Facts About Solvability and Nilpotence

Solvability is an interesting property of a Lie algebra , in that it tends to “infect” many related algebras. For one thing, all subalgebras and quotient algebras of are also solvable. For the first count, it should be clear that if then . On the other hand, if is a quotient epimorphism then any element in has a representative in , so if the derived series of bottoms out at then so must the derived series of .

As a sort of converse, suppose that is a solvable quotient of by a solvable ideal ; then is itself solvable. Indeed, if and is the quotient epimorphism then , as we saw above. That is, , but since is solvable this means that — as a subalgebra — is solvable, and thus is as well.

Finally, if and are solvable ideals of then so is . Here, we can use the third isomorphism theorem to establish an isomorphism . The right hand side is a quotient of , and so it’s solvable, which makes a solvable quotient by a solvable ideal, meaning that is itself solvable.

As an application, let be any Lie algebra and let be a maximal solvable ideal, contained in no larger solvable ideal. If is any other solvable ideal, then is solvable as well, and it obviously contains . But maximality then tells us that , from which we conclude that . Thus we conclude that the maximal solvable ideal is unique; we call it the “radical” of , written .

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

In general, the quotient is semisimple, since if it had a solvable ideal it would have to be of the form for some containing . But if is a solvable quotient of by a solvable ideal, then must be solvable, which means it must be contained in the radical of . Thus the only solvable ideal of is , 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 — where is the center of — is nilpotent then is as well. Indeed, to say that is to say that for some . But then .

Finally, if is nilpotent, then . To see this, note that if is the first term of the descending central series that equals zero, then , since the brackets of everything in with anything in are all zero.