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Let $G$ be a simply connected semisimple algebraic $K$-group and $K$ be a finite extension of $k$. Is $R_{K/k}G$ still a simply connected algebraic group?

We say $G$ is simply connected if for any central isogeny $G'\to G$ is in fact an isomorphism of algebraic groups.

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As pointed out by Jeff Adler, it may be more useful here to characterize "simply connected" in terms of the relationship between the full weight lattice and the character group of a maximal torus in $G$. In any case it's important to indicate whether you need any extra assumptions about the fields or the field extension involved. –  Jim Humphreys Feb 26 '11 at 13:19
    
I think I would assume $K/k$ is separable. Are there any convenient reference for this full weight lattice characterization? –  ronggang Feb 26 '11 at 14:06
    
It's important to realize that the characterization is intended for connected semisimple algebraic groups, where it agrees over $\mathbb{C}$ with the topological characterization. The notion comes up in many books and papers, such as the papers by Borel-Tits on reductive groups available at NUMDAM numdam.org:80/?lang=en. Or see 31.1 in my 1975 Springer GTM21 on linear algebraic groups. Some of the standard online reference sources are not too helpful here. –  Jim Humphreys Feb 26 '11 at 14:44

1 Answer 1

up vote 3 down vote accepted

May I assume that $K/k$ is separable?

Let $T$ be a maximal torus in $G$. Since $G$ is simply connected, the weight lattice and character lattice for $T$ are the same. This remains true if we replace $G$ by a direct product of $[K:k]$ copies of $G$, and $T$ by a corresponding product of tori. Over the algebraic closure, our direct product is isomorphic to $R_{K/k}G$. Our condition on the lattices doesn't depend on the rational structure of an algebraic group, so this latter group is also simply connected.

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In fact, the $k$-group $R_{K/k}G$ already becomes isomorphic to the indicated direct product after extending scalars to $K$. –  George McNinch Feb 26 '11 at 12:21
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@George. Yes, under the extra condition that K/k is Galois. –  Peter McNamara Feb 27 '11 at 0:12
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@Peter: Oops, yes you are of course correct-- thanks! In part the point of my comment was just that "one doesn't have to go all the way to an alg. closure". Probably what I should written is that $R_{K/k}G$ becomes isomorphic to the indicated direct product over any Galois extension $L/k$ with $K \subset L$. –  George McNinch Feb 27 '11 at 2:31

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