# The Unapologetic Mathematician

## 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 | Algebra, Lie Algebras | 7 Comments