## Kernel of adjoint representation of a Lie group

Let $G$ be a matrix Lie group with associated Lie algebra $\mathfrak{g}$. $\mathrm{Aut}(\mathfrak{g})$ denotes the group of Lie algebra automorphisms of $\mathfrak{g}$ under composition. Recall we have the adjoint representation

$\mathrm{Ad} : G \to \mathrm{Aut}(\mathfrak{g})$

of $G$, given by

$g \mapsto (x \mapsto gxg^{-1})$.

Of course, $gxg^{-1}$ is always in the Lie algebra, essentially due to the fact that conjugation “passes through” the exponential unmolested.

It was remarked in the PMATH 763 lectures that if $Z(G)$ denotes the centre of $G$, then $\ker \mathrm{Ad} = Z(G)$. However, after doing some reading I found this was not quite true and that indeed $\ker \mathrm{Ad} = C_G(G_0)$, where $G_0$ denotes the connected component of the identity, and $C_G(S)$ denotes the centralizer of $S$ in $G$. I couldn’t find a proof anywhere, but luckily it wasn’t too difficult to solve.

The proof of this gives us a bit of insight into the imperfect but palatable relationship between commutativity in $G$ and commutativity in $\mathfrak{g}$.

If $g \in C_G(G_0)$, to show $g \in \ker \mathrm{Ad}$ the idea is that ${\log} \circ {\exp} = \mathrm{id}$ in a sufficiently small neighbourhood of the identity. Using this we can prove (by scaling) that $gXg^{-1} = X$ for any $X \in \mathfrak{g}$, which is what we wanted. The details are a quick exercise.

Conversely, if $g \in \ker \mathrm{Ad}$, then to show $g \in C_G(G_0)$ the idea is to use the fact that $G_0$ is generated by ${\exp}(\mathfrak{g})$. Again I leave the details for you.

Of course, if $G$ is connected, then $G_0 = G$ and so

$C_G(G_0) = C_G(G) = Z(G)$,

the centre of $G$.