[proofplan]
We use the structural decomposition of a finite-dimensional semisimple Lie algebra into a finite direct sum of simple ideals and the classification of finite-dimensional simple Lie algebras over an algebraically closed field of characteristic $0$ by connected finite type Dynkin diagrams. The map from semisimple Lie algebras to diagrams is obtained by taking the disjoint union of the diagrams of the simple summands. We prove that this assignment is well-defined, construct its inverse by taking direct sums of the simple Lie algebras attached to each connected component, and then verify that the two constructions are inverse to each other using uniqueness of the simple-ideal decomposition and the simple classification theorem.
[/proofplan]
[step:Decompose each semisimple Lie algebra into simple ideals]
Let $\mathfrak{g}$ be a finite-dimensional semisimple Lie algebra over $k$. By the decomposition theorem for finite-dimensional semisimple Lie algebras into simple ideals (citing a result not yet in the wiki: complete reducibility and simple-ideal decomposition of semisimple Lie algebras), there exist simple ideals
\begin{align*}
\mathfrak{g}_1,\dots,\mathfrak{g}_r \subset \mathfrak{g}
\end{align*}
for some integer $r \geq 0$ such that
\begin{align*}
\mathfrak{g} = \mathfrak{g}_1 \oplus \cdots \oplus \mathfrak{g}_r
\end{align*}
as Lie algebras. Here $r = 0$ means $\mathfrak{g} = 0$, the empty direct sum.
Moreover, the same decomposition theorem states that the unordered list of isomorphism classes
\begin{align*}
[\mathfrak{g}_1],\dots,[\mathfrak{g}_r]
\end{align*}
is uniquely determined by the isomorphism class of $\mathfrak{g}$, including multiplicities.
[guided]
Let $\mathfrak{g}$ be a finite-dimensional semisimple Lie algebra over $k$. The first structural input is that semisimple Lie algebras split into their simple building blocks. More precisely, by the decomposition theorem for finite-dimensional semisimple Lie algebras into simple ideals (citing a result not yet in the wiki: complete reducibility and simple-ideal decomposition of semisimple Lie algebras), there are simple 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*}
The direct sum here is a direct sum of Lie algebras: each $\mathfrak{g}_i$ is an ideal, the [vector space](/page/Vector%20Space) sum is direct, and for $i \neq j$ one has $[\mathfrak{g}_i,\mathfrak{g}_j] = 0$ because distinct simple ideals in such a direct-sum decomposition commute.
The theorem also gives uniqueness in the form needed here: if
\begin{align*}
\mathfrak{g} = \mathfrak{h}_1 \oplus \cdots \oplus \mathfrak{h}_s
\end{align*}
is another decomposition into simple ideals, then $r = s$ and, after reordering, $\mathfrak{g}_i \cong \mathfrak{h}_i$ for each $i$. Thus the unordered list of simple isomorphism classes occurring in $\mathfrak{g}$ is intrinsic to $\mathfrak{g}$.
[/guided]
[/step]
[step:Attach a finite disjoint union of Dynkin diagrams to a semisimple Lie algebra]
For each simple ideal $\mathfrak{g}_i$, let $\Delta_i$ denote its Dynkin diagram. By the classification of finite-dimensional simple Lie algebras over an algebraically closed field of characteristic $0$ by connected finite type Dynkin diagrams (citing a result not yet in the wiki: Cartan-Killing classification of finite-dimensional simple Lie algebras), each $\Delta_i$ is a connected finite type Dynkin diagram, and the isomorphism class of $\mathfrak{g}_i$ determines $\Delta_i$ uniquely.
Define
\begin{align*}
\Phi([\mathfrak{g}]) := \Delta_1 \sqcup \cdots \sqcup \Delta_r,
\end{align*}
where $\sqcup$ denotes disjoint union of diagrams and $[\mathfrak{g}]$ denotes the isomorphism class of $\mathfrak{g}$. If $r = 0$, define $\Phi([0])$ to be the empty disjoint union. The uniqueness of the simple-ideal decomposition shows that $\Phi([\mathfrak{g}])$ is independent of the chosen decomposition of $\mathfrak{g}$.
[guided]
For every simple summand $\mathfrak{g}_i$, the simple classification theorem assigns a connected finite type Dynkin diagram. We denote this diagram by $\Delta_i$. The classification theorem says two things needed here: first, every finite-dimensional simple Lie algebra over $k$ has a connected finite type Dynkin diagram; second, two such simple Lie algebras are isomorphic exactly when their Dynkin diagrams are isomorphic.
We therefore define a map
\begin{align*}
\Phi:
\left\{
\begin{array}{c}
\text{isomorphism classes of finite-dimensional} \\
\text{semisimple Lie algebras over } k
\end{array}
\right\}
&\to
\left\{
\begin{array}{c}
\text{finite disjoint unions of connected} \\
\text{finite type Dynkin diagrams}
\end{array}
\right\}
\end{align*}
by
\begin{align*}
\Phi([\mathfrak{g}]) := \Delta_1 \sqcup \cdots \sqcup \Delta_r.
\end{align*}
The symbol $\sqcup$ means disjoint union of graphs, so the connected components of $\Phi([\mathfrak{g}])$ are exactly the diagrams attached to the simple ideals of $\mathfrak{g}$.
We must check that this is well-defined. The possible ambiguity is the choice of simple-ideal decomposition. But the previous step says that any two decompositions of $\mathfrak{g}$ into simple ideals have the same simple factors up to reordering and isomorphism. Since the simple classification theorem identifies isomorphic simple Lie algebras with the same Dynkin diagram, the resulting disjoint union of diagrams is unchanged up to isomorphism of finite disjoint unions.
[/guided]
[/step]
[step:Construct a semisimple Lie algebra from a finite disjoint union of diagrams]
Let
\begin{align*}
\Delta = \Delta_1 \sqcup \cdots \sqcup \Delta_r
\end{align*}
be a finite disjoint union of connected finite type Dynkin diagrams. By the simple classification theorem, for each $i \in \{1,\dots,r\}$ there exists a finite-dimensional simple Lie algebra $\mathfrak{s}_i$ over $k$ whose Dynkin diagram is $\Delta_i$, and $\mathfrak{s}_i$ is unique up to isomorphism.
Define
\begin{align*}
\Psi(\Delta) := [\mathfrak{s}_1 \oplus \cdots \oplus \mathfrak{s}_r].
\end{align*}
The direct sum of simple Lie algebras is semisimple, so $\Psi(\Delta)$ is an isomorphism class of finite-dimensional semisimple Lie algebras over $k$. If $\Delta$ is the empty disjoint union, define $\Psi(\Delta) := [0]$. The uniqueness part of the simple classification theorem shows that $\Psi(\Delta)$ is independent of the chosen representatives $\mathfrak{s}_i$.
[guided]
Now start from the diagram side. Let
\begin{align*}
\Delta = \Delta_1 \sqcup \cdots \sqcup \Delta_r
\end{align*}
be a finite disjoint union of connected finite type Dynkin diagrams. The connected components are listed as $\Delta_1,\dots,\Delta_r$; if the disjoint union is empty, then $r = 0$.
For each connected component $\Delta_i$, the simple classification theorem gives a finite-dimensional simple Lie algebra $\mathfrak{s}_i$ over $k$ with Dynkin diagram $\Delta_i$. The same theorem also says that this $\mathfrak{s}_i$ is unique up to isomorphism. We then form the direct sum Lie algebra
\begin{align*}
\mathfrak{s}_1 \oplus \cdots \oplus \mathfrak{s}_r,
\end{align*}
whose bracket is defined componentwise:
\begin{align*}
[(x_1,\dots,x_r),(y_1,\dots,y_r)]
=
([x_1,y_1],\dots,[x_r,y_r])
\end{align*}
for $x_i,y_i \in \mathfrak{s}_i$.
Because each $\mathfrak{s}_i$ is simple, each $\mathfrak{s}_i$ is semisimple. A finite direct sum of semisimple Lie algebras is semisimple, since its radical is the direct sum of the radicals of the summands, and each of those radicals is $0$. Thus
\begin{align*}
\mathfrak{s}_1 \oplus \cdots \oplus \mathfrak{s}_r
\end{align*}
is a finite-dimensional semisimple Lie algebra over $k$.
Define
\begin{align*}
\Psi(\Delta) := [\mathfrak{s}_1 \oplus \cdots \oplus \mathfrak{s}_r].
\end{align*}
This is independent of the chosen representatives because replacing any $\mathfrak{s}_i$ by an isomorphic simple Lie algebra gives an isomorphic direct sum.
[/guided]
[/step]
[step:Verify that the two constructions are inverse maps]
Let $\mathfrak{g}$ be a finite-dimensional semisimple Lie algebra over $k$, and choose its decomposition into simple ideals
\begin{align*}
\mathfrak{g} = \mathfrak{g}_1 \oplus \cdots \oplus \mathfrak{g}_r.
\end{align*}
If $\Delta_i$ is the Dynkin diagram of $\mathfrak{g}_i$, then
\begin{align*}
\Phi([\mathfrak{g}]) = \Delta_1 \sqcup \cdots \sqcup \Delta_r.
\end{align*}
Applying $\Psi$ chooses, for each $\Delta_i$, a simple Lie algebra isomorphic to $\mathfrak{g}_i$. Therefore
\begin{align*}
\Psi(\Phi([\mathfrak{g}]))
=
[\mathfrak{g}_1 \oplus \cdots \oplus \mathfrak{g}_r]
=
[\mathfrak{g}].
\end{align*}
Conversely, let
\begin{align*}
\Delta = \Delta_1 \sqcup \cdots \sqcup \Delta_r
\end{align*}
be a finite disjoint union of connected finite type Dynkin diagrams. By construction,
\begin{align*}
\Psi(\Delta) = [\mathfrak{s}_1 \oplus \cdots \oplus \mathfrak{s}_r],
\end{align*}
where the Dynkin diagram of $\mathfrak{s}_i$ is $\Delta_i$. Applying $\Phi$ gives
\begin{align*}
\Phi(\Psi(\Delta)) = \Delta_1 \sqcup \cdots \sqcup \Delta_r = \Delta.
\end{align*}
Thus $\Phi$ and $\Psi$ are inverse bijections.
[guided]
We now check both compositions.
First let $\mathfrak{g}$ be semisimple, and write
\begin{align*}
\mathfrak{g} = \mathfrak{g}_1 \oplus \cdots \oplus \mathfrak{g}_r
\end{align*}
as a direct sum of simple ideals. Let $\Delta_i$ be the Dynkin diagram of $\mathfrak{g}_i$. The map $\Phi$ sends $[\mathfrak{g}]$ to
\begin{align*}
\Delta_1 \sqcup \cdots \sqcup \Delta_r.
\end{align*}
When we apply $\Psi$, each connected component $\Delta_i$ is replaced by the unique simple Lie algebra with that diagram, up to isomorphism. Since $\mathfrak{g}_i$ itself has diagram $\Delta_i$, the chosen simple Lie algebra is isomorphic to $\mathfrak{g}_i$. Hence the resulting semisimple Lie algebra is isomorphic to
\begin{align*}
\mathfrak{g}_1 \oplus \cdots \oplus \mathfrak{g}_r,
\end{align*}
which is $\mathfrak{g}$. Therefore
\begin{align*}
\Psi(\Phi([\mathfrak{g}])) = [\mathfrak{g}].
\end{align*}
Conversely, begin with a finite disjoint union
\begin{align*}
\Delta = \Delta_1 \sqcup \cdots \sqcup \Delta_r.
\end{align*}
The map $\Psi$ assigns to it the isomorphism class of
\begin{align*}
\mathfrak{s}_1 \oplus \cdots \oplus \mathfrak{s}_r,
\end{align*}
where each $\mathfrak{s}_i$ is simple with Dynkin diagram $\Delta_i$. Applying $\Phi$ to this direct sum records the Dynkin diagrams of precisely those simple summands, so it returns
\begin{align*}
\Delta_1 \sqcup \cdots \sqcup \Delta_r.
\end{align*}
Thus
\begin{align*}
\Phi(\Psi(\Delta)) = \Delta.
\end{align*}
The two assignments are inverse to one another, so the desired bijection follows.
[/guided]
[/step]
[step:Translate the bijection into the isomorphism criterion]
The bijection just proved implies that finite-dimensional semisimple Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$ over $k$ satisfy
\begin{align*}
\mathfrak{g} \cong \mathfrak{h}
\end{align*}
if and only if
\begin{align*}
\Phi([\mathfrak{g}]) \cong \Phi([\mathfrak{h}])
\end{align*}
as finite disjoint unions of connected finite type Dynkin diagrams. This is exactly the stated classification of isomorphism classes of finite-dimensional semisimple Lie algebras over $k$ by finite disjoint unions of connected finite type Dynkin diagrams.
[/step]
