3
$\begingroup$

$\newcommand{\dmod}{\text{-}\mathrm{mod}}$ Let $A$ be a finite-dimensional $k$-algebra, $A\dmod$ be a category of finite-dimensional A-modules and $\mathrm{U}_A:A\dmod \to \textbf{Vect}_k$ be a forgetful functor. We can reconstruct $A$ as $\mathrm{End}(\mathrm{U}_A)$ by using Tannaka reconstruction thorem.

Question : Does the claim hold even if the assumption of "finite-dimensional" is excluded?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

Yes, of course. You still have a natural homomorphism $A\rightarrow END(U_A)$. Since ${}_AA$ is a free $A$-module, an endomorphism $x\in END(U_A)$ is determined by its value $x_A$ on ${}_AA$. This proves that the natural homomorphism is an isomorphism: $$x_A \in End (A_{End_{{}_AA}})= End (A_{{A}})=A.$$

Improve this answer
$\endgroup$
4
  • $\begingroup$ Is $\mathrm{End}(U_A)$ a set? If F and G are functors, I think $\mathrm{Nat}(F,G)$ is not a set in general. $\endgroup$
    – yohei ohta
    Commented Nov 26, 2021 at 0:41
  • 1
    $\begingroup$ It is a set because your category is "presentable". It has a generator ${}_AA$, a value on which determines a natural endomorphism. $\endgroup$
    – Bugs Bunny
    Commented Nov 26, 2021 at 8:25
  • $\begingroup$ Thx! I would like to ask another related question. I think Tannaka reconstruction theorem holds for an arbitary complete closed monoidal category $\mathcal{C}$. Let A be a monoid in $\mathcal{C}$ ,$A\text{-}\mathrm{Mod}$ be a category of left A-modules in $\mathcal{C}$ and $U_A:A\text{-}\mathrm{Mod} \to \mathcal{C}$ be a forgetful functor. Is $\mathrm{Nat}(U_A,U_A)$ a object of $\mathcal{C}$? $\endgroup$
    – yohei ohta
    Commented Nov 26, 2021 at 9:55
  • $\begingroup$ No, it is not. Take ${\mathcal C}$ to be vector space. Then $Nat(U_A,U_A)$ is typically not a vector space. $\endgroup$
    – Bugs Bunny
    Commented Nov 29, 2021 at 8:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .