# Tannaka Duality

I only know two theorems for neutral Tannaka categories.

(1) One states that the set of $k$-group scheme homomorphisms $m:G_1\to G_2$ is in one to one correspondence with the set of $k$-linear tensor functors $m^*:{\rm Rep}_k(G_2)\to {\rm Rep}_k(G_1)$ which commute with the forgetful fiber functors.

(2) The other states that for any neutral Tannaka category $(C,\omega)$ there is a canonical $k$-linear tensor equivalence $\alpha_C:C\to {\rm Rep}_k(G_C)$ for some affine group scheme $G_C$, and the equivalence commutes with the fiber functors.

Question: Is it true that if I have a $k$-linear tensor functor $b:C\to C'$ which commutes with the fiber functors, then there is a unique $k$-group scheme homomorphism $m_b:G_{C'}\to G_C$ which satisfies $\alpha_{C'}b=m_b^*\alpha_C$. Here "=" means strictly equal not an isomorphism of functors.

My problem is that (1) is a correspondence between group homomorphism and "strict"functors. There are different morphisms $m_1,m_2$ which induce two isomorphic functors i.e. $m_1^*\cong m_2^*$. For example, if $G$ is an affine $k$-group scheme which admits a non-trivial $k$-rational point $g\in G(k)$. Let $m_1: G\to G$ be the identity and $m_2: G\to G$ be the conjugation: $a\to gag^{-1}$. Then for any $V\in{\rm Rep}_k(G)$ we define an isomorphism $m_1^*(V)\to m_2^*(V)$ by the isomorphism $V\xrightarrow{ g\cdot}V$.

The reason I ask this is that if I have functors between neutral Tannakian categories $a_{12}:(C_1,\omega_1)\to (C_2,\omega_2)$, $a_{23}:(C_2,\omega_2)\to (C_3,\omega_3)$, and $a_{13}:(C_1,\omega_1)\to (C_3,\omega_3)$, and if we know $a_{13}=a_{23}a_{12}$, then I want to have also a commutative diagram of the corresponding Tannakian group schemes. If we only have $a_{13}\cong a_{23}a_{12}$ this is already false becasuse of the example I provided, but I would hope this could be true when we have a strict equality.

• If $b$ is not a strict functor, how can it be strictly equal to something? – Will Sawin Mar 8 '13 at 17:52
• I assume $b$ is a functor not an isomorphism class of functors. – stefan Mar 8 '13 at 17:59
• Then why isn't the question answered by (1)? – Will Sawin Mar 8 '13 at 18:14
• Because (2) only provides an equivalence, not an isomorphism of categories. Because of this we have to choose a quasi-inverse which is very non-canonical. – stefan Mar 8 '13 at 18:31
• Nothing fails. In your example, the functor sending a representation to its conjugate by an non-unit element does not commute with the fibre functor. – Xandi Tuni Mar 8 '13 at 21:00

## 1 Answer

I think you're confused. The issue in (1) is not about strict functors. Instead, it is about what it means for a trio of functors to be commutative. If $f: A \to B$, $g: B \to C$, and $h: A \to C$ form a commutative triangle, that means $g \circ f = h$ - they are the same functor. But even if these are strictly equivalent functors, there are many different choices for the isomorphism between $g \circ f$ and $h$. Indeed, there are many different choices for the isomorphism between a functor and itself. In that case you can choose the identity, but that won't have meaning here.

A commutative triangle of functors should be a seen as trio $f,g,h$, with an isomorphism $g \circ f = h$. This is what gets you a morphism of groups, and this is the appropriate notion of a morphism of Tannakian categories.

In fact, the different functors from $Rep_G$ to itself are all the same strict functor, because they act the same on the set of representations (assuming we take the set of representations to have one element for each isomorphism class.)

• The point is how you explain my example. If you think (1) is a one to one correspondence between the set of homomorphisms from $G_1$ to $G_2$ and the set of "isomorphism class" of functors then my example is a counterexample for you. – stefan Mar 9 '13 at 15:15
• The set of isomorphism classes of functors with a choice of isomorphism to make the triangle with the fibers commutative is what you're looking for. In your case, the action on the fibers is by multiplication by $g$. – Will Sawin Mar 9 '13 at 17:36