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Stream: theory: category theory

Topic: enchilada categories

John Baez (Mar 24 2020 at 20:37):

Wow! Lawvere had his Hegelian taco, and we all know that a monad is like a burrito, but now there are enchilada categories:

John Baez (Mar 24 2020 at 20:38):

Exact sequences in the enchilada category

M. Eryüzlü, S. Kaliszewski, and John Quigg

We define exact sequences in the enchilada category of C*-algebras and
correspondences, and prove that the reduced-crossed-product functor is not
exact for the enchilada categories. Our motivation was to determine
whether we can have a better understanding of the Baum-Connes conjecture
by using enchilada categories. Along the way we prove numerous results
showing that the enchilada category is rather strange.

Theory and Applications of Categories, Vol. 35, 2020, No. 12, pp 350-370.

http://www.tac.mta.ca/tac/volumes/35/12/35-12.pdf

Reed Mullanix (Mar 24 2020 at 20:43):

I always thought that burritos more comonadic than monadic. Perhaps the tortilla endofunctor has some deeper underlying structure

David Michael Roberts (Mar 25 2020 at 06:39):

Many other people have studied this under a less hip, but more informative name. Also, they actually look at the bicategory, not the homotopy category, which is better. Also, I believe the 1-arrows they use (apart from committing the error of truncating) are not general enough. I can come up with easy maps between spaces such that the induced maps between C*-algebras aren't in their category.

David Michael Roberts (Mar 25 2020 at 06:39):

Other than that, I do want to understand what they do, because I have a long-incubating project on a related (bi)category

Nicolas Blanco (Mar 25 2020 at 10:47):

David Michael Roberts said:

Many other people have studied this under a less hip, but more informative name. Also, they actually look at the bicategory, not the homotopy category, which is better. Also, I believe the 1-arrows they use (apart from committing the error of truncating) are not general enough. I can come up with easy maps between spaces such that the induced maps between C*-algebras aren't in their category.

A naive question: You said that looking at the bicategory is better. Would it make sense to look at the double category instead (with *-homomorphisms as vertical arrows and correspondences as horizontal arrows)?

John Baez (Mar 25 2020 at 20:17):

Many other people have studied this under a less hip, but more informative name. Also, they actually look at the bicategory, not the homotopy category, which is better.

Can you sketch what bicategory you're talking about? (I have no idea what this enchilada category is; I just found the name amusing.)

(=_=) (Mar 26 2020 at 01:27):

@John Baez said:

Many other people have studied this under a less hip, but more informative name. Also, they actually look at the bicategory, not the homotopy category, which is better.

Can you sketch what bicategory you're talking about? (I have no idea what this enchilada category is; I just found the name amusing.)

I believe it would have C *-algebras as objects, C *-correspondences as morphisms and Morita-Rieffel equivalences intertwiners as 2-morphisms. The more informative name would then be a correspondence bicategory. You might be interested in this because C *-correspondences are spans. Ralf Meyer and his school have written a lot about this.

(=_=) (Mar 26 2020 at 01:32):

Paging @Min Ro: you might find this interesting.

(=_=) (Mar 26 2020 at 01:34):

Nicolas Blanco said:

A naive question: You said that looking at the bicategory is better. Would it make sense to look at the double category instead (with *-homomorphisms as vertical arrows and correspondences as horizontal arrows)?

Not David, but I get the feeling people know how to work with bicategories better than they do with double categories. By "better", I think David meant that it's better than merely looking at the homotopy category.

Min Ro (Mar 26 2020 at 01:36):

Rongmin Lu said:

Paging Min Ro: you might find this interesting.

Thanks for the invitation.

David Michael Roberts (Mar 26 2020 at 09:13):

@Nicolas Blanco that's even better: it would then be some kind of pseudo-double category. I have my own views on using bimodules rather than a certain class of cospans, but the result should be biequivalent

David Michael Roberts (Mar 26 2020 at 09:15):

@Rongmin Lu @John Baez $C^*$-algebras as objects, $C^*$-correspondences (which are a certain type of bimodule with an inner product valued in the codomain algebra,+axioms), and intertwiners. Morita–Rieffel equivalences are the quasi-invertible 1-arrows.

Nicolas Blanco (Mar 26 2020 at 09:49):

David Michael Roberts said:

Nicolas Blanco that's even better: it would then be some kind of pseudo-double category. I have my own views on using bimodules rather than a certain class of cospans, but the result should be biequivalent

Yes, I had in mind this rough analogy with the (pseudo-?)double category of rings, ring homomorphisms and bimodules. Or more generally with the pseudo-double category of enriched categories, enriched functors and enriched profunctors.
I am not sure if this analogy can be made more precise though.

(=_=) (Mar 26 2020 at 12:37):

Nicolas Blanco said:

Yes, I had in mind this rough analogy with the (pseudo-?)double category of rings, ring homomorphisms and bimodules.

This is precisely the analogy pursued by Meyer, although he's only interested in it as a bicategory AFAIK.

(=_=) (Mar 26 2020 at 12:43):

David Michael Roberts said:

Rongmin Lu John Baez $C^*$-algebras as objects, $C^*$-correspondences (which are a certain type of bimodule with an inner product valued in the codomain algebra,+axioms), and intertwiners. Morita–Rieffel equivalences are the quasi-invertible 1-arrows.

Thanks for the correction! :sweat_smile:

Nicolas Blanco (Mar 26 2020 at 13:55):

In the ring case, the bicategory (and it should work for the double category also) can be thought of as a subbicategory of the bicategory of Ab-enriched categories and Ab-enriched profunctors - where you think of a ring as a one object Ab-category, so a monoid internal to Ab. Do you know if there is something similar for $C^\ast$-algebras and their bimodules?

John Baez (Mar 27 2020 at 19:30):

Thanks!

David Michael Roberts (Mar 30 2020 at 01:21):

@Nicolas Blanco there are indeed such things as $C^*$-categories, but in general one needs them to be non-unital (!!!)

Nicolas Blanco (Mar 30 2020 at 09:11):

David Michael Roberts said:

Nicolas Blanco there are indeed such things as $C^*$-categories, but in general one needs them to be non-unital (!!!)

Thanks, I have heard about the existence of these things before. From the nLab page it seems that any $C^*$-algebra is a one-object $C^*$-category. In the ring case, this correspondence can be extended to a correspondence between modules and presheaves and bimodules and profunctors. Do you know if there is an analogous thing for $C^*$-algebra/$C^*$-category. Is Hilbert $C^*$-module the notion that is considered or something else?
Also can you provide some sort of intuition why the $C^*$-algebra needs to be non-unital?

David Michael Roberts (Mar 30 2020 at 09:53):

Take a non-compact space: its $C^*$-algebra is non-unital. The compact operators on a Hilbert space are also a non-unital algebra, and both of these are fundamental examples.

John Baez (Mar 30 2020 at 14:37):

Hmm, I hoped you were going to give a stronger argument for non-unital C-algebras. For any non-unital C-algebra you can just adjoint a unit, which in the topological case corresponds to taking the one-point compactification. So you have to weigh the damage this does against the convenience of requiring a unit.

But it's true, precisely for this reason, that tons of results about unital C*-algebras generalize to nonunital ones.

My thesis advisor invented C*-algebras! :+1:

Min Ro (Mar 30 2020 at 14:59):

Wasn't it your advisor who said "Back in the good old days, every Cstar-algebra was unital. If you found one that wasn't, you immediately unitized it. Nowadays, every Cstar-algebra is non-unital, if you find one that is, then you immediately tensor it with K."?