Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is a follow-up question to this coend computation. Here's an attempt at a slightly simpler computation:

$\int^{a \in A} \mbox{hom_A(a,a)}$

This should be similar to the trace operator. In attempting to follow the derivation

$\begin{array}{l}\mbox{Set}(\int^{b \in B}\mbox{hom}(a, b) \times F(b), S)\\ \cong \int_b \mbox{Set}(\mbox{hom}(a,b) \times F(b), S)\\ \cong \int_b \mbox{Set}(\mbox{hom}(a,b), \mbox{Set}(F(b), S)) \\ \cong \mbox{Nat}(\mbox{hom}(a,-), \mbox{Set}(F(-), S) \\ \cong \mbox{Set}(F(a), S),\end{array}$

I get

$\begin{array}{l}\mbox{Set}(\int^{a \in A} \mbox{hom}_A(a,a), S) \\ \cong \int_{a \in A} \mbox{Set}(\mbox{hom}_A(a,a), S) \\ \cong \mbox{Nat}(\mbox{hom}_A(-,-), S)\end{array}$

So here I guess we have the set of natural transformations from the hom functor to the constant functor $S$. For any first parameter $a$, we have the set of natural transformations from hom$(a,-)$ to $S(a,-)$, which by Yoneda's lemma is isomorphic to $S(a,a) = S$. So I think it goes

$\begin{array}{l}\cong \displaystyle \prod_a \mbox{Nat}(\mbox{hom}_A(a,-), S(a,-)) \\ \cong \prod_a S\\ \cong S^{\mbox{Ob}(A)} \\ \cong \mbox{Set}(\mbox{Ob}(A), S).\end{array}$

So $\int^{a \in A} \mbox{hom_A(a,a)} \cong \mbox{Ob}(A).$ Is that right?

share|improve this question
    
Who is "you"? Presumably one of the commenters on your previous question. But please revise to make this slightly more stand-alone? –  Theo Johnson-Freyd Apr 7 '10 at 4:03
1  
I copied over the answer from the previous question; hopefully it's clearer now. –  Mike Stay Apr 7 '10 at 4:42

2 Answers 2

up vote 5 down vote accepted

I agree with Reid's answer, but I want to add a bit more.

Putting Reid's calculation into a more general setting, if $A$ is any category then $$ \int^{a \in A} \mathrm{hom}_A (a, a) = (\mathrm{endomorphisms\ in\ } A)/\sim $$ where $\sim$ is the (rather nontrivial) equivalence relation generated by $gh \sim hg$ whenever $g$ and $h$ are arrows for which these composites are defined.

You can see confirmation there that your instinct about traces was right. If we wanted to define a 'trace map' on the endomorphisms in $A$, it should presumably satisfy $\mathrm{tr}(gh) = \mathrm{tr}(hg)$, i.e. it should factor through $\int^a \mathrm{hom}(a, a)$.

In fact, Simon Willerton has done work on 2-traces in which exactly this coend appears. See for instance these slides, especially the last one.

You can see in that slide something about the dual formula, the end $$ \int_{a \in A} \mathrm{hom}_A(a, a). $$ By the "fundamental fact" I mentioned before, this is the set $$ {}[A, A](\mathrm{id}, \mathrm{id}) = \mathrm{Nat}(\mathrm{id}, \mathrm{id}) $$ of natural transformations from the identity functor on $A$ to itself. Evidently this is a monoid, and it's known as the centre of $A$. For example, when $G$ is a group construed as a one-object category, it's the centre in the sense of group theory. So your set might, I suppose, be called the co-centre of $A$.

share|improve this answer
1  
(endomorphisms in A)/~. Ie. closed loops of arrows without regard to start/endpoint. So this is telling us something about the topology of A considered as a directed graph. Is that right? –  Dan Piponi Apr 7 '10 at 18:07
    
Yes, I think it must be. At the moment I can't see quite what it's telling us about the topology of A, but it must be telling us something. I hadn't thought of it that way before. –  Tom Leinster Apr 7 '10 at 22:02

The step $\int_{a \in A} \mathrm{Set}(\mathrm{hom}_A(a, a), S) = \mathrm{Nat}(\mathrm{hom}_A(-,-),S)$ does not really make sense, because $a \mapsto \mathrm{hom}_A(a,a)$ is not a functor. And $\int^{a \in A} \mathrm{hom}_A(a,a)$ is not equal to $\mathrm{Ob}(A)$ in general. For instance, let $G$ be a group and let $A = BG$ be the groupoid with a single object with automorphism group $G$. Then $\int^{a \in A} \mathrm{hom}_A(a,a)$ can be identified with the abelianization set of conjugacy classes of $G$. In general, $\int^{a \in A} \mathrm{hom}_A(a,a)$ is the "Hochschild homology" of the category $A$.


Edit: I was originally thinking of the case of $G$ abelian, and generalized wrongly to the nonabelian case. As atonement, let me write out the computation directly from the definition of coend:

$$\int^{a \in A} \mathrm{hom}_A(a, a) = \operatorname{colim} \left[ \coprod_{f:a \to b} \mathrm{hom}_A(b,a)\rightrightarrows \coprod_{a\in A} \mathrm{hom}_A(a,a)\right] = \operatorname{colim} [G \times G \rightrightarrows G]$$

where the two maps send $(g, h)$ to $gh$ and $hg$ respectively. So the coend is the quotient of $G$ in the category of sets by the relation $gh = hg$, or $g = hgh^{-1}$, i.e., it is the set of conjugacy classes of elements of $G$.

share|improve this answer
    
Yeah, I clearly got that wrong. At best I've got a dinatural transformation from a constant functor to hom that picks out the identity. –  Mike Stay Apr 7 '10 at 5:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.