# Is Tannaka theory easy?

$$\def\A{\mathcal A}\require{AMScd}$$

Disclaimer 1: "Tannaka theory" is an umbrella term referring to a family of results; I might have cherry-picked a version of the theorem that is particularly easier to prove abstractly, but it does not seem different -in generality if not in spirit- from the one in [1, Thm 7.9] or [2, 2.6.1]; correct me if I'm wrong!

Disclaimer 2: I do not claim to have reinvented the wheel. :-) just curious about the reason why the technology to prove the theorem existed 6 years before its claim, and yet nobody attempted to write down a similar proof (or did someone?).

Theorem. Let $$k$$ be a ring, $$F : \A \to mod(k)$$ a $$k$$-linear, faithful, strong monoidal functor with domain a $$k$$-linear rigid monoidal category. The codomain $$mod(k)$$ is finitely-generated projective $$k$$-modules.

Then there is a bialgebra $$B\in Mod(k)$$ (not necessarily fin.gen) such that $$\A$$ is monoidally equivalent to the category of $$B$$-modules $$Mod(B)$$.

I do category theory, I'm a simple man: allow me to prove this theorem with what I know.

1. I can Kan extend $$F$$ along itself on the right: $$Ran_FF : Mod(k) \to Mod(k)$$ is a monad, the codensity monad of $$F$$.

2. $$Ran_FF(k)$$ corresponds to the $$k$$-module $$\int_A \hom_k(\hom_k(k,FA),FA) = \int_A \hom_k(FA,FA)$$ i.e. to the monoid of endo-natural transformations $$Nat(F,F)$$ (the monoid operation is vertical composition). I claim this is the $$B$$ we are looking for.

3. I can Kan extend $$F$$ on the left: $$Lan_FF$$ is a comonad, the density comonad of $$F$$.

4. $$Lan_FF(k)$$ is an object of $$Mod(k)$$, not very far from $$B$$: \begin{align*} Lan_FF(k) &\cong \int^A \hom_k(FA, k)\otimes FA \\ &\cong \int^A (FA)^*\otimes FA\\ &\cong \int^A (FA\otimes FA^*)^*\\ &\cong \left(\int_A FA\otimes FA^*\right)^* = B^*\\ \end{align*}

5. "every object of $$\A$$ is a $$B$$-module" in the following sense: the universal wedge of the end $$\int_A \hom_k(FA,FA)$$ is made by maps $$\epsilon_A : B \to \hom_k(FA,FA) = End(FA)$$ and this is a ring map giving $$FA$$ a structure of $$B$$-module; a morphism $$\phi : A\to A'$$ in $$\A$$ now fits in the commutative square $$\begin{CD} B @>\epsilon_A>> \hom_k(FA,FA) \\ @V\epsilon_{A'}VV @VV(F\phi)_* V \\ \hom_k(FA',FA') @>>(F\phi)^*> \hom_k(FA,FA')\end{CD}$$ which means that for every $$b\in B$$, $$F\phi(b.x)=b.F\phi(x)$$ where $$b.\_ = \epsilon(b)$$; thus it is a homomorphism of $$B$$-modules.

I feel this is enough to define a functor $$\tilde F : \A \to Mod(B)$$ (just corestrict $$F$$) in such a way that it is an equivalence of categories; it is full and strictly surjectve on objects, and strong monoidal & fatihful by assumption. Point 4 above entails that the multiplication of $$B$$ given by vertical composition of natural transformation is compatible with a comultiplication on $$B^*$$, precisely $$\begin{array}{c} (Nat(F,F)\otimes Nat(F,F) \to Nat(F,F))^* \\ \hline Nat(F,F)^*\otimes Nat(F,F)^* \leftarrow Nat(F,F)^* \end{array}$$ (and $$B\cong B^*$$ because as $$B$$-module it is of course 1-dimensional).

So the theorem is proved.

Question 1: am I missing something?

Question 2: If not, why this is not the standard proof of the theorem above? It is streamlined, allows for countless generalizations, and uses nothing but elementary coend calculus. And basic algebra (but this is to be expected).

[1]:Tørris Koløen Bakke, Hopf algebras and monoidal categories (2007)

[2]: Rivano, Neantro Saavedra. "Catégories tannakiennes." Bulletin de la Société Mathématique de France 100 (1972): 417-430.

• Why is it essentially surjective on objects and full ? Oct 25 '19 at 15:03
• I think Tannakian reconstruction is a monoidal upgrade of the Barr-Beck theorem (wikipedia says that this was first observed by Deligne, and is now the standard approach). And Barr-Beck can be proved very easily using ingredients that are similar to the ones in your argument (which I haven't read carefully). So the answer is probably: yes, Tannaka duality is easy and should admit formal proofs using co-ends. Oct 25 '19 at 15:24
• Just because something is easy now certainly doesn't make it easy half a century ago. Oct 26 '19 at 2:07
• I agree! That's why we teach things the simple way as soon as we can. Is it the case of Tannaka here? Oct 26 '19 at 10:08
• I think most people who learn Tannaka-Krein duality do not know Kan extensions, Barr-Beck, or "elementary coend calculus." Which isn't to say that this might not be a great approach, just that it's not easy for everyone. Nov 4 '19 at 19:17

This is not an answer, but a collection of thoughts that make me doubt about the correctness of the proof. I am a bit familiar with Tannaka-Krein reconstruction but I am not familiar at all with Kan extensions, hence I am not able to say if the proof is correct and I am wrong, or to find a mistake in it. Therefore, I hope somebody will help me understanding what's going on either by commenting or by answering Fosco's question.

1. The category of modules over a bialgebra is an abelian category, whence I imagine that one has to assume $$\mathcal{A}$$ abelian to have the equivalence at the end;
2. The category of modules over a bialgebra is not rigid in general, not even the category of those that are finitely generated and projective over $$k$$, whence I imagine that the conclusion should sound like: there exists a Hopf algebra $$H$$ over $$k$$ such that $$\mathcal{A}$$ is monoidally equivalent to the category of $$H$$-modules whose underlying $$k$$-module structure is finitely generated and projective.
3. Last but not the least: if I didn't misunderstand the content of the paper, Alain Bruguières proved in Théorie tannakienne non commutative that the non-commutative (in the sense of: $$k$$ non-commutative) analogue of Tannaka-Krein should be: there exists a Hopf algebroid $$\mathcal{H}$$ over $$k$$ (i.e. an algebraic groupoid) such that $$\mathcal{A}$$ is monoidally equivalent to the category of $$\mathcal{H}$$-modules whose underlying $$k$$-module structure is finitely generated and projective.

Can somebody see if I misinterpreted Fosco's argument or if there is a way to fix the whole picture?