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Stream: learning: questions

Topic: Cotangent bundle vs comonad ?

Peva Blanchard (Aug 22 2024 at 22:12):

I've been re-reading stuff in differential geometry, and I noticed an analogy. Let $M$ be a smooth manifold. One can build the cotangent bundle $T^*M$ with the projection

$$\pi_M : T^*M \to M$$

Interestingly, there is also a tautological 1-form on $T^*M$

$$\theta_M : T^*M \to T^*T^*M$$

It really looks like $T^*$ is a comonad, with the counit $\pi$ and the comultiplication $\theta$.

I'm not exactly sure which ambient category should be used. This MO question seems to suggest that $T^*$ is an endofunctor on the category of smooth manifolds with étale maps.

My question is: is $T^*$ a comonad in some appropriate category?

Tim Hosgood (Aug 22 2024 at 23:11):

one reason you would struggle to get a comonad (or even a monad) from the cotangent functor is that it is contravariant — I don't know if there actually even is a good definition of (co)monads for a contravariant functor

Tim Hosgood (Aug 22 2024 at 23:11):

however, the answer to your question is almost yes, if you talk about the tangent bundle instead: see e.g. https://arxiv.org/abs/1401.0940

Tim Hosgood (Aug 22 2024 at 23:12):

actually that paper also mentions the issue with the cotangent functor (e.g. the bottom of page 12)

fosco (Aug 23 2024 at 08:18):

Tim Hosgood said:

one reason you would struggle to get a comonad (or even a monad) from the cotangent functor is that it is contravariant — I don't know if there actually even is a good definition of (co)monads for a contravariant functor

Yes, contravariant (co)monads exist. The original definition was due to Guitart:

• Les monades involutives en théorie élémentaire des ensembles, CRAS, Paris, t. 277, 1973, 935-937
• Construction de monades involutives, CRAS Paris, t. 279, 1974, 491-493

These references are somewhat difficult to access. To a large extent, the definition is used and expanded in
http://www.numdam.org/item/CTGDC_1982__23_2_115_0.pdf
https://eudml.org/doc/91146

I only know about these because I was about to re-discover them, just to be gaslighted into believing that the notion doesn't make sense ("a monad... but it's contravariant? Foolish, preposterous!"). Good that I didn't give up.

A "contramonad" is, instead, a very natural notion, if anything Guitart makes it a pain in the ass to parse (there are no diagrams in his paper...)

fosco (Aug 23 2024 at 08:25):

I have been on and off the task of making Guitart's paper more readable (most of his results can be proved more swiftly in 2024, and a large part of his theorems are fragments of a "formal theory of contravariant monads"). At some point in the future a short survey on contravariant monads will appear. I still have to decide in what form...

For Guitart the motivating example is the contravariant powerset functor, or taking everything up a notch (=dimension) the presheaf construction on Cat. But there are other examples.

(Fun fact, related to this story because it appears in one of the papers of this family: Guitart was the first to notice and put on paper that profunctors are just(TM) the Kleisli bicategory of the presheaf construction.)

Tim Hosgood (Aug 23 2024 at 10:46):

oh that's exciting!

fosco (Aug 23 2024 at 10:50):

Tim Hosgood said:

oh that's exciting!

I agree!

Tim Hosgood (Aug 23 2024 at 15:53):

so is the answer to the original question actually "yes"?

John Baez (Aug 23 2024 at 16:18):

Peva mentioned etale maps. I usually consider $T^\ast$ to be a contravariant endofunctor on the category of smooth manifolds and all smooth maps. I don't know if $T^\ast$ is a contramonad on this category, just because I don't know what a contramonad is. Is the definition on the nLab somewhere?

Tim Hosgood (Aug 23 2024 at 16:40):

the MO question linked says that you can't consider it as a covariant endofunctor when you take all maps, because you don't get a morphism $T^*M\to T^*N$ unless $M\to N$ is étale

Tim Hosgood (Aug 23 2024 at 16:42):

but I think there's also the subtle point that it's not even a contravariant endofunctor for arbitrary maps, right? my guess (though it's usually bad to guess what somebody is thinking...) is that you're thinking that a morphism $f\colon M\to N$ induces a pullback map $f^*\colon T^*N\to T^*M$, but what it actually induces is a pullback map on sections $f^*\colon \Gamma(N,T^*N)\to\Gamma(M,T^*M)$, and this isn't the same as inducing a map on the actual underlying spaces of the bundles themselves! in other words, you get a map $f^*(T^*N)\to T^*M$, not $T^*N\to T^*M$

Tim Hosgood (Aug 23 2024 at 16:43):

the answer to the MO question says that étale maps are exactly the class of maps for which you do actually get a pushforward (because they care about the covariant endofunctor) on points, not sections

Tim Hosgood (Aug 23 2024 at 16:43):

I haven't read the reference, so I have no idea if this is also true for the pullback story (in order to get a contravariant endofunctor, to which you could then apply what fosco was talking about)

Tim Hosgood (Aug 23 2024 at 16:45):

essentially, I derailed the whole conversation at the very beginning by trying to say that the cotangent bundle functor should be contravariant, because the whole point of the original question was that actually for the class of étale maps it defines a perfectly good covariant endofunctor!

Cole Comfort (Aug 23 2024 at 18:03):

John Baez said:

Peva mentioned etale maps. I usually consider $T^\ast$ to be a contravariant endofunctor on the category of smooth manifolds and all smooth maps. I don't know if $T^\ast$ is a contramonad on this category, just because I don't know what a contramonad is. Is the definition on the nLab somewhere?

Surely a contramonad is just a relative monad between $\mathcal C$ and ${\mathcal C}^{\mathsf op}$ .

Nathanael Arkor (Aug 23 2024 at 18:37):

A relative monad needs to be relative to a functor/distributor, but there isn't one in general betweeen $\mathcal C$ and $\mathcal C^{\text{op}}$.

fosco (Aug 23 2024 at 19:37):

Cole Comfort said:

John Baez said:

Peva mentioned etale maps. I usually consider $T^\ast$ to be a contravariant endofunctor on the category of smooth manifolds and all smooth maps. I don't know if $T^\ast$ is a contramonad on this category, just because I don't know what a contramonad is. Is the definition on the nLab somewhere?

Surely a contramonad is just a relative monad between $\mathcal C$ and ${\mathcal C}^{\mathsf op}$ .

It is not, but there is a nice description: a contramonad is precisely a monad the Kleisli category of which is equipped with a duality $Kl(T)\cong Kl(T)^\text{op}$

fosco (Aug 23 2024 at 19:40):

(precisely stated: given a monad T such that Kl(T) is anti-equivalent to itself, $C \to Kl(T)\cong Kl(T)^\text{op} \to C^\text{op}$ is a contramonad; conversely, given a contramonad, there is a certain, non-immediate way to cook a monad in the usual sense, plus an equivalence...

John Baez (Aug 23 2024 at 22:15):

Tim Hosgood said:

but I think there's also the subtle point that it's not even a contravariant endofunctor for arbitrary maps, right?

Aargh, you're right, $T^\ast$ is neither a covariant nor contravariant functor from the category of smooth manifolds to itself.

my guess (though it's usually bad to guess what somebody is thinking...) is that you're thinking that a morphism $f\colon M\to N$ induces a pullback map $f^*\colon T^*N\to T^*M$, but what it actually induces is a pullback map on sections $f^*\colon \Gamma(N,T^*N)\to\Gamma(M,T^*M)$, and this isn't the same as inducing a map on the actual underlying spaces of the bundles themselves!

You are flattering me to say that's what I was thinking; unfortunately I was thinking some nonsense. If I were thinking straight, I might have said that. And I might have said this: $T^*$ gives a pullback map on fibers, i.e. given $f \colon M \to N$ with $f(x) = y$ we get $f^* \colon T^*_y N \to T^*_x M$, which is simply the linear adjoint of the map between tangent spaces $f_* \colon T_x M \to T_y N$.

Peva Blanchard (Aug 24 2024 at 15:22):

Thank you all for the answers!

Peva Blanchard (Aug 24 2024 at 15:45):

So I understand that the statement "$T^*$ is a comonad as a ..." is not an known obvious fact; and possibly not a fact at all. It was a sort of on-the-fly question for me, so I'm not sure I will dig more into it.

I notice, however that it relates to the interplay between étale space vs sheaf of sections, which, quite fortunately, we have been discussing on @David Egolf's thread on topos.

John Baez (Aug 24 2024 at 17:52):

Since I like differential geometry and physics, I'm supposed to know all sorts of stuff about manifolds like $T(TX)$ and $T(T^\ast X)$ and $T^\ast (T^\ast X)$ and maps between these. I know that $T$ is a monad on the category of smooth manifolds. Thus, I should know if $T^\ast$ is a comonad or contramonad in some sense or other. But I don't.

If $f: X \to Y$ is an etale map between smooth manifolds, i.e. a local diffeomorphism, then every point $x \in X$ has a neighborhood such that $f|_U$ is a diffeomorphism, so $f|_U$ induces an isomorphism of tangent bundles and thus of cotangent bundles. So the category of smooth manifolds and etale maps may be a good category in which to study $T^\ast$. But that's as far as I've gotten in that direction.

fosco (Aug 24 2024 at 18:32):

Does it help if I write down the definition for a contramonad? It would be nice to have another example

Tim Hosgood (Aug 24 2024 at 19:42):

that would be cool! since we're deciding to work with étale maps, which give an iso of both tangent and cotangent bundles, it would be interesting to see if the cotangent bundle has any "preference" for being a monad or a contramonad

fosco (Aug 24 2024 at 20:28):

image.png

fosco (Aug 24 2024 at 20:28):

image.png

fosco (Aug 24 2024 at 20:28):

image.png

fosco (Aug 24 2024 at 20:29):

image.png

fosco (Aug 24 2024 at 20:29):

this is most of Guitart's paper

John Baez (Aug 24 2024 at 23:09):

Thanks! The concept of involutive monad, and the theorem you mention relating them to contramonads, should make it quicker to decide whether the cotangent bundle construction gives a contramonad on some category or not.

Tim Hosgood (Aug 24 2024 at 23:53):

fosco said:

image.png

I'm confused about the definition here: is the equivalence between the category and its op, or between the Kleisli category and its op?

edit: oh, are you just writing $\mathcal{C}$ to mean both the category and the Kleisli category?

John Baez (Aug 25 2024 at 07:41):

I think he should write $\mathcal{C}_T$ instead of $\mathcal{C}$ the last two times.

fosco (Aug 25 2024 at 08:41):

yeah, sorry, typo!