The n-Category Café

You should think of the suspense as a good thing — math classes are often short on suspense.

But, the answer is explained in the next few lectures. So, if the tension is really unbearable, you can peek ahead to see lectures that haven’t been ‘officially released’ yet. Right now it’s the last week of the Fall quarter at UCR, but I plan to dole out the lectures up to lecture 20 at a slow rate throughout the Christmas break, to keep everyone happily entertained until classes resume on January 4th.

Good question, David. Believe it or not, we haven’t thought too much about the $n$-groupoidal or $\infty$-groupoidal generalizations of what we’ve done so far. But, if you think of a complex algebraic variety as giving a particularly nice $\infty$-groupoid, our ideas are related to something well-known!

In algebraic geometry people like to ponder correspondences between algebraic varieties. A correspondence from a variety $X$ to a variety $Y$ is a very nice binary relation: a Zariski-closed subset of $X \times Y$. If gives a span of algebraic varieties

$$X \stackrel{p}{\leftarrow} S \stackrel{q}{\rightarrow} Y$$

and this span is sufficiently well-behaved that we get a map on homology groups

$$H_*(X) \to H_*(Y)$$

using a ‘pull–push’ strategy:

$$H_*(X) \stackrel{p^!}{\to} H_*(S) \stackrel{q_*}{\to} H_*(Y)$$

Here $q_*$ is the usual ‘pushforward’ on homology, while $p^!$ is the sneaky ‘pullback’ part — often called the ‘transfer map’.

The pushforward is easy to define, but we can only define the transfer map

$$p^! : H_*(X) \to H_*(S)$$

when the map

$$p: S \to X$$

is sufficiently well-behaved. The simplest case is when $p$ is a covering map… if you know about singular homology, you can easily guess how this works just by drawing a few pictures! A simplex down in $X$ lifts up to a bunch of simplices in $S$. Apparently branched covers also work.

I’d love it if an algebraic geometer would step up and say in general what kind of map between varieties (or schemes) admits a well-behaved transfer. I don’t understand this stuff very well, and some of the things I said above could be missing some technical hypotheses.

Anyway: there’s a close relation between this stuff and the concept of ‘Euler characteristic’. What Jim and I seem to have (unintentionally) done is to generalize this idea to a different context using homotopy cardinality instead of Euler characteristic!

Euler characteristic tends to be well-defined for algebraic varieties. It doesn’t tend to be well-defined for $K(G,1)$’s when $G$ is a finite group, or groupoid. But homotopy cardinality is well-defined for such spaces, and it serves as a substitute for Euler characteristic. It shows up in Jim’s definition of the ‘transfer’ map

$$p^! : H_0(X) \to H_0(S)$$

for a functor

$$p: S \to X$$

between finite groupoids. This definition will show up in lecture 19.

The question has a homological flavor to it. I think $Mat(R)$ need not be finitely complete even if $R$ is a ring; counterexamples occur for at least some rings of homological dimension greater than 1.

For a rig $R$, define an $R$-module in the expected way: as a commutative monoid $A$ equipped with a rig homomorphism $R \to hom(A, A)$. (In other words, a rig $R$ is nothing but a one-object $CMon$-enriched category, and a module is a enriched functor $R \to CMon$.) If $R$ is a ring, then an $R$-module in this sense agrees with the usual notion: the underlying commutative monoid is an abelian group, since $-1 \in R$ acts by additive inverse.

The free $R$-module functor takes a set $X$ to an $X$-indexed coproduct of copies of $R$. By the universal property of free modules, the category $Mat(R)$ is equivalent to the category of finitely generated free $R$-modules, $Free_R$. So the question is whether $Free_R$ admits equalizers.

But take $R = \mathbb{Z}_6$. The map $m_2: R \to R$ given by multiplication by 2 has a kernel which is not free, so there is no equalizer of the pair of maps

$$m_2, 0: R \to R$$

in $Free_R$. (One should convince one’s self that the inclusion $Free_R \hookrightarrow Mod_R$ preserves those equalizers which happen to exist, so that in $Free_R$ one may calculate equalizers in the “usual” way. I feel pretty sure the claim is true…)

Can I make the question even more complicated by pointing out that, to a constructive mathematician, the answer may be negative even when R is a field? This is because (for example), if you want to calculate the dimension of the kernel of x ↦ ax, then you must determine whether or not a is zero.

(Here I take the correct constructive notion of ‘field’ which is satisfied by, for example, the Dedekind reals: given any element a of the commutaive ring R, a = 0 iff a is not invertible. If you use the condition that every element is either invertible xor equal to 0 —which is a stronger condition in intuitionistic logic but is not satisfied by the ring of Dedekind real numbers in constructive analysis—, then Mat(R) does have equalisers.)

If you don’t believe in constructive mathematics, then just ask the question internal to a nonBoolean topos. But my intuition is that this is further evidence that the answer is likely to be complicated, even in the classical case. If I may be so bold, it even suggests to me that Mat(R) is simply the wrong category to use.