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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 $A\to B \otimes M$?

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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.

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  • $\begingroup$ 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 $B\otimes M \cong \hom(M^\vee,B)$? $\endgroup$ – 57Jimmy Feb 22 at 14:03
  • $\begingroup$ 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. $\endgroup$ – Claudius Feb 22 at 14:22
  • $\begingroup$ Ok, I suspected that. Thanks! :) $\endgroup$ – 57Jimmy Feb 22 at 14:33
  • $\begingroup$ 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 $-\otimes M$ $\endgroup$ – 57Jimmy Feb 22 at 15:31
  • $\begingroup$ 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 $x = 0$. $\endgroup$ – Claudius Feb 22 at 22:33

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