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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!

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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 $\mathfrak h\hookrightarrow\mathfrak g$.

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  • $\begingroup$ 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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