Facts About Solvability and Nilpotence

Solvability is an interesting property of a Lie algebra $L$ , in that it tends to “infect” many related algebras. For one thing, all subalgebras and quotient algebras of $L$ are also solvable. For the first count, it should be clear that if $K\subseteq L$ then $K^{(i)}\subseteq L^{(i)}$ . On the other hand, if $L\to L/I$ is a quotient epimorphism then any element in $(L/I)^{(i)}$ has a representative in $L^{(i)}$ , so if the derived series of $L$ bottoms out at $0$ then so must the derived series of $L/I$ .

As a sort of converse, suppose that $L/I$ is a solvable quotient of $L$ by a solvable ideal $I$ ; then $L$ is itself solvable. Indeed, if $(L/I)^{(n)}=0$ and $\pi:L\to L/I$ is the quotient epimorphism then $\phi(L^{(n)})=0$ , as we saw above. That is, $L^{(n)}\subseteq I$ , but since $I$ is solvable this means that $L^{(n)}$ — as a subalgebra — is solvable, and thus $L$ is as well.

Finally, if $I$ and $J$ are solvable ideals of $L$ then so is $I+J$ . Here, we can use the third isomorphism theorem to establish an isomorphism $(I+J)/J\cong I/(I\cap J)$ . The right hand side is a quotient of $I$ , and so it’s solvable, which makes $(I+J)/J$ a solvable quotient by a solvable ideal, meaning that $I+J$ is itself solvable.

As an application, let $L$ be any Lie algebra and let $S$ be a maximal solvable ideal, contained in no larger solvable ideal. If $I$ is any other solvable ideal, then $S+I$ is solvable as well, and it obviously contains $S$ . But maximality then tells us that $S+I=S$ , from which we conclude that $I\subseteq S$ . Thus we conclude that the maximal solvable ideal $S$ is unique; we call it the “radical” of $L$ , written $\mathrm{Rad}(L)$ .

In the case that the radical of $L$ is zero, we say that $L$ is “semisimple”. In particular, a simple Lie algebra is semisimple, since the only ideals of $L$ are itself and $0$ , and $L$ is not solvable.

In general, the quotient $L/\mathrm{Rad}(L)$ is semisimple, since if it had a solvable ideal it would have to be of the form $I/\mathrm{Rad}(L)$ for some $I\subseteq L$ containing $\mathrm{Rad}(L)$ . But if $I/\mathrm{Rad}(L)$ is a solvable quotient of $I$ by a solvable ideal, then $I$ must be solvable, which means it must be contained in the radical of $L$ . Thus the only solvable ideal of $L/\mathrm{Rad}(L)$ is $0$ , as we said.

We also have some useful facts about nilpotent algebras. First off, just as for solvable algebras all subalgebras and quotient algebras of a nilpotent algebra are nilpotent. Even the proof is all but identical.

Next, if $L/Z(L)$ — where $Z(L)$ is the center of $L$ — is nilpotent then $L$ is as well. Indeed, to say that $(L/Z(L))^n=0$ is to say that $L^n\subseteq Z(L)$ for some $n$ . But then $L^{n+1}=[L,L^n]\subseteq[L,Z(L)]=0$ .

Finally, if $L\neq0$ is nilpotent, then $Z(L)\neq0$ . To see this, note that if $L^{n+1}$ is the first term of the descending central series that equals zero, then $0\neq L^n\subseteq Z(L)$ , since the brackets of everything in $L^n$ with anything in $L$ are all zero.

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August 21, 2012 - Posted by John Armstrong | Algebra, Lie Algebras

7 Comments »

[...] elements. By induction on the dimension of we assume that is actually nilpotent, which proves that itself is nilpotent. Share this:StumbleUponDiggRedditLike this:LikeBe the first to like [...]

Pingback by Engel’s Theorem « The Unapologetic Mathematician | August 22, 2012 | Reply
[...] is a subalgebra of , so we know it’s also solvable, so induction tells us that there’s a common eigenvector for the [...]

Pingback by Lie’s Theorem « The Unapologetic Mathematician | August 25, 2012 | Reply
[...] We let this span the one-dimensional subspace and consider the action of on the quotient . Since we know that the image of in will again be solvable, we get a common eigenvector . Choosing a pre-image [...]

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[...] thus automatically solvable. The image is thus the solvable quotient of by a solvable kernel, so we know that itself is solvable. Share this:StumbleUponDiggRedditLike this:LikeBe the first to like [...]

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[...] recall that there was another “radical” we’ve mentioned: the radical of a Lie algebra is its maximal solvable ideal. This is not necessarily the same as the radical of [...]

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[...] if is semisimple then there is a collection of ideals, each of which is simple as a Lie algebra in its own right, [...]

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[...] turns out that all the derivations on a semisimple Lie algebra are inner derivations. That is, they’re all of the form for some . We know that [...]

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