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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeMay 12th 2015
• (edited May 13th 2015)

It seems to me that the category of bundles over some base space, with morphisms the differential operators on spaces of sections of these bundles, is equivalently the co-Kleisli category of the Jet bundle comonad. Is this known? (It seems to be a different statement than that discussed by Blute-Cockett-Seely, as far as I see.)

More in detail, consider differential cohesion with infinitesimal shape modality $\Im$. For a given base space $X$, write

$i \colon X \longrightarrow \Im X$

for the $X$-component of the unit of the $\Im$-monad. Then the operation of forming jet bundles is the comonad given by the base change adjoint triple $(i_! \dashv i^\ast \dashv i_\ast)$:

$Jet \coloneqq i^\ast i_\ast \;\colon\; \mathbf{H}_{/X} \to \mathbf{H}_{/X} \,.$

Now, it is a standard fact that given two bundles $E_1, E_2$ over $X$, then differential operators

$D \colon \Gamma(E_1) \to \Gamma(E_2)$

are equivalently bundle maps

$\tilde D \;\colon\; Jet(E_1) \longrightarrow E_2 \,,$

where the equivalence is given by

$D(\phi) = \tilde D \circ j^\infty(\phi)$

with $j^\infty \phi \in \Gamma(Jet(E_1))$ the jet bundle prolongation of $\phi$.

Combining this with the information that $Jet$ is a comonad, we have the impulse to say that $Jet(E_1) \to E_2$ is to be regarded as a morphism in its co-Kleisli category, and hence that under the above equivalence the composition of differential operators $D_2 \circ D_1$ corresponds to the composite

$Jet(E_1) \to Jet (Jet(E_1)) \stackrel{Jet(\tilde D_1)}{\longrightarrow} Jet(E_2) \stackrel{\tilde D_2}{\longrightarrow} E_3 \,,$

where the first map is the co-product of the $Jet$ co-monad.

And it seems to me that this is actually true.

• CommentRowNumber2.
• CommentAuthorMichael_Bachtold
• CommentTimeMay 12th 2015
• (edited May 12th 2015)

In Verbovetsky Krasilsh’chik you find the co-product $Jet(E_1) \to Jet (Jet(E_2))$, on page 13 where they call it the co-glueing. They don’t really use the terminology of monads and the Kleisil category, but I think they do use the terminology of adjoints and representative objects. Or at least I recall that from other writings (there is a book by Lychagin, Krasilsh’chik, Vinogradov and some lecture notes which I’ll reference here as soon as I can)

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeMay 12th 2015

Thanks! And on p. 17 they check that the co-gluing is indeed a co-product.

1. Right. Here are the lecture notes I recalled. Around page 25. There they call it the co-composition and check the co-associativity.

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeMay 12th 2015

Thanks again!

Now, do they ever say explicitly that the “Jet-associated” (as they call it) map corresponding to the composite of two differential operators is given by (in whatever notation) the formula I gave at the bottom of #1 above?

I think it follows from what they say, but for citation purposes I wonder if they (or anyone else) says it explicitly.

2. I don’t see that they state it explicitly.

But it also seems to me, that it follows from what they say, by writing out a commutative diagram (which I don’t know how to display here).

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeMay 12th 2015
• (edited May 12th 2015)

You mean this commuting diagram here, I suppose:

$\array{ P_1 &\stackrel{id}{\longrightarrow}& P_1 &\stackrel{\Delta_1}{\longrightarrow}& P_2 &\stackrel{\Delta_2}{\longrightarrow}& P_3 \\ \downarrow^{\mathrlap{id}} && \downarrow^{\mathrlap{j_\infty}} && \downarrow^{\mathrlap{j_\infty}} && \downarrow^{\mathrlap{id}} \\ P_1 &\stackrel{j_\infty}{\longrightarrow}& J^\infty(P_1) &\stackrel{J^\infty(\Delta_1)}{\longrightarrow}& J^\infty(P_2) \\ \downarrow^{\mathrlap{j_\infty}} && \downarrow^{\mathrlap{j_\infty}} && \downarrow^{\mathrlap{id}} && \downarrow^{\mathrlap{id}} \\ J^\infty (P_1) &\stackrel{c^{\infty,\infty}}{\longrightarrow}& J^\infty(J^\infty(P_1)) &\underoverset{= J^\infty (\psi^{\Delta_1})}{\psi^{\Delta_1}_\infty}{\longrightarrow}& J^\infty(P_2) &\stackrel{\psi^{\Delta_2}}{\longrightarrow}& P_3 }$
• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeMay 12th 2015
• (edited May 12th 2015)

Here the key point, for the purpose of #1, is the middle rectangle (in #7). Regarding the full middle rectangle identifies the bottom middle morphism as $\psi_\infty^{\Delta_1}$, while factoring the middle rectangle, as shown, through the $J^\infty$-image of the square defining $\psi^{\Delta_1}$ identifies the bottom middle morphism as $J^\infty(\psi^{\Delta_1})$.

• CommentRowNumber9.
• CommentAuthorUrs
• CommentTimeMay 12th 2015

I have turned that into a statement in the entry on differential operator, see this defnition/proposition with consecutive proof.

Let me know what you think.

Strictly speaking, what is still missing is a precise proof that the co-product $c^{\infty,\infty}$ used there is really the coproduct of the Jet comonad as defined more abstractly.

• CommentRowNumber10.
• CommentAuthorUrs
• CommentTimeMay 12th 2015

the statement is explicit in Marvan 93, section 1.1

• CommentRowNumber11.
• CommentAuthorUrs
• CommentTimeMay 12th 2015
• (edited May 12th 2015)

in Marvan 89 at least the statement about the Jet operation itself being a comonad is made in passing. It seems that Marvan says that he introduced the terminology “jet comonad”.

• CommentRowNumber12.
• CommentAuthorDavid_Corfield
• CommentTimeMay 13th 2015
• (edited May 13th 2015)

So we now have two similar constructions

$Jet \coloneqq i^\ast i_\ast \;\colon\; \mathbf{H}_{/X} \to \mathbf{H}_{/X} \,.$

and

$\Box_W: \mathbf{H}_{/W} \to \mathbf{H}_{/W} \,.$

Both arise as dependent product then context extension for the units of modalities, $\ast$ and $\Im$, in the Aufhebung table.

Would it be worth looking at this for the other modalities, so $\sharp$, ʃ, $R$ and $\rightrightarrows$? Then also possibilty, reader, writer, randomness.

Then what about these for the counits of comodalities, e.g., along $\Re X \to X$?

3. Re 7&8: Yes, thanks! I woke up noticing that in your first post you wrote $Jet(Jet(E_2))$ instead of $Jet(Jet(E_1))$, thats a typo right?

Thanks for the references of Marvan, I had not seen them.

Strictly speaking, what is still missing is a precise proof that the co-product c∞,∞ used there is really the coproduct of the Jet comonad as defined more abstractly.

I see your point. I first need to catch up with your writings on differential cohesion etc. to say more. I keep wanting to do that. Maybe you could help me gain some intuition starting with

$i \colon X \longrightarrow \Im X$

For concreteness, suppose $X$ is the affine line $R$ from SDG, can I think of $\Im R$ as a quotient of $R$, where infinitesimally close points have been identified? Although that does not make sense with what is written in jet bundle (“inclusion of constant paths into all infinitesimal paths”).

• CommentRowNumber14.
• CommentAuthorMichael_Bachtold
• CommentTimeMay 13th 2015
• (edited May 13th 2015)

Oh, I see that my interpretation is basically written in de Rham space, under Properties, As a quotient. But then the comment “inclusion of constant paths into all infinitesimal paths” below $i \colon X \longrightarrow \Im X$ in jet bundle is still somewhat confusing for me. Shouldn’t it read: “collapsing infinitesimal paths to constant paths”?

• CommentRowNumber15.
• CommentAuthorDavid_Corfield
• CommentTimeMay 13th 2015
• (edited May 13th 2015)

Michael I was going to point to the page infinitesimal path ∞-groupoid, but there needs to be some fixing of typos first.

It says

$(Red \dashv \Pi_{inf}) \colon \mathbf{H}_{th} \stackrel{\overset{i_*}{\leftarrow}}{\underset{i^*}{\to}} \mathbf{H} \stackrel{\overset{i_!}{\leftarrow}}{\underset{i_*}{\to}} \mathbf{H} \,.$

and should be

$(Red \dashv \Pi_{inf}) \colon \mathbf{H}_{th} \stackrel{\overset{i_!}{\leftarrow}}{\underset{i^*}{\to}} \mathbf{H} \stackrel{\overset{i^*}{\leftarrow}}{\underset{i_*}{\to}} \mathbf{H}_{th} \, ,$

I think.

Note that $\Pi_{inf}$ is now written $\Im$, and $Red$ is written $\Re$.

• CommentRowNumber16.
• CommentAuthorDavid_Corfield
• CommentTimeMay 13th 2015

Re #14,

Shouldn’t it read: “collapsing infinitesimal paths to constant paths”

I think Urs would point out that identifying infinitesimally close points does not mean collapsing them as in a strict quotient. As with the shape modality, identifying path connected points in a space, one retains the information of different identifications, and identifications between identifications, etc.

• CommentRowNumber17.
• CommentAuthorUrs
• CommentTimeMay 13th 2015
• (edited May 13th 2015)

Michael, David,

thanks for catching typos! Have fixed them now.

Regarding the path inclusion: think of $\Pi_{inf}X \coloneqq \Im(X)$ as the result of forming a new groupoid from the groupoid $X$ by adding in further isomorphisms between all objects (points of $X$) that are infinitesimal neighbours. These isomorphisms are hence infinitesimal paths. The canonical map $X \to \Pi_{inf}(X)$ sends all the existing objects and morphsims of $X$ “to themselves” in $\Pi_{inf}X$ and doesn’t hit any of these new non-trivial isomorphisms between infinitesimal neighbours.

This is a standard story from algebraic geometry. I think this is where the word “crystal” in “crystalline cohomology” comes from, in that one visualizes $\Pi_{inf}X$ as looking like a crystalline version of $X$, where all these new infinitesimal paths are little edges of the crystal.

I don’t know for sure if this is the historical visualization behind Grothendieck’s “crystalline”, but certainly texts on crystalline cohomology draw these kinds of pictures.

• CommentRowNumber18.
• CommentAuthorUrs
• CommentTimeMay 13th 2015

David, re #12,

that’s a neat analogy that you are observing there. I’ll think about this.

• CommentRowNumber19.
• CommentAuthorUrs
• CommentTimeMay 13th 2015
• (edited May 13th 2015)

Michael, re #13,

I believe I see how to prove that the abstractly defined Jet comonad has the coproduct operation that it is supposed to have.

There is now a remark 5.3.88 in the dcct pdf which indicates how, dually, the product of the left adjoint monad works. I think from this it is pretty clear how it goes, but of course this remark is not yet a proof of anything.

• CommentRowNumber20.
• CommentAuthorDavid_Corfield
• CommentTimeMay 13th 2015

Re #18, so it’s a process of taking something bundle like and returning another bundle where the fibre at a point is now sections through fibres of ’adjoining’ points. So for necessity, a necessary man at this world picks out a man under variation across worlds, such as ’the tallest man’. So they’re like germs. With shape, you’d have at a point all sections for the component of the point?

For $Rh$ this would send a fibre of entities to a fibre of entities varying over points which only differ in the super directions. Could they be called super-jets?

I see talking about an ’infinite jet superbundle’, p. 17 of On the (non)removability of spectral parameters in Z2-graded zero-curvature representations and its applications. But that’s just over a real manifold. I guess we’d want it over a superspace.

• CommentRowNumber21.
• CommentAuthorUrs
• CommentTimeMay 13th 2015
• (edited May 13th 2015)

Regarding your first paragraph: right, so by adjunction, the sections of $i^\ast i_\ast E$ over some base $X$ are bundles maps $i^\ast i_! X \to E$, hence are certain “generalized sections” of the original $E$.

When $i \colon X \to \int X$ is the shape unit, then $i^\ast i_! X = X \times_{\int X} X$ is a kind of bundle over $X$ whose fibers are the homotopy types of smooth collections of based path spaces of paths emanating at that point and smoothly parameterized by their endpoint. For whatever that’s worth. That tells one in principle what the sections of any $i^\ast i_\ast E$ are in this case, though right now I don’t recognize it as anything. But possibly this is some important concept and we should think about it.

Regarding the article that you point to: thanks for the reference. Yes, those jet bundles of superbundles which these authors dicuss should be precisely what the Jet comonad in super formal smooth homotopy types produces when applied to these bundles.

• CommentRowNumber22.
• CommentAuthorUrs
• CommentTimeMay 13th 2015

I have contacted Michal Marvan, and he has sent me a scan of his original article (I am in the process of checking if I may share it here). In there he in fact proves that the Eilenberg-Moore category of coalgebras over the jet comonad is equivalently the category of differential equations with variables in the given base manifold.

• CommentRowNumber23.
• CommentAuthorUrs
• CommentTimeMay 13th 2015

for the moment the link is here: Marvan 86, a nice article.

(my understanding at the moment is that it is okay to link to the pdf)

• CommentRowNumber24.
• CommentAuthorDavid_Corfield
• CommentTimeMay 13th 2015

Marvan’s name appears 18 times in that paper I mentioned in #20.

• CommentRowNumber25.
• CommentAuthorUrs
• CommentTimeMay 13th 2015

chasing references, I found Kock 80 which has not the comonad structure on $Jet$, but does have the statement that $Jet(-)$ is right adjoint to forming infinitesimal disk bundles (and I have created this entry now).

• CommentRowNumber26.
• CommentAuthorMichael_Bachtold
• CommentTimeMay 13th 2015
• (edited May 13th 2015)

Re #15-18: Thanks David and Urs! That helped. Although I still have some basic “obstructions” in my understanding of cohesive types, which I might ask in a separate thread.

Concerning #25: it seems that infinitesimal disk bundle are basically the same as what Kock calls the “bundle of k-monads” in Synthetic Geometry of Manifolds p. 39. (I don’t know why he calls them monads though.)

• CommentRowNumber27.
• CommentAuthorUrs
• CommentTimeMay 13th 2015

Yes, I have just added this to infinitesimal disk bundle.

• CommentRowNumber28.
• CommentAuthorUrs
• CommentTimeMay 13th 2015
• (edited May 13th 2015)

I don’t know why he calls them monads though.

I believe this is to refer to something like Leibniz’s monads, but not, in any case, to monads in the sense of category theory. There is a lot in the Science of Logic about the issue of Leibniz’s monads and infinitesimal units, but I’ll spare us the details.