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Let $k$ be an algebraically closed field of characteristic zero. and let $$\sigma: SL_n(k)\rightarrow SL_n(k)$$
be an involution. My questions are:

  • How could one calculate the fundamental group of $SL_n(k)^\sigma$ ? (the invariant subgroup)

  • In particular, what is $\pi_1(SO_n(k))$ and $\pi_1(Sp_{2n}(k))$?

thanks

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    $\begingroup$ Are you thinking of these as algebraic groups? Over $\mathbb{C}$ the fundamental group of $SO(n)$ is $\mathbb{Z}/2 \mathbb{Z}$ for $n > 2,$ $\mathbb{Z}$ when $n=2,$ while the symplectic group is simply connected. $\endgroup$
    – Igor Rivin
    Commented Oct 9, 2015 at 8:53
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    $\begingroup$ Yes I know that, but I need to know it over an algebraicaly closed field of characteristic zero, not necessarily $\mathbb C$ $\endgroup$
    – Gest2015
    Commented Oct 9, 2015 at 9:07
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    $\begingroup$ Does the question makes sense? the fundamental group is defined for an algebraic group, not for the group of $k$-points of an algebraic $k$-group. $\endgroup$
    – YCor
    Commented Oct 9, 2015 at 11:38
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    $\begingroup$ It's important to make explicit what you mean by "fundamental group" in this context. Beyond the usual notion in algebraic topology, there is an algebraic notion introduced by Chevalley in the study of semisimple (or reductive) algebraic groups over an arbitrary algebraically closed field: the quotient of the abstract group of weights of a maximal torus by the actual character group of the torus. (This turns out to be equivalent to the topological definition in characteristic 0.) $\endgroup$ Commented Oct 9, 2015 at 12:31
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    $\begingroup$ I agree with Matthias Wendt. I will post an answer tomorrow. $\endgroup$ Commented Oct 9, 2015 at 20:33

1 Answer 1

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Let $k$ be an algebraically closed field of characteristic 0. Let $G$ be a connected reductive group over $k$. The notion of the algebraic fundamental group of $\pi_1(G)$ was introduced in my memoir here and generalized to arbitrary characteristic here and to reductive group schemes here.

Let $G^{\rm ss}=[G,G]$ denote the commutator subgroup of $G$ (which is semisimple). Let $G^{\rm sc}\twoheadrightarrow G^{\rm ss}$ denote the universal covering of $G^{\rm ss}$ (then $G^{\rm sc}$ is simply connected). We consider the composite homomorphism $$ \rho\colon G^{\rm sc} \twoheadrightarrow G^{\rm ss} \hookrightarrow G.$$

Let $T\subset G$ be a maximal torus. By abuse of notation, we write $T^{\rm sc}$ for the preimage of $T$ in $G^{\rm sc}$. We have a homomorphism $$\rho\colon T^{\rm sc}\to T,$$ which in general is neither surjective nor injective.

Let $X_*(T)=\{\chi\colon \mathbf{G}_{m,k}\to T\}$ denote the cocharacter group of $T$. We obtain a homomorphism $$ \rho_*\colon X_*(T^{\rm sc})\to X_*(T). $$

Definition. $\pi_1(G)=X_*(T)/\rho_* X_*(T^{\rm sc}). $

This algebraic fundamental group $\pi_1(G)$ does not depend on the choice of $T$ (up to a canonical isomorphism). Further, if $K$ is an algebraically closed field extension of $k$, then clearly $$ X_*(T)=X_*(T\times_k K)$$ and $$ \pi_1(G)=\pi_1(G\times_k K).$$

Let $k={\mathbb{C}}$. A cocharacter $\chi\colon \mathbf{G}_{m,{\mathbb{C}}}\to T$ induces a continuous homomorphism ${\mathbb{C}}^\times\to T({\mathbb{C}})$ and a homomorphism of topological fundamental groups $$ \pi_1^{\mathrm{top}}({\mathbb{C}}^\times)\to\pi_1^{\mathrm{top}}(T({\mathbb{C}}))\to\pi_1^{\mathrm{top}}(G({\mathbb{C}})).$$ By Proposition 11.1 of the memoir, in this way we obtain a canonical isomorphism $$ \pi_1(G)\overset{\sim}{\to}\mathrm{Hom}\left[\pi_1^{\mathrm{top}}({\mathbb{C}}^\times) \to\pi_1^{\mathrm{top}}(G({\mathbb{C}}))\right]. $$ After we choose one of the two generators of $\pi_*^{\mathrm{top}}({\mathbb{C}}^\times)$, we obtain a noncanonical isomorphism $$ \pi_1(G)\overset{\sim}{\to}\pi_1^{\mathrm{top}}(G({\mathbb{C}})). $$

Now reducing to the case $k={\mathbb{C}}$, one can easily see that Igor Rivin's comment works over any algebraically closed field $k$ of characteristic 0. I show below how to see this without reducing to ${\mathbb{C}}$, by elaborating on the comment of Matthias Wendt.

First assume that $G$ is a simply connected semisimple group. Then $G^{\rm sc}=G^{\rm ss}=G$, hence $T^{\rm sc}=T$ and $\pi_1(G)=0$ (as one should expect!). Since $\mathrm{Sp}_{2n}$ is simply connected, we conclude that $\pi_1(\mathrm{Sp}_{2n})=0$.

Then assume that $G$ is a torus. Then $G^{\rm sc}=1$, $T=G$, $T^{\rm sc}=1$, hence $\pi_1(G)=X_*(G)$. Since $\mathrm{SO}_2$ is a 1-dimensional torus, we conclude that $\pi_1(\mathrm{SO}_2)\simeq\mathbb{Z}$.

Now let $G$ be a semisimple group over $k$. Note that $\ker\rho$ is always a finite abelian group. By Example 1.6(3) in the memoir, we have a canonical isomorphism $$\pi_1(G)\cong \mathrm{Hom}(\mathrm{Hom}_k(\ker\rho,\mathbf{G}_{m,k}),\mathbb{Q}/\mathbb{Z}).$$ It is well known that for $\mathrm{SO}_n$ for $n>2$ we have $\ker\rho\simeq\mu_2$, hence $\pi_1(G)\simeq\mathbb{Z}/2\mathbb{Z}$ (again, as one should expect).

Unfortunately, Example 1.6(3) was given without proof. For a proof see, e.g., this preprint, Lemma 15.2.

The point of introducing the algebraic fundamental groups was as follows. If $G$ is actually defined over a nonclosed field $k_0$ such that $k$ is an algebraic closure of $k_0$, then the Galois group $\mathrm{Gal}(k/k_0)$ acts on $\pi_1(G)$. Then, following an idea of Kottwitz, from the Galois module $\pi_1(G)$ one can compute arithmetic invariants of $G$ over $k_0$, such as the Galois cohomology $H^1(k_0,G)$ when $k_0$ is a $p$-adic field, and the Tate-Shafarevich kernel and the defect of weak approximation for $G$ when $k_0$ is a number field.

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    $\begingroup$ oh that's a wonderful and complete answer, by the way I was reading you memoir this morning! thanks. $\endgroup$
    – Gest2015
    Commented Oct 10, 2015 at 17:05
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    $\begingroup$ You ask about $\pi_1(G)$, where $G=(\mathrm{SL}_n)^\sigma$ and $\sigma$ is an involution. Such $\sigma$ and $G$ are classified, see the book by Onishchik and Vinberg (1990), Table 7. They are $\mathrm{Sp}_{2m}$ for $2m=n$, $\mathrm{O}_n$ (not $\mathrm{SO}_n$), and $\mathrm{S}(\mathrm{GL}_p\times \mathrm{GL}_{n-p})$. For the last group we have $\pi_1(G)=\mathbb{Z}$. $\endgroup$ Commented Oct 10, 2015 at 17:44

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