[proofplan] We attach to a finite-dimensional simple Lie algebra $\mathfrak g$ its root system relative to a Cartan subalgebra and then pass from the root system to its Dynkin diagram. Simplicity forces the root system, hence the Dynkin diagram, to be irreducible, so the diagram is connected and of finite type. Injectivity is the Cartan matrix isomorphism criterion: the Dynkin diagram determines the Cartan matrix, and the Cartan matrix determines the simple Lie algebra. Surjectivity is Serre's construction theorem together with the classification of irreducible finite crystallographic root systems. [/proofplan]

[step:Associate a connected finite type Dynkin diagram to a simple Lie algebra]Let $\mathfrak g$ be a finite-dimensional simple Lie algebra over $k$. Since $k$ is algebraically closed of characteristic zero and $\mathfrak g$ is simple, $\mathfrak g$ is semisimple. Choose a Cartan subalgebra $\mathfrak h \subset \mathfrak g$. Let \begin{align*} \Phi_{\mathfrak g}: \mathfrak h^* \setminus \{0\} &\to \{\text{root spaces of }\mathfrak g\} \\ \alpha &\mapsto \mathfrak g_\alpha := \{x \in \mathfrak g : [h,x] = \alpha(h)x \text{ for every } h \in \mathfrak h\} \end{align*} and define the root set \begin{align*} R(\mathfrak g,\mathfrak h) := \{\alpha \in \mathfrak h^* \setminus \{0\} : \mathfrak g_\alpha \ne 0\}. \end{align*} By the root space decomposition theorem for finite-dimensional semisimple Lie algebras over an algebraically closed field of characteristic zero, $R(\mathfrak g,\mathfrak h)$ is a reduced crystallographic finite root system in the real [vector space](/page/Vector%20Space) spanned by $R(\mathfrak g,\mathfrak h)$. Choose a base of simple roots \begin{align*} \Pi = \{\alpha_1,\dots,\alpha_r\} \subset R(\mathfrak g,\mathfrak h), \end{align*} and let $A(\mathfrak g,\mathfrak h,\Pi) = (a_{ij})_{1 \le i,j \le r}$ be the associated Cartan matrix, where \begin{align*} a_{ij} := \frac{2(\alpha_i,\alpha_j)}{(\alpha_j,\alpha_j)}. \end{align*} This proof uses the column-normalized Cartan matrix convention: the denominator is $(\alpha_j,\alpha_j)$, so the arrows in non-simply-laced Dynkin diagrams are encoded by the ordered pair $(a_{ij},a_{ji})$ with this convention fixed. Let $V_{\mathbb R}$ be the real vector space spanned by $R(\mathfrak g,\mathfrak h)$, and let $W = W(R(\mathfrak g,\mathfrak h))$ be the Weyl group generated by the root reflections on $V_{\mathbb R}$. The Killing form $\kappa_{\mathfrak g}: \mathfrak g \times \mathfrak g \to k$ is nondegenerate on $\mathfrak h$ because $\mathfrak g$ is semisimple; it identifies $\mathfrak h^*$ with $\mathfrak h$, and the standard real form $V_{\mathbb R} \subset \mathfrak h^*$ inherits a positive definite $W$-invariant inner product $(\cdot,\cdot)$ from this identification. The Dynkin diagram $\Delta(\mathfrak g)$ is the diagram encoded by this Cartan matrix. The Dynkin diagram is independent of the chosen Cartan subalgebra and base up to diagram isomorphism: Cartan subalgebras are conjugate by an automorphism of $\mathfrak g$, and different bases of the same finite root system are related by the Weyl group, which preserves the Cartan matrix up to simultaneous relabelling of rows and columns. Since $\mathfrak g$ is simple, the root system cannot decompose as an orthogonal disjoint union $R_1 \sqcup R_2$ with both $R_i$ nonempty; such a decomposition would give a direct sum decomposition of $\mathfrak g$ into two nonzero ideals. Therefore $R(\mathfrak g,\mathfrak h)$ is irreducible, and $\Delta(\mathfrak g)$ is connected. Since the root system is finite crystallographic, its diagram is of finite type.[/step]

[guided]We first define the object that the classification map assigns to $\mathfrak g$. Let $\mathfrak h \subset \mathfrak g$ be a Cartan subalgebra. For each nonzero linear functional $\alpha \in \mathfrak h^*$, define the root space \begin{align*} \mathfrak g_\alpha := \{x \in \mathfrak g : [h,x] = \alpha(h)x \text{ for every } h \in \mathfrak h\}. \end{align*} This is a $k$-linear subspace of $\mathfrak g$. Define the root set \begin{align*} R(\mathfrak g,\mathfrak h) := \{\alpha \in \mathfrak h^* \setminus \{0\} : \mathfrak g_\alpha \ne 0\}. \end{align*} The hypotheses on $k$ and $\mathfrak g$ are used here: over an algebraically closed field of characteristic zero, a finite-dimensional simple Lie algebra is semisimple, so the root space decomposition theorem applies. It gives a finite reduced crystallographic root system $R(\mathfrak g,\mathfrak h)$. Choose a base of simple roots \begin{align*} \Pi = \{\alpha_1,\dots,\alpha_r\} \subset R(\mathfrak g,\mathfrak h). \end{align*} The Cartan matrix attached to this base is the integer matrix $A(\mathfrak g,\mathfrak h,\Pi) = (a_{ij})$ defined by \begin{align*} a_{ij} := \frac{2(\alpha_i,\alpha_j)}{(\alpha_j,\alpha_j)}, \qquad 1 \le i,j \le r. \end{align*} We use the column-normalized convention, so the denominator is the squared length of the root indexed by the column. Let $V_{\mathbb R}$ be the real span of the roots, and let $W = W(R(\mathfrak g,\mathfrak h))$ be the Weyl group generated by the reflections in the roots. The Killing form on $\mathfrak g$ is nondegenerate on $\mathfrak h$ because $\mathfrak g$ is semisimple; via the resulting identification of $\mathfrak h^*$ with $\mathfrak h$, the standard real root space $V_{\mathbb R}$ carries a positive definite $W$-invariant inner product $(\cdot,\cdot)$. The Dynkin diagram $\Delta(\mathfrak g)$ is precisely the labelled graph encoded by this Cartan matrix. We must check that this construction depends only on the isomorphism class of $\mathfrak g$. Cartan subalgebras of a finite-dimensional semisimple Lie algebra over an algebraically closed field of characteristic zero are conjugate under automorphisms of $\mathfrak g$, so replacing $\mathfrak h$ transports the root system by a linear isomorphism. Replacing the base $\Pi$ by another base applies an element of the Weyl group, which relabels the simple roots and preserves the Cartan integers. Hence the diagram is well-defined up to graph isomorphism. Finally, simplicity forces connectedness. If $R(\mathfrak g,\mathfrak h)$ decomposed as an orthogonal disjoint union $R_1 \sqcup R_2$ with both parts nonempty, then the subalgebras generated by the root spaces for $R_1$ and $R_2$ would be commuting nonzero ideals whose direct sum is $\mathfrak g$. This contradicts the simplicity of $\mathfrak g$. Therefore the root system is irreducible, and the corresponding Dynkin diagram is connected. Because the root system is finite and crystallographic, the diagram is of finite type.[/guided]

[step:Show that isomorphic simple Lie algebras have the same diagram] Let $\mathfrak g$ and $\mathfrak g'$ be finite-dimensional simple Lie algebras over $k$, and let \begin{align*} \varphi: \mathfrak g &\to \mathfrak g' \end{align*} be a Lie algebra isomorphism. If $\mathfrak h \subset \mathfrak g$ is a Cartan subalgebra, then $\varphi(\mathfrak h) \subset \mathfrak g'$ is a Cartan subalgebra. For every root $\alpha \in R(\mathfrak g,\mathfrak h)$, define \begin{align*} \alpha': \varphi(\mathfrak h) &\to k \\ \varphi(h) &\mapsto \alpha(h). \end{align*} Then $\varphi(\mathfrak g_\alpha) = \mathfrak g'_{\alpha'}$, because \begin{align*} [\varphi(h),\varphi(x)] = \varphi([h,x]) = \varphi(\alpha(h)x) = \alpha(h)\varphi(x) = \alpha'(\varphi(h))\varphi(x). \end{align*} Thus $\varphi$ induces an isomorphism of root systems and hence an isomorphism of Dynkin diagrams. Therefore the map $[\mathfrak g] \mapsto \Delta(\mathfrak g)$ is well-defined on isomorphism classes. [/step]

[step:Use the Cartan matrix criterion to prove injectivity]Suppose $\mathfrak g$ and $\mathfrak g'$ are finite-dimensional simple Lie algebras over $k$ with isomorphic Dynkin diagrams. Choose Cartan subalgebras $\mathfrak h \subset \mathfrak g$ and $\mathfrak h' \subset \mathfrak g'$, and choose bases of simple roots \begin{align*} \Pi = \{\alpha_1, \dots,\alpha_r\}, \qquad \Pi' = \{\alpha'_1,\dots,\alpha'_r\} \end{align*} whose diagrams are matched by the given diagram isomorphism. After relabelling $\Pi'$, the Cartan matrices agree: \begin{align*} A(\mathfrak g,\mathfrak h,\Pi) = A(\mathfrak g',\mathfrak h',\Pi'). \end{align*} By the finite-type Cartan matrix classification theorem for semisimple Lie algebras, a finite-dimensional semisimple Lie algebra over an algebraically closed field of characteristic zero is determined up to isomorphism by its finite type Cartan matrix, equivalently by its reduced crystallographic root system with chosen simple roots, up to simultaneous relabelling of the simple roots. Since both algebras are simple and their Cartan matrices agree after relabelling, this theorem gives a Lie algebra isomorphism \begin{align*} \mathfrak g \cong \mathfrak g'. \end{align*} Thus the assignment is injective on isomorphism classes.[/step]

[guided]The injectivity question asks whether the diagram has lost information. A Dynkin diagram is not just an unlabelled graph: its edges and arrows encode the Cartan integers \begin{align*} a_{ij} = \frac{2(\alpha_i,\alpha_j)}{(\alpha_j,\alpha_j)}. \end{align*} So if $\mathfrak g$ and $\mathfrak g'$ have isomorphic Dynkin diagrams, we may relabel the chosen simple roots of $\mathfrak g'$ so that their Cartan matrices are identical: \begin{align*} A(\mathfrak g,\mathfrak h,\Pi) = A(\mathfrak g',\mathfrak h',\Pi'). \end{align*} Now we apply the finite-type Cartan matrix classification theorem for semisimple Lie algebras. Its hypotheses are satisfied: both $\mathfrak g$ and $\mathfrak g'$ are finite-dimensional semisimple Lie algebras because they are finite-dimensional simple Lie algebras over an algebraically closed field of characteristic zero; the chosen Cartan subalgebras and simple root bases define finite type Cartan matrices using the same column-normalized convention; and those matrices agree after relabelling. The theorem states that a finite-dimensional semisimple Lie algebra over such a field is determined up to isomorphism by its finite type Cartan matrix, equivalently by its reduced crystallographic root system with chosen simple roots, up to simultaneous relabelling. Since each algebra here is simple, the conclusion is \begin{align*} \mathfrak g \cong \mathfrak g'. \end{align*} Therefore two simple Lie algebras with the same Dynkin diagram represent the same isomorphism class, which proves injectivity.[/guided]

[step:Construct a simple Lie algebra from each connected finite type diagram]Let $\Delta$ be a connected finite type Dynkin diagram, and let $A_\Delta = (a_{ij})_{1 \le i,j \le r}$ be its Cartan matrix. Because $\Delta$ is of finite type, $A_\Delta$ is a finite type generalized Cartan matrix. By Serre's construction theorem, there exists a finite-dimensional semisimple Lie algebra $\mathfrak g(A_\Delta)$ over $k$ generated by elements \begin{align*} e_i, f_i, h_i \in \mathfrak g(A_\Delta), \qquad 1 \le i \le r, \end{align*} subject to the Serre relations determined by $A_\Delta$, and its Cartan matrix is exactly $A_\Delta$. Since $\Delta$ is connected, the matrix $A_\Delta$ is indecomposable. The finite-type Serre construction theorem further says that the semisimple Lie algebra obtained from an indecomposable finite type generalized Cartan matrix has irreducible root system, and a finite-dimensional semisimple Lie algebra has irreducible root system exactly when it has one simple ideal. Hence $\mathfrak g(A_\Delta)$ is simple. Its Dynkin diagram is $\Delta$, so every connected finite type Dynkin diagram occurs.[/step]

[guided]For surjectivity, begin with a connected finite type Dynkin diagram $\Delta$. The diagram determines a Cartan matrix \begin{align*} A_\Delta = (a_{ij})_{1 \le i,j \le r}. \end{align*} The finite type assumption is exactly the condition needed for the Serre construction to produce a finite-dimensional semisimple Lie algebra rather than an infinite-dimensional Kac-Moody algebra. Apply Serre's construction theorem to $A_\Delta$. The theorem requires a finite type generalized Cartan matrix over an algebraically closed field of characteristic zero. These hypotheses are satisfied by construction of $A_\Delta$ and by the hypothesis on $k$. The theorem produces a finite-dimensional semisimple Lie algebra $\mathfrak g(A_\Delta)$ over $k$ generated by elements \begin{align*} e_i, f_i, h_i \in \mathfrak g(A_\Delta), \qquad 1 \le i \le r, \end{align*} with Serre relations determined by $A_\Delta$, and the resulting root system has Cartan matrix $A_\Delta$. It remains to check simplicity. Connectedness of $\Delta$ means that $A_\Delta$ is indecomposable: after any simultaneous permutation of rows and columns, it cannot be written as a block diagonal matrix with two nonempty blocks. The finite-type Serre construction theorem includes the corresponding structural statement: the semisimple Lie algebra obtained from an indecomposable finite type generalized Cartan matrix has irreducible root system. For a finite-dimensional semisimple Lie algebra, the root system decomposes as the orthogonal disjoint union of the root systems of its simple ideals; therefore irreducibility of the root system is equivalent to having exactly one simple ideal. Applying this to $\mathfrak g(A_\Delta)$, we conclude that $\mathfrak g(A_\Delta)$ has one simple ideal, hence is simple. Its Dynkin diagram is the original diagram $\Delta$, so $\Delta$ lies in the image of the classification map.[/guided]

[step:List the connected finite type diagrams and conclude the bijection] The classification of irreducible finite crystallographic root systems says that every irreducible finite crystallographic root system has exactly one of the types \begin{align*} A_n \ (n \ge 1),\quad B_n \ (n \ge 2),\quad C_n \ (n \ge 3),\quad D_n \ (n \ge 4),\quad E_6,E_7,E_8,F_4,G_2, \end{align*} and that each listed type occurs. Passing from each root system to its Dynkin diagram gives precisely the connected finite type Dynkin diagrams. The first step shows that every finite-dimensional simple Lie algebra determines such a diagram, the injectivity step shows that this diagram determines the isomorphism class of the Lie algebra, and the construction step shows that every connected finite type diagram occurs. Hence the assignment is a bijection, with the listed types as its image. [/step]