Timeline for How can this argument calculating the Haar measure on a parabolic subgroup be generalized to the non-split case?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Feb 14, 2017 at 8:47 | comment | added | user1688 | What you really do compute here is the modular function. This computation carries over to any parabolic $P=LN$, since it has a finite index subgroup of the form $MAN$, where $A$ is a split torus. | |
| Feb 14, 2017 at 7:43 | comment | added | user1688 | In order to compute the Haar measure on $P$, these computations are not necessary. If $dm$ is a Haar measure on $M$ and $dn$ one on $N$, you get a Haar integral on $P$ by $\int_M\int_N f(mn)\,dn\,dm$. You only need an extra factor in the reverse order $nm$. | |
| Feb 14, 2017 at 4:59 | comment | added | user94041 | Uhm I see. No, $P(F)$ is not open in $P(F')$. | |
| Feb 14, 2017 at 0:28 | comment | added | D_S | I'm not sure. Is $P(F)$ open in $P(F')$? In general, if you take a Haar measure on a group $G$, and you restrict the measure to the Borel sets of a closed subgroup $H$, you don't get a Haar measure on $H$. | |
| Feb 14, 2017 at 0:22 | comment | added | user94041 | Can't you just run the above argument in some finite extension $F' / F$ splitting $G$, obtain a (say, left) Haar measure for $P(F')$, and give $P(F)$ the induced measure as a subgroup? | |
| Feb 13, 2017 at 23:11 | history | asked | D_S | CC BY-SA 3.0 |