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We get a "nice" Haar measure on $G=SL(2,R)$ in Iwasawa coordinates $G=NAK$ as follows: $dg=dx {dy\over y^2} dk$. Here $N=\{ n_x\}$, $A=\{a_y\}$ and $K=SO(2)$. Note that $dg=dn\, da\, dk$ is a product of Haar measures on the three subgroups.

Question: In what generality does this happen? If I have some nice Lie group $G$, and I have, say, a local bijection $H\times K\to G$ (in a neighborhood of $e$), is $dg=dh\, dk$? (No in general; e.g. $\bar N\times A\times N\to G=SL(2,R)$. Haar measure is a mess involving all three variables.)

Thanks!

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  • $\begingroup$ I'm not sure what you mean by writing $\mathrm dg = \mathrm dx(\mathrm dy/y^2)\mathrm dk$ and $\mathrm dg = \mathrm dn\,\mathrm da\,\mathrm dk$; $\mathrm da$ is $\mathrm dy/y$, not $\mathrm dy/y^2$ (assuming $a = \begin{pmatrix} y \\ & y^{-1} \end{pmatrix}$). $\endgroup$
    – LSpice
    Commented Apr 12, 2021 at 20:40
  • $\begingroup$ But, to the general question, I suspect that, absent the obvious case where conjugation by $H$ acts by measure-preserving isomorphisms on $K$, or vice versa, there is no very good general answer for when this happens. $\endgroup$
    – LSpice
    Commented Apr 12, 2021 at 20:42
  • $\begingroup$ Without thinking at all deeply about this, my suspicion is that in the first example one is really getting a product of Haar measures on P and K where P=NA; and then it so happens that the parabolic P has a nice semidirect product decomposition. I also suspect that for this particular example the compactness of K is important, but this is just a tentative hunch and I could easily be shown wrong here $\endgroup$
    – Yemon Choi
    Commented Apr 12, 2021 at 23:23

1 Answer 1

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FYI, the answer to my question is this Stack Exchange post (h/t Stephen D Miller):

https://math.stackexchange.com/a/2325533

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