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Let $G$ be a (connected) reductive group over a field $k$. Then there is a natural functor from the category of representations of $G$ to the category of representations of $G(k)$.

Under which circumstances can one say that this functor is a) full and b) faithful?

Edited for clarity:

Here we think of algebraic representations of $G$ over the field $k$ (so, for example, morphisms of $G$ to $GL(V)$ for $k$-vector spaces $V$, defined over $k$).

Given such a morphism $\phi : G \rightarrow GL(V)$, we take $k$ points to obtain a $k$-linear representation of the group $G(k)$ on the $k$-vector space $V$ (thus a homomorphism $\phi(k) : G(k) \rightarrow GL(V)(k)$).

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It would be helpful to make the language more precise: what is meant here by a "representation" of $G$ (rational?) and does that relate to $k$. Are you identifying a group scheme $G$ with its group of rational points over an algebraic closure of $k$? And is that group connected? Also, are the representations of $G(k)$ taken over $k$ or its algebraic closure or some other field? The extreme case when $k$ is finite shows some of the problems that can come up with the restriction functor. Even going down from $\mathbb{C}$ to $\mathbb{R}$ requires some care. –  Jim Humphreys May 27 '10 at 17:14
    
Thanks - I edited to hopefully make the question clearer. –  user1594 May 27 '10 at 17:24
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Assuming you use finite-dimensional $k$-rational representations and connected $G$, faithfulness holds for infinite $k$ because conn'd reductive groups are unirational over any field (so $G(k)$ is Zar.-dense in $G$ when $k$ is infinite). Must assume connectedness, as otherwise if $k$ isn't sep. closed then $G$ could correspond to nontrivial finite Galois module with no nonzero Gal-fixed points, so $G(k) = \{1\}$. Should assume $G$ semisimple, or else nontrivial torus quotients will make a mess, and even simply connected or else $G(k)$ can have nontrivial finite commutative quotients. –  BCnrd May 27 '10 at 17:32
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To give a concrete example of the problems with non-s.c. $G$ in a "classical" setting, consider the degree-$n$ central isogeny $f:{\rm{SL}}_ n \rightarrow {\rm{PGL}}_ n$ over $k = \mathbf{R}$ with $n$ odd. Since the kernel is $\mu_ n$, the induced map on $k$-points is a real-analytic isomorphism, so the inverse map on $k$-points defines a real-analytic linear representation of ${\rm{PGL}}_ n(k)$ which is certainly not "algebraic" (since $f$ has no algebraic inverse, as its degree is $> 1$). Note here that ${\rm{PGL}}_ n$ is not simply connected. –  BCnrd May 27 '10 at 17:40

2 Answers 2

up vote 1 down vote accepted

A bit of Tannakian formalism clarifies the situation. Recall that for every abstract group $\Gamma$ there is a notion of "algebraic hull" $\Gamma^{alg}$ constructed as follows: Consider pairs $(\varphi,H)$ where $H$ is an algebraic group over $k$ and $\varphi:\Gamma\to H(k)$ a homomorphism of groups with Zariski-dense image. Maps between such pairs are defined in the obvious was, and one defines $$\Gamma^{alg} := \lim_{(\varphi,H)}H$$ So this is an affine group scheme over $k$. The group $\Gamma^{alg}$ has a universal property, namely that for any algebraic group $H$ over $k$, any map $\Gamma \to H(k)$ factors over $\Gamma^{alg}$. From Tannakian formalism we know:

The canonical map $\Gamma\to\Gamma^{alg}$ induces an equivalnence of monoidal categories between the category of finite dimensional algebraic $\Gamma^{alg}$-representations and finite dimensional $k$-representations of $\Gamma$.

Now, let's apply this to $\Gamma = G(k)$ for a fixed algebraic, not necessarily reductive group $G$ over $k$. The universal property of the algebraic hull yields a canonical morphism $\Gamma^{alg}\to G$ whose image is the Zariski-closure of $\Gamma$ in $G$. We have then results like this:

Proposition (Deligne, LNM900, p.139): Let $f: H\to G$ be a morphism of affine group schemes over $k$ and let $\omega$ be the induced functor $Rep_k(G) \to Rep_k(H)$ between categories of finite dimensional representations. The morphism $f$ is faithfully flat if and only if $\omega$ is fully faithful and if for every representation $V$ of $G$ every subrepresentation of $\omega(V)$ is isomorphic to the image of a subrepresentation of $V$.

For instance, faithful flatness of $f$ implies surjectivity of $f$, which is therefore a necessary condition for the induced functor of representations to be fully faithful - this is what's behind Brian's comment. If you assume $G$ to be reductive then the category $Rep(G)$ is semisimple (in characteristic zero) and the last condition in Deligne's proposition can, maybe, be rephrased in a simpler way.

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@Xandi: I suppose "full flatness" is meant to be "faithful flatness", and "surjectivity" is meant to be "implies" in the last paragraph (only equivalent in the reduced case, such as in char. 0)? Also, the final sentence is only true in char. 0. Does Deligne's result require assuming $k$ has char. 0? –  BCnrd May 28 '10 at 15:47
    
I have corrected the misspellings. No, in Delignes results there is no hypothesis on the characteristic of $k$. –  Xandi Tuni May 28 '10 at 20:29
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Actually, "Deligne's result" is already in Saavedra, and the reference given is to an article by Deligne and somebody else. –  JS Milne May 29 '10 at 2:25
    
My apologies. I hope you trust that in a research article I would have gotten this right. –  Xandi Tuni May 29 '10 at 11:04

Edit Sorry: I just realized that I conflated "essentially surjective" with "full" in my head this afternoon. So this is mostly an answer to a question that wasn't asked.

The functor is not essentially surjective, already when $G = T$ is a split torus.

Take for example the 1 dimensional torus $T = \mathbf{G}_m$; in this case, the 1 dimensional algebraic representations of $T$ are parametrized by $m \in \mathbf{Z} = X^*(T)$; to an $m$ corresponds the 1 dimensional representation $k_m = k$ on which $T$ acts with weight $1$ (so in particular an element $t$ of $T(k) = k^\times$ acts by multiplication with $t^m$).

Now let $V$ be the representation $k_1$, let $\sigma$ be a non-trivial automorphism of the field $k$, and let $\ ^\sigma V$ be the representation of $T(k) = k^\times$ obtained from $V$ by "twisting with $\sigma$"; thus an element $t \in T(k) = k^\times$ acts by multiplication with $\sigma(t)$.

If $k$ has characteristic 0, or if $\sigma$ is not a power of the Frobenius automorphism, the representation $\ ^\sigma V$ is not isomorphic as $T(k)$-representation to $k_m$ for any $m$. Thus, the functor isn't essentially surjective.

When $k$ is finite, this construction shows that the indicated functor is not full. Indeed, if $k = \mathbf{F}_q$ then $\operatorname{Hom}_T(k_1,k_q) = 0$ but $\operatorname{Hom}_{T(k)}(k_1,k_q) = k$ since $k_1$ and $k_q$ are isomorphic when restricted to $T(k)$.

This isn't special to tori: e.g. if $G = \operatorname{SL}(V)_{/\mathbf{F}_p}$ then $V$ and the first Frobenius twist $V^{[1]}$ (defined by twisting the action of $\operatorname{SL}(V)$ on $V$ by the Frobenius map) are distinct simple algebraic $G$-modules, but they give isomorphic representations of $G(\mathbf{F}_p)$.

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