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1. • CommentRowNumber1.
   • CommentAuthorMike Shulman
   • CommentTimeMar 28th 2014
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   Author: Mike Shulman
   Format: MarkdownItexOnce we decide there's no reason to require differential forms to be "linear", I think it's about as easy to define cojet forms as it was to define cogerm forms. Namely, a cojet form on $X$ is just a (perhaps smooth) function $J X \to \mathbb{R}$, where $J X$ is the space of jets in $X$. Of course, we need to define $J X$. I think the following definition is sensible; is it well-known? Let $T$ be the tanget-bundle functor; the projections $T X \to X$ make $T$ into a copointed endofunctor. Now $J X$ is the cofree $T$-coalgebra on $X$. We can construct $J X$ by the usual sort of sequential limit: $$ \cdots \to J_3 X \to J_2 X \to J_1 X \to J_0 X = X $$ where $J_0 X = X$ and $J_1 X = T X$, while $J_{n+1} X$ for $n\ge 1$ is the equalizer of $T (J_n X) \to J_n X \to J_{n-1} X$ and $T(J_n X) \to T(J_{n-1} X) \to J_{n-1} X$. So $J_2 X$, for instance, consists of a tangent vector to $T X$ at $(x,v)$, say $(u,w)$, such that $u=v$. Thus, for instance, $J (\mathbb{R}^m) \cong (\mathbb{R}^m)^\omega$, consisting of a point $x$, a tangent vector at that point (= 1-jet), a 2-jet at that 1-jet, etc. Of course, $J X$ is infinite-dimensional, so we'd have to be working in some category of generalized smooth spaces. For the limit to produce a terminal coalgebra, we need $T$ to preserve the sequential limit, but this shouldn't be a problem; e.g. in SDG, $T$ is a right adjoint $(-)^D$ so it preserves all limits. Note that a $T$-coalgebra in general is just a smooth space $Y$ equipped with a vector field $\xi:Y\to T Y$. The universal property of $J X$ is that given such a $(Y,\xi)$ and any smooth map $f:Y\to X$, there is a unique extension $E(f,\xi) : Y \to J X$. The 1-jet part of $E(f,\xi)$ is the composite $Y \xrightarrow{\xi} T Y \xrightarrow{T f} T X$, and so on. Now I'm saying we can define a **cojet differential form** on $X$ to be a function $\omega:J X \to \mathbb{R}$. The **differential** of such an $\omega$ is the composite $J X \to T(J X) \to \mathbb{R}$, where the first map is the canonical vector field on $J X$ (the terminal $T$-coalgebra structure) and the second is just the ordinary differential of $\omega$ regarded as a function on a smooth space. In coordinates on $X=\mathbb{R}$, this means $\omega$ is a function of countably many variables $x$, $\mathrm{d}x$, $\mathrm{d}^2x$, etc. To take its cojet differential, we take its ordinary differential regarded as a function of countably many variables, and then we set $\mathrm{d}(\mathrm{d}^n x) = \mathrm{d}^{n+1} x$ for all $n$. We can also integrate cojet forms in essentially the way that I suggested in the other thread for cogerm forms. Any closed interval $[a,b]$ comes with a family of canonical vector fields $\xi_h$ which (under the identification $T\mathbb{R} \cong \mathbb{R}$) is constant at $h$, so given a curve $c:[a,b] \to X$ we have an induced $E(c,{\xi_h}):[a,b] \to J X$ and thus $\omega \circ E(c,{\xi_h}) : [a,b] \to \mathbb{R}$. Now a tagged partition $a=x_0 \lt x_1 \lt \dots \lt x_n=b$ with tags $t_i$ and widths $\Delta x_i = x_i - x_{i-1}$ yields a Riemann sum $$ \sum_{i=1}^n \omega(E(c,\xi_{\Delta x_i})(t_i)) $$ and we can take the limit.

   Once we decide there’s no reason to require differential forms to be “linear”, I think it’s about as easy to define cojet forms as it was to define cogerm forms. Namely, a cojet form on is just a (perhaps smooth) function , where is the space of jets in .

   Of course, we need to define . I think the following definition is sensible; is it well-known? Let be the tanget-bundle functor; the projections make into a copointed endofunctor. Now is the cofree -coalgebra on .

   We can construct by the usual sort of sequential limit:

   where and , while for is the equalizer of and . So , for instance, consists of a tangent vector to at , say , such that . Thus, for instance, , consisting of a point , a tangent vector at that point (= 1-jet), a 2-jet at that 1-jet, etc. Of course, is infinite-dimensional, so we’d have to be working in some category of generalized smooth spaces. For the limit to produce a terminal coalgebra, we need to preserve the sequential limit, but this shouldn’t be a problem; e.g. in SDG, is a right adjoint so it preserves all limits.

   Note that a -coalgebra in general is just a smooth space equipped with a vector field . The universal property of is that given such a and any smooth map , there is a unique extension . The 1-jet part of is the composite , and so on.

   Now I’m saying we can define a cojet differential form on to be a function . The differential of such an is the composite , where the first map is the canonical vector field on (the terminal -coalgebra structure) and the second is just the ordinary differential of regarded as a function on a smooth space.

   In coordinates on , this means is a function of countably many variables , , , etc. To take its cojet differential, we take its ordinary differential regarded as a function of countably many variables, and then we set for all .

   We can also integrate cojet forms in essentially the way that I suggested in the other thread for cogerm forms. Any closed interval comes with a family of canonical vector fields which (under the identification ) is constant at , so given a curve we have an induced and thus . Now a tagged partition with tags and widths yields a Riemann sum

   and we can take the limit.

2. • CommentRowNumber2.
   • CommentAuthorFosco
   • CommentTimeMar 28th 2014
   • (edited Mar 28th 2014)
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   Author: Fosco
   Format: MarkdownItexNot an answer, but some times ago I was extremely interested in expanding the description of the jet space $J X$ in a more intrinsic way (trying to understand the nlab page about the [variational bicomplex](http://ncatlab.org/nlab/show/variational+bicomplex)). I had a vague idea on the same spirit (consider the tangent space functor $T : \mathbf{GoodSpaces}\to\mathbf{GoodSpaces}$ as copointed). I would like to go deeper (and maybe learn better the topic to give a true answer!). Any advice?

   Not an answer, but some times ago I was extremely interested in expanding the description of the jet space in a more intrinsic way (trying to understand the nlab page about the variational bicomplex). I had a vague idea on the same spirit (consider the tangent space functor as copointed).

   I would like to go deeper (and maybe learn better the topic to give a true answer!). Any advice?

3. • CommentRowNumber3.
   • CommentAuthorTobyBartels
   • CommentTimeMar 29th 2014
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   Author: TobyBartels
   Format: MarkdownItexThat sequential-limit construction is so obvious that I\'ve always taken it for granted that this is what anybody would mean by the space of jets on $X$. I don\'t know that I ever bothered to check.

   That sequential-limit construction is so obvious that I've always taken it for granted that this is what anybody would mean by the space of jets on . I don't know that I ever bothered to check.

4. • CommentRowNumber4.
   • CommentAuthorMichael_Bachtold
   • CommentTimeMar 29th 2014
   • (edited Mar 29th 2014)
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   Author: Michael_Bachtold
   Format: MarkdownItexCould there be a mistake in the sequential limit construction as stated? I'm assuming that that the first map in $T (J_n X) \to J_n X \to J_{n-1} X$ is just the standard projection from the tangent space, while the first one in $T(J_n X) \to T(J_{n-1} X) \to J_{n-1} X$ is the tangent map of $J_n X \to J_{n-1} X$. But then these two compositions give the same map. I would suggest the construction of $J_{n+1} X$ as the equalizer of $T (J_n X ) \to J_n X \to T ( J_{n-1} X)$ and $T (J_n X) \to T (J_{n-1} X)$, where the first arrow in $T (J_n X) \to J_n X \to T (J_{n-1} X)$ is the standard projection and the second one is the previously obtained equalizer (starting the induction with $J_1 X \to T X$ the identity), while the arrow $T(J_n X ) \to T( J_{n-1} X)$ is the tangent map of $J_n X \to J_{n-1} X$. I haven't seen the construction of jets formulated in these categorical terms (in particular as cofree coalgebras) which I find very nice. The inductive definition though is probably well know, I think I have seen it in papers of Spencer.

   Could there be a mistake in the sequential limit construction as stated? I’m assuming that that the first map in is just the standard projection from the tangent space, while the first one in is the tangent map of . But then these two compositions give the same map. I would suggest the construction of as the equalizer of and , where the first arrow in is the standard projection and the second one is the previously obtained equalizer (starting the induction with the identity), while the arrow is the tangent map of .

   I haven’t seen the construction of jets formulated in these categorical terms (in particular as cofree coalgebras) which I find very nice. The inductive definition though is probably well know, I think I have seen it in papers of Spencer.

5. • CommentRowNumber5.
   • CommentAuthorMike Shulman
   • CommentTimeMar 29th 2014
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   Author: Mike Shulman
   Format: MarkdownItexOh, sorry about the typo. The construction should be dual to the one [here](http://nlab.mathforge.org/nlab/show/transfinite+construction+of+free+algebras); I don't know whether that's the same as what you suggested.

   Oh, sorry about the typo. The construction should be dual to the one here; I don’t know whether that’s the same as what you suggested.

6. • CommentRowNumber6.
   • CommentAuthorMike Shulman
   • CommentTimeApr 3rd 2014
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   Author: Mike Shulman
   Format: MarkdownItexLet's try to attack the question of higher forms from this perspective. We can of course replace the tangent-bundle functor $T$ in the construction of jet space by any copointed endofunctor, and there are a couple natural choices to try for 2-forms. A first try might be the functor $\lambda X. \bigwedge^2(T X)$, the exterior square of the tangent bundle. Then a coalgebra is a space equipped with a bivector field, and we can construct the cofree coalgebra $J^{\wedge 2} X$ and consider forms $\omega : J^{\wedge 2} X \to \mathbb{R}$. Since any domain in $\mathbb{R}^2$ has a canonical bivector field, I think we can integrate any such form over a parametrized surface in a similar way to what we did above for 1-forms. However, since every bivector on $\mathbb{R}$ is zero, we can't define a cojet differential analogously to how we did it for 1-forms above, and I don't have ideas about how to define an exterior product or exterior derivative. Another option would be the functor $\lambda X. T X \times_X T X$, for which a coalgebra is a space equipped with two vector fields. Again, $\mathbb{R}^2$ has two canonical vector fields on it, so we can integrate any form $\omega : J^{\times 2} X \to \mathbb{R}$ over any parametrized surface in $X$. (Unlike for the bivector case, the integral will not in general be invariant under rotational reparametrization, but for "nice" forms it can be.) Now the two canonical vector fields on $J^{\times 2}X$ give us *two* differentials on forms: we can take the usual differential $T(J^{\times 2} X) \to \mathbb{R}$ of $\omega$ as a scalar function and compose it with either vector field, to obtain two new forms $\mathrm{d}_1\omega$ and $\mathrm{d}_2\omega$. In particular, the coordinates of $J^{\times 2} X$ should be things like $\mathrm{d}_1\mathrm{d}_2^2 \mathrm{d}_1 x$: a "2-dimensional jet" in this sense is like an infinite binary tree whose branches are higher-order changes at each step. The two vector fields on $J^{\times 2} X$ also give us two maps $J^{\times 2} X \to J X$, and thus two ways to regard a cojet 1-form $\omega : J X \to \mathbb{R}$ as a cojet 2-form; let's denote them by $\omega_1$ and $\omega_2$. We then have $\mathrm{d}_1 \omega_1 = (\mathrm{d}\omega)_1$ and similarly. I think it's natural to write $\omega \otimes \eta = \omega_1 \eta_2$, and $$ \omega\wedge \eta = \omega \otimes \eta - \eta \otimes \omega. $$ This wedge product ought to behave like the ordinary one on linear forms (but I don't think that squaring will distribute over it). The obvious definition of an exterior derivative is $\mathrm{d}\wedge \omega = \mathrm{d}_1\omega_2 - \mathrm{d}_2 \omega_1$. This seems *almost* right, but not quite. For instance, if $X=\mathbb{R}^2$ and $\omega = f\, \mathrm{d}x$ for some function (0-form) $f$, then we have $$ \begin{aligned} \mathrm{d}\wedge \omega &= \mathrm{d}_1(f \, \mathrm{d}_2 x) - \mathrm{d}_2 (f \, \mathrm{d}_1 x)\\ &= \frac{\partial f}{\partial x} \mathrm{d}_1 x \,\mathrm{d}_2 x + \frac{\partial f}{\partial y} \mathrm{d}_1 y \,\mathrm{d}_2 x + f \mathrm{d}_1 \mathrm{d}_2 x - \frac{\partial f}{\partial x} \mathrm{d}_2 x \,\mathrm{d}_1 x - \frac{\partial f}{\partial y} \mathrm{d}_2 y \,\mathrm{d}_1 x - f \mathrm{d}_2 \mathrm{d}_1 x\\ &= \frac{\partial f}{\partial y} (\mathrm{d}_1 y \wedge \mathrm{d}_2 x) + f (\mathrm{d}_1 \mathrm{d}_2 x - \mathrm{d}_2 \mathrm{d}_1 x) \end{aligned} $$ when it ought to be only the first term. I haven't yet managed to think of a way to modify the definition of $J^{\times 2} X$ so as to make $\mathrm{d}_1 \mathrm{d}_2 = \mathrm{d}_2 \mathrm{d}_1$ without $\mathrm{d}_1 = \mathrm{d}_2$.

   Let’s try to attack the question of higher forms from this perspective. We can of course replace the tangent-bundle functor in the construction of jet space by any copointed endofunctor, and there are a couple natural choices to try for 2-forms.

   A first try might be the functor , the exterior square of the tangent bundle. Then a coalgebra is a space equipped with a bivector field, and we can construct the cofree coalgebra and consider forms . Since any domain in has a canonical bivector field, I think we can integrate any such form over a parametrized surface in a similar way to what we did above for 1-forms. However, since every bivector on is zero, we can’t define a cojet differential analogously to how we did it for 1-forms above, and I don’t have ideas about how to define an exterior product or exterior derivative.

   Another option would be the functor , for which a coalgebra is a space equipped with two vector fields. Again, has two canonical vector fields on it, so we can integrate any form over any parametrized surface in . (Unlike for the bivector case, the integral will not in general be invariant under rotational reparametrization, but for “nice” forms it can be.)

   Now the two canonical vector fields on give us two differentials on forms: we can take the usual differential of as a scalar function and compose it with either vector field, to obtain two new forms and . In particular, the coordinates of should be things like : a “2-dimensional jet” in this sense is like an infinite binary tree whose branches are higher-order changes at each step.

   The two vector fields on also give us two maps , and thus two ways to regard a cojet 1-form as a cojet 2-form; let’s denote them by and . We then have and similarly. I think it’s natural to write , and

   This wedge product ought to behave like the ordinary one on linear forms (but I don’t think that squaring will distribute over it).

   The obvious definition of an exterior derivative is . This seems almost right, but not quite. For instance, if and for some function (0-form) , then we have

   when it ought to be only the first term. I haven’t yet managed to think of a way to modify the definition of so as to make without .

7. • CommentRowNumber7.
   • CommentAuthorMike Shulman
   • CommentTimeApr 3rd 2014
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   Author: Mike Shulman
   Format: MarkdownItexOn the other hand, it might be that because of the equality of mixed partials, $\mathrm{d}_1\mathrm{d}_2 x$ and $\mathrm{d}_2\mathrm{d}_1 x$ have the same integral over any smooth surface, so that the extra term at least wouldn't interfere with Stokes' theorem. (I'm not 100% sure and I don't have time to write it out carefully right now.)

   On the other hand, it might be that because of the equality of mixed partials, and have the same integral over any smooth surface, so that the extra term at least wouldn’t interfere with Stokes’ theorem. (I’m not 100% sure and I don’t have time to write it out carefully right now.)

8. • CommentRowNumber8.
   • CommentAuthorMike Shulman
   • CommentTimeApr 4th 2014
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   Author: Mike Shulman
   Format: MarkdownItexI think what I want is the cofree space with two *commuting* vector fields cogenerated by $X$; that should ensure that $\mathrm{d}_1 \mathrm{d}_2 = \mathrm{d}_2 \mathrm{d}_1$. Spaces with two commuting vector fields aren't the coalgebras for a single copointed endofunctor, but I think we can construct cofree ones using limits of comonads, dually to the construction of higher inductive types. If a space $X$ has two vector fields $v$ and $w$, then I think that $v$ and $w$ commute iff the two composites $X \xrightarrow{v} T X \xrightarrow{T w} T^2 X$ and $X \xrightarrow{w} T X \xrightarrow{T v} T^2 X \xrightarrow{swap} T^2 X$ are equal. These maps define two natural transformations $J^{\times 2} \to T^2$ over the identity, hence two maps of comonads $J^{\times 2} \to C(T^2)$, where $C(T^2)$ is the cofree comonad generated by $T^2$. The equalizer of these, in the category of comonads, should be a comonad whose coalgebras are spaces equipped with two commuting vector fields. And we should be able to construct that equalizer using a sequential limit, either out of $J^{\times 2}$ and $C(T^2)$ dually to [here](http://ncatlab.org/nlab/show/transfinite+construction+of+free+algebras#colimits_of_monads), or by mixing these equalizers into the sequential-limit construction of $J^{\times 2}$.

   I think what I want is the cofree space with two commuting vector fields cogenerated by ; that should ensure that . Spaces with two commuting vector fields aren’t the coalgebras for a single copointed endofunctor, but I think we can construct cofree ones using limits of comonads, dually to the construction of higher inductive types. If a space has two vector fields and , then I think that and commute iff the two composites and are equal. These maps define two natural transformations over the identity, hence two maps of comonads , where is the cofree comonad generated by . The equalizer of these, in the category of comonads, should be a comonad whose coalgebras are spaces equipped with two commuting vector fields. And we should be able to construct that equalizer using a sequential limit, either out of and dually to here, or by mixing these equalizers into the sequential-limit construction of .