Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose that $F$ is a nonarchimedean local field, $G_1$ and $G_2$ are connected (linear) algebraic groups over $F$, and $\phi:G_1\to G_2$ is a surjective homomorphism of algebraic groups. Suppose $H$ is a hyperspecial maximal compact subgroup of $G_1$. Is the image $\phi(H)$ necessarily a hyperspecial maximal compact subgroup of $G_2$?

share|improve this question

1 Answer 1

up vote 4 down vote accepted

If by "surjective" you mean surjective in the usual sense (for example on $\overline{F}$-points) then maybe you have a problem, because $G_1(F)$ may not surject onto $G_2(F)$. So for example $SL(2)$ surjects onto $PGL(2)$ but if $R$ is the integers of $F$ then $SL(2,R)$ is hyperspecial max compact but its image in $PGL(2,F)$ isn't (it's not even maximal, as $PGL(2,R)$ strictly contains the image of $SL(2,R)$).

However if $G_1\to G_2$ is, say, a $z$-extension, then (by definition) the kernel is central in $G_1$ and has no $H^1$, so the long exact sequence shows $G_1(F)\to G_2(F)$ is surjective. Moreover, if I've got things right, then I think that $G_1$ unramified forces the kernel to be unramified, and if you take a smooth integral model of $G_1$ with $G_1(R)$ equal to the hyperspecial you thought of, then the quotient of this model of $G_1$ by the Zariski closure of the kernel will also be unramified, and the same cohomology argument shows that $G_1(R)$ surjects onto $G_2(R)$, so in this case you win.

share|improve this answer

Your Answer

 
discard

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.