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.

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About mlbaker

just another guy trying to make the diagrams commute.
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