What is the internal Hom object in the category $\mathcal{C} = \mathbf{Rep}_k(G)$?

Question

Here $k$ is a field and $\mathcal{C}$ is the category of representation of an affine $k$-group scheme $G$ over the finite-dimensional $k$-vector spaces. Supposedly, the internal Hom object is the representation of $G$ on $\mathrm{Hom}_{k-\text{lin}}(X, Y)$, but I'm not sure how to prove that this is the case.

Answer 1

$\DeclareMathOperator{\Hom}{Hom}$ So from the comments, this has been reduced to showing that if $\varphi: \Hom_k(Z\otimes X, Y)\to \Hom_k(Z,\Hom_k(X,Y))$ is given by $\varphi(f)(z)(x) = f(z\otimes x)$ then this is a homomorphism of representations of $G$.

I will abuse notation somewhat here to make everything more readable. Technically, $G$ is a functor, so everything are really natural transformations, and we ought to pick a $k$-algebra $A$ and work with $G(A)$ and with the natural transformations on this level. Just assume that all of this has been fixed (and I will also ignore all questions of these things being natural). For $g\in G$ and some element $x$ in a representation of $G$, I will write $g.x$ for the action of $g$ on $x$, to better distinguish the action of $G$ from applications of functions.

So given $f\in \Hom_k(Z\otimes X, Y)$ and $g\in G$ we have $g.f$ defined by $(g.f)(z\otimes x) = g.(f(g^{-1}.z \otimes g^{-1}.x))$ (extended linearly to all of $Z\otimes X$), so $\varphi(g.f)(z)(x) = (g.f)(z\otimes x) = g.(f(g^{-1}.z\otimes g^{-1}.x))$.

On the other hand, for $h\in \Hom_k(Z,\Hom_k(X,Y))$ we have $g.h$ defined by $(g.h)(z) = g.(h(g^{-1}.z))$, but now $h(g^{-1}.z)\in \Hom_k(X,Y)$ so by definition $(g.h)(z)(x) = (g.(h(g^{-1}.z))(x) = g.(h(g^{-1}.z)(g^{-1}.x))$.

Now, if $h = \varphi(f)$ then this becomes $g.(\varphi(f)(g^{-1}.z)(g^{-1}.x)) = g.(f(g^{-1}.z\otimes g^{-1}.x)) = \varphi(g.f)$ by the above, which shows that $\varphi$ is a homomorphism of representations.