A faithful completely reducible Lie algebra representation implies reductivity

Question

Suppose $\mathfrak g$ is a finite-dimensional Lie algebra over a field $k$, which we can assume of characteristic zero. In Milne's LAG, Proposition 6.4 claims that $\mathfrak g$ is a reductive Lie algebra if and only if there exists a faithful and completely reducible finite-dimensional representation of $\mathfrak g$.

I understood the proof in the book saying that if $\mathfrak g$ is reductive, we can take the direct sum of the adjoint representation for $Z(\mathfrak g) \oplus \mathcal D \mathfrak g = \mathfrak g$ given by the adjoint representation for $\mathcal D\mathfrak g$ and a direct sum of $1$-dimensional faithful representations for $Z(\mathfrak g)$ given by adding morphisms of the form $k \simeq \mathfrak{gl}(k)$. What I don't understand is the converse, for which nothing is mentioned.

The existence of this faithful completely reducible representation of $\mathfrak g$ implies that we have an inclusion $\mathfrak g \subseteq \mathfrak{gl}(V)$ for some finite-dimensional $k$-vector space $V$. Since $\mathcal D(\mathfrak{gl}(V)) = \mathfrak{sl}(V)$ is semisimple, $\mathrm{rad}(\mathfrak{sl}(V)) = 0$, thus $\mathcal D(\mathfrak g) \subseteq \mathfrak{sl}(V)$ is also semisimple. If $x \in Z(\mathfrak g) \cap \mathcal D \mathfrak g$, this also means $x \in Z(\mathcal D \mathfrak g) = 0$, so we know that $Z(\mathfrak g) \oplus \mathcal D \mathfrak g$ is an ideal of $\mathfrak g$. That's what I managed to do so far.

Two questions (which are kind of linked) :

• Why doesn't Milne mention anything about this direction? Did he forget or am I missing something obvious?
• Does anyone have a proof?

Answer 1

Let ${\cal G}$ be a Lie algebra defined over a field of characteristic $0$. Suppose that $D{\cal G}$ is semi-simple, write the Levi Decomposition ${\cal G}=S\oplus rad({\cal G})$, where $S$ is semi-simple and $rad({\cal G})$ solvable. We have $[S,S]=S$, this implies that $S\subset [{\cal G},{\cal G}]$ and $[{\cal G},{\cal G}]=S\oplus U$ where $U\subset rad({\cal G})$ thus $U$ is solvable. We deduce that $[{\cal G},{\cal G}]$ is semi-simple, if and only if $[{\cal G},{\cal G}]=S$. This implies that $[rad({\cal G},rad({\cal G})]=0$ and $rad({\cal G})$ is commutative.

But I don't understand your incomplete argument, you seem to assume that a subalgebra of a semi-simple Lie algebra is also semi-simple and this is not true.

Let $\phi:{\cal G}\rightarrow sl(V)$ be a complete reducible representation of ${\cal G}$, you can write $V=\oplus_iV_i$ as a sum of irreducible module.

Suppose that $dim V_i>1$, let $\phi_i:{\cal G}\rightarrow gl(V_i)$ be the representation induced by $\phi$ on $V_i$. Remark that $[rad({\cal G}),rad({\cal G})]$ is a nilpotent ideal. There exists $x\in V_i$ such that $[rad({\cal G}),rad({\cal G})]x=0$, since $[rad({\cal G}),rad({\cal G})]$ is an ideal, you deduce that $V=\{ x\in V_i, [rad({\cal G}),rad({\cal G})]x=0\}$ is a submodule thus it is $V_i$. This implies that $\phi_i(rad({\cal G})$ is commutative for $dim(V_i)>1$. Since $gl(k)$ is commutative, you deduce that $\phi(rad({\cal G})$ is commutative and $rad({\cal G})$ is commutative since $\phi$ is faithful.

Answer 2

You can actually find a proof of this claim in Milne, though you have to read it carefully. Right after proposition 6.4, he defines the nilpotent radical $\mathfrak{s}$ of $\mathfrak{g}$, which is the intersection of the kernels of all simple representations of $\mathfrak{g}$. Since $\mathfrak{s}$ is contained in the kernel of any semisimple representation of $\mathfrak{g}$, the hypothesis that $\mathfrak{g}$ has a faithful semisimple representation implies that $\mathfrak{s}=0$.

Milne then proves the following theorem (theorem 6.9):

Let $\mathfrak{g}$ be a Lie algebra, $\mathfrak{r}$ its radical and $\mathfrak{s}$ its nilpotent radical. Then $\mathfrak{s} = [\mathfrak{g}, \mathfrak{r}]$.

Looking at the proof, we see that the full theorem depends on proposition 6.4. However, the inclusion $[\mathfrak{g}, \mathfrak{r}] \subset \mathfrak{s}$ does not. It only depends on lemmas 6.11 and 6.12, which are elementary.

Since $\mathfrak{s}=0$, we conclude that $[\mathfrak{g}, \mathfrak{r}]=0$, so the radical $\mathfrak{r}$ is contained in the center $Z(\mathfrak{g})$. Note that the inclusion $Z(\mathfrak{g}) \subset \mathfrak{r}$ always holds, so we conclude $Z(\mathfrak{g})= \mathfrak{r}$. This is Milne's definition of an algebra being reductive.

In a sense, the $[\mathfrak{g}, \mathfrak{r}] \subset \mathfrak{s}$ part of the theorem should have been stated and proved before proposition 6.4. Proposition 6.4 then follows quite easily.

Answer 3

Here is a possibly different way of doing it:

Claim: Let $\mathfrak g$ be Lie algebra over $\mathbb{R}$, and $I,J$ be two ideals of it such that $I \cap J=0$ and such that $\mathfrak g/I$,$\mathfrak g/J$ are reductive then $\mathfrak g$ is reductive as well. Proof: Let $p:\mathfrak g\rightarrow \mathfrak g /I,q:\mathfrak g\rightarrow \mathfrak g/J$ be the natural quotient maps. Let $A$ be an abelian ideal of $\mathfrak g$, then $p[A]$ is abelian ideal $\mathfrak g/I$ and by reductivity of $\mathfrak g/I$ we get that $p[A]\subseteq Z(\mathfrak g/I)$. Hence: $$p([A,\mathfrak g])=[p(A),p(\mathfrak g)]=[p(A),\mathfrak g/I]=0 \text{ (As $p[A]\subseteq Z(\mathfrak g$/I) ) }$$ So $[A,\mathfrak g]\subseteq I$, similarly we get $[A,\mathfrak g]\subseteq J $. Hence, $[A,\mathfrak g]\subseteq I\cap J=0$, so $A\subseteq Z(\mathfrak g)$. So every abelian ideal of $\mathfrak g$ lives in its center. We also have $$p(Z(\mathfrak g) \cap {\mathfrak g}^2)\subseteq Z(\mathfrak g/I) \cap (\mathfrak g/I)^2=0 \text{ , (By reductivity of $\mathfrak g/I$) } $$ Thus, $Z(\mathfrak g)\cap \mathfrak g^2\subseteq I$, a similar argument gives $Z(\mathfrak g)\cap \mathfrak g^2\subseteq J$, so $Z(\mathfrak g)\cap \mathfrak g^2\subseteq I \cap J=0$, and this completes the proof of reductivity of $\mathfrak g$ $\square$

Now, we prove the statement: "Let $\mathfrak g$ be a finite dimensional $\mathbb{R}$- Lie algebra that admits a faithful completely reducible $\mathfrak g$-module, then $\mathfrak g$ is reductive" by induction on $\dim(\mathfrak g)$.

Proof: Let $V$ be the faithful completely reducible $\mathfrak g$-module. If this representation is irreducible, then we are done because irreducible linear lie algebras are reductive. So assume that the representation is not irreducible. Let $S$ be a proper nontrivial $\mathfrak g$-submodule of $V$. By complete reducibility, we have a $\mathfrak g$-submodule $T$ of $V$ such that $V=S\oplus T$. $V$ is completely reducible , thus so are the $\mathfrak g$-submodules $S,T$. It follows that $S$ is a faithful completely reducible $\mathfrak g/Ann(S)$-module and $T$ is a faithful completely reducible $\mathfrak g/Ann(T)$-module. Hence, by induction we have that $\mathfrak g/Ann(S)$, $\mathfrak g/Ann(T)$ are reductive. By faithfulness, we have $Ann(S)\cap Ann(T)=Ann(V)=0$ . So by previous claim, we are done. $\square$