# The rigid additive tensor category freely generated by an object

I've found myself to be absolutely mystified by something in Deligne and Milne's notes on Tannakian categories. Namely, on p. 16 they are showing that there is a rigid additive tensor category $\mathsf{C}$ with a distinguished object $T$ such that for every other rigid additive tensor category $\mathsf{D}$ there is an equivalence of categories

$$\text{Hom}^{\otimes} (\mathsf{C}, \mathsf{D})\xrightarrow{\sim}\mathsf{D}, \quad F\mapsto F(T)$$

They call this pair $(\mathsf{C}, T)$ the rigid additive tensor category freely generated by $T$, and its UMP is like a categorified version of representing the "identity functor" on the $2$-category of all rigid additive tensor categories (I don't really know anything about $2$-categories so I'm not sure how technically accurate this is!). I find it quite an interesting object, and would like to see how it exists, but the construction has left me completely confused. I will describe what they say in the notes for completeness, although the notes do not have many more details.

## Setup in the notes

They start by looking at a construction on modules. Namely, if $V$ is a free $R$-module of finite rank they set $T^{a,b}(V) = T^{a,b} = V^{\otimes a} \otimes (V^\vee) ^{\otimes b}$ as the $R$-module of tensors of covariant degree $a$ and contravariant degree $b$.A map $f:T^{a,b}\to T^{c,d}$ can be identified with an element $"f"\in T^{b+c, a+d}$. The identity map $\text{id}: V \to V$ gives a distinguished element $"\text{id}" \in T^{1,1}$ and the higher tensor powers of this element are in $T^{b+c,a+d}$ whenever $a+d = b+c$. We can permute the contravariant components of this special element via elements of the symmetric group $S_{a+d}$, giving a map

$$\epsilon: S_{a+d} \to T^{b+c, a+d} = \text{Hom} (T^{a,b}, T^{c,d})$$

which induces a map $R[S_{a+d}]\to \text{Hom} (T^{a,b}, T^{c,d})$ that is injective provided $\text{rank} (V)\geq a+d$. If we have maps

$$T^{a,b}\xrightarrow{\epsilon(\sigma)} T^{c,d} \xrightarrow{\epsilon(\tau)} T^{e,f}$$

then apparently $\epsilon (\tau) \circ \epsilon(\sigma)$ is given by a universal formula

$$\epsilon (\tau) \circ \epsilon(\sigma) = \text{rank} (V)^N \epsilon (\rho).$$

Then a general category is constructed by using this case of modules as a model.

## Questions

I know this is a lot to ask, but I would really appreciate it if anyone who has understood this construction could help me with some basic questions. Concretely, some of the main things I don't understand are:

• If $"\text{id}"\in T^{1,1}$ is obtained from the identity map $V\to V$, then we can write $"\text{id}" = v\otimes f \in V\otimes V^\vee$. Then, to me, the higher tensor powers of $"\text{id}"$ are just elements of the form

$$"\text{id}"^{\otimes n} = a\otimes \dots a \otimes f\otimes \dots \otimes f$$

But surely, then, permuting the contravariant components of this element by elements of $S_n$ does nothing, since they are all the same. So what have I misinterpreted, and also is there a reason for permuting the contravariant components and not the covariant ones?

• I suspect this depends highly on my understanding of the map $R[S_{a+d}]\to \text{Hom}(T^{a,b}, T^{c,d})$ (which is very poor without an answer to my first question), but how can one go about calculating the universal formula?
• Finally, are there other descriptions of this construction in the literature, possibly going by different names? I have looked but I've been unable to find anything. It would be helpful for my understanding to see it done in multiple ways.

Edit: I have accepted the answer below as it most relevantly answered my question. I have also added some thoughts of a different, written as another answer although they are far from a complete understanding. I would certainly still welcome other answers or comments, even if they are also incomplete!

Let $V$ have basis $e_1, \ldots, e_n$. There is a basis $\delta_, \ldots, \delta_n$ of $V^\vee$ called the dual basis characterized by the property $\delta_i(e_j) = \begin{cases}1 & \text{if }i=j \\ 0 & \text{otherwise}\end{cases}$.

The element $"\mathrm{id}" \in T \otimes T^\vee$ corresponding to the identity $V \to V$ is then $\sum_{i=1}^n e_i \otimes \delta_i$. As you can see the tensor powers of $"\mathrm{id}"$ are fairly complicated and permuting the contravariant components is definitely not the identity.

So, why does that element correspond to the identity? Let's go backwards: given $\sum_i v_i \otimes \alpha_i \in V \otimes V^\vee$, the linear transformation $V \to V$ corresponding to it is given by $F(x) = \sum_i \alpha_i(x)v_i$. If you try this on the candidate for $"\mathrm{id}"$, you see that it corresponds to $x \mapsto \sum_{i=1}^n \delta_i(x) e_i$. Since the $\delta_i$ give the coordinates with respect to the basis $e_1, \ldots, e_n$, that sum is precisely $x$.

EDIT: About the correspondence between maps $T^{a,b} \to T^{c,d}$ and elements of $T^{b+c,a+d}$. The correspondence described above $V\otimes V^\vee \to \mathrm{Hom}(V,V)$ generalizes straightforwardly to give a map $V \otimes W^\vee \to \mathrm{Hom}(W,V)$ ---just use the same formula! (Warning: this map is an isomorphism only if $W$ is finite dimensional.) Then we have $T^{b+c,a+d} \cong T^{c,d} \otimes T^{b,a} \cong T^{c,d} \otimes (T^{a,b})^\vee \to \mathrm{Hom}(T^{a,b},T^{c,d})$.

• Dear Omar, thanks for this very clear and helpful answer! It seems I had forgotten how one actually transfers elements across the isomorphism. I will try calculating the universal formula from this now that I know how it works. – Alex Saad May 16 '16 at 20:37
• My apologies - I'm struggling to write down the right relationship between considering $\epsilon(\sigma)$ as an element of $T^{b+c, a+d}$ and as a map $T^{a,b}\to T^{c,d}$. What's confusing me is how your map generalises to the possibility of dual factors (i.e. $b>0$). What's the correspondence in this case? – Alex Saad May 16 '16 at 22:53
• I added a bit about $T^{a,b}$ to the answer, @AlexSaad. I didn't add an explicit formula, but I think this might actually be clearer than one. – Omar Antolín-Camarena May 18 '16 at 18:05
• I managed to figure this out before you edited it in, but it certainly is clearer than an actual formula. Thanks for adding it - it's been a good lesson in actually developing my understanding of the tensor-hom adjunction. The universality of the category Deligne constructs eventually follows because we're allowed to use this trick - everything in a "rigid" tensor category is "finite dimensional" i.e. isomorphic to its double dual. – Alex Saad May 19 '16 at 11:54

Not really an answer, but I wanted to post this here in case anyone else ends up thinking about this thing. It might help set you on the right track.

Anyway, after browsing through this collection of slides (see "Ingredients of Construction" slide) I now know that the morphisms in this category can be described using walled Brauer diagrams (see e.g. page 6-7 here). I haven't completely tied up all the holes in my understanding but it has helped quite a lot.

The rule for going from an element $\sigma \in S_{a+d}$ to a map $\epsilon(\sigma):T^{a,b}\to T^{c,d}$ is as follows: draw the element $\sigma$ as in my diagram below as a collection of strands joining the $a+d = b+c$ points along two lines from top to bottom. Additionally, there is a green "wall" between the $a$th and $(a+1)$st points on the top connected to the point between the $c$th and $(c+1)$st on the bottom (this wall represents the "split" between covariant and contravariant factors in the map). Then "flip" the right (contravariant) part of the diagram along the green wall to get a pictorial description of a map $T^{a,b}\to T^{c,d}$. One obtains a similar diagram, but with different numbers of nodes on the top and bottom (the top has $a+b$ nodes, the bottom has $c+d$ nodes). Lines that crossed the green wall then become "semi-loops" attached to one edge, which I have taken to mean "evaluation" and "coevaluation". The other lines show which factors are sent to which other factors in the corresponding map

The universal formula mentioned above feels within reach when you follow the formal rule

$$\text{closed loop in a Brauer diagram} = \text{factor of } \text{rank}(V) \text{ multiplying the diagram without the loop.}$$

Assuming this rule works in this vague interpretation, I managed to demonstrate the formula in the image below for a particular element $(2,3)$ of the symmetric group $S_5$. Why one must follow this rule is still unclear to me, but I guess it is something like this: in a composition $$T^{a,b}\xrightarrow{\epsilon(\sigma)} T^{c,d}\xrightarrow{\epsilon(\tau)}T^{e,f}$$ where $a+d = b+c$ and $c+f = e+d$, there is some number $N$ of evaluations and coevaluations occuring between elements of $V$ and $V^\vee$, which can be seen as closed loops in the centre of the Brauer diagram. When each of these pairings between coevaluations and evaluations occurs, a sum of basis vectors of the form $\delta_i(e_i)$ (notation in the accepted answer to this question) multiplies the morphism by a factor of the rank of $V$. Therefore every time a loop occurs in the diagram, we thus pick up another factor of the rank.

When you calculate what the corresponding map $\epsilon(\tau\sigma): T^{a,b}\to T^{e,f}$ does, some "double-overlapping" of the wall kills off certain strands in the diagrams, which would get mapped to closed loops after "flipping" the diagram. Multiplying by the number of strands that get killed by this gives you the same thing: hence $$\epsilon(\tau)\circ\epsilon(\sigma) = \text{rank} (V)^N \epsilon(\tau\sigma).$$

This is quite a hard procedure to describe, and after reading about it there seems to be lots of extremely deep maths related to this construction including links to supersymmetry in physics. I hope someone else may find this to be a helpful and/or interesting note. Of course, if anyone would like to correct me or add to what I have written in this answer I would be very happy!