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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.

J. Doe
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  • Where do you get stuck? I assume that you already know how to define the structure of a $G$-module on that Hom-space? – Tobias Kildetoft Feb 05 '18 at 10:50
  • @TobiasKildetoft Sorry, I took some time to respond to your comment. Yes I am familiar with how $X$ is given the structure of a $G$-module, but I'm struggling to see the connection here. This isn't where I was stuck, though. I was not sure if the result was right in the first place and was looking for a way to directly prove that $\mathrm{Hom}(- \otimes X, Y) \cong \mathrm{Hom}(-, \mathrm{Hom}(X, Y))$, assuming it was true (and trying to do so seem a bit foolish now). – J. Doe Feb 05 '18 at 11:17
  • You need to be very careful with the notation here, as you have two different Hom's in play, one being the Hom of modules and the other being the Hom of vector spaces. What you need to show is that the isomorphism holds when the outside Hom is for modules, but one way to do this is to show it for the one for vector spaces and then see that the $G$-actions coincide – Tobias Kildetoft Feb 05 '18 at 11:26
  • @TobiasKildetoft The notation is a problem, yes. I'll write $\mathcal{X}$ for the representation on $X$ of $G$ from here on. That is, $\mathcal{X}=(X, \rho : G \to GL_X)$. I need to prove this: $\mathrm{Hom}(- \otimes\mathcal{X}, \mathcal{Y}) \cong \mathrm{Hom}(-, (\mathrm{Hom}_{k-\text{lin}}(X, Y), \rho))$.

    My understanding is that you're saying I can equivalently prove that $\mathrm{Hom}{k-\text{lin}}(- \otimes X, Y) \cong \mathrm{Hom}{k-\text{lin}}(-, \mathrm{Hom}_{k-\text{lin}}(X, Y))$, which I just figured how to do. I still don't know how to see that the $G$-actions coincide, tho.

     – J. Doe Feb 05 '18 at 12:30
  • Just write up what the $G$-action is on either side and check that the isomorphism is compatible with that. – Tobias Kildetoft Feb 05 '18 at 12:33
  • @TobiasKildetoft I thought about this for some time now. I'm still getting no where. Reading about this problem on the web, there's also a connection with the representation $(X^{\vee} \otimes Y, \rho)$. I haven't moved much further than where I started at, however, and I'd appreciate if you can give me some hints. – J. Doe Feb 05 '18 at 19:14
  • Did you find an explicit isomorphism between the spaces of linear maps? In one direction it is easy to write up explicitly an isomorphism, and using that makes it a lot easier. – Tobias Kildetoft Feb 06 '18 at 07:33
  • @TobiasKildetoft Yes, the isomorphism $\mathrm{Hom}{k-\text{lin}}(Z⊗X,Y)≅\mathrm{Hom}{k-\text{lin}}(Z,\mathrm{Hom}_{k-\text{lin}}(X,Y))$ is given by $(f :Z\otimes X \to Y) \mapsto(g : z \mapsto (h:x \mapsto f(z \otimes x)))$. I don't understand how you mean I should be using this isomorphism. Your comments are a bit 'cryptic', if you don't mind my saying so. – J. Doe Feb 06 '18 at 18:41
  • Sorry for being cryptic, that was not quite intended. It was, however, intended that I wanted to give away as little as possible at a time, since I think it is important to work through this in detail at least once on your own. Now you have the isomorphism, let's call it $\varphi$. You want to show that for any $g\in G$ and any linear map $f: Z\otimes X\to Y$ you have $g\varphi(f) = \varphi(gf)$. This should just be a matter of writing up what the action is in terms of the actions on each of $X$, $Y$ and $Z$. – Tobias Kildetoft Feb 06 '18 at 19:47
  • @TobiasKildetoft I understand what you mean. I want to know, however, what do you mean by taking $g \in G$? To me $G$ is a representable group-valued functor from the $k$-algebras. Are we first treating $G$ only as an 'ordinary' group. Can you help me first see what the bigger picture is, since at this point I'm a bit confused and don't even understand clearly why we can make a switch of the Hom sets that you mentioned in your second comment. Thank you. – J. Doe Feb 07 '18 at 11:52
  • Right, so by $g\in G$ I am being very sloppy with the notation (for convenience, but I really should not be). What I really mean is to fix a $k$-algebra $A$ and take $g\in G(A)$, then do everything "at the level of $A$". One then needs to show that everything done like this really is functorial in $A$. – Tobias Kildetoft Feb 07 '18 at 11:56
  • @TobiasKildetoft I can understand if you're not interested in answering this question now. However, if you would like to help me, can you maybe explain how to prove $g\varphi(f) = \varphi(gf)$? I'm still stuck with this problem for so long now and no one that I know of could help me either with this. I've been so far reluctant to comment here because I've made zero progress on this problem due to my inability to define an action that would prove the equality. – J. Doe Mar 06 '18 at 14:17

1 Answers1

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$\DeclareMathOperator{\Hom}{Hom}$ 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.

Tobias Kildetoft
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