# The Unapologetic Mathematician

## 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

$\displaystyle\phi\left([x,y]\right)=\left[\phi(x),\phi(y)\right]=\left[\phi(x),0\right]=0$

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 $Z(L)=C_L(L)$.

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Posted by John Armstrong | Algebra, Lie Algebras

## 4 Comments	»

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 [...]

Pingback by | August 14, 2012 | Reply

2. [...] if — where is the center of — is nilpotent then is as well. Indeed, to say that is to say that for some . But then [...]

Pingback by | August 22, 2012 | Reply

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

Pingback by | August 22, 2012 | Reply

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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