# Does the notion of a simply connected algebraic group depend on the base field?

Let $$G$$ be a connected semisimple algebraic group over a field $$k$$ with $$\text{char}(k) = 0$$.

One usually defines $$G$$ to be simply connected if every isogeny $$G' \to G$$ for a connected algebraic group $$G'$$ is an isomorphism. As far as I understand, the isogeny and $$G'$$ need not be defined over $$k$$.

Question: Sometimes people require in this definition that instead every $$k$$-isogeny $$G' \to G$$ for a connected $$k$$-group $$G'$$ is a $$k$$-isomorphism. Does this give the same notion of a simply connected algebraic group and if so, why?

$$\newcommand{\sc}{\mathrm{sc}}\newcommand{\Gal}{\mathrm{Gal}}$$You don't even need $$\mathrm{char}(k)=0$$.

Theorem: Let $$k$$ be a perfect field and let $$G$$ be a semisimple algebraic group over $$k$$. Then, there exists some semisimple algebraic group $$G^\sc$$ and a central isogeny $$G^\sc\to G$$ with the following universal property: if $$H$$ is a semisimple group over $$k$$ and $$H\to G$$ is a central isogeny, then the map $$G^\sc\to G$$ factors uniquely through $$H$$.

Proof: If $$G$$ is split this is well-known. Suppose now that $$G$$ is arbitrary. Then, we can find some finite Galois extension $$K/k$$ such that $$G_K$$ becomes split. Consider then $$U:=(G_K)^\sc$$. Note then that for every $$\sigma\in\Gal(K/k)$$ we have that by pulling back along $$\sigma$$ gives us a map

$$U^\sigma \to (G_K)^\sigma\cong G_K$$

which is evidently a central isogeny. Then, by definition, there exists a unique central isogeny $$U\to U^\sigma$$ factorizing this map. Note though that since the kernel of $$U^\sigma\to G_K$$ is the same degree as $$U\to G$$ we have that $$U\to U^\sigma$$ is an isomorphism. It's not hard to see that this gives a $$\Gal(K/k)$$-descent datum on $$U$$ and thus $$U$$ descends uniquely to some $$\widetilde{G}$$ over $$k$$ equipped with a map $$\widetilde{G}\to G$$.

Suppose now that $$H\to G$$ is a central isogeny. Note then that we get a central isogeny $$H_K\to G_K$$ and thus a unique factorization $$U\to H_K$$. Note then that by construction, and the fact that $$(H_K)^\sigma\cong H$$ naturally, shows that this descends uniquely to a map $$\widetilde{G}\to H$$. Thus, $$\widetilde{G}=G^\sc$$ as desired. $$\blacksquare$$

Note then that saying that $$G$$ is simply-connected is merely the same as saying that $$G^\sc=G$$. By the above construction it's clear that $$(G^\sc)_K=(G_K)^\sc$$ so that this statement is independent of which extension of $$k$$ you consider.

• Thank you, nice writeup. One question: Do I see it right that in the theorem, one can also leave out the condition that $H$ is semisimple (because it should be so automatically when it is isogenous to $G$) and instead require only that $H$ is connected? – abenthy Apr 21 at 9:24
• @abenthy If you contain $\mathbb{G}_m$ then your universal cover would have to be infinite degree over you, and so wouldn't be finite type. – Alex Youcis Apr 21 at 16:03
• @abenthy Oops. I misread your question. Yes. If $H\to G$ is an isogeny then $H$ is automatically semisimple. – Alex Youcis Apr 21 at 17:52