=E_{21},\qquad e_{-\alpha_2}=E_{32},\qquad e_{-(\alpha_1+\alpha_2)}=E_{31}. \end{align*} Using $E_{ij}E_{kl}=\delta_{jk}E_{il}$, the positive bracket is \begin{align*} [E_{12},E_{23}] &=E_{12}E_{23}-E_{23}E_{12}\\ &=\delta_{2,2}E_{13}-\delta_{3,1}E_{22}\\ &=E_{13}-0\\ &=E_{13}. \end{align*} Thus $[e_{\alpha_1},e_{\alpha_2}]=e_{\alpha_1+\alpha_2}$, so the corresponding structure constant is $1$. For the opposite roots, \begin{align*} [E_{21},E_{32}] &=E_{21}E_{32}-E_{32}E_{21}\\ &=\delta_{1,3}E_{22}-\delta_{2,2}E_{31}\\ &=0-E_{31}\\ &=-E_{31}. \end{align*} Thus $[e_{-\alpha_1},e_{-\alpha_2}]=-e_{-(\alpha_1+\alpha_2)}$, so the corresponding structure constant is $-1$. In this concrete Chevalley basis for $\mathfrak{sl}_3(k)$, the nonzero constants appearing in these root-vector brackets are integers, and reversing to the opposite root spaces changes the sign. [/example] The theorem is stronger than a convenient basis result. It says that the Lie bracket remembers only the root-string combinatorics, up to signs, once the rank-one normalisations are fixed. ## Serre Generators and Relations The next problem is external: if no ambient Lie algebra has been built yet, what relations should be imposed on symbols $e_i,f_i,h_i$ so that they reproduce the same rank-one and root-string behaviour? The Cartan matrix contains exactly the integers needed to write those relations. [definition: Generalised Cartan Matrix] An $n\times n$ integer matrix $A=(a_{ij})$ is a generalised Cartan matrix if \begin{align*} a_{ii} &= 2,\\ a_{ij} &\le 0 \quad (i\ne j),\\ a_{ij}=0 &\iff a_{ji}=0. \end{align*} It is of finite type if it is the Cartan matrix of a finite crystallographic root system. [/definition] For a base $\Delta=\{\alpha_1,\dots,\alpha_n\}$, the associated Cartan matrix is \begin{align*} a_{ij}=\langle \alpha_j,\alpha_i^\vee\rangle. \end{align*} The entry $a_{ij}$ measures how the $i$th rank-one subalgebra acts on the root vector corresponding to $\alpha_j$. We need a presentation that turns these numbers into generators and relations before any ambient semisimple Lie algebra has been constructed. [definition: Serre Presentation] Given a finite type Cartan matrix $A=(a_{ij})$, the Serre presentation is the Lie algebra generated by \begin{align*} e_i,\\ f_i,\\ h_i \qquad (1\le i\le n) \end{align*} subject to the relations \begin{align*} [h_i,h_j] &= 0,\\ [h_i,e_j] &= a_{ij}e_j,\\ [h_i,f_j] &= -a_{ij}f_j,\\ [e_i,f_j] &= \delta_{ij}h_i,\\ (\operatorname{ad} e_i)^{1-a_{ij}}(e_j) &= 0 \quad (i\ne j),\\ (\operatorname{ad} f_i)^{1-a_{ij}}(f_j) &= 0 \quad (i\ne j). \end{align*} [/definition] The first four lines say that each triple $(e_i,h_i,f_i)$ behaves like the standard generators of $\mathfrak{sl}_2$ and that the different triples see each other through the Cartan matrix. We need the theorem below because the Serre relations should be exactly sufficient, not merely necessary, for rebuilding the finite-dimensional semisimple Lie algebra. [quotetheorem:4721] [citeproof:4721] The finite-dimensional semisimple hypothesis ensures that the root strings are finite and that the integers $a_{ij}$ come from an actual crystallographic root system. If one starts from arbitrary generators satisfying only the first four relations, there is no reason for the iterated brackets $(\operatorname{ad} e_i)^m(e_j)$ to stop, so the positive part can become much too large. The Serre relations are therefore not decorative relations but exactly the truncation rules forced by the geometry of roots. This prepares the external construction, where the same relations will be imposed before any ambient Lie algebra exists. [example: The A Two Relations] For type $A_2$ the Cartan matrix is \begin{align*} A=\begin{pmatrix}2&-1\\-1&2\end{pmatrix}, \end{align*} so $a_{12}=-1$ and $a_{21}=-1$. The Serre exponent for $e_i$ acting on $e_j$ is $1-a_{ij}$ when $i\ne j$, hence \begin{align*} 1-a_{12}&=1-(-1)=2,\\ 1-a_{21}&=1-(-1)=2. \end{align*} Therefore the positive Serre relations are \begin{align*} (\operatorname{ad}e_1)^2(e_2) &=(\operatorname{ad}e_1)([e_1,e_2])\\ &=[e_1,[e_1,e_2]]\\ &=0, \end{align*} and \begin{align*} (\operatorname{ad}e_2)^2(e_1) &=(\operatorname{ad}e_2)([e_2,e_1])\\ &=[e_2,[e_2,e_1]]\\ &=0. \end{align*} The same calculation for the negative generators gives \begin{align*} (\operatorname{ad}f_1)^2(f_2)=[f_1,[f_1,f_2]]=0, \end{align*} and \begin{align*} (\operatorname{ad}f_2)^2(f_1)=[f_2,[f_2,f_1]]=0. \end{align*} Thus the positive part starts with $e_1$ and $e_2$, creates the mixed bracket $[e_1,e_2]$, and then stops because the next possible brackets $[e_1,[e_1,e_2]]$ and $[e_2,[e_2,e_1]]$ are forced to vanish. The three positive root vectors are therefore represented by $e_1$, $e_2$, and $[e_1,e_2]$, matching the three positive roots $\alpha_1$, $\alpha_2$, and $\alpha_1+\alpha_2$ of type $A_2$. [/example] This example is the template for the general construction: the Cartan matrix says how far commutators can be continued, and the Serre relations force them to stop at exactly the root-system boundary. ## Constructing the Lie Algebra from a Cartan Matrix We now face the main reconstruction question. Given a finite type Cartan matrix $A$, does the Serre presentation produce a finite-dimensional semisimple Lie algebra whose root system has Cartan matrix $A$? The existence theorem says yes, and it is the formal inverse to the earlier process of extracting roots from a semisimple Lie algebra. The finite type condition is crucial: for a generalised Cartan matrix outside finite type, the same generators and Serre relations lead to Kac-Moody algebras, which are usually infinite-dimensional. Thus the theorem is not a statement about all generalised Cartan matrices, but about the special matrices whose root combinatorics are finite. [quotetheorem:4722] [citeproof:4722] The theorem depends on finite type in the strongest possible way. If $A$ is affine or indefinite, the same presentation generally produces an infinite-dimensional Kac-Moody algebra rather than a finite-dimensional semisimple Lie algebra. The complex setting is also part of the classification statement: over non-algebraically closed fields, forms of the same complex Lie algebra can have different arithmetic behaviour. Within its hypotheses, the theorem supplies the missing existence half of classification, while uniqueness follows by comparing any other algebra with the same Chevalley generators. [remark: Uniqueness from the Cartan Matrix] For finite type matrices, the semisimple Lie algebra produced by the Serre presentation is unique up to isomorphism once the Cartan matrix is fixed. This is because any semisimple Lie algebra with that Cartan matrix receives a homomorphism from the Serre-presented algebra sending generators to Chevalley generators, and the root-space decomposition forces this map to be an isomorphism. [/remark] The smallest nontrivial test case is type $A_2$, where all of the abstract presentation can be matched directly with matrices. This example also shows how the dimension predicted by roots plus rank appears from the generators and relations. [example: Recovering Sl Three from A Two] Starting with \begin{align*} A=\begin{pmatrix}2&-1\\-1&2\end{pmatrix}, \end{align*} the Serre presentation has generators $e_1,e_2,f_1,f_2,h_1,h_2$. Since $a_{12}=a_{21}=-1$, the positive Serre relations are \begin{align*} (\operatorname{ad}e_1)^2(e_2)&=[e_1,[e_1,e_2]]=0,\\ (\operatorname{ad}e_2)^2(e_1)&=[e_2,[e_2,e_1]]=0. \end{align*} Put $x=[e_1,e_2]$. Then \begin{align*} [e_1,x]&=[e_1,[e_1,e_2]]=0,\\ [e_2,x]&=[e_2,[e_1,e_2]] =[e_2,-[e_2,e_1]] =-[e_2,[e_2,e_1]] =0. \end{align*} Thus the positive part is spanned by \begin{align*} e_1,\qquad e_2,\qquad x=[e_1,e_2]. \end{align*} Similarly, put $y=[f_2,f_1]$. The negative Serre relations give \begin{align*} [f_2,y]&=[f_2,[f_2,f_1]]=0,\\ [f_1,y]&=[f_1,[f_2,f_1]] =[f_1,-[f_1,f_2]] =-[f_1,[f_1,f_2]] =0, \end{align*} so the negative part is spanned by \begin{align*} f_1,\qquad f_2,\qquad y=[f_2,f_1]. \end{align*} Together with $h_1,h_2$, this gives \begin{align*} 3+2+3=8 \end{align*} basis elements, matching $\dim \mathfrak{sl}_3(k)=3^2-1=8$. The concrete identification in $\mathfrak{sl}_3(k)$ is \begin{align*} e_1&=E_{12}, & e_2&=E_{23},\\ f_1&=E_{21}, & f_2&=E_{32},\\ h_1&=E_{11}-E_{22}, & h_2&=E_{22}-E_{33}. \end{align*} Using $E_{ij}E_{kl}=\delta_{jk}E_{il}$, the remaining positive root vector is \begin{align*} [e_1,e_2] &=[E_{12},E_{23}]\\ &=E_{12}E_{23}-E_{23}E_{12}\\ &=\delta_{2,2}E_{13}-\delta_{3,1}E_{22}\\ &=E_{13}. \end{align*} The corresponding negative root vector is \begin{align*} [f_2,f_1] &=[E_{32},E_{21}]\\ &=E_{32}E_{21}-E_{21}E_{32}\\ &=\delta_{2,2}E_{31}-\delta_{1,3}E_{22}\\ &=E_{31}. \end{align*} Therefore the eight presented generators and brackets map to \begin{align*} E_{12},\ E_{23},\ E_{13},\ E_{21},\ E_{32},\ E_{31},\ E_{11}-E_{22},\ E_{22}-E_{33}, \end{align*} which are the standard root-vector and Cartan basis elements of $\mathfrak{sl}_3(k)$. Thus the Serre presentation for type $A_2$ reconstructs $\mathfrak{sl}_3(k)$, with the three positive root vectors corresponding to $\alpha_1$, $\alpha_2$, and $\alpha_1+\alpha_2$. [/example] The dimension count is not an accident: type $A_2$ has six roots and rank two, so reconstruction predicts dimension $6+2=8$. ## Rank-Two Reconstructions: B Two and C Two Rank two is where the relations are still visible by hand but already distinguish root lengths. The main question is how the asymmetric Cartan matrix entries control which iterated brackets survive, and how this distinguishes $B_2$ and $C_2$ as labelled root systems. [example: Recovering So Five from B Two] For type $B_2$ choose the convention in which $\alpha_1$ is short, $\alpha_2$ is long, so the Cartan matrix is \begin{align*} A=\begin{pmatrix}2&-2\\-1&2\end{pmatrix}. \end{align*} Thus $a_{12}=-2$ and $a_{21}=-1$, and the positive Serre exponents are \begin{align*} 1-a_{12}&=1-(-2)=3,\\ 1-a_{21}&=1-(-1)=2. \end{align*} Therefore the positive Serre relations are \begin{align*} (\operatorname{ad}e_1)^3(e_2) &=(\operatorname{ad}e_1)^2([e_1,e_2])\\ &=(\operatorname{ad}e_1)([e_1,[e_1,e_2]])\\ &=[e_1,[e_1,[e_1,e_2]]]\\ &=0, \end{align*} and \begin{align*} (\operatorname{ad}e_2)^2(e_1) &=(\operatorname{ad}e_2)([e_2,e_1])\\ &=[e_2,[e_2,e_1]]\\ &=0. \end{align*} Put \begin{align*} x&=[e_1,e_2],\\ z&=[e_1,x]=[e_1,[e_1,e_2]]. \end{align*} Then the first Serre relation gives \begin{align*} [e_1,z]=[e_1,[e_1,[e_1,e_2]]]=0, \end{align*} while the second gives \begin{align*} [e_2,x] &=[e_2,[e_1,e_2]]\\ &=[e_2,-[e_2,e_1]]\\ &=-[e_2,[e_2,e_1]]\\ &=0. \end{align*} Thus the positive part is generated by the four root-vector candidates \begin{align*} e_1,\qquad e_2,\qquad x=[e_1,e_2],\qquad z=[e_1,[e_1,e_2]], \end{align*} and the Serre relations stop the two possible continuations in the simple directions: \begin{align*} [e_2,x]=0,\qquad [e_1,z]=0. \end{align*} For the negative generators the same Cartan entries give \begin{align*} (\operatorname{ad}f_1)^3(f_2)&=[f_1,[f_1,[f_1,f_2]]]=0,\\ (\operatorname{ad}f_2)^2(f_1)&=[f_2,[f_2,f_1]]=0. \end{align*} Writing \begin{align*} u&=[f_1,f_2],\\ v&=[f_1,u]=[f_1,[f_1,f_2]], \end{align*} these become \begin{align*} [f_1,v]&=0,\\ [f_2,u] &=[f_2,[f_1,f_2]]\\ &=[f_2,-[f_2,f_1]]\\ &=-[f_2,[f_2,f_1]]\\ &=0. \end{align*} So the negative part is represented by \begin{align*} f_1,\qquad f_2,\qquad [f_1,f_2],\qquad [f_1,[f_1,f_2]]. \end{align*} Together with the two Cartan generators $h_1,h_2$, the reconstructed algebra has \begin{align*} 4+4+2=10 \end{align*} basis elements. This is the finite-dimensional simple Lie algebra of type $B_2$, conventionally identified over $k$ with $\mathfrak{so}_5(k)$. [/example] The asymmetric exponents in this example are the first place where root lengths affect reconstruction. Keeping track of which simple root is short determines which iterated bracket survives for one additional step. [illustration:b2-positive-root-brackets] [example: Comparing B Two and C Two] The root systems $B_2$ and $C_2$ have the same underlying Weyl group and the same number of roots, but the roles of long and short roots are interchanged. In the preceding $B_2$ convention, $\alpha_1$ was short and $\alpha_2$ was long, giving \begin{align*} \begin{pmatrix}2&-2\\-1&2\end{pmatrix}. \end{align*} If the labels are reversed so that $\alpha_1$ is long and $\alpha_2$ is short, the Cartan matrix becomes \begin{align*} A=\begin{pmatrix}2&-1\\-2&2\end{pmatrix}. \end{align*} Thus \begin{align*} a_{12}&=-1,\\ a_{21}&=-2. \end{align*} The positive Serre exponents are therefore \begin{align*} 1-a_{12}&=1-(-1)=2,\\ 1-a_{21}&=1-(-2)=3. \end{align*} So the positive Serre relations are \begin{align*} (\operatorname{ad}e_1)^2(e_2) &=(\operatorname{ad}e_1)([e_1,e_2])\\ &=[e_1,[e_1,e_2]]\\ &=0, \end{align*} and \begin{align*} (\operatorname{ad}e_2)^3(e_1) &=(\operatorname{ad}e_2)^2([e_2,e_1])\\ &=(\operatorname{ad}e_2)([e_2,[e_2,e_1]])\\ &=[e_2,[e_2,[e_2,e_1]]]\\ &=0. \end{align*} Put \begin{align*} x&=[e_2,e_1],\\ z&=[e_2,x]=[e_2,[e_2,e_1]]. \end{align*} Then the second Serre relation gives \begin{align*} [e_2,z]=[e_2,[e_2,[e_2,e_1]]]=0. \end{align*} The first Serre relation also says \begin{align*} [e_1,[e_1,e_2]]=0. \end{align*} Since $[e_2,e_1]=-[e_1,e_2]$ by skew-symmetry of the Lie bracket, this is the same stopping rule in the other order: \begin{align*} [e_1,x] &=[e_1,[e_2,e_1]]\\ &=[e_1,-[e_1,e_2]]\\ &=-[e_1,[e_1,e_2]]\\ &=0. \end{align*} Thus, under this $C_2$ labelling, the positive root-vector candidates are \begin{align*} e_1,\qquad e_2,\qquad [e_2,e_1],\qquad [e_2,[e_2,e_1]], \end{align*} whereas in the preceding $B_2$ labelling they were \begin{align*} e_1,\qquad e_2,\qquad [e_1,e_2],\qquad [e_1,[e_1,e_2]]. \end{align*} The surviving double bracket has moved from the $e_1$-direction to the $e_2$-direction. The negative Serre relations swap in the same way: \begin{align*} (\operatorname{ad}f_1)^2(f_2)&=[f_1,[f_1,f_2]]=0,\\ (\operatorname{ad}f_2)^3(f_1)&=[f_2,[f_2,[f_2,f_1]]]=0. \end{align*} So the labelled root data remember which simple root is long and which is short, even though over the complex numbers the reconstructed Lie algebras of types $B_2$ and $C_2$ are isomorphic by the exceptional low-rank isomorphism $\mathfrak{so}_5(k)\cong\mathfrak{sp}_4(k)$. [/example] This comparison is a useful warning about classification data. A Dynkin diagram without labels may hide the choice of root lengths, while the Cartan matrix records both the incidence and the direction of the multiple edge. ## The Classification Loop The reconstruction theorem completes the central argument of the course. Starting from a finite-dimensional semisimple Lie algebra, one chooses a Cartan subalgebra, obtains a crystallographic root system, chooses simple roots, and forms a finite type Cartan matrix. Starting from that Cartan matrix, the Serre presentation reconstructs a semisimple Lie algebra with the same root data. [explanation: From Geometry to Presentation] The Chevalley basis theorem translates geometric root data into integral bracket constants. The Serre relations then compress the bracket table further: instead of listing all brackets between all root spaces, it is enough to keep the simple generators and the string-length relations imposed by the Cartan matrix. This is why Dynkin diagrams classify semisimple Lie algebras over the complex numbers. The diagram determines the Cartan matrix up to the standard conventions on arrows and root lengths; the Cartan matrix determines the Serre presentation; and the presentation determines the Lie algebra. [/explanation] The final theorem packages the two directions into a single classification statement. Its force is that no extra continuous parameter remains after the Dynkin diagram has been fixed. [quotetheorem:4723] [citeproof:4723] The restrictions in the statement mark the boundary of the theorem. If the Lie algebra is reductive rather than semisimple, an additional central torus is present and the Dynkin diagram records only the derived semisimple part; if the base field is not algebraically closed, real and arithmetic forms are not distinguished by the finite type diagram alone. If the Cartan matrix is not finite type, the reconstruction process leaves the finite-dimensional classification and enters Kac-Moody theory. The forward connection is that Chevalley bases and integral structure constants make it possible to build integral forms and Chevalley groups, so the same reconstruction data later reappears in representation theory, algebraic groups, and arithmetic geometry. The practical outcome is that the classification has no remaining hidden analytic or geometric input. All finite-dimensional semisimple Lie algebras over $k$ arise from the finite type Cartan matrices, and the exceptional types appear because their Cartan matrices satisfy the same finite root-system axioms as the classical families. This completes the existence direction of the classification loop. The final chapter revisits the same data from the uniqueness side, making explicit why the Cartan matrix and root data determine the isomorphism class rather than merely producing examples. # 10. Uniqueness and the Classification Theorem The preceding chapters built the classification in pieces: extract a root system from a semisimple Lie algebra, classify the finite root systems by Dynkin diagrams, and reconstruct an algebra from finite type Cartan data. The final task is to check that these pieces really give a classification theorem rather than a list of examples. This chapter explains why the Cartan matrix determines the isomorphism class, why connected diagrams correspond to simple algebras, and how the classical and exceptional families exhaust the finite-dimensional semisimple case. ## Recovering a Lie Algebra from Its Cartan Matrix The first problem is uniqueness: if two semisimple Lie algebras have the same root-theoretic data, why must they be isomorphic? The roots record weights for a chosen Cartan subalgebra, but the Lie bracket also involves constants between root spaces. The content of the uniqueness theorem is that these constants are forced, up to choices of root vectors, by the Cartan matrix. Let $L$ be a finite-dimensional semisimple Lie algebra over $k=\mathbb C$, let $H \subseteq L$ be a Cartan subalgebra, and let $\Phi \subseteq H^*$ be the corresponding root system. Choose a base of simple roots $\Delta = \{\alpha_1,\dots,\alpha_\ell\}$. [definition: Cartan Matrix of a Root System] The Cartan matrix associated to $\Delta$ is the integer matrix $A = (a_{ij})_{1 \le i,j \le \ell}$ defined by \begin{align*} a_{ij} = \langle \alpha_j, \alpha_i^\vee \rangle = \frac{2(\alpha_j,\alpha_i)}{(\alpha_i,\alpha_i)}. \end{align*} [/definition] The ordering convention matters: the row indexed by $i$ records pairings with the coroot $\alpha_i^\vee$. The matrix has diagonal entries $2$, off-diagonal entries in $\mathbb Z_{\le 0}$, and $a_{ij}=0$ iff $a_{ji}=0$. [example: Type A Two Cartan Matrix] For the root system $A_2$, choose simple roots $\alpha_1,\alpha_2$ with equal squared length $(\alpha_1,\alpha_1)=(\alpha_2,\alpha_2)=r^2$ and angle $2\pi/3$. Since $\cos(2\pi/3)=-1/2$, their inner product is \begin{align*} (\alpha_1,\alpha_2)=|\alpha_1||\alpha_2|\cos(2\pi/3)=r^2\left(-\frac12\right)=-\frac{r^2}{2}. \end{align*} Using the convention $a_{ij}=2(\alpha_j,\alpha_i)/(\alpha_i,\alpha_i)$, the four entries are \begin{align*} a_{11}&=\frac{2(\alpha_1,\alpha_1)}{(\alpha_1,\alpha_1)}=2,\\ a_{12}&=\frac{2(\alpha_2,\alpha_1)}{(\alpha_1,\alpha_1)} =\frac{2(-r^2/2)}{r^2}=-1,\\ a_{21}&=\frac{2(\alpha_1,\alpha_2)}{(\alpha_2,\alpha_2)} =\frac{2(-r^2/2)}{r^2}=-1,\\ a_{22}&=\frac{2(\alpha_2,\alpha_2)}{(\alpha_2,\alpha_2)}=2. \end{align*} Hence \begin{align*} A=\begin{pmatrix}2&-1\\-1&2\end{pmatrix}. \end{align*} The $\alpha_1$-string through $\alpha_2$ is $\alpha_2,\alpha_2+\alpha_1$: the element $\alpha_2-\alpha_1$ is not a root, while $\alpha_1+\alpha_2$ is the third positive root of $A_2$. Similarly, the $\alpha_2$-string through $\alpha_1$ is $\alpha_1,\alpha_1+\alpha_2$. Thus each simple root acts on the other by a string of two roots, and this is the Cartan matrix that later appears in the Serre presentation of $\mathfrak{sl}_3$. [/example] To pass from a Cartan matrix to a diagram, one suppresses redundant numerical information while remembering the relative root lengths and angles. We need this compressed object because the classification is stated graphically, and the graph must still retain enough data to recover the Cartan matrix. [definition: Dynkin Diagram] The Dynkin diagram of a base $\Delta = \{\alpha_1,\dots,\alpha_\ell\}$ is the graph with one vertex for each $\alpha_i$, with $a_{ij}a_{ji}$ edges between vertices $i$ and $j$ when $i \ne j$, and with an arrow pointing toward the shorter root when $a_{ij}a_{ji} > 1$. [/definition] Thus no edge means orthogonality, a single edge means angle $2\pi/3$ with equal lengths, a double edge means a $B_2/C_2$ pair, and a triple edge means a $G_2$ pair. We need the reconstruction theorem because, after the convention on arrow direction is fixed, the diagram determines the matrix and should determine the Lie algebra. [quotetheorem:4724] [citeproof:4724] The theorem explains why the Cartan matrix is stronger than an invariant: it is a presentation datum. Once the simple-root generators are fixed, all higher root spaces are generated by iterated brackets, and the Serre relations prescribe precisely when these iterated brackets vanish. [remark: Choice of Cartan Subalgebra] The criterion is independent of the Cartan subalgebra and base of simple roots. Cartan subalgebras are conjugate, and different bases of the same root system are related by the Weyl group. Hence the Dynkin diagram attached to a semisimple Lie algebra is well-defined up to graph isomorphism. [/remark] ## Simplicity and Irreducibility The next problem is to recognise simple Lie algebras from their diagrams. Semisimple Lie algebras split as direct sums of simple ideals, while root systems may split as orthogonal unions. The key compatibility is that these two decompositions are the same phenomenon seen on different sides of the Cartan decomposition. [definition: Reducible Root System] A root system $\Phi$ in a Euclidean space $E$ is reducible if there exist nonempty root systems $\Phi_1$ and $\Phi_2$ in orthogonal subspaces of $E$ such that \begin{align*} \Phi = \Phi_1 \sqcup \Phi_2. \end{align*} It is irreducible if it is not reducible. [/definition] After the definition, the role of orthogonality is worth emphasising: roots in different components have zero Cartan integers against each other, so their root spaces commute in the semisimple Lie algebra. We need the theorem below to identify this diagrammatic disconnectedness with the algebraic decomposition into simple ideals. [quotetheorem:4725] [citeproof:4725] This theorem turns the classification of simple Lie algebras into the classification of connected Dynkin diagrams. For semisimple algebras, disconnected diagrams are allowed; each connected component corresponds to one simple ideal. [example: Direct Sum and Disconnected Diagram] Let the two summands have standard basis $e_i,f_i,h_i$ with $[h_i,e_i]=2e_i$, $[h_i,f_i]=-2f_i$, and $[e_i,f_i]=h_i$ for $i=1,2$. For the Cartan subalgebra $H=kh_1\oplus kh_2$, define $\alpha,\beta\in H^*$ by \begin{align*} \alpha(t h_1+s h_2)=2t,\qquad \beta(t h_1+s h_2)=2s. \end{align*} Then \begin{align*} [t h_1+s h_2,e_1]&=2t e_1=\alpha(t h_1+s h_2)e_1,\\ [t h_1+s h_2,f_1]&=-2t f_1=-\alpha(t h_1+s h_2)f_1,\\ [t h_1+s h_2,e_2]&=2s e_2=\beta(t h_1+s h_2)e_2,\\ [t h_1+s h_2,f_2]&=-2s f_2=-\beta(t h_1+s h_2)f_2. \end{align*} Thus the roots are $\{\pm\alpha,\pm\beta\}$, with one $A_1$ pair coming from each copy of $\mathfrak{sl}_2$. The cross-brackets vanish because the bracket in a direct sum is componentwise: \begin{align*} [(x,0),(0,y)]=([x,0],[0,y])=(0,0). \end{align*} Also $\alpha(h_2)=0$ and $\beta(h_1)=0$, so the Cartan matrix for the simple roots $\alpha,\beta$ is \begin{align*} \begin{pmatrix} 2&0\\ 0&2 \end{pmatrix}. \end{align*} Hence the Dynkin diagram has two isolated vertices, so the root system is $A_1\sqcup A_1$. The algebra is semisimple as a direct sum of two simple summands, but it is not simple because either summand is a nonzero proper ideal. [/example] The product example shows how disconnected diagrams encode direct sums. The next computation contrasts it with a connected chain, where the same root-space bookkeeping detects simplicity rather than decomposition. [example: Type A Three Is Irreducible] For $L=\mathfrak{sl}_4(k)$, let \begin{align*} H=\{\operatorname{diag}(t_1,t_2,t_3,t_4):t_1+t_2+t_3+t_4=0\}. \end{align*} Write $\varepsilon_i(\operatorname{diag}(t_1,t_2,t_3,t_4))=t_i$. If $E_{ij}$ is the matrix with $1$ in the $(i,j)$ entry and $0$ elsewhere, then for $h=\operatorname{diag}(t_1,t_2,t_3,t_4)$ with trace zero, \begin{align*} [h,E_{ij}] &=hE_{ij}-E_{ij}h\\ &=t_iE_{ij}-t_jE_{ij}\\ &=(t_i-t_j)E_{ij}\\ &=(\varepsilon_i-\varepsilon_j)(h)E_{ij}. \end{align*} Thus the root space spanned by $E_{ij}$ has root $\varepsilon_i-\varepsilon_j$, so \begin{align*} \Phi=\{\varepsilon_i-\varepsilon_j:i\ne j\}. \end{align*} Choose \begin{align*} \alpha_1=\varepsilon_1-\varepsilon_2,\qquad \alpha_2=\varepsilon_2-\varepsilon_3,\qquad \alpha_3=\varepsilon_3-\varepsilon_4. \end{align*} Using the standard inner product on the hyperplane $\sum_i x_i=0$, the squared lengths and adjacent inner products are \begin{align*} (\alpha_i,\alpha_i)&=(\varepsilon_i-\varepsilon_{i+1},\varepsilon_i-\varepsilon_{i+1})=1+1=2,\\ (\alpha_1,\alpha_2)&=(\varepsilon_1-\varepsilon_2,\varepsilon_2-\varepsilon_3)=-1,\\ (\alpha_2,\alpha_3)&=(\varepsilon_2-\varepsilon_3,\varepsilon_3-\varepsilon_4)=-1,\\ (\alpha_1,\alpha_3)&=(\varepsilon_1-\varepsilon_2,\varepsilon_3-\varepsilon_4)=0. \end{align*} Therefore the Cartan entries $a_{ij}=2(\alpha_j,\alpha_i)/(\alpha_i,\alpha_i)$ give \begin{align*} A= \begin{pmatrix} 2&-1&0\\ -1&2&-1\\ 0&-1&2 \end{pmatrix}. \end{align*} The nonzero off-diagonal entries connect $\alpha_1$ to $\alpha_2$ and $\alpha_2$ to $\alpha_3$, while $\alpha_1$ and $\alpha_3$ are not connected. Hence the Dynkin diagram is the connected chain $A_3$, so the root system is irreducible. By *Simple Lie Algebras Have Irreducible Root Systems*, $\mathfrak{sl}_4(k)$ is simple. [/example] ## The Finite Type Dynkin Diagrams Having reduced simplicity to connectedness, the remaining combinatorial question is which connected diagrams actually come from finite root systems. The rank two restrictions already show that multiple edges are tightly controlled, and positive definiteness rules out cycles and most branching. The final list is finite in each rank but includes four infinite families. [quotetheorem:4719] [citeproof:4719] This theorem is the combinatorial spine of the classification. It does not yet assert that every diagram is realised by a Lie algebra, nor that the realisation is unique; those are supplied by existence through Serre relations and the uniqueness theorem above. [example: Rank Two Possibilities] Let $\alpha_1,\alpha_2$ be simple roots in a rank two root system, and write $\theta$ for the angle between them. From the Cartan matrix convention, \begin{align*} a_{12}a_{21} &=\frac{2(\alpha_2,\alpha_1)}{(\alpha_1,\alpha_1)} \frac{2(\alpha_1,\alpha_2)}{(\alpha_2,\alpha_2)}\\ &=\frac{4(\alpha_1,\alpha_2)^2}{(\alpha_1,\alpha_1)(\alpha_2,\alpha_2)}\\ &=4\cos^2\theta. \end{align*} Since $a_{12},a_{21}\in\mathbb Z_{\le 0}$, the product $a_{12}a_{21}$ is a nonnegative integer. The Gram matrix of $\alpha_1,\alpha_2$ is \begin{align*} G= \begin{pmatrix} (\alpha_1,\alpha_1)&(\alpha_1,\alpha_2)\\ (\alpha_2,\alpha_1)&(\alpha_2,\alpha_2) \end{pmatrix}, \end{align*} so positive definiteness gives \begin{align*} 0<\det G &=(\alpha_1,\alpha_1)(\alpha_2,\alpha_2)-(\alpha_1,\alpha_2)^2\\ &=(\alpha_1,\alpha_1)(\alpha_2,\alpha_2)\left(1-\frac{a_{12}a_{21}}{4}\right). \end{align*} Because $(\alpha_i,\alpha_i)>0$, this implies $a_{12}a_{21}<4$. Hence the only possible products are $0,1,2,3$. These products give the four rank two finite possibilities: \begin{align*} a_{12}a_{21}=0&:\quad \begin{pmatrix}2&0\\0&2\end{pmatrix} \quad\text{of type }A_1\times A_1,\\ a_{12}a_{21}=1&:\quad \begin{pmatrix}2&-1\\-1&2\end{pmatrix} \quad\text{of type }A_2,\\ a_{12}a_{21}=2&:\quad \begin{pmatrix}2&-2\\-1&2\end{pmatrix} \quad\text{or its transpose, of type }B_2=C_2,\\ a_{12}a_{21}=3&:\quad \begin{pmatrix}2&-3\\-1&2\end{pmatrix} \quad\text{or its transpose, of type }G_2. \end{align*} Thus rank two already forces the allowed edge multiplicities in a finite Dynkin diagram: no edge, single edge, double edge, or triple edge, and nothing larger. [/example] The example also points to a boundary of the idea. The following remark, Why Cycles Do Not Occur, records that interpretation before the construction is used again. [remark: Why Cycles Do Not Occur] A finite type Dynkin diagram has an associated positive definite quadratic form. If a simply-laced diagram contained a cycle, assigning the same nonzero coefficient to every vertex around the cycle and zero elsewhere would produce a nonpositive value for the corresponding form. Multiple edges make the form even more restrictive, so finite type diagrams are trees. [/remark] ## The Classification Theorem We can now state the endpoint of the course. Starting with a semisimple Lie algebra, Chapters 2 through 8 give a Dynkin diagram; starting with a finite type Dynkin diagram, Chapter 9's Serre construction gives a semisimple Lie algebra; the uniqueness theorem says that returning to root data loses no information. [quotetheorem:4727] [citeproof:4727] For semisimple algebras, the same statement is obtained by allowing finite disjoint unions of finite type connected diagrams. A semisimple Lie algebra is then the direct sum of the simple Lie algebras attached to the connected components, with repetitions allowed. This passage from connected diagrams to disjoint unions is not a new construction but the diagrammatic shadow of direct sums: roots from different components are orthogonal and their root spaces commute. It also explains why the simple classification is the essential case; semisimple classification adds only multiplicities and component bookkeeping. [quotetheorem:4728] [citeproof:4728] This is the promised complex classification. The hypotheses are part of the statement: over $\mathbb C$, Cartan subalgebras split, root-space decompositions are available, complete reducibility holds, and the Serre construction produces the required finite-dimensional algebras. The result is also a uniqueness statement, not merely an existence list: two semisimple Lie algebras with the same finite Dynkin diagram have the same simple summands and hence are isomorphic. Conversely, every diagram on the list is realized by the construction from its Cartan matrix, so the combinatorial classification and the Lie-algebra classification match exactly. ## Classical Lie Algebras in the Classification The abstract list becomes more recognisable when matched with the classical matrix Lie algebras. These examples also fix the indexing conventions: type $A_{n-1}$ corresponds to $\mathfrak{sl}_n$, while the orthogonal and symplectic families split into the $B$, $C$, and $D$ series. [example: Special Linear Algebras Have Type A] Let $L=\mathfrak{sl}_n(k)$ for $n\ge 2$, and let \begin{align*} H=\{\operatorname{diag}(t_1,\dots,t_n):t_1+\cdots+t_n=0\}. \end{align*} Write $\varepsilon_i(\operatorname{diag}(t_1,\dots,t_n))=t_i$. If $E_{ij}$ is the matrix with $1$ in the $(i,j)$ entry and $0$ elsewhere, then for $h=\operatorname{diag}(t_1,\dots,t_n)\in H$ and $i\ne j$, \begin{align*} [h,E_{ij}] &=hE_{ij}-E_{ij}h\\ &=t_iE_{ij}-t_jE_{ij}\\ &=(t_i-t_j)E_{ij}\\ &=(\varepsilon_i-\varepsilon_j)(h)E_{ij}. \end{align*} Thus $kE_{ij}$ is the root space of root $\varepsilon_i-\varepsilon_j$, so \begin{align*} \Phi=\{\varepsilon_i-\varepsilon_j:i\ne j\}. \end{align*} Choose simple roots \begin{align*} \alpha_i=\varepsilon_i-\varepsilon_{i+1}\qquad 1\le i\le n-1. \end{align*} Using the standard inner product on the hyperplane $\sum_i x_i=0$, we have \begin{align*} (\alpha_i,\alpha_i) &=(\varepsilon_i-\varepsilon_{i+1},\varepsilon_i-\varepsilon_{i+1})\\ &=(\varepsilon_i,\varepsilon_i)-2(\varepsilon_i,\varepsilon_{i+1})+(\varepsilon_{i+1},\varepsilon_{i+1})\\ &=1-0+1\\ &=2. \end{align*} For adjacent simple roots, \begin{align*} (\alpha_i,\alpha_{i+1}) &=(\varepsilon_i-\varepsilon_{i+1},\varepsilon_{i+1}-\varepsilon_{i+2})\\ &=(\varepsilon_i,\varepsilon_{i+1})-(\varepsilon_i,\varepsilon_{i+2}) -(\varepsilon_{i+1},\varepsilon_{i+1})+(\varepsilon_{i+1},\varepsilon_{i+2})\\ &=0-0-1+0\\ &=-1. \end{align*} If $|i-j|>1$, then the indices $i,i+1,j,j+1$ are distinct, so each pairing between one $\varepsilon$ from $\alpha_i$ and one $\varepsilon$ from $\alpha_j$ is zero, and hence \begin{align*} (\alpha_i,\alpha_j)=0. \end{align*} Therefore the Cartan entries $a_{ij}=2(\alpha_j,\alpha_i)/(\alpha_i,\alpha_i)$ are \begin{align*} a_{ii}&=\frac{2(\alpha_i,\alpha_i)}{(\alpha_i,\alpha_i)}=2,\\ a_{i,i+1}&=\frac{2(\alpha_{i+1},\alpha_i)}{(\alpha_i,\alpha_i)} =\frac{2(-1)}{2}=-1,\\ a_{i+1,i}&=\frac{2(\alpha_i,\alpha_{i+1})}{(\alpha_{i+1},\alpha_{i+1})} =\frac{2(-1)}{2}=-1, \end{align*} and $a_{ij}=0$ when $|i-j|>1$. Thus the Dynkin diagram has one single edge between consecutive vertices $\alpha_i$ and $\alpha_{i+1}$, and no edge between nonconsecutive vertices. This is the chain with $n-1$ vertices, so $\mathfrak{sl}_n(k)$ has type $A_{n-1}$. [/example] The special linear family gives the baseline chain diagrams. The symplectic family keeps a chain but changes the final Cartan integers, so it is the first place where the arrowed double bond appears. [example: Symplectic Algebras Have Type C] Let $L=\mathfrak{sp}_{2n}(k)$ for $n\ge 2$, and take the standard Cartan subalgebra \begin{align*} H=\{\operatorname{diag}(t_1,\dots,t_n,-t_1,\dots,-t_n):t_i\in k\}. \end{align*} Write $\varepsilon_i(h)=t_i$. For a diagonal matrix $h=\operatorname{diag}(d_1,\dots,d_{2n})$ and a matrix unit $E_{ab}$, \begin{align*} [h,E_{ab}] &=hE_{ab}-E_{ab}h\\ &=d_aE_{ab}-d_bE_{ab}\\ &=(d_a-d_b)E_{ab}. \end{align*} Since the diagonal entries are $t_1,\dots,t_n,-t_1,\dots,-t_n$, the possible nonzero weights occurring in $\mathfrak{sp}_{2n}(k)$ are \begin{align*} \Phi=\{\pm\varepsilon_i\pm\varepsilon_j:i\ne j\}\cup\{\pm2\varepsilon_i:1\le i\le n\}. \end{align*} Choose simple roots \begin{align*} \alpha_i=\varepsilon_i-\varepsilon_{i+1}\quad(1\le i<n), \qquad \alpha_n=2\varepsilon_n. \end{align*} Using the standard inner product with $(\varepsilon_i,\varepsilon_j)=\delta_{ij}$, we have \begin{align*} (\alpha_i,\alpha_i) &=(\varepsilon_i-\varepsilon_{i+1},\varepsilon_i-\varepsilon_{i+1})\\ &=1-0-0+1\\ &=2 \end{align*} for $1\le i<n$, while \begin{align*} (\alpha_n,\alpha_n)=(2\varepsilon_n,2\varepsilon_n)=4. \end{align*} For $1\le i<n-1$, \begin{align*} (\alpha_i,\alpha_{i+1}) &=(\varepsilon_i-\varepsilon_{i+1},\varepsilon_{i+1}-\varepsilon_{i+2})\\ &=0-0-1+0\\ &=-1. \end{align*} At the final edge, \begin{align*} (\alpha_{n-1},\alpha_n) &=(\varepsilon_{n-1}-\varepsilon_n,2\varepsilon_n)\\ &=0-2\\ &=-2. \end{align*} Nonconsecutive simple roots have disjoint indices, except for no shared $\varepsilon$ term, so their inner product is $0$. Therefore the Cartan entries $a_{ij}=2(\alpha_j,\alpha_i)/(\alpha_i,\alpha_i)$ satisfy \begin{align*} a_{ii}&=2,\\ a_{i,i+1}&=\frac{2(-1)}{2}=-1, \qquad a_{i+1,i}=\frac{2(-1)}{2}=-1 \quad(1\le i<n-1),\\ a_{n-1,n}&=\frac{2(-2)}{2}=-2,\\ a_{n,n-1}&=\frac{2(-2)}{4}=-1, \end{align*} and all other off-diagonal entries are $0$. Thus the Dynkin diagram is a chain: all edges are single except the final edge, where $a_{n-1,n}a_{n,n-1}=2$. Since $(\alpha_{n-1},\alpha_{n-1})=2$ and $(\alpha_n,\alpha_n)=4$, the shorter final-adjacent root is $\alpha_{n-1}$, so the double edge points toward $\alpha_{n-1}$. Hence $\mathfrak{sp}_{2n}(k)$ has type $C_n$. [/example] The symplectic computation identified type $C_n$ by the direction of its double bond. The odd orthogonal computation gives the complementary type $B_n$, where the same coordinate method produces the opposite length convention. [example: Odd Orthogonal Algebras Have Type B] Let $L=\mathfrak{so}_{2n+1}(k)$ for $n\ge 2$, with respect to a nondegenerate symmetric form, and let $H$ be the standard diagonal Cartan subalgebra. Write $\varepsilon_i\in H^*$ for the functional extracting the $i$th diagonal parameter. The roots are \begin{align*} \Phi = \{\pm \varepsilon_i \pm \varepsilon_j : i\ne j\}\cup\{\pm \varepsilon_i : 1\le i\le n\}. \end{align*} Choose \begin{align*} \alpha_i=\varepsilon_i-\varepsilon_{i+1}\quad(1\le i<n), \qquad \alpha_n=\varepsilon_n. \end{align*} These are simple roots for the usual positive system: for $i<j$, \begin{align*} \varepsilon_i-\varepsilon_j &=(\varepsilon_i-\varepsilon_{i+1})+(\varepsilon_{i+1}-\varepsilon_{i+2})+\cdots+(\varepsilon_{j-1}-\varepsilon_j)\\ &=\alpha_i+\alpha_{i+1}+\cdots+\alpha_{j-1}, \end{align*} and \begin{align*} \varepsilon_i &=(\varepsilon_i-\varepsilon_{i+1})+\cdots+(\varepsilon_{n-1}-\varepsilon_n)+\varepsilon_n\\ &=\alpha_i+\cdots+\alpha_{n-1}+\alpha_n. \end{align*} Therefore \begin{align*} \varepsilon_i+\varepsilon_j &=(\varepsilon_i-\varepsilon_j)+2\varepsilon_j\\ &=(\alpha_i+\cdots+\alpha_{j-1})+ 2(\alpha_j+\cdots+\alpha_{n-1}+\alpha_n), \end{align*} so every positive root is a nonnegative integer combination of the chosen $\alpha_i$. Using the standard inner product $(\varepsilon_i,\varepsilon_j)=\delta_{ij}$, the squared lengths are \begin{align*} (\alpha_i,\alpha_i) &=(\varepsilon_i-\varepsilon_{i+1},\varepsilon_i-\varepsilon_{i+1})\\ &=1-0-0+1\\ &=2 \end{align*} for $1\le i<n$, while \begin{align*} (\alpha_n,\alpha_n)=(\varepsilon_n,\varepsilon_n)=1. \end{align*} For $1\le i<n-1$, \begin{align*} (\alpha_i,\alpha_{i+1}) &=(\varepsilon_i-\varepsilon_{i+1},\varepsilon_{i+1}-\varepsilon_{i+2})\\ &=0-0-1+0\\ &=-1. \end{align*} At the final edge, \begin{align*} (\alpha_{n-1},\alpha_n) &=(\varepsilon_{n-1}-\varepsilon_n,\varepsilon_n)\\ &=0-1\\ &=-1. \end{align*} If $|i-j|>1$ and neither root is the final pair just computed, the two expressions involve disjoint $\varepsilon$-indices, so every term in the expansion has inner product $0$ and hence $(\alpha_i,\alpha_j)=0$. Now compute the Cartan entries $a_{ij}=2(\alpha_j,\alpha_i)/(\alpha_i,\alpha_i)$. The diagonal entries are \begin{align*} a_{ii}=\frac{2(\alpha_i,\alpha_i)}{(\alpha_i,\alpha_i)}=2. \end{align*} For $1\le i<n-1$, \begin{align*} a_{i,i+1} &=\frac{2(\alpha_{i+1},\alpha_i)}{(\alpha_i,\alpha_i)} =\frac{2(-1)}{2} =-1,\\ a_{i+1,i} &=\frac{2(\alpha_i,\alpha_{i+1})}{(\alpha_{i+1},\alpha_{i+1})} =\frac{2(-1)}{2} =-1. \end{align*} For the final adjacent pair, \begin{align*} a_{n-1,n} &=\frac{2(\alpha_n,\alpha_{n-1})}{(\alpha_{n-1},\alpha_{n-1})} =\frac{2(-1)}{2} =-1,\\ a_{n,n-1} &=\frac{2(\alpha_{n-1},\alpha_n)}{(\alpha_n,\alpha_n)} =\frac{2(-1)}{1} =-2. \end{align*} All other off-diagonal entries are $0$. Thus the Dynkin diagram is a chain with single edges until the final pair, where \begin{align*} a_{n-1,n}a_{n,n-1}=(-1)(-2)=2. \end{align*} Since $(\alpha_n,\alpha_n)=1<2=(\alpha_{n-1},\alpha_{n-1})$, the final root $\alpha_n$ is shorter, so the double edge points toward $\alpha_n$. Hence $\mathfrak{so}_{2n+1}(k)$ has type $B_n$. [/example] Odd orthogonal algebras produce the non-simply-laced $B_n$ chain. Even orthogonal algebras instead produce a simply laced fork, so the final comparison separates the two orthogonal families diagrammatically. [example: Even Orthogonal Algebras Have Type D] Let $L=\mathfrak{so}_{2n}(k)$ for $n\ge 4$, with its standard diagonal Cartan subalgebra. Write $\varepsilon_i$ for the functional extracting the $i$th diagonal parameter. The roots are \begin{align*} \Phi=\{\pm\varepsilon_i\pm\varepsilon_j:i\ne j\}. \end{align*} Choose \begin{align*} \alpha_i=\varepsilon_i-\varepsilon_{i+1}\quad(1\le i\le n-1), \qquad \alpha_n=\varepsilon_{n-1}+\varepsilon_n. \end{align*} These are the usual simple roots: for $i<j$, \begin{align*} \varepsilon_i-\varepsilon_j &=(\varepsilon_i-\varepsilon_{i+1})+(\varepsilon_{i+1}-\varepsilon_{i+2})+\cdots+(\varepsilon_{j-1}-\varepsilon_j)\\ &=\alpha_i+\alpha_{i+1}+\cdots+\alpha_{j-1}, \end{align*} while for $i<j<n$, \begin{align*} \varepsilon_i+\varepsilon_j &=(\varepsilon_i-\varepsilon_{i+1})+\cdots+(\varepsilon_{j-1}-\varepsilon_j)+2\varepsilon_j\\ &=\alpha_i+\cdots+\alpha_{j-1}+2(\alpha_j+\cdots+\alpha_{n-2})+\alpha_{n-1}+\alpha_n. \end{align*} For $j=n$, this becomes \begin{align*} \varepsilon_i+\varepsilon_n &=(\varepsilon_i-\varepsilon_{i+1})+\cdots+(\varepsilon_{n-2}-\varepsilon_{n-1})+(\varepsilon_{n-1}+\varepsilon_n)\\ &=\alpha_i+\cdots+\alpha_{n-2}+\alpha_n. \end{align*} Thus every positive root is a nonnegative integer combination of the chosen $\alpha_i$. Using the standard inner product $(\varepsilon_i,\varepsilon_j)=\delta_{ij}$, every simple root has squared length $2$: \begin{align*} (\alpha_i,\alpha_i) &=(\varepsilon_i-\varepsilon_{i+1},\varepsilon_i-\varepsilon_{i+1})\\ &=1-0-0+1\\ &=2 \end{align*} for $1\le i\le n-1$, and \begin{align*} (\alpha_n,\alpha_n) &=(\varepsilon_{n-1}+\varepsilon_n,\varepsilon_{n-1}+\varepsilon_n)\\ &=1+0+0+1\\ &=2. \end{align*} For $1\le i<n-2$, \begin{align*} (\alpha_i,\alpha_{i+1}) &=(\varepsilon_i-\varepsilon_{i+1},\varepsilon_{i+1}-\varepsilon_{i+2})\\ &=0-0-1+0\\ &=-1. \end{align*} The two final attachments to $\alpha_{n-2}$ are \begin{align*} (\alpha_{n-2},\alpha_{n-1}) &=(\varepsilon_{n-2}-\varepsilon_{n-1},\varepsilon_{n-1}-\varepsilon_n)\\ &=0-0-1+0\\ &=-1,\\ (\alpha_{n-2},\alpha_n) &=(\varepsilon_{n-2}-\varepsilon_{n-1},\varepsilon_{n-1}+\varepsilon_n)\\ &=0+0-1-0\\ &=-1. \end{align*} The last two simple roots are orthogonal: \begin{align*} (\alpha_{n-1},\alpha_n) &=(\varepsilon_{n-1}-\varepsilon_n,\varepsilon_{n-1}+\varepsilon_n)\\ &=1+0-0-1\\ &=0. \end{align*} All other nonconsecutive pairs have disjoint $\varepsilon$-indices, so each term in their inner product expansion is $0$. Now compute $a_{ij}=2(\alpha_j,\alpha_i)/(\alpha_i,\alpha_i)$. Since every $(\alpha_i,\alpha_i)=2$, we have \begin{align*} a_{ii}&=2,\\ a_{ij}&=(\alpha_j,\alpha_i) \end{align*} for $i\ne j$. Hence $a_{ij}=-1$ exactly for the adjacent pairs \begin{align*} (1,2),(2,3),\dots,(n-3,n-2),(n-2,n-1),(n-2,n), \end{align*} and $a_{ij}=0$ for all other off-diagonal pairs. Therefore the Dynkin diagram is a chain up to $\alpha_{n-2}$, with two single edges leaving $\alpha_{n-2}$ toward $\alpha_{n-1}$ and $\alpha_n$, and no edge between $\alpha_{n-1}$ and $\alpha_n$. This is the forked diagram $D_n$, so $\mathfrak{so}_{2n}(k)$ has type $D_n$. [/example] The exceptional types $E_6,E_7,E_8,F_4,G_2$ are not realised as the four standard infinite matrix families, although they can be constructed by Serre generators and relations or by more geometric models. Their existence is not an anomaly: it is forced by the same finite-type diagram classification. ## Reading the Theorem Backwards The classification theorem is often most useful in reverse. Instead of beginning with a Lie algebra and computing its diagram, one may begin with a diagram and know the rank, root lengths, possible subalgebras generated by pairs of simple roots, and the shape of the Weyl group action. [example: Weyl Group of Type B Three] For the root system $B_3$ in $\mathbb R^3$ with standard orthonormal basis $\varepsilon_1,\varepsilon_2,\varepsilon_3$, the roots are \begin{align*} \Phi=\{\pm\varepsilon_i:1\le i\le 3\}\cup\{\pm\varepsilon_i\pm\varepsilon_j:i\ne j\}. \end{align*} The short roots have squared length \begin{align*} (\pm\varepsilon_i,\pm\varepsilon_i)=(\varepsilon_i,\varepsilon_i)=1, \end{align*} while the long roots have squared length \begin{align*} (\pm\varepsilon_i\pm\varepsilon_j,\pm\varepsilon_i\pm\varepsilon_j) =(\varepsilon_i,\varepsilon_i)+(\varepsilon_j,\varepsilon_j)=2 \end{align*} for $i\ne j$, since $(\varepsilon_i,\varepsilon_j)=0$. The reflection in the root $\alpha$ is \begin{align*} s_\alpha(v)=v-\frac{2(v,\alpha)}{(\alpha,\alpha)}\alpha. \end{align*} For a short root $\alpha=\varepsilon_i$ and $v=x_1\varepsilon_1+x_2\varepsilon_2+x_3\varepsilon_3$, we have $(v,\varepsilon_i)=x_i$ and $(\varepsilon_i,\varepsilon_i)=1$, so \begin{align*} s_{\varepsilon_i}(v) &=v-2x_i\varepsilon_i\\ &=x_1\varepsilon_1+\cdots+x_i\varepsilon_i+\cdots+x_3\varepsilon_3-2x_i\varepsilon_i\\ &=x_1\varepsilon_1+\cdots-x_i\varepsilon_i+\cdots+x_3\varepsilon_3. \end{align*} Thus $s_{\varepsilon_i}$ changes the sign of the $i$th coordinate and fixes the other two coordinates. For a long root $\alpha=\varepsilon_i-\varepsilon_j$, where $i\ne j$, we have $(\alpha,\alpha)=2$ and \begin{align*} (v,\varepsilon_i-\varepsilon_j)=x_i-x_j. \end{align*} Therefore \begin{align*} s_{\varepsilon_i-\varepsilon_j}(v) &=v-\frac{2(x_i-x_j)}{2}(\varepsilon_i-\varepsilon_j)\\ &=v-(x_i-x_j)\varepsilon_i+(x_i-x_j)\varepsilon_j. \end{align*} The $i$th coordinate becomes \begin{align*} x_i-(x_i-x_j)=x_j, \end{align*} and the $j$th coordinate becomes \begin{align*} x_j+(x_i-x_j)=x_i, \end{align*} while every other coordinate is unchanged. Hence $s_{\varepsilon_i-\varepsilon_j}$ swaps the $i$th and $j$th coordinates. The sign changes $s_{\varepsilon_i}$ generate $(\mathbb Z/2\mathbb Z)^3$, and the coordinate swaps $s_{\varepsilon_i-\varepsilon_j}$ generate $S_3$. Together they generate all signed permutation matrices, so the Weyl group is \begin{align*} W(B_3)\cong (\mathbb Z/2\mathbb Z)^3\rtimes S_3. \end{align*} Given short roots $\sigma\varepsilon_i$ and $\tau\varepsilon_j$, with $\sigma,\tau\in\{\pm1\}$, first permute coordinates to send $\varepsilon_i$ to $\varepsilon_j$, then change the sign of the $j$th coordinate if $\sigma\ne\tau$. Thus all short roots lie in one Weyl group orbit. Given long roots $\sigma\varepsilon_i+\tau\varepsilon_j$ and $\sigma'\varepsilon_p+\tau'\varepsilon_q$, with $i\ne j$ and $p\ne q$, first permute coordinates to send the unordered pair $\{i,j\}$ to $\{p,q\}$, then change signs in the $p$ and $q$ coordinates as needed. Thus all long roots lie in one Weyl group orbit. The two orbits remain distinct because every signed permutation preserves the standard squared length. [/example] The classical families show how the main diagrams arise from matrix algebras. Highest roots add a different kind of data: instead of naming the diagram, they measure how the positive roots sit above the chosen simple basis. [example: Highest Roots in Small Types] For $A_3$, use the standard realization with simple roots \begin{align*} \alpha_1=\varepsilon_1-\varepsilon_2,\qquad \alpha_2=\varepsilon_2-\varepsilon_3,\qquad \alpha_3=\varepsilon_3-\varepsilon_4. \end{align*} Then the positive roots are $\varepsilon_i-\varepsilon_j$ for $i<j$, and the largest one in the usual order is \begin{align*} \varepsilon_1-\varepsilon_4 &=(\varepsilon_1-\varepsilon_2)+(\varepsilon_2-\varepsilon_3)+(\varepsilon_3-\varepsilon_4)\\ &=\alpha_1+\alpha_2+\alpha_3. \end{align*} Thus the highest root of $A_3$ is $\alpha_1+\alpha_2+\alpha_3$. For $C_3$, take \begin{align*} \alpha_1=\varepsilon_1-\varepsilon_2,\qquad \alpha_2=\varepsilon_2-\varepsilon_3,\qquad \alpha_3=2\varepsilon_3. \end{align*} The long positive roots are $2\varepsilon_1,2\varepsilon_2,2\varepsilon_3$, and among them $2\varepsilon_1$ has the largest expansion in the simple-root basis. Indeed, \begin{align*} 2\alpha_1+2\alpha_2+\alpha_3 &=2(\varepsilon_1-\varepsilon_2)+2(\varepsilon_2-\varepsilon_3)+2\varepsilon_3\\ &=2\varepsilon_1-2\varepsilon_2+2\varepsilon_2-2\varepsilon_3+2\varepsilon_3\\ &=2\varepsilon_1. \end{align*} So the highest root of $C_3$ is $2\alpha_1+2\alpha_2+\alpha_3$. For $G_2$, with $\alpha_1$ short, $\alpha_2$ long, and Cartan matrix \begin{align*} \begin{pmatrix}2&-3\\-1&2\end{pmatrix}, \end{align*} the positive roots are \begin{align*} \alpha_1,\quad \alpha_2,\quad \alpha_1+\alpha_2,\quad 2\alpha_1+\alpha_2,\quad 3\alpha_1+\alpha_2,\quad 3\alpha_1+2\alpha_2. \end{align*} Their heights are respectively \begin{align*} 1,\quad 1,\quad 2,\quad 3,\quad 4,\quad 5. \end{align*} The unique positive root of largest height is therefore $3\alpha_1+2\alpha_2$, so this is the highest root of $G_2$. [/example] The example also points to a boundary of the idea. The following remark, What Has Been Classified, records that interpretation before the construction is used again. [remark: What Has Been Classified] The theorem classifies finite-dimensional simple Lie algebras over $\mathbb C$ up to Lie algebra isomorphism. It does not classify real forms, algebraic groups, Lie groups with a fixed Lie algebra, or representations of the simple Lie algebras. Those are later classification problems built on this one. [/remark] The course has now completed its main arc. Semisimplicity made the Killing form nondegenerate; Cartan subalgebras produced roots; root strings produced Cartan integers; Cartan matrices produced Dynkin diagrams; and Serre theorem together with uniqueness turned the diagrams into a complete classification.