The distribution of the trace of a random element of $SU_2$ is the Sato-Tate distribution.

The analogue of the Gaussian distribution in free probability theory is the Wigner semicircle distribution.

The Sato-Tate distribution and the Wigner semicircle distribution are the same.

Combinatorially, this comes from the fact that the moments of both those two distributions are the Catalan numbers.

We can categorify this as follows: If $A$ is a self-dual element of a monoidal abelian category over a field, then $Hom(1, A^{\otimes n})$ is a vector space. (By self-duality, this is also equal to $Hom(A^{\otimes a}, A^{\otimes b} )$ for any $a,b$ such that $a+b=n$.)

This categorifies the $n$th moment in the following way: If $G$ is a compact Lie group, and $A$ is a self-dual representation in the category of representations of $A$, then the dimension of $Hom(1, A^{\otimes n})$ is the $n$th moment of the trace of a random element of $G$ acting on $A$. So in the case $G=SU_2$, $A$ the standard representation, the dimension is the Catalan numbers.

On the other hand, if you take a free monoidal abelian category on one self-dual generator, the moments are the Catalan numbers. (One has to take dimensions, not just over the base field, but over the ring generated over the base field by the element of $Hom(1,1)$ defined by the canonical morphism $1 \to A \otimes A \to 1$.)

Since the category of representations of $SU_2$ is certainly a monoidal abelian category, we have a functor from the second category to the first that sends the free generator to the standard representation. So we have a morphism from the free moment vector spaces to the moment vector spaces of $SU_2$.

I believe I can show by an explicit combinatorial argument that this is an isomorphism, categorifying this identity between the two definitions of Catalan numbers.

Hence the free monoidal abelian category on one self-dual generator is equivalent to the category of finite-dimensional representations of $SU_2$.

Is there some conceptual reason why this is so?

Note that the limit of the category of representations of $USP_{2n}$ as $n$ goes to $\infty$ is the free symmetric monoidal abelian category on one self-dual generator.

  • 2
    $\begingroup$ Qiaochu asked a similar question a while ago mathoverflow.net/questions/17197/… and then blogged about this at some point. I can't find all the posts, but one of them is qchu.wordpress.com/2010/03/07/… $\endgroup$ Nov 26, 2014 at 2:58
  • $\begingroup$ Unrelated to the main thrust of the question, but here's an idle thought I had once: the above construction categorifies moments. How should we categorify cumulants? There's a diagram I can write down that I was tempted to turn into a chain complex whose class in the Grothendieck group might be a candidate, but I couldn't figure out anything that I could do with this construction so I stopped worrying about it. $\endgroup$ Nov 26, 2014 at 5:44
  • $\begingroup$ Where can I learn about the symmetric case alluded to in the last line of the question? $\endgroup$
    – Tim Campion
    Nov 26, 2014 at 14:18

1 Answer 1


I think what you said is not quite right; let me explain why.

The free monoidal category on a self-dual object (ignoring abelian for now) has a description involving a version of the cobordism hypothesis (namely the "tangle hypothesis": see Theorem 4.4.4 in Lurie's On the classification of topological field theories). What it works out to is that this gadget has

  • Objects finite sets of points in $\mathbb{R}$ and
  • Morphisms ambient isotopy classes of cobordisms in $\mathbb{R} \times [0, 1]$ between objects (living in $\mathbb{R} \times \{ 0 \}$ and $\mathbb{R} \times \{ 1 \}$ respectively).

Here the monoidal structure is stacking sets of points on top of each other. This category deserves to be called something like the category of planar tangles. As you know, it's a fun exercise to work out that, letting $X$ denote the object generating this category, there are exactly $C_n$ elements of $\text{Hom}(1, X^{\otimes 2n})$ all of whose connected components intersect the boundary at least once. This is essentially a slightly disguised version of the interpretation of the Catalan numbers in terms of parentheses. In fact, $\text{End}(1)$ is freely generated by a circle, and as an $\text{End}(1)$-module $\text{Hom}(1, X^{\otimes 2n})$ is freely generated by those $C_n$ elements.

Starting from here you can get something more linear by first asking for the free monoidal $k$-linear category on a self-dual object, and you can get that just by taking the free $k$-linear category on the above. Now $\text{End}(1)$ is a polynomial ring $k[x]$ but everything else I just said is still true. This category is a version of the Temperley-Lieb category.

Specializing to $k = \mathbb{C}$ you get something that looks an awful lot like the monoidal subcategory of $\text{Rep}(\text{SU}(2))$ generated by the defining representation $V$, but not quite: the circle, which corresponds to $x$ in our free category, gets sent to $-2$ in $\text{Rep}(\text{SU}(2))$ (this is due to the fact that $V$ is self-dual but a map $V \otimes V \to 1$ exhibiting that self-duality can't be chosen to be symmetric; in other words, $V$ is quaternionic).

If you want to get every object in $\text{Rep}(\text{SU}(2))$, it turns out you don't have to go anywhere near as far as abelian: it suffices to first require the existence of finite direct sums and second to require that idempotents split. In other words, it suffices to require Cauchy completeness. The corresponding free object gets us every object in $\text{Rep}(\text{SU}(2))$ but again there's that subtlety with $x \in \mathbb{C}[x]$ being sent to $-2$.

At this point I want to convince you that you cannot recover $\text{SU}(2)$ itself from this object. There are two ways I know of to try something like this, one involving Tannaka-Krein reconstruction using a fiber functor and one involving Doplicher-Roberts reconstruction. In both situations it's crucial that you have access to not only the monoidal structure on $\text{Rep}(\text{SU}(2))$ but the braided monoidal structure, which this construction doesn't provide (and trying the corresponding free braided monoidal construction gets you extra morphisms coming from tangles in $\mathbb{R}^3$). In fact you can choose various braided monoidal structures that don't get you $\text{SU}(2)$ but that will instead get you the quantum groups $U_q(\mathfrak{su}(2))$. Here $x$ will be sent to something like $ - q^2 - q^{-2}$ depending on your conventions.

So we haven't found $\text{SU}(2)$ itself but some kind of combinatorial / topological shadow of both it and its quantum group relatives. Incidentally, I have no idea what this has to do with free probability; it would be cool if someone knew.

  • $\begingroup$ There are some subtleties here involving framings I can't claim to have a 100% grasp on, and it's possible I've made a few incorrect statements above because of them. $\endgroup$ Nov 26, 2014 at 6:10
  • $\begingroup$ Thanks for clearing up the details of the category theory. That all looks right. When I wrote down the equations that make a homomorphism $A \otimes A \to A \otimes A$ into a symmetric structure on the category, I got $x^2=4$, so I think plus and minus $2$ both work. $\endgroup$
    – Will Sawin
    Nov 26, 2014 at 15:01
  • $\begingroup$ Yes, certainly you need the braided structure to reconstuct $SU_2$. But what interests me is that you can in fact find a braided structure. $\endgroup$
    – Will Sawin
    Nov 26, 2014 at 15:07
  • $\begingroup$ For me, this came up in a context where I have a monoidal category and an explicit probability distribution associted to each object where the moments were exactly this. Some subcategories look like the representation theory of a group, and some, I suspect look like "random matrices", i.e. free monoidal categories. I wanted to see if you could distinguish the two by categorifying, instead of looking at their moments, but it turns out you can't. So this provides some insight into the relation between category theory and probability. $\endgroup$
    – Will Sawin
    Nov 26, 2014 at 15:13
  • $\begingroup$ I think in general, the $K$-theory of a rigid monoidal category is a $C^*$ algebra, and counting the multiplicity of the unit is a state, which sends each element/observable to a (quantum) probability distribution. Then a free category corresponds to a free $C^*$ algebra as studied in free probability. So I think that's where it comes from. $\endgroup$
    – Will Sawin
    Nov 26, 2014 at 15:16

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