[proofplan] We attach to each finite-dimensional semisimple Lie algebra a root system relative to a Cartan subalgebra, choose a simple root basis, and record the associated finite type Dynkin diagram; the zero Lie algebra is assigned the empty diagram. The [decomposition of a semisimple Lie algebra into simple ideals](/theorems/3821) matches the decomposition of its root system into irreducible components, so direct sums correspond to disjoint unions of diagrams. Conversely, each finite type Dynkin diagram determines a generalized Cartan matrix of finite type, and the Serre presentation constructs the corresponding semisimple Lie algebra, with the empty diagram producing the zero Lie algebra. Finally, the uniqueness part of the Serre construction shows that equal diagrams give isomorphic Lie algebras, and connectedness is exactly the condition for nonzero simplicity. [/proofplan] [step:Attach a finite type Dynkin diagram to a semisimple Lie algebra] Let $\mathfrak g$ be a finite-dimensional [semisimple Lie algebra](/page/Semisimple%20Lie%20Algebra) over $k$. If $\mathfrak g = 0$, define its Dynkin diagram to be the empty diagram and skip the root construction below. Otherwise, choose a [Cartan subalgebra](/page/Cartan%20Subalgebra) $\mathfrak h \subset \mathfrak g$. By the root space [decomposition theorem for semisimple Lie algebras](/theorems/3816) over an algebraically closed field of characteristic zero, applied to the pair $(\mathfrak g,\mathfrak h)$, there is a finite reduced crystallographic [root system](/page/Root%20System) $\Phi \subset \mathfrak h^*$ and a decomposition \begin{align*} \mathfrak g = \mathfrak h \oplus \bigoplus_{\alpha \in \Phi} \mathfrak g_\alpha, \end{align*} where \begin{align*} \mathfrak g_\alpha := \{x \in \mathfrak g : [h,x] = \alpha(h)x \text{ for all } h \in \mathfrak h\}. \end{align*} Here we are citing a result not yet in the wiki: root space decomposition and root system theorem for semisimple Lie algebras. Choose a base of simple roots \begin{align*} \Delta = \{\alpha_1,\dots,\alpha_n\} \subset \Phi. \end{align*} The associated [Cartan matrix](/page/Cartan%20Matrix) is the integer matrix \begin{align*} A = (a_{ij})_{1 \le i,j \le n}, \qquad a_{ij} := \langle \alpha_j,\alpha_i^\vee\rangle, \end{align*} where $\alpha_i^\vee$ denotes the [coroot](/page/Coroot) of $\alpha_i$. Since $\Phi$ is a finite crystallographic root system, $A$ is a Cartan matrix of finite type. The [Dynkin diagram](/page/Dynkin%20Diagram) $\Gamma(\mathfrak g,\mathfrak h,\Delta)$ is the diagram whose vertices are the simple roots $\alpha_1,\dots,\alpha_n$ and whose edges are determined by the off-diagonal entries $a_{ij}$ and $a_{ji}$. Changing the Cartan subalgebra or the simple root basis changes the diagram only by an isomorphism of diagrams, by the conjugacy of Cartan subalgebras and the Weyl group transitivity on bases of the root system. Here we are citing a result not yet in the wiki: conjugacy of Cartan subalgebras and Weyl chamber transitivity. Hence the isomorphism class of the finite type Dynkin diagram is an invariant of the isomorphism class of $\mathfrak g$. [/step] [step:Relate simple ideal decompositions to connected components] By the decomposition theorem for finite-dimensional semisimple Lie algebras, there are [simple Lie algebra](/page/Simple%20Lie%20Algebra) ideals \begin{align*} \mathfrak g_1,\dots,\mathfrak g_r \subset \mathfrak g \end{align*} such that \begin{align*} \mathfrak g = \mathfrak g_1 \oplus \cdots \oplus \mathfrak g_r. \end{align*} Here $r \geq 0$; when $r=0$, the direct sum is the zero Lie algebra and its diagram is the empty diagram by definition. Here we are citing a result not yet in the wiki: decomposition of semisimple Lie algebras into simple ideals. For each $m \in \{1,\dots,r\}$, choose a Cartan subalgebra $\mathfrak h_m \subset \mathfrak g_m$. Then \begin{align*} \mathfrak h := \mathfrak h_1 \oplus \cdots \oplus \mathfrak h_r \end{align*} is a Cartan subalgebra of $\mathfrak g$. Let $\Phi_m \subset \mathfrak h_m^*$ be the root system of $\mathfrak g_m$ relative to $\mathfrak h_m$. Under the natural inclusion $\mathfrak h_m^* \subset \mathfrak h^*$ obtained by extending functionals by zero on the other summands, the root system of $\mathfrak g$ is \begin{align*} \Phi = \Phi_1 \sqcup \cdots \sqcup \Phi_r. \end{align*} Roots belonging to different summands are orthogonal, because $[\mathfrak g_m,\mathfrak g_\ell]=0$ for $m \ne \ell$. Therefore the Cartan matrix of $\Phi$ is block diagonal, with blocks the Cartan matrices of $\Phi_1,\dots,\Phi_r$. It follows that the Dynkin diagram of $\mathfrak g$ is the disjoint union \begin{align*} \Gamma(\mathfrak g) = \Gamma(\mathfrak g_1) \sqcup \cdots \sqcup \Gamma(\mathfrak g_r). \end{align*} Moreover, each $\Gamma(\mathfrak g_m)$ is connected, since a disconnected diagram would split the root system of $\mathfrak g_m$ into two orthogonal nonempty root subsystems, which would split $\mathfrak g_m$ into two nonzero commuting ideals, contradicting simplicity. [/step] [step:Construct a semisimple Lie algebra from each finite type diagram] Let $\Gamma$ be a finite disjoint union of connected finite type Dynkin diagrams. Write its connected components as \begin{align*} \Gamma = \Gamma_1 \sqcup \cdots \sqcup \Gamma_r. \end{align*} Here $r \geq 0$. If $r=0$, then $\Gamma$ is the empty diagram and we define $\mathfrak g_\Gamma := 0$ with the zero Lie bracket; this Lie algebra is semisimple because it has no nonzero solvable ideal, and its Dynkin diagram is empty. For each component $\Gamma_m$, let \begin{align*} A_m = ((a_m)_{ij}) \end{align*} be the Cartan matrix encoded by $\Gamma_m$. By the Serre existence theorem for finite type Cartan matrices, applied to $A_m$, there exists a finite-dimensional simple Lie algebra $\mathfrak s_m$ over $k$ whose Cartan matrix is $A_m$. Here we are citing a result not yet in the wiki: Serre existence and simplicity theorem for finite type Cartan matrices. For $r>0$, define the Lie algebra \begin{align*} \mathfrak g_\Gamma := \mathfrak s_1 \oplus \cdots \oplus \mathfrak s_r \end{align*} with componentwise Lie bracket. Since each $\mathfrak s_m$ is simple, each $\mathfrak s_m$ is semisimple, and a finite direct sum of semisimple Lie algebras is semisimple. The Dynkin diagram of $\mathfrak g_\Gamma$ is the disjoint union of the Dynkin diagrams of the $\mathfrak s_m$, hence is precisely $\Gamma$. [/step] [step:Prove that equal diagrams give isomorphic Lie algebras] Let $\mathfrak g$ and $\mathfrak g'$ be finite-dimensional semisimple Lie algebras over $k$ with isomorphic Dynkin diagrams. If the common diagram is empty, then the preceding construction assigns both Lie algebras no simple summands, so $\mathfrak g=0=\mathfrak g'$ and they are isomorphic. We now assume the common diagram is nonempty. Choose decompositions into simple ideals \begin{align*} \mathfrak g = \mathfrak g_1 \oplus \cdots \oplus \mathfrak g_r, \qquad \mathfrak g' = \mathfrak g'_1 \oplus \cdots \oplus \mathfrak g'_r \end{align*} so that, after relabelling the simple ideals of $\mathfrak g'$, the connected Dynkin diagram of $\mathfrak g_m$ is isomorphic to the connected Dynkin diagram of $\mathfrak g'_m$ for every $m \in \{1,\dots,r\}$. For each $m$, choose simple root bases for $\mathfrak g_m$ and $\mathfrak g'_m$. The diagram isomorphism relabels the simple roots and identifies their Cartan matrices. By the uniqueness part of the Serre presentation theorem for finite type Cartan matrices, the simple Lie algebras determined by the same Cartan matrix are isomorphic. Hence there is a Lie algebra isomorphism \begin{align*} \varphi_m : \mathfrak g_m \to \mathfrak g'_m \end{align*} for each $m$. Define \begin{align*} \varphi : \mathfrak g \to \mathfrak g', \qquad \varphi(x_1+\cdots+x_r) := \varphi_1(x_1)+\cdots+\varphi_r(x_r), \end{align*} where $x_m \in \mathfrak g_m$. Since each $\varphi_m$ is a Lie algebra isomorphism and the brackets between distinct direct summands vanish, $\varphi$ is a Lie algebra isomorphism. Therefore two semisimple Lie algebras with the same finite disjoint union of Dynkin diagrams are isomorphic. [/step] [step:Identify the simple Lie algebras with connected diagrams] If $\mathfrak g$ is simple, then its decomposition into simple ideals has one summand, namely $\mathfrak g$ itself. By the direct-sum argument above, its Dynkin diagram has exactly one connected component, so it is connected. Conversely, suppose $\mathfrak g$ is semisimple and its Dynkin diagram is connected. Write \begin{align*} \mathfrak g = \mathfrak g_1 \oplus \cdots \oplus \mathfrak g_r \end{align*} as a direct sum of simple ideals. The Dynkin diagram of $\mathfrak g$ is the disjoint union of the connected diagrams of the $\mathfrak g_m$. Since the diagram is connected, this disjoint union has only one component, so $r=1$. Hence $\mathfrak g=\mathfrak g_1$ is simple. Combining existence, uniqueness, and the connectedness criterion proves the asserted bijection between semisimple Lie algebras and finite disjoint unions of finite type Dynkin diagrams, with the zero Lie algebra corresponding to the empty diagram and nonzero simple Lie algebras corresponding exactly to connected finite type Dynkin diagrams. [/step]