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Let $\mathfrak{g}$ be a compact Lie algebra and let $\mathfrak{t}$ be a Cartan algebra on it. Consider the adjoint representation of $\mathfrak{g}$ : $ad: \mathfrak{g}\rightarrow End(\mathfrak{g})$.

Let $H \in \mathfrak{t}$. What is the kernel of $ad(H)$? I believe it is $\mathfrak{t}$ since I've come across a statement which say that

$ad(H)$ is invertible on $\mathfrak{g}/ \mathfrak{t}$ because the adjoint representation of $\mathfrak{t}$ on $\mathfrak{g}/\mathfrak{t}$ has only non-zero roots.

But I don't know how to prove it, because the thing which confuse me is the fact that I see a lot while I'm learning about representation theory the space of kernel of roots, if it is trivial they wouldn't work with it !

Any help about the kernel of $ad(H), H \in \mathfrak{t} $ and why is $ ad(H): \mathfrak{g/t} \rightarrow \mathfrak{g/t}$ invertible would be greatly appreciated!

Mira
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    Your title sounds as if you are looking for the kernel of the map $ad: \mathfrak g \rightarrow End(\mathfrak g)$, but in the question body you are actually asking about the kernel of $ad(H)$ for specific $H$. Those are different things, so please clarify. In any case, the kernel of $ad$ is the centre of $\mathfrak g$, while for any element $x \in \mathfrak g$, the kernel of $ad(x)$ is the centralizer of $x$ in $\mathfrak g$. For elements from a maximal torus, the centraliser is often studied. Look up "regular (semisimple) elements". – Torsten Schoeneberg May 16 '21 at 02:01
  • @Torsten, thank you for your comment! I'm interested in the kernel of $ ad(H), H \in \mathfrak{t}$, I'll fix the title . Could you please elaborate more in this. – Mira May 16 '21 at 16:54

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If $\mathfrak g$ is any Lie algebra and $x \in \mathfrak g$ is any element, the kernel of the map $ad(x): \mathfrak g \rightarrow \mathfrak g, y \mapsto [x,y]$ is the centraliser of $x$ in $\mathfrak g$, $C_\mathfrak{g}(x):= \lbrace y \in \mathfrak g: [x,y]=0 \rbrace$.

If $\mathfrak g$ is any semisimple Lie algebra and $x \in \mathfrak g$ is any semisimple element, the centraliser $C_\mathfrak{g}(x)$ is a reductive subalgebra of $\mathfrak g$, see Centralizer of semi-simple element in semi-simple Lie algebra. Such elements are called regular if their centraliser is a Cartan subalgebra; equivalently, they lie in exactly one CSA, cf. If $x\in L$ lies in exactly one Cartan subalgebra, then $x$ is regular. If they don't, their centralisers are non-abelian i.e. contain a semisimple part.

In general, "most" semisimple elements are regular (vaguely said, the irregular ones lie in subspaces of lower dimension).

Examples: Let $\mathfrak g = \mathfrak{su}_n$, i.e. the skew-hermitian $n \times n$-matrices of trace $0$ viewed as real Lie algebra (of dimension $n^2-1$). Let's look at the standard Cartan $\mathfrak t$ given by the diagonal matrices in there (these are the diagonal matrices whose entires are all purely imaginary and add up to $0$).

For $n=2$, every element $\neq 0$ in the one-dimensional $\mathfrak t$ is regular.

For $n=3$, the irregular elements in the two-dimensional $\mathfrak t$ are of the form $\pmatrix{xi &0&0 \\ 0&x i&0 \\ 0&0&-2xi}$ or $\pmatrix{xi &0&0 \\ 0&-2x i&0 \\ 0&0&xi}$ or $\pmatrix{-2xi &0&0 \\ 0&x i&0 \\ 0&0&xi}$ i.e. the union of three one-dimensional subspaces. (Exercise: Find the centralisers of each.)

For $n=4$, e.g. the element $\pmatrix{i&0&0&0\\0&-i&0&0\\0&0&0&0\\0&0&0&0}$ has centraliser $\pmatrix{\ast&0&0&0\\0&\ast&0&0\\0&0&\ast&\ast\\0&0&\ast&\ast}$. Exercise: What other elements in $\mathfrak t$ are irregular?

Note that although the examples are compact Lie algebras, nowhere have we really used that. Indeed, the only interesting thing here would be that in a compact semisimple Lie algebra, all elements are semisimple and thus belong to some CSA. Thus your question is actually asking for the centraliser of an arbitrary element in such a Lie algebra.

Torsten Schoeneberg
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