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

`$G$`

(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 17:14connected$G$, 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