Ytukyg

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 $mathcal Dmathfrak g$ 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?

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 ${cal G}=Soplus rad({cal G})$, where $S$ is semi-simple and $rad({cal G})$ solvable. We have $[S,S]=S$, this implies that $Ssubset [{cal G},{cal G}]$ and $[{cal G},{cal G}]=Soplus U$ where $Usubset rad({cal G})$ 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 $xin V_i$ such that $[rad({cal G}),rad({cal G})]x=0$, since $[rad({cal G}),rad({cal G})]$ is an ideal, you deduce that $V={ xin V_i, [rad({cal G}),rad({cal G})]x=0}$ 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.

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.

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 $p:mathfrak grightarrow mathfrak g /I,q:mathfrak grightarrow mathfrak g/J$ 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, $[A,mathfrak g]subseteq Icap J=0$, so $Asubseteq Z(mathfrak g)$. 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, $Z(mathfrak g)cap mathfrak g^2subseteq I$, a similar argument gives $Z(mathfrak g)cap mathfrak g^2subseteq J$, so $Z(mathfrak g)cap mathfrak g^2subseteq I cap J=0$, 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 $V=Soplus T$. $V$ 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. $square$