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.

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 […]

Pingback by Flags « The Unapologetic Mathematician | August 25, 2012 | Reply
[…] 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 […]

Pingback by Cartan’s Criterion « The Unapologetic Mathematician | September 1, 2012 | Reply
[…] 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 […]

Pingback by The Radical of the Killing Form « The Unapologetic Mathematician | September 6, 2012 | Reply
[…] if is semisimple then there is a collection of ideals, each of which is simple as a Lie algebra in its own right, […]

Pingback by Decomposition of Semisimple Lie Algebras « The Unapologetic Mathematician | September 8, 2012 | Reply
[…] 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 […]

Pingback by All Derivations of Semisimple Lie Algebras are Inner « The Unapologetic Mathematician | September 11, 2012 | Reply

About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

RSS Feeds

RSS - Posts
RSS - Comments
Feedback

Got something to say? Anonymous questions, comments, and suggestions at Formspring.me!
Subjects

Subjects
Archives

August 2012

M T W T F S S

1 2 3 4 5

6 7 8 9 10 11 12

13 14 15 16 17 18 19

20 21 22 23 24 25 26

27 28 29 30 31

« Jul Sep »
Search for:

The Unapologetic Mathematician

Mathematics for the interested outsider