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I would like to know a natural categorification of the rig of integers $\mathbb{Z}$. This should be a $2$-rig. Among the various notions of 2-rigs, we obviously have to exclude those where $+$ is a coproduct (since otherwise $a+b \cong 0 \Rightarrow a \cong b \cong 0$). So let's take rig categories instead.

Question. What is a natural and non-discrete rig category whose rig of isomorphism classes is $\mathbb{Z}$?

I've read MO/3476, but the answers are not really satisfactory. Both the category of tangles and Schnauel's categories of polyhedra don't qualify.

Here is my approach. Notice that $\mathbb{N}$ is the initial rig, so its categorification should be the initial rig category, which turns out to be the groupoid of finite sets and bijections, which is equivalent to the permutation groupoid $\mathbb{P}$. The rig $\mathbb{Z}$ is the free rig on one generator $x$ subject to the relations $x^2=1$ and $x+1=0$. For short, $\mathbb{Z} = \mathbb{N}[x]/(x^2=1,x+1=0)$. This suggests that our categorification is $\mathbb{P}[X]/(X^2 \cong 1,X+1 \cong 0)$, where these isomorphisms probably should satisfy some coherence condition (which are not visible in the decategorification $\mathbb{Z}$) in order to "flatten" the rig category. Namely, if $e : X^2 \to 1$ and $f : X+1 \to 0$ are the isomorphisms, we could require (here I omit the coherence isomorphisms of the rig category) that $eX = Xe : X^3 \to X$ and $fX = Xf =f \circ (e+X): X^2+X \to 0$.

I think that $\mathbb{P}[X,X^{-1}]$ should be the rig category of vector bundles on the projective space $\mathbb{P}^1_{\mathbb{F}_0}$, where $\mathbb{F}_0$ is the "field with no element" in the sense of Durov. Here, the Serre twist $\mathcal{O}(1)$ is not inverse to itsself, but this changes when we consider the Möbius strip $H$ on $S^1$. But vector bundles in topology have too much morphisms. For the same reason we cannot take $\mathbb{P}^1_{\mathbb{C}}$.

Question. What is a natural realization of the rig category $\mathbb{P}[X]/(X^2 \cong 1,X+1 \cong 0)$?

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    $\begingroup$ Very disappointing that this question did not get an answer... $\endgroup$ May 1, 2014 at 2:17
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    $\begingroup$ Has someone access to Joyal's paper "Règle des signes en algèbre combinatoire"? It seems to deal with this question. $\endgroup$ May 28, 2014 at 13:12
  • $\begingroup$ You can deduce that $X^2=1$ from $X+1=0$, so perhaps it would be enough to think about $f$. $\endgroup$ Aug 11, 2015 at 18:22
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    $\begingroup$ No, $X+1=0$ does not imply $X^2=1$. We are in a rig. $\endgroup$ Aug 12, 2015 at 15:13
  • $\begingroup$ I think that mathoverflow.net/questions/214767 will answer my question. $\endgroup$ Aug 14, 2015 at 9:18

2 Answers 2

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It should definitely be mentioned here that one often-used categorification of the integers is the sphere spectrum, $\mathbb S$, i.e. the infinite loop space $\mathbb S = \varinjlim \Omega^n S^n$. I've heard this idea attributed to Waldhausen. There's a fun exposition in TWF 102 by John Baez.

The idea is to think of $\mathbb Z$ is the (additive) group completion of the rig $\mathbb N$, and to categorify the group completion process. A natural categorification of the rig $\mathbb N$ is the 2-rig $\mathsf{FinBij}$ of finite sets and bijections, with $\times$ distributing over $\amalg$. So we group complete $\mathsf{FinBij}$ the $\amalg$ part. The thing is, rather than creating a 1-groupoid with two monoidal structures and the appropriate universal property, it's more natural to regard $\mathsf{FinBij}$ as an $\infty$-groupoid with two monoidal structures, and to construct a group completion in the world of $\infty$-groupoids. But by the homotopy hypothesis, an $\infty$-groupoid is the same thing as a space. So we end up with a space that has two monoidal structures, one distributing over the other. The "additive" monoidal structure is appropriately commutative, so this structure can be regarded as the structure of an infinite loop space, otherwise known as a connective spectrum. And the multiplicative structure makes it a "multiplicative infinite loop space", otherwise known as a ring spectrum.

Note that $\pi_0(\mathbb S) = \mathbb Z$, so this is a categorification in the most basic sense.

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  • $\begingroup$ I only saw your answer now. Thank you! $\endgroup$ Feb 15, 2020 at 18:18
  • $\begingroup$ Still I am very interested in the 2-rig interpretation of my question. $\endgroup$ Feb 15, 2020 at 18:19
  • $\begingroup$ Some other words which should appear on this answer: the theorem that the group completion of $FinBij$ is $\mathbb S$ is originally to Barratt, Priddy, and Quillen. $\endgroup$
    – Tim Campion
    Feb 10, 2021 at 4:05
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    $\begingroup$ Perhaps it should be pointed out that a grouplike rig ($\infty$-)category is automatically a(n $\infty-$)groupoid. In particular, a rig category which categorifies $\mathbb Z$, since it is grouplike, must be a groupoid. So the answer is the same as the answer for groupoids which categorify $\mathbb Z$, and for those the most natural candidate does seem to be $\mathbb S$ (or its $1$-truncation if you want to think about $1$-groupoids). $\endgroup$ Jul 24, 2021 at 20:07
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I think the answer is given by the free Picard groupoid $\mathbb{S}$ on one object. This is the $1$-truncation of the sphere spectrum, keeping only its first two homotopy groups $\mathbb{Z}$ and $\mathbb{Z}_{2}$, which is defined and studied in Section 3 of

Niles Johnson, Angélica M. Osorno, Modeling stable one-types. Theory Appl. Categ. 26 (2012), No. 20, 520–537; arXiv:1201.2686.

The Picard groupoid $\mathbb{S}$ is defined as follows:

  • The objects of $\mathbb{S}$ are the integers;
  • For $m,n\in\mathrm{Obj}(\mathbb{S})$, we have $$ \mathrm{Hom}_{\mathbb{S}}(m,n) \overset{\mathrm{def}}{=} \begin{cases} \emptyset &\text{if $m\neq n$,}\\ \mathbb{Z}_{2} &\text{if $m=n$;} \end{cases} $$
  • The monoidal structure $\oplus$ on $\mathbb{S}$ is given by addition of integers;
  • The symmetry of $\mathbb{S}$ at $(A,B)$ is the morphism $\beta^{\mathbb{S},\oplus}_{m,n}\colon m\oplus n\to n\oplus m$ defined by $$ \beta^{\mathbb{S},\oplus}_{m,n} \overset{\mathrm{def}}{=} \begin{cases} 0 &\text{if $mn$ is even,}\\ \eta_{m+n} &\text{if $mn$ is odd,} \end{cases} $$ where $\eta_{m+n}$ is the unique non-zero element of $\mathrm{Hom}_{\mathbb{S}}(n,n)$.

There are at least three senses in which this is the right answer:

  • The first one is that it truly is the $1$-truncation of the sphere spectrum:

    Proposition 3.3 of Johnson–Osorno. There is a map of $\mathbb{E}_{\infty}$-ring spaces

    $$\overline{\mathrm{B}\xi}\colon QS^{0}\longrightarrow\mathrm{B}\mathbb{S}$$ witnessing $\mathrm{B}\mathbb{S}$ as the Postnikov $1$-truncation of $QS^{0}$.

  • There second one relates to $2$-rigs and $2$-rings: the statement

    Rigs and rings are monoids in $(\mathsf{CMon},\otimes_{\mathbb{N}},\mathbb{N})$ and $(\mathsf{Ab},\otimes_{\mathbb{Z}},\mathbb{Z})$.

    categorifies to

    $2$-rigs and $2$-rings are pseudomonoids in $(\mathsf{SymMonCats},\otimes_{\mathbb{F}},\mathbb{F})$ and $(\mathsf{2Ab},\otimes_{\mathbb{S}},\mathbb{S})$,

    where $\mathbb{F}$ is the symmetric monoidal category of finite sets and bijections, the monoidal unit of the tensor product $\otimes_{\mathbb{F}}$ of symmetric monoidal categories, and $\mathbb{S}$ is the Picard groupoid of Johnson–Osorno, which is again the monoidal unit of the tensor product $\otimes_{\mathbb{S}}$ of Picard groupoids.

  • Finally, we have an analogue of Baez's conjecture for $\mathbb{S}$. In Remark 3.2 of their paper, Johnson–Osorno note that $\mathbb{S}$ admits also a second monoidal structure given by \begin{align*} (m,n) &\mapsto mn,\\ (m\xrightarrow{f}m,n\xrightarrow{g}n) &\mapsto nf+mg, \end{align*} with symmetry isomorphism $\beta^{\mathbb{S},\otimes}_{m,n}\colon m\otimes n \to n\otimes m$ defined by $$ \beta^{\mathbb{S},\otimes}_{m,n} \overset{\mathrm{def}}{=} \begin{cases} 0 &\text{if ${m\choose 2}{n\choose 2}$ is even,}\\ \eta_{nm} &\text{if ${m\choose 2}{n\choose 2}$ is odd.} \end{cases} $$ We then have:

Just as $\mathbb{N}$ is the initial rig and $\mathbb{F}$ is the initial $2$-rig, the ring $\mathbb{Z}$ is the initial ring and $\mathbb{S}$ is the initial $2$-ring.

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    $\begingroup$ I don't understand how the conjecture doesn't follow from the description of 2-rings as pseudomonoids in the category you indicated. The unit is always the initial pseudomonoid, isn't it ? But the conjecture also follows from the fact that $\mathbb S$ is the truncation of the sphere spectrum. Am I missing something ? $\endgroup$ Jul 24, 2021 at 20:09
  • $\begingroup$ @MaximeRamzi It's not a conjecture then! Thanks! $\endgroup$
    – Emily
    Jul 24, 2021 at 20:39
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    $\begingroup$ The recent big trilogy of books by Johnson and Yau prove, among many other things, Baez's conjecture in greater generality, including the case of braided bimonoidal categories. I think the original version was proved by Elgueta. $\endgroup$
    – David Roberts
    Jul 24, 2021 at 21:39

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