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Stream: theory: category theory

Topic: Isbell duality, functor tensor products, and traces

Emily (Apr 20 2024 at 18:19):

The Yoneda lemma says that the assignments

$$\begin{align*} \mathcal{F} &\mapsto [X\mapsto\mathrm{Nat}(h_{X},\mathcal{F})],\\ F &\mapsto [X\mapsto\mathrm{Nat}(h^{X},F)] \end{align*}$$

are naturally isomorphic to the identity functors on $\mathsf{PSh}(\mathcal{C})$ and $\mathsf{CoPSh}(\mathcal{C})$. Considering then the assignments

$$\begin{align*} \mathcal{F} &\mapsto [X\mapsto\mathrm{Nat}(\mathcal{F},h_{X})],\\ F &\mapsto [X\mapsto\mathrm{Nat}(F,h^{X})] \end{align*}$$

leads to functors $\mathsf{O}$ and $\mathsf{Spec}$ forming an adjunction

$$(\mathsf{O}\dashv\mathsf{Spec})\colon\mathsf{PSh}(\mathcal{C})\rightleftarrows\mathsf{CoPSh}(\mathcal{C})^{\mathsf{op}},$$

called the Isbell duality adjunction.

As it turns out, $\mathsf{O}$ can be viewed as right Kan lift in $\mathsf{Prof}$,

rift.png

and the natural transformation in this diagram has components

$$\theta_{A,B} \colon \underbrace{\mathcal{F}(A)\times[\mathsf{O}(\mathcal{F})](B)]}_{\overset{\mathrm{def}}{=}\mathcal{F}(A)\times\mathrm{Nat}(\mathcal{F},h_{B})} \to \mathrm{Hom}_{\mathcal{C}}(A,B)$$

given by

$$\theta_{A,B}(\phi,\alpha) \mathbin{\overset{\mathrm{def}}{=}} \alpha_A(\phi).$$

Now, if we take the coend of both sides, i.e. the coends of the functors $(A,B)\mapsto\mathcal{F}(A)\times[\mathsf{O}(\mathcal{F})](B)$ and $(A,B)\mapsto\mathrm{Hom}_{\mathcal{C}}(A,B)$, we get a map

$$\theta_{\mathcal{F}} \colon \mathcal{F}\boxtimes\mathsf{O}(\mathcal{F}) \to \mathrm{Tr}(\mathcal{C})$$

from the functor tensor product of $\mathcal{F}$ with $\mathsf{O}(\mathcal{F})$ to the trace of $\mathcal{C}$.

For example:

1. When $\mathcal{C}$ is the one-object category associated to a group $G$, this map takes the form of a map from a certain quotient of $X\times\mathsf{Sets}^{\mathrm{L}}_{G}(X,G)$ to the set of conjugacy classes of $G$, with $\mathcal{F}$ now corresponding to a left $G$-set $(X,\lambda_{X})$.
2. Working on the $\mathsf{Mod}_{R}$-enriched setting and taking $\mathcal{C}$ to be the one-object category associated to a ring $R$, a presheaf $\mathcal{F}$ on $\mathcal{C}$ is now given by a left $R$-module $M$, and this map goes from a certain quotient of $M\otimes_{R}M^*$ to the zeroth Hochschild homology $\mathrm{HH}_{0}(R)$ of $R$, where $M^*=\mathrm{Hom}_{R}(M,R)$ is the dual module of $M$.

This seems like a very natural and neat map; has anyone here ever seen it appear somewhere in the literature, even if only for the special cases above? Is there any use of it?

John Baez (Apr 20 2024 at 18:36):

So are you implying there's a map from $M \otimes_R M^*$ to the zeroth Hochschild homology of $R$? This reminds me a lot of Simon Willerton's work on Hochschild homology as a "2-trace", but I'm forgetting some details of that so I can't instantly tell if it's another viewpoint on what you're talking about.

Emily (Apr 20 2024 at 18:51):

John Baez said:

So are you implying there's a map from $M \otimes_R M^*$ to the zeroth Hochschild homology of $R$? This reminds me a lot of Simon Willerton's work on Hochschild homology as a "2-trace", but I'm forgetting some details of that so I can't instantly tell if it's another viewpoint on what you're talking about.

It's only a map from a quotient of $M \otimes_R M^*$ to the zeroth Hochschild homology.

I think the quotient is given by quotienting out $M\otimes_RM^*$ by something like the submodule generated by $(rm,\phi)-(m,\phi r)$ with $\phi r$ the map $m\mapsto\phi(m)r$, but I'd have to check it to be sure/work out all details carefully.

dusko (Apr 20 2024 at 20:32):

Emily said:

the natural transformation in this diagram has components

$$\theta_{A,B} \colon \underbrace{\mathcal{F}(A)\times[\mathsf{O}(\mathcal{F})](B)]}_{\overset{\mathrm{def}}{=}\mathcal{F}(A)\times\mathrm{Nat}(\mathcal{F},h_{B})} \to \mathrm{Hom}_{\mathcal{C}}(A,B)$$

given by

$$\theta_{A,B}(\phi,\alpha) \mathbin{\overset{\mathrm{def}}{=}} \alpha_A(\phi).$$

Now, if we take the coend of both sides, i.e. the coends of the functors $(A,B)\mapsto\mathcal{F}(A)\times[\mathsf{O}(\mathcal{F})](B)$ and $(A,B)\mapsto\mathrm{Hom}_{\mathcal{C}}(A,B)$, we get a map

$$\theta_{\mathcal{F}} \colon \mathcal{F}\boxtimes\mathsf{O}(\mathcal{F}) \to \mathrm{Tr}(\mathcal{C})$$

from the functor tensor product of $\mathcal{F}$ with $\mathsf{O}(\mathcal{F})$ to the trace of $\mathcal{C}$.

this natural transformation is the categorical version of george mackey's dual pairings of TVS. the theory of such pairings was spelled out in grothendieck's thesis. the theory of the categorical pairings will presumably first retract that path, and then depart onto the obviously much higher ground where the coends will take us.

i don't think anyone worked all of this out, but lawvere for one was fully aware of it.

old people have a story about everything so here is a quick one. peter freyd asked at some point in the 90s for a common denominator of adjoint operators in hilbert spaces and the categorical adjunctions. i presented both as suitable Chu-objects. then bill lawvere called me and spent 2 hours explaining (to a very thick postdoc) how it was all envisioned in george mackey's thesis, and how grothendieck missed the categorical content of the dual pairs that he worked out...

so lots of people philosophized about this structure and maybe it is finally time for someone to really work it out :)

John Baez (Apr 20 2024 at 22:33):

Emily said:

John Baez said:

So are you implying there's a map from $M \otimes_R M^*$ to the zeroth Hochschild homology of $R$?

It's only a map from a quotient of $M \otimes_R M^*$ to the zeroth Hochschild homology.

But a map from a quotient of something gives a map from that something. So a map from $M \otimes_R M^*$ to the zeroth Hochschild homology gives a map from $M \otimes_R M^*$ to Hochschild homology.

Emily (Apr 20 2024 at 22:43):

John Baez said:

But a map from a quotient of something gives a map from that something. So a map from $M \otimes_R M^*$ to the zeroth Hochschild homology gives a map from $M \otimes_R M^*$ to Hochschild homology.

Yep, I was being overly dense/afraid of saying too much without actually writing and checking things down. As Duško let me know, this map is actually given by $(m,\phi)\mapsto\phi(m)$.