The Unapologetic Mathematician

Mathematics for the interested outsider

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