How to show that $\int_G f(t) dt = \int_G f(t^{-1}) dt$?

Question

I am reading the lecture notes. On page 34, line 13, it is said that $\int_G f(t) dt = \int_G f(t^{-1}) dt$. How to prove this identity? I think that if we let $s=t^{-1}$, then $\int_Gf(t^{-1})dt=\int_{G^{-1}}f(s)d(s^{-1}) = \int_G f(s)\frac{-1}{s^2}ds$. But I didn't get $\int_G f(t) dt = \int_G f(t^{-1}) dt$. Thank you very much.

Answer 1

As mentioned in the comments of the main thread and FelxCQ's answer, the lecture notes state that $t$ is a right invariant Haar measure on a compact Lie group $G$. It is demonstrated in Haar measure on a compact group that for any subset $A$ of $G$: $$t(A) = t(A^{-1})$$

(Note that your the notes also specify that $t$ has been normalized, i.e. $\int_G dt = 1$)

where $A$ is the set of inverses of group elements. i.e. $t$ is invariant under the inverse operation.

Now $$\int_G f(t^{-1}) dt = \int_{G^{-1}} f(t) dt^{-1} = \int_G f(t) dt$$

Answer 2

I believe this notation denotes the group inverse, not the real involution function.

I don't know how to prove this, but intuitively, as $g$ describes the whole group $G$, $g^{-1}$ also describes $G$ "at the same rate", so both integrals should be equal.

Answer 3

Because G is compact, so is unimodular. Therefore Haar modulus of G is constant function 1. So the equality is clear.