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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 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
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
@George. Yes, under the extra condition that K/k is Galois. – Peter McNamara Feb 27 '11 at 0:12
@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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