# What's the internal hom of linear representations of categories?

The category of $\mathbb F$-linear representations of a group $G$ is closed, with internal hom given by the underlying internal hom of vector spaces, and $G$-action defined by $$f(v)\mapsto g\cdot f(g^{-1}\cdot v).$$

1. How to see this is the correct formula (giving a right adjoint), and not e.g $f(v)\mapsto g\cdot f(g\cdot v)$?

I think this carries over to linear representations $F,G:\mathsf G\to \mathsf{Vect_\mathbb{F}}$ of a groupoid using the formulas $$\begin{gathered}\mathsf{hom}(F,G)(A)=\mathsf{hom}(FA,GA), \\ \mathsf{hom}(F,G)(f:A\to B)=\mathsf{hom}(Ff^{-1},Gf).\end{gathered}$$ But what happens for categories? By abstract nonsense the functor category should still inherit the closed structure.

Theorem. Let $\mathscr V$ be a complete monoidal closed category and $\mathsf C$ a small category. Then the functor category $[\mathsf C,\mathscr V]$ is monoidal closed with pointwise tensor product.

Unfortunately the proof is a bit terse for me and I don't know how to read off what the internal hom should be in the functor category $[\mathsf C,\mathsf{Vect_\mathbb{F}}]$ of $\mathbb F$-linear representations of $\mathsf C$.

1. What's the internal hom in the category of representations of a fixed small category $\mathsf C$?
• For 2, IIRC you can adapt the Yoneda lemma. Then, since a functor is determined by its values on objects and morphisms, you can simply compute the values of $[F,G]$ on objects and morphisms by expressing them as natural transformations to which you can apply the hom-tensor adjunction.
– user14972
Jun 24 '17 at 12:58
• @Hurkyl your comment is a bit terse for me, sorry. I don't understand what the formula for the internal hom should be. Jun 24 '17 at 18:55
• $f(v) \mapsto gf(gv)$ is not an action at all (check this). Jun 25 '17 at 0:32

It might be helpful to take a significantly different approach, here, with apologies if I just add to the confusion. It's a bit unnatural to think of $C$ as a plain category in putting a closed structure on $[C,V]$, since the internal hom has $F^G$ has to send $c$ to an object of $V$, rather than just a set. Here's the intuition: if $V=\mathbf{Set}$, then $F^G(c)=\mathrm{Nat}(\hat c\times G,F)$ is some kind of set of natural transformations from $G$ to $F$ weighted by $c$. (In particular, if $c$ is initial, then we recover the ordinary set of natural transformations.) Now for more general $V$, $F^G(c)$ has to be an object of $V$. Perhaps you'll buy that this should feel like an object of (weighted) $V$-natural transformations between $G$ and $F$; but that doesn't make sense if $F,G$ are ordinary functors, rather than $V$-functors.

However, if we assume $V$ is cocomplete, then we can construct the free $V$-category $V(C)$ on $C$, which has the same objects $C$ and as hom-objects $V(C)(c_1,c_2)=\coprod_{C(c_1,c_2)} I$, where $I$ is the monoidal unit of $V$. In case this construction seems unfamiliar, work it out in the case $C$ is a group and $V$ is abelian groups-you should just get the group ring! Now $V$-functors from $V(C)$ to $V$ are the same as functors from $C$ to $V$ (recovering the fact that $G$-representations are modules over the group ring) and we now have easier access to $V$-objects of ($V$-)natural tranformations between functors $C\to V$. Recall the definition of $V-\mathrm{Nat}(F,G)$ as the end $\int_C G(c)^{F(c)}$, which can be calculated as the equalizer of the two maps $\prod_C G(c)^{F(c)}$ to $\prod_{f:c_1\to c_2} G(c_2)^{F(c_1)}$ whose components at $f$ are, respectively, $G(f)^{F(c_1)}$ and $G(c_2)^{F(f)}$. Thus again in case $C$ is a group and $V$ is abelian groups, the $V$-natural transformations between two $C$-modules $F$ and $G$ are those abelian group homomorphisms $a:F\to G$ such that, for every $f$ in $C$, $G(f)\circ a=a\circ F(f)$. In more traditional notation, $a(fx)=f(ax)$, for every $x\in F$: these are just the equivariant group homomorphisms, as expected, and indeed the points of the object of $V$-natural transformations between $V$-functors out of free $V$-categories are just ordinary natural transformations, so that over something like abelian groups we don't notice much difference. (Over something with a more dramatic forgetful functor, more would already have happened: the category structure on the natural transformations between functors from a category to the monoidal category of categories invents for us the notion of modification.)

Now we'll try to get the internal hom in $V^J$ for any small $V$-category $J$ in a way familiar from the case $V=\mathrm{Set}$. We have an enriched Yoneda lemma guaranteeing that, for $F,G:J\to V$, we must have $G^F(j)=V^J(\hat j,G^F)=V^J(\hat j\otimes F,G)$, for every $j\in J$; the action of $G^F$ on morphism objects, $J(j_1,j_2)\to V(V^J(\hat j_1\otimes F,G),V^J(\hat j_2\otimes F,G))$, is defined by the composition $$J(j_1,j_2)\otimes V^J(\hat j_1\otimes F,G)\to V^J(\hat j_2\otimes F,\hat j_1\otimes F)\otimes V^J(\hat j_1\otimes F,G)\to V^J(\hat j_2\otimes F,G)$$ The first map there applies Yoneda, $J(j_1,j_2)\to V^J(\hat j_2,\hat j_1)$, and then tensors with $F$; the second map is just composition. Using the enriched co-Yoneda lemma, that makes every functor a colimit of representables, one can show that this really does give a Cartesian closed structure on $V^J$.

Hopefully that's a less obscure, if ludicrously wordier, approach to the Cartesian closure of the general functor categories. Anyway, what's this all mean in the case $J=V(C)$ with $C$ a group and $V$ abelian groups? Given $F$ and $G$, we know $G^F(c)$, where $c\in C$ is the unique object, is supposed to be $V^C(F\otimes \hat c, G)$, the abelian group of equivariant homomorphisms into $G$ from $F$ tensored with the regular representation (with the left action.) Now comes a trick: equivariant homomorphisms $\bar h:F\otimes \hat c\to G$ are naturally identified with arbitrary homomorphisms $h: F\to G$! This is because $\hat c$ is free and transitive: we can choose $\bar h$ arbitrarily on the summand of $\hat c(c)$ corresponding to the identity morphism, and everything else is uniquely determined.

So, we've finally found that $G^F(c)$ is the group of non-equivariant homomorphisms $F\to G$, and we have only to determine the $C$-action. Well, we worked out the formula above: a morphism/element of the group $f:c\to c$ acts on $h:F\to G$ by viewing $h$ as an equivariant map $\bar h: F\otimes \hat c\to G$ and acting by $f$ on the $\hat c$ coordinate. But note that the contravariance of the Yoneda we're using here makes $f$ act on the right (otherwise it also wouldn't be a module map.) So, abusing notation, we can say $f\bar h(x\otimes f')=\bar h(x\otimes f'f)$. Now to turn $f\bar h$ back into a non-equivariant map $F\to G$, we evaluate on the identity. So $$(fh)(x)=f\bar h(x\otimes e)=\bar h(x\otimes f)=f \bar h(f^{-1} x\otimes e)=f(h(f^{-1}x))$$

And there you have it! Notice that the appearance of an inverse here suggests things aren't so simple for monoids (or categories!) as indeed they aren't-if $C$ is a monoid that's not a group then an equivariant map out of $F\otimes \hat c$ is not uniquely determined by its restriction to the identity component. However, this internal hom is still not always bad to calculate. I highly recommend trying it for functors out of the arrow category as an exercise.

Notice also that I could (should?) have told this story a lot faster by sticking with $V=\mathrm{Set}$-you get the same internal hom formula for $G$-sets as from $G$-representations.

• Thank you very much for this answer. It will probably take me a while to digest it. I have also asked this related question. Is there an internal tensor-hom adjunction hiding anywhere in your answer? Also, I still don't understand how a category representation - even a monoid representation - should act on arrows. Jun 25 '17 at 13:25
• I'm not sure what you're distinguishing by mentioning "internal" adjunction. For the other question, this can be calculated using the formula I gave, so that $F^G(c)$ is the equivariant maps $G\otimes\hat c\to F$, with action in the $\hat c$ variable. So say $C$ is the monoid freely generated by an idempotent element $a$; then we have functions $f_1:G\otimes 1\to F$ and $f_2:G\otimes a\to F$ such that $f_2(ax\otimes a)=af_1(x)$. And $a$ acts on the hom representation by sending $(f_1,f_2)$ to $(f_2,f_2)$. Notice that the fixed points are still the equivariant maps $G\to F$. Jun 26 '17 at 2:47
• By internal adjunction I mean natural isomorphisms $\mathsf{hom}(A\otimes B,C)\cong \mathsf{hom}(A,\mathsf{hom}(B,C))$ where all homs are internal, not just the inner one on the right. For the rest, I will have to digest for a while :) Jun 26 '17 at 10:10
• Ah, got it. No, I didn't mention anything about this in my answer. But it should also be true, for formal reasons coming from $V$. Jun 26 '17 at 10:48

About question 1 I guess that there are many possible different answers but the most easy, at least to me, is that with that actions the classical unit and counit of the adjunction $$-\otimes W \dashv [W,-]$$ in $\mathbb K$-vect that is the mappings $$\epsilon_{V} \colon [V,W] \otimes V \longrightarrow W$$ $$\epsilon_{V} (f \otimes v) = f(v)$$ and $$\eta_V \colon V \longrightarrow [W,V \otimes W]$$ $$\eta_V(v)(w) = v \otimes w$$ become morphisms of $G$-representations. In this way you basically prove that the whole adjunction $(-\otimes W,[W,-],\epsilon,\eta)$ lift to an adjunction in $\mathbf{GRep}$ (the category of $G$-representations).

About question 2 I am a little bit short in the adjoint lifting theorem, so I'll need some time to think it through.