The Unapologetic Mathematician

Mathematics for the interested outsider

Ideals of Lie Algebras

As we said, a homomorphism of Lie algebras is simply a linear mapping between them that preserves the bracket. I want to check, though, that this behaves in certain nice ways.

First off, there is a Lie algebra 0. That is, the trivial vector space can be given a (unique) Lie algebra structure, and every Lie algebra has a unique homomorphism L\to0 and a unique homomorphism 0\to L. This is easy.

Also pretty easy is the fact that we have kernels. That is, if \phi:L\to L' is a homomorphism, then the set I=\left\{x\in L\vert\phi(x)=0\in L'\right\} is a subalgebra of L. Indeed, it’s actually an “ideal” in pretty much the same sense as for rings. That is, if x\in L and y\in I then [x,y]\in I. And we can check that


proving that \mathrm{Ker}(\phi)\subseteq L is an ideal, and thus a Lie algebra in its own right.

Every Lie algebra has two trivial ideals: 0\subseteq L and L\subseteq L. Another example is the “center” — in analogy with the center of a group — which is the collection Z(L)\subseteq L of all z\in L such that [x,z]=0 for all x\in L. That is, those for which the adjoint action \mathrm{ad}(z) is the zero derivation — the kernel of \mathrm{ad}:L\to\mathrm{Der}(L) — which is clearly an ideal.

If Z(L)=L we say — again in analogy with groups — that L is abelian; this is the case for the diagonal algebra \mathfrak{d}(n,\mathbb{F}), for instance. Abelian Lie algebras are rather boring; they’re just vector spaces with trivial brackets, so we can always decompose them by picking a basis — any basis — and getting a direct sum of one-dimensional abelian Lie algebras.

On the other hand, if the only ideals of L are the trivial ones, and if L is not abelian, then we say that L is “simple”. These are very interesting, indeed.

As usual for rings, we can construct quotient algebras. If I\subseteq L is an ideal, then we can define a Lie algebra structure on the quotient space L/I. Indeed, if x+I and y+I are equivalence classes modulo I, then we define

\displaystyle [x+I,y+I]=[x,y]+I

which is unambiguous since if x' and y' are two other representatives then x'=x+i and y'=y+j, and we calculate

\displaystyle [x',y']=[x+i,y+j]=[x,y]+\left([x,j]+[i,y]+[i,j]\right)

and everything in the parens on the right is in I.

Two last constructions in analogy with groups: the “normalizer” of a subspace K\subseteq L is the subalgebra N_L(K)=\left\{x\in L\vert[x,K]\in K\right\}. This is the largest subalgebra of L which contains K as an ideal; if K already is an ideal of L then N_L(K)=L; if N_L(K)=K we say that K is “self-normalizing”.

The “centralizer” of a subset X\subseteq L is the subalgebra C_L(X)=\left\{x\in L\vert[x,X]=0\right\}. This is a subalgebra, and in particular we can see that Z(L)=C_L(L).

August 13, 2012 - Posted by | Algebra, Lie Algebras


  1. […] seen that the category of Lie algebras has a zero object and kernels; now we need cokernels. It would be nice to just say that if is a homomorphism then is the […]

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  3. […] killed by every . But this means that for all , while . That is, is strictly contained in the normalizer […]

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  4. Reblogged this on Peter's ruminations and commented:
    in Turing’s on permutations paper, he refers (upon editing) to normalizers, idealizers etc. We can get a feel for what these are:- (the ideal is a bit like the 0, in Hs = 0 for LDPCs)

    Comment by | August 27, 2012 | Reply

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