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Murat Güngör
Murat Güngör
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How do these two Haar measures on $\mathrm{SL}SL(2,\mathbb{R})$ compare?

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Murat Güngör
Murat Güngör
  • 486
  • 3
  • 11

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}a\,\mathrm{d}n\,\mathrm{d}k$$\mathrm{d}_1 x=\mathrm{d}k\,\mathrm{d}n\,\mathrm{d}a$ so that $$\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?

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 $\mathrm{d}_1 x=\mathrm{d}a\,\mathrm{d}n\,\mathrm{d}k$ 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?

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?

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Murat Güngör
Murat Güngör
  • 486
  • 3
  • 11

How do these two Haar measures on $\mathrm{SL}(2,\mathbb{R})$ compare?

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 $\mathrm{d}_1 x=\mathrm{d}a\,\mathrm{d}n\,\mathrm{d}k$ 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?