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

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