# Categorification of $\pi$?

Is there a categorification of $\pi$?

I have to admit that this is a very vague question. Somehow it is motivated by this recent MO question, which made me stare at some digits and somehow forgot my animosity about this branch of mathematics, wondering if there is a connection to the branches I love so much.

$\pi$ is the area of the unit circle, so perhaps we have to categorify the unit circle (using the projective line?) and the area of such an object. Areas are values of integrals, and there are some kind of integrals in category theory (ends), but this is really just a wild guess.

For some great examples of categorification see this list on MO, and for the meaning of categorification see this MO question or that article by Baez/Dolan. Inspired by the answers of Todd Trimble, we may consider a categorified sinus function

$$\sin(X) = \sum_{n \geq 0} (-1)^{\otimes n} X^{\otimes (2n+1)} / (2n+1)!$$

in any complete symmetric monoidal category (if we can make sense of $-1$).

Edit: Perhaps $(-1)$ should be the universal invertible object $\mathcal{L}$ such that the symmetry on $\mathcal{L} \otimes \mathcal{L}$ equals $-1$. In the theory of $k$-linear cocomplete symmetric monoidal categories, this is the category of super vector spaces over $k$, i.e. $\mathbb{Z}/2$-graded vector spaces, with a twisted symmetry. Here, $\mathcal{L}$ is $1$-dimensional concentrated in degree $1$. Thus, we have a sine function for super vector spaces, namely $\sin(V) = \oplus_{n \geq 0} \mathcal{L}^{\otimes n} \otimes V^{\otimes (2n+1)} / \Sigma_{2n+1}$. How does it look like, and can we extract something which resembles $\pi$?

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Pi is a period integral, right? My understanding is that period integrals are supposed to be some kind of decategorification of motives. – Qiaochu Yuan Jan 12 '11 at 23:37
I can't figure out what "-1" should be, because it is not clear if such an object should exist in ANY complete symmetric monoidal category or not (and in that case, sin(-) of an object makes sense only if I can find it). In the second case, consider the monoidal structure on SETS given by the set-theoretic product and having the singleton {} as "neutral element"... There are definitely no way to have a set such that AxA = {}... – Fosco Loregian Jan 13 '11 at 14:26
I mean that I want to recover the "obvious" (?) relation $(-1)\otimes(-1)\cong I$, where $I$ is the identity wrt the tensor $\otimes$. – Fosco Loregian Jan 13 '11 at 14:27
Given the plethora of infinite series representations of $\pi$, I sort of expect that there are many categorifications of $\pi$, none of which is necessarily canonical. – deoxygerbe Apr 25 '12 at 23:21
@deoxygerbe: The question is: Is there a series representation which doesn't involve negatives and whose summands admit a categorification? – Martin Brandenburg Dec 19 '13 at 18:57

One approach to questions of this general nature is to find a groupoid whose groupoid cardinality is equal to the number in question; this is a form of groupoidification. For example, the groupoid cardinality of the groupoid $\text{FinSet}_0$ of finite sets and bijections is $\sum \frac{1}{n!} = e$, so this is a reasonable categorification of $e$; in fact arguably this is "the" categorification of $e$ and goes a long way towards explaining its prevalence in mathematics. Similarly, for any finite set $X$, the groupoid cardinality of the groupoid of $X$-colored finite sets and color-preserving bijections is $\sum \frac{|X|^n}{n!} = e^{|X|}$. Note that this groupoid is the coproduct of $|X|$ copies of $\text{FinSet}_0$; see also this math.SE question.
In a TWF which I can't find (because John Baez made them very hard to search through for some reason!) John Baez mentions that he looked into the problem of finding a "natural" groupoid whose cardinality is $\pi$, but that he (and possibly some collaborators) weren't able to come up with any nice examples. So possibly this is the wrong direction to go in the particular case of $\pi$.
Can you make sense of the «grupoid of $(-1)$-colored finite sets and finite color-presrving bijections» so as to categorify $e^{-1}$ (which comes up in a sensible combinatorial way, of course) – Mariano Suárez-Alvarez Jan 13 '11 at 3:30
Nice. Now do $e\cdot e^{-1}=1$ :) – Mariano Suárez-Alvarez Jan 13 '11 at 3:47