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Let $G$ be a linear reductive algebraic over an algebraically closed field of characteristic $0$. I know that there is a surjective map with finite Kernel $$G' \times T \to G$$ where $G'$ is semisimple and $T$ is a torus.

When I talk about semisimple/simply connected groups, I refer to their characterization based on their root system as in Springer book "Linear algebraic group".

I would like to prove that for every semisimple algebraic groups there is a finite covering $$G'' \to G'$$ such that $G''$ is simply connected, but I really do not know how to do.

Also ,if it is possible, I would like to see a reference for the proof of the fact that over $\mathbb{C}$ a simply connected group is simply connected in the standard topological sense.

Tommaso Scognamiglio
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    For the first one, you may want to check out chapter 20 in Milne's notes https://www.jmilne.org/math/CourseNotes/iAG200.pdf or A.4.10 and A.4.11 of B. Conrad's 'Pseudo-reductive groups'. A proof of the fact that $G''(\mathbb{C})$ is simply connected with the usual topology is sketched here: https://math.stackexchange.com/questions/3183/is-the-universal-cover-of-an-algebraic-group-an-algebraic-group – Alvaro Martinez Apr 19 '19 at 10:43
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    As Alvaro pointed out, this is a very standard fact that you should just read in a textbook. – Alex Youcis Apr 20 '19 at 02:20

1 Answers1

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Just to get this off the unanswered list.

For a proof that every semisimple group has a universal cover, at least in the split case, one can see [Mil, §18.d]. The general existence case can be handled by Galois descent.

Suppose now that $G$ is a semisimple group over $\mathbb{C}$. You want to show that the obvious map

$$\pi_1(G^\mathrm{an},e)\to \pi_1(G,e)$$

is an isomorphism. To do this we note that this map is induced by a map of Galois categories

$$\left\{\begin{matrix}\text{Central isogenies}\\ \text{over }G\end{matrix}\right\}\to \left\{\begin{matrix}\text{Finite topological covers}\\ \text{of }G^\mathrm{an}\end{matrix}\right\}$$

given by $G'\to G$ maps to $(G')^\mathrm{an}\to G^\mathrm{an}$. By abstract theory (e.g. see [Stacks, Tag0BN6]) it suffices to note that if $G'$ is connected then so is $(G')^\mathrm{an}$, but this is well-known (e.g. see [3, XII Proposition 2.4]).

From this, we see that it suffices to show that $\pi_1(G^\mathrm{an},e)$ and $\pi_1(G,e)$ have the same finite cardinality. But, to do this it suffices to show that the universal cover $\widetilde{G}$ of $G$ has simply connected analytification, since then

$$\pi_1(G,e)=\ker(\widetilde{G}\to G)(\mathbb{C})=\ker(\widetilde{G}^\mathrm{an}\to\widetilde{G})=\pi_1(G^\mathrm{an},e)$$

For this, see [Con, Proposition D.4.1] and note that the root data of semisimple gorup being simply connected is equivalent to the group itself being simply connected by the isogeny theorem (e.g. see [Mil, Theorem 23.9]).

[Con] Conrad, B., 2014. Reductive group schemes. Autour des schémas en groupes, 1, pp.93-444.

[Mil] Milne, J.S., 2017. Algebraic groups: The theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.

[Stacks] Various authors, 2020. Stacks project. https://stacks.math.columbia.edu/

Alex Youcis
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