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Let $p:X\rightarrow Y$ be a double cover of curves, denote by $$SU_n:=(p_*SL_n(\mathcal O_X))^{\tilde{\sigma}}$$ i.e. the $\tilde{\sigma}-$invariant part, the action of $\tilde{\sigma}$ is given by $$\tilde{\sigma}(g)=\,^t(g\circ\sigma)^{-1}$$ where $\sigma$ is the involution induced by the double cover. $SU_n$ is well knowing to be a parahoric group shceme in the sens of Bruhat-Tits.

My question: What is $\pi_1(({SU_n})_\eta)$ ? (the algebraic fundamental group). where $\eta$ is the generic point of $Y$.

Thanks

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  • $\begingroup$ Do you really mean to take $g \circ \sigma$ and not $\sigma(g)$? I don't see why $\sigma$ should lie in $p_* SL_n$. $\endgroup$
    – Will Sawin
    Commented Nov 23, 2015 at 17:24
  • $\begingroup$ I see $g$ as an automorphism of $SL_r\times X$ and that's why I used the notation $g\circ \sigma$, $\endgroup$
    – Z.A.Z.Z
    Commented Nov 23, 2015 at 17:45

1 Answer 1

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There is an exact sequence

$$ \pi_1 ((SU_n)_{\overline{\eta}}) \to \pi_1((SU_n)_{\eta}) \to \operatorname{Gal}(\overline{\eta}|\eta)$$

Over $\overline{\eta}$, $p_* SL_n$ is just $SL_n \times SL_n$. The involution $\tilde{\sigma}$ acts by switching the two $SL_n$s and then doing an performing some automorphism, so the involution invariants $SU_n$ are just the diagonal $SL_n$ embedded by the graph of the automorphism.

In characteristic $0$, $\pi_1(SL_n)$ is trivial, so the fundamental group is just the Galois group of the base field. In characteristic $p$ things are more complicated as the fundamental group is infinite, as witnessed by the family of etale covers $g \to g^{-1} \operatorname{Frob}_q(g)$ which is Galois of group $SL_n(\mathbb F_q)$ for each $q$.

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  • $\begingroup$ I don't see why over $\bar \eta$ the fiber shoud be two copies of $SL_r$? what is $X_{\bar\eta}$? $\endgroup$
    – Z.A.Z.Z
    Commented Nov 23, 2015 at 18:01
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    $\begingroup$ @Z.h $X_{\overline{\eta}}$ is just the pullback of $X$ along the map $\overline{\eta} \to X$, which is just $\overline{\eta}$ - a single point. $Y_{\overline{\eta}}$ is the pullback of $Y$ along the same map, which is two points. Thus the pushforward of $SL_r$ is two cope of $SL_r$. $\endgroup$
    – Will Sawin
    Commented Nov 24, 2015 at 1:48
  • $\begingroup$ I considred the 2:1 cover as $\pi:X\rightarrow Y$, so I think you mean $Y_{\bar\eta}$ is one point and $X_{\bar\eta}$ is two points? right? $\endgroup$
    – Z.A.Z.Z
    Commented Nov 24, 2015 at 19:14
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    $\begingroup$ @Z.h yes, precisely that. $\endgroup$
    – Will Sawin
    Commented Nov 24, 2015 at 21:34
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    $\begingroup$ @Z.h For the other map, use the nontrivial automorphism of $K(X)$ over $K(Y)$ and then apply the embedding into $\overline{K(X)}$. Alternately, use that $X_{\overline{\eta} } = \operatorname{Spec} K(X) \otimes_{K(Y)} \overline{K(X)} = \operatorname{Spec} \overline{K(X)} \times \overline{K(X)}$ $\endgroup$
    – Will Sawin
    Commented Nov 28, 2015 at 13:08

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