Points of reductive groups

Question

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 $\phi(k) : G(k) \rightarrow GL(V)(k)$).

Answer 1

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.

Answer 2

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 $G(\mathbf{F}_p)$.