# Category of representations is Tannakian

Tannakian categories are modeled on categories of linear representations. I am trying to learn about both subjects, but I cannot see why a category of representations should be a rigid tensor category. Specifically: for each object $$M$$ in a rigid tensor category $$\mathcal{T}$$, the axioms require that the dual object $$M^\vee$$ should satisfy the property that the functor $$-\otimes M^\vee$$ is left-adjoint to the functor $$-\otimes M$$. But why does this property hold for representations? If I have a morphism of vector spaces (representations) $$A\otimes M^\vee \to B$$, how do I get a morphism $$A\to B \otimes M$$?

This only works for finite-dimensional vector spaces. In that case the canonical map $$M\to M^{\vee\vee}$$, $$x\mapsto \widehat{x}$$, where $$\widehat{x}$$ is given by $$\widehat{x}(\xi) = \xi(x)$$ for $$\xi\in M^\vee$$, is an isomorphism.
Hence, we have $$B\otimes M\cong B\otimes M^{\vee\vee} \cong \hom(M^\vee, B)$$. But then we have natural isomorphisms $$\hom(A\otimes M^\vee, B) \cong \hom(A, \hom(M^\vee, B)) \cong \hom(A, B\otimes M)$$ via the tensor-hom adjunction.
• Thank you. Is the isomorphism $B\otimes M^{\vee\vee} \cong \hom(M^\vee,B)$ given by $b\otimes \overline{m} \mapsto (\xi \mapsto \xi(m) \cdot b)$ ? Is there an explicit inverse? And why does this not work in infinite imension, since I can avoid using $M^{\vee\vee}$ and I can just use the same map to see that $B\otimes M \cong \hom(M^\vee,B)$? – 57Jimmy Feb 22 at 14:03
• The map you suggest is correct. As far as I know, there is no explicit inverse (except, maybe, if you choose bases). If $M$ is not finite-dimensional then the canonical map $B\otimes M\to \hom(M^\vee, B)$ is not surjective. – Claudius Feb 22 at 14:22
• And I imagine that if $M$ is infinite dimensional, then also the map $B \otimes M^\vee \to \hom(M,B)$ is not an isomorphism because it is not injective? Whereas in the finite dimensional case it is, and $-\otimes M^\vee$ is also right-adjoint to $-\otimes M$ – 57Jimmy Feb 22 at 15:31
• This map is always injective. Choose a basis $\{e_i\}_i$ of $B$. If some $x = \sum_{i=1}^n e_i\otimes \xi_i$ maps to the zero homomorphism, then $\sum_{i=1}^n \xi_i(m) e_i = 0$ for all $m\in M$, hence $\xi_i(m) = 0$ for all $m\in M$, all $i$, by linear independence of the $e_i$. But this means $\xi_i = 0$ for all $i$ and thus $x = 0$. – Claudius Feb 22 at 22:33