Let F be a function field, and A its adele ring. I want to consider U(A)/U(F), where U(A) is the space of strictly upper triangular matrices with entries from A, and U(F) is the same with entries from F. I was wondering anyone has an idea or a source that proves this space is compact so that we get a finite measure.
6
-
1$\begingroup$ Isn't such a group a repeated extension of copies of the adele class group (which is compact)? $\endgroup$– Pete L. ClarkCommented Apr 15, 2010 at 18:23
-
2$\begingroup$ If $G$ is any solvable smooth connected affine gp over global $F$ and has no nontrivial $F$-rat'l character (e.g., unip.) then $G(A)/G(F)$ is compact. Over number fields, due to Godement. Argument generalized to function fields by Oesterle; see Thm 1.3 in Ch. IV of Oesterle's Inv. paper on Tamagawa numbers. For case you ask about, just use comp. series with successive qts as vector gps, and lots of Hilb. 90 for exactness on adelic, local, global pts. In fact, for any smooth conn'd affine $G$ over $F$ with no nontriv $F$-rat'l characters, $G(A)/G(F)$ has finite vol; lies deeper in char > 0. $\endgroup$– BCnrdCommented Apr 15, 2010 at 18:31
-
$\begingroup$ @Brian: I know that in general there are more unipotent groups in positive characteristic, e.g. Witt vector groups. But do they arise in this case? $\endgroup$– Pete L. ClarkCommented Apr 15, 2010 at 18:33
-
$\begingroup$ @Pete: Even Witt groups have filtration by copies of the additive group, so there are far worse unipotent groups out there (over imperfect fields). That said, they don't arise for the question as posed, for the reason mentioned in your first comment and mine (I have the bad habit to say "Hilbert 90" for the additive version too, which I suppose must have another name.) $\endgroup$– BCnrdCommented Apr 15, 2010 at 19:22
-
$\begingroup$ @Brian: right, there's this "k-wound" nastiness that I read about in BLR. (That's why I said e.g.) Thanks for your response. $\endgroup$– Pete L. ClarkCommented Apr 15, 2010 at 19:25
|
Show 1 more comment