Just a simply question. I came across the following statement which is used for deriving Weyl's integral formula:

''$\text{Ad}_G(h)|_{\mathfrak{h}} = \text{Ad}_H(h)$ due to functoriality in the Lie group for the adjoint representation''.

Therefore we have $\text{Ad}_G(t_0^{-1})$ mod $\mathfrak{t}$ = Ad$_{G/T}(t_0^{-1})$.

Could somebody define for me the ''functoriality'' in the context of adjoint representation?

Thank you very much for your attention!


What is meant here is that the adjoint representation is nicely compatible with homomorphisms of Lie groups. For a homomorphism $\phi:H\to G$, let $\phi':\mathfrak h\to\mathfrak g$ be the derivative. Then it is a basic fact of Lie theory that $\phi(exp(tX))=\exp(t\phi'(X))$ for all $X\in\mathfrak h$. Now for $h\in H$ and $X\in\mathfrak h$, you get Ad$(h)(X)$ as the derivative at $t=0$ of the curve $h\exp(tX)h^{-1}$. Using the above fact, you get $\phi(h\exp(tX)h^{-1})=\phi(h)\exp(t\phi'(X))\phi(h)^{-1}$. Differentiating at $t=0$, we obtain $\phi'(Ad(h)(X))=Ad(\phi(h))(\phi'(X))$, which is the functorial property the argument refers to. You just have to apply this to the inclusion $i:H\to G$ whose derivative $i'$ is the inclusion $\mathfrak h\hookrightarrow\mathfrak g$.

  • $\begingroup$ Thanks a lot Andreas! That is very clear explanation! But it seems like it is better called the consequence of the local diffeomorphism property of exponential map! $\endgroup$ – PhysicsMath May 2 '16 at 17:40
  • $\begingroup$ The result certainly is a consequence of the fact that the exponential map is a local diffeomorphism, but this is true for large parts of Lie theory. It really is the compatibility of the adjoint representation with homomorphisms. $\endgroup$ – Andreas Cap May 3 '16 at 6:43
  • $\begingroup$ Got you Andreas! You are emphasizing the compatibility of the adjoint representation with homomorphisms over the local diffeomorphism property of the exponential map. $\endgroup$ – PhysicsMath May 9 '16 at 2:10

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