MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

By using the Iwasawa decomposition, one obtains a (bi-invariant) Haar measure on $G:=\mathrm{SL}(2,\mathbb{R})$ which can be symbolically written as $\mathrm{d}x=\mathrm{d}a\,\mathrm{d}n\,\mathrm{d}k$, the measures appearing on the right-hand side being the usual ones. This actually means $$\int_G f(x)\,\mathrm{d}x=\int_K \int_N \int_A f(ank)\,\mathrm{d}a\,\mathrm{d}n\,\mathrm{d}k$$ for suitable $f$. (Here the order in $ank$, which is expressed by the notation $\mathrm{d}x=\mathrm{d}a\,\mathrm{d}n\,\mathrm{d}k$, is important, whereas the order of the triple integral is immaterial.) Analogously, one can define another Haar measure $\mathrm{d}_1 x=\mathrm{d}k\,\mathrm{d}n\,\mathrm{d}a$ so that $$\int_G f(x)\,\mathrm{d}_1 x=\int_A \int_N \int_K f(kna)\,\mathrm{d}k\,\mathrm{d}n\,\mathrm{d}a.$$ We must have $\mathrm{d}_1 x=c\cdot\mathrm{d}x$ for some $c>0$. Is $c$ equal to 1?

share|cite|improve this question
The modular function is $\Delta$ is constant one for $SL_2(\mathbb{R})$ and any other reductive group over a local field. – Marc Palm May 10 '13 at 11:45
up vote 4 down vote accepted

Yes, since you get one from the other by applying the inversion and there is a general rule for the inversion that says $\int_Gf(x^{-1})dx=\int_Gf(x)dx\Delta(x)$, where $\Delta(x)$ is the modular function.

share|cite|improve this answer

Use Fubini to get:

$I:=\int_G f(x) \mathrm{d}_1x = \int_A \int_N \int_K f(kna) \mathrm{d}k\mathrm{d}n\mathrm{d}a=\int_K \int_N \int_A f(kna) \mathrm{d}a\mathrm{d}n\mathrm{d}k$.

Use unimodularity of K,N,A to get:

$I=\int_K \int_N \int_A f(k^{-1}n^{-1}a^{-1}) \mathrm{d}a\mathrm{d}n\mathrm{d}k=\int_K \int_N \int_A f((ank)^{-1}) \mathrm{d}a\mathrm{d}n\mathrm{d}k$.

Now use the unimodularity of $SL(2,\mathbb{R})$:

$I=\int_G f(x^{-1}) \mathrm{d}x=\int_G f(x) \mathrm{d}x$.

“Using unimodularity” means using the identity given by anton: $\int_G f(x^{-1}) \mathrm{d}x = \int_G f(x)\Delta(x) \mathrm{d}x$ where you use the left Haar measure (integrating $f(x^{-1})$ with respect to the left Haar measure is the same as integrating $f$ with respect to the right Haar measure). As you noticed for these groups there exists a bi-invariant Haar measure, thus the groups are unimodular (i. e. $\Delta(x)=1$).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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