Let $G$ be a locally compact topological group and $H$ a closed subgroup. Choose a left Haar measure $d\zeta$ for $H$, and let $d\mu$ be any measure for $G$. Also let $f$ and $g$ be continuous compactly supported real functions on $G$. I'm stumped by a step in a proof where the following equality is asserted :

$$\int_H \Delta_H(\zeta^{-1}) \Bigg[\int_G f(x\zeta^{-1})g(x)d\mu(x)\Bigg]d\zeta = \int_G g(x)\Bigg[ \int_H f(x\zeta) d\zeta\Bigg] d\mu (x).$$

I can only make the left hand side look like $$\int_G g(x) \Bigg[\int_H \Delta_H(\zeta^{-1})f(x\zeta^{-1})d\zeta \Bigg]d\mu (x)$$

Clearly I'm supposed to use the properties of the modular function $\Delta_H$, but I don't know what to do when the argument for $\Delta_H$ is the variable I'm integrating over. Any help would be greatly appreciated. Thanks!

  • $\begingroup$ Is $\Delta_H(\zeta^{-1})d\zeta$ not a right Haar measure on H, or am I wrong? $\endgroup$ – Lucien Nov 21 '13 at 18:13
  • $\begingroup$ @Lucien Oh interesting! Looks like it is right invariant. What does that do though? $\endgroup$ – Tim kinsella Nov 21 '13 at 19:23

Define a measure $d\zeta^{-1}$ on $H$ by the condition $$ \int f(\zeta)\,d\zeta^{-1}=\int f(\zeta^{-1})\,d\zeta. $$ I claim that there is an equality of measures $$ \Delta_H(\zeta)\,d\zeta^{-1} = d\zeta. $$ Integrating $f(x\zeta)$ against both these measures shows the inner integral on the right hand side of the original equality is equal to the inner integral in your transformation of the left hand side.

To prove the equality of measures: first one checks that $\Delta_H(\zeta)\,d\zeta^{-1}$ is a left Haar measure (this is basically equivalent to Lucien's observation that $\Delta_H(\zeta^{-1})\,d\zeta$ is a right Haar measure). This means there is a constant $C>0$ such that $\Delta_H(\zeta)\,d\zeta^{-1}=C\,d\zeta$. Making the substitution $\zeta\leftrightarrow\zeta^{-1}$ shows that $C^2=1$, so that $C=1$.


Let us call $d\lambda(\zeta)$ your left Haar measure on $H$. Notice that, as already pointed out by Lucien, $\Delta_H(\zeta^{-1})d\lambda(\zeta)$ is a right Haar measure on $H$. Actually, it is exactly the measure $d\lambda(\zeta^{-1})$ (e.g. see Folland, "A course in abstract harmonic analysis", p.48), i.e. the measure $\rho$ defined by $\rho(E)=\lambda(E^{-1})$. So your left hand side square bracket becomes

$$\int_H f(x\zeta^{-1})d\lambda(\zeta^{-1})$$

Now apply the general formula for the push-forward measure under the map $(\cdot)^{-1}\colon H\to H$ to get

$$\int_H f(x\zeta^{-1})d\lambda(\zeta^{-1})=\int_H f(x(\zeta^{-1})^{-1})d\lambda (\zeta)=\int_H f(x\zeta)d\lambda(\zeta)$$ and we're done.

  • $\begingroup$ For $f(\zeta)=1-\zeta$ on $0<\zeta\leq 1$ and 0 otherwise, and $x>0$, I compute $\int_0^\infty f(x\zeta)\,\frac{d\zeta}{\zeta}=+\infty$. $\endgroup$ – Julian Rosen Dec 13 '13 at 17:11
  • $\begingroup$ I stand corrected. I'm going to edit. $\endgroup$ – johndoe Dec 14 '13 at 0:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.