Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let ‎$‎G‎$ ‎be a‎ ‎locally ‎compact ‎group‎, ‎$‎H‎$ ‎be a‎ ‎closed ‎subgroup ‎and ‎‎$‎N‎$ ‎be a‎ ‎normal ‎subgroup ‎of ‎‎$‎G‎$ ‎such ‎that ‎‎$‎H‎\subseteq ‎N‎$‎. ‎How ‎can ‎we get $$\int_{G/H} ‎f(xH)d‎\mu_{G/H}‎(xH)=‎\mu_{N/H}‎(N/H)‎‎‎‎‎‎‎\int_{G/N}f(xN)d‎\mu_{G/N}(xN)\ \ ?$$

share|improve this question
add comment

closed as too localized by Dan Petersen, Bill Johnson, Suvrit, Andreas Blass, Mark Meckes Sep 7 '12 at 1:31

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

1 Answer

First, in order for an invariant measure on $G/H$ to exist, you need for the modular functions the equality $$ \Delta_G|_H=\Delta_H. $$ Under these circumstances the left hand side if the equation is defined, provided the function $f$ is $H$-invariant. If in addition $f$ is even $N$-invariant the right hand side is defined as well (the modular condition is automatic, as $N$ is normal). If the measure $\mu_{N/H}(N/H)$ is finite, it then is possible to choose the invariant measures such that the claimed equality holds for every measurable $N$-invariant function which is such that either side of the equality exists. A proof can be found in Deitmar/Echterhoff: Principles of Harmonic Analysis.

share|improve this answer
add comment

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