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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 $X$ 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.

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 $J X$ 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 $T : \mathbf{GoodSpaces}\to\mathbf{GoodSpaces}$ as copointed).

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

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 $X$. 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 $T (J_n X) \to J_n X \to J_{n-1} X$ 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.

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 $T$ 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 $\mathrm{d}_1 = \mathrm{d}_2$.

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, $\mathrm{d}_1\mathrm{d}_2 x$ 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.)

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 $X$; 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, or by mixing these equalizers into the sequential-limit construction of $J^{\times 2}$.