I'm going through a proof of the theorem, any finite dimensional module of a semisimple Lie algebra is semisimple, from page 10 of these notes (pdf). I am having a difficult time understanding few parts of it and I would appreciate any explanations. I labeled the parts I would like to have some explanations (1), (2), (3) in the proof below. Thank you very much.

The proof goes as follows: Let $L$ be a semisimple Lie algebra where the base field $K$. It suffices to show that any submodule $A$ of a finite dimensional $L$-module $V$ has a complement which is invariant under $L$. First the case when $A$ is of codimension $1$ is proved (which I am fine with).

The general case (which I paraphrase the proof): We are looking for a projection $\pi: V \rightarrow A$, which commutes with the action of $L$. If we can find such a projection, then we are done because $V = A \oplus \ker(\pi)$. Let $\mathcal{V}$ and $\mathcal{A}$ denote the spaces of linear maps $V \rightarrow A$ whose restriction to $A$ is a homothety respectively $0$. $\mathcal{A}$ is a submodule of codimension $1$ in $\mathcal{V}$ (1).

Any element of $\mathcal{V}$ with non-zero restriction to $A$ is a scalar multiple of a projection from $V$ onto $A$ (2). The space $\mathcal{V}$ is mapped into $\mathcal{A}$ under the action of $L$ on $Hom(V,A)$, since $$ (x \cdot \phi)(a) = x \cdot \phi(a)- \phi(x \cdot a) = 0 $$
for any $a \in A$ and any $\phi$ which is a homothery on $A$. The last identity also shows that those $\phi \in \mathcal{V}$ such that $K \phi$ is stable under $L$ and their restriction to $A$ are not zero are exactly the multiples of the projections $V \rightarrow A$ which commutes with $L$ (3). Hence, by our assumption we can find a submodule $K \phi$ such that $\mathcal{V} = \mathcal{A} \oplus K \phi$. Since the restriction of $\phi$ to non-zeoro we are done.


Regarding your question (1), this follows from the way you defined $\mathcal{A}$ and $\mathcal{V}$, as the subspaces of $\hom(V,A)$ of maps $T:V\to A$ such that $T\in \mathcal{A}$ iff $T\mid_A=0$ and $T\in\mathcal{V}$ iff $T\mid_A$ is a homothety. Since $0$ is a homothety, it holds that $\mathcal{A}\subseteq\mathcal{V}$. Also, we know that $\mathcal{A}$ is a proper subspace of $\mathcal{V}$. For example, if we take $\lbrace{v_1,\ldots,v_m}\rbrace$ to be a basis for $A$, and extend it to a basis $\lbrace{v_1,\ldots,v_n}\rbrace$ of $V$, then the projection map $\phi:V\to A$, defined by $$\phi(v_i)=\begin{cases}v_i&i\le m\\0 &i>m\end{cases}$$ is an element of $\mathcal{V}\setminus \mathcal{A}$.

To see that $\mathcal{A}$ has codimension one, it suffices to prove that $\mathcal{V}=K\cdot \phi+\mathcal{A}$. The inclusion $\supseteq$ here is trivial. Coversely, given $\psi\in\mathcal{V}$, there exists $\lambda\in K$ such that $\psi(a)=\lambda a$ for all $a\in A$. In particular, applying $\psi$ to the basis above, we have that $\psi-\lambda\phi$ vanishes on $A$, and hence is an element of $\mathcal{A}$. Thus $\psi=\lambda\phi+(\psi-\lambda\phi)\in K\phi+\mathcal{A}$.

Regarding point (2), I'm actually unsure what the author meant here- the definition I know for projection is of a map $\phi:V\to V$ such that $\phi^2=1_V$, but perhaps he meant a map $\phi:V\to A$ such that $\phi\mid_A=1_A$. Given the latter interpretation, the previous paragraph shows that this is indeed the case.

For point (3), let's make sure that we understand the displayed equation and what it means here. If $\psi\in\mathcal{V}$ is given, then there exists $\lambda\in K$ such that $\psi(a)=\lambda a$ for all $a\in A$. Taking $x\in L$, and recalling that $A$ is an $L$-module, we have that $\psi(x\cdot a)=\lambda x\cdot a$, just because $x\cdot a\in A$ as well. So the equation holds because $$(x\phi)(a)=x\phi(a)-\phi(xa)=\lambda x a-\lambda x a=0. $$ In particular $x\phi\in\mathcal{A}$. Note that this equation does not tell you anything about how $\phi$ acts on elements of $V\setminus A$.

Now, to understand (3), let $\phi\in\mathcal{V}$ be such that $K\phi$ is $L$-stable. In particular, the assumption that $K\phi$ is $L$-stable means that there exists $\mu\in K$ such that $(x\cdot\phi)(v)=\mu\phi(v)$ for all $v\in V$. Since this necessarily holds for $v\in A$, it implies that $\mu\cdot \phi(a)=0$ for all $a\in A$. Thus, either $\phi$ itself vanishes on $A$, or $\mu=0$. In the latter case, by the definition of the action of $L$ on $\hom(V,A)$, it follows that any such $\phi\in\mathcal{V}$ which does not vanish on $A$ must satisfy $(x\cdot \phi)=0$, for all $x\in L$, and hence commute with the action of $L$.

Hope that clarifies things.

  • $\begingroup$ Thank you very much for your help!! It is very clear! I guess one of the things I was confused with was that when it says "any element of $\mathcal{V}$ with non-zero restriction to $A$ is a scalar multiple of a projection from $V$ onto $A$" it actually means that the restriction to $A$ is a scalar multiple of the identity map on $A$, and that it shouldn't be taken literally... $\endgroup$ – Johnny T. Feb 26 at 19:37

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