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    • CommentRowNumber1.
    • CommentAuthorMike Shulman
    • CommentTimeMar 28th 2014

    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 XX is just a (perhaps smooth) function JXJ X \to \mathbb{R}, where JXJ X is the space of jets in XX.

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

    We can construct JXJ X by the usual sort of sequential limit:

    J 3XJ 2XJ 1XJ 0X=X \cdots \to J_3 X \to J_2 X \to J_1 X \to J_0 X = X

    where J 0X=XJ_0 X = X and J 1X=TXJ_1 X = T X, while J n+1XJ_{n+1} X for n1n\ge 1 is the equalizer of T(J nX)J nXJ n1XT (J_n X) \to J_n X \to J_{n-1} X and T(J nX)T(J n1X)J n1XT(J_n X) \to T(J_{n-1} X) \to J_{n-1} X. So J 2XJ_2 X, for instance, consists of a tangent vector to TXT X at (x,v)(x,v), say (u,w)(u,w), such that u=vu=v. Thus, for instance, J( m)( m) ωJ (\mathbb{R}^m) \cong (\mathbb{R}^m)^\omega, consisting of a point xx, a tangent vector at that point (= 1-jet), a 2-jet at that 1-jet, etc. Of course, JXJ 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 TT to preserve the sequential limit, but this shouldn’t be a problem; e.g. in SDG, TT is a right adjoint () D(-)^D so it preserves all limits.

    Note that a TT-coalgebra in general is just a smooth space YY equipped with a vector field ξ:YTY\xi:Y\to T Y. The universal property of JXJ X is that given such a (Y,ξ)(Y,\xi) and any smooth map f:YXf:Y\to X, there is a unique extension E(f,ξ):YJXE(f,\xi) : Y \to J X. The 1-jet part of E(f,ξ)E(f,\xi) is the composite YξTYTfTXY \xrightarrow{\xi} T Y \xrightarrow{T f} T X, and so on.

    Now I’m saying we can define a cojet differential form on XX to be a function ω:JX\omega:J X \to \mathbb{R}. The differential of such an ω\omega is the composite JXT(JX)J X \to T(J X) \to \mathbb{R}, where the first map is the canonical vector field on JXJ X (the terminal TT-coalgebra structure) and the second is just the ordinary differential of ω\omega regarded as a function on a smooth space.

    In coordinates on X=X=\mathbb{R}, this means ω\omega is a function of countably many variables xx, dx\mathrm{d}x, d 2x\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 d(d nx)=d n+1x\mathrm{d}(\mathrm{d}^n x) = \mathrm{d}^{n+1} x for all nn.

    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][a,b] comes with a family of canonical vector fields ξ h\xi_h which (under the identification TT\mathbb{R} \cong \mathbb{R}) is constant at hh, so given a curve c:[a,b]Xc:[a,b] \to X we have an induced E(c,ξ h):[a,b]JXE(c,{\xi_h}):[a,b] \to J X and thus ωE(c,ξ h):[a,b]\omega \circ E(c,{\xi_h}) : [a,b] \to \mathbb{R}. Now a tagged partition a=x 0<x 1<<x n=ba=x_0 \lt x_1 \lt \dots \lt x_n=b with tags t it_i and widths Δx i=x ix i1\Delta x_i = x_i - x_{i-1} yields a Riemann sum

    i=1 nω(E(c,ξ Δx i)(t i)) \sum_{i=1}^n \omega(E(c,\xi_{\Delta x_i})(t_i))

    and we can take the limit.

    • CommentRowNumber2.
    • CommentAuthorFosco
    • CommentTimeMar 28th 2014
    • (edited Mar 28th 2014)

    Not an answer, but some times ago I was extremely interested in expanding the description of the jet space JXJ 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:GoodSpacesGoodSpacesT : \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?

    • CommentRowNumber3.
    • CommentAuthorTobyBartels
    • CommentTimeMar 29th 2014

    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 XX. I don't know that I ever bothered to check.

    • CommentRowNumber4.
    • CommentAuthorMichael_Bachtold
    • CommentTimeMar 29th 2014
    • (edited Mar 29th 2014)

    Could there be a mistake in the sequential limit construction as stated? I’m assuming that that the first map in T(J nX)J nXJ n1XT (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 nX)T(J n1X)J n1XT(J_n X) \to T(J_{n-1} X) \to J_{n-1} X is the tangent map of J nXJ n1XJ_n X \to J_{n-1} X. But then these two compositions give the same map. I would suggest the construction of J n+1XJ_{n+1} X as the equalizer of T(J nX)J nXT(J n1X)T (J_n X ) \to J_n X \to T ( J_{n-1} X) and T(J nX)T(J n1X)T (J_n X) \to T (J_{n-1} X), where the first arrow in T(J nX)J nXT(J n1X)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 1XTXJ_1 X \to T X the identity), while the arrow T(J nX)T(J n1X)T(J_n X ) \to T( J_{n-1} X) is the tangent map of J nXJ n1XJ_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.

    • CommentRowNumber5.
    • CommentAuthorMike Shulman
    • CommentTimeMar 29th 2014

    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.

    • CommentRowNumber6.
    • CommentAuthorMike Shulman
    • CommentTimeApr 3rd 2014

    Let’s try to attack the question of higher forms from this perspective. We can of course replace the tangent-bundle functor TT 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 λX. 2(TX)\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 2XJ^{\wedge 2} X and consider forms ω:J 2X\omega : J^{\wedge 2} X \to \mathbb{R}. Since any domain in 2\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 λX.TX× XTX\lambda X. T X \times_X T X, for which a coalgebra is a space equipped with two vector fields. Again, 2\mathbb{R}^2 has two canonical vector fields on it, so we can integrate any form ω:J ×2X\omega : J^{\times 2} X \to \mathbb{R} over any parametrized surface in XX. (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 ×2XJ^{\times 2}X give us two differentials on forms: we can take the usual differential T(J ×2X)T(J^{\times 2} X) \to \mathbb{R} of ω\omega as a scalar function and compose it with either vector field, to obtain two new forms d 1ω\mathrm{d}_1\omega and d 2ω\mathrm{d}_2\omega. In particular, the coordinates of J ×2XJ^{\times 2} X should be things like d 1d 2 2d 1x\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 ×2XJ^{\times 2} X also give us two maps J ×2XJXJ^{\times 2} X \to J X, and thus two ways to regard a cojet 1-form ω:JX\omega : J X \to \mathbb{R} as a cojet 2-form; let’s denote them by ω 1\omega_1 and ω 2\omega_2. We then have d 1ω 1=(dω) 1\mathrm{d}_1 \omega_1 = (\mathrm{d}\omega)_1 and similarly. I think it’s natural to write ωη=ω 1η 2\omega \otimes \eta = \omega_1 \eta_2, and

    ωη=ωηηω. \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 dω=d 1ω 2d 2ω 1\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= 2X=\mathbb{R}^2 and ω=fdx\omega = f\, \mathrm{d}x for some function (0-form) ff, then we have

    dω =d 1(fd 2x)d 2(fd 1x) =fxd 1xd 2x+fyd 1yd 2x+fd 1d 2xfxd 2xd 1xfyd 2yd 1xfd 2d 1x =fy(d 1yd 2x)+f(d 1d 2xd 2d 1x) \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 ×2XJ^{\times 2} X so as to make d 1d 2=d 2d 1\mathrm{d}_1 \mathrm{d}_2 = \mathrm{d}_2 \mathrm{d}_1 without d 1=d 2\mathrm{d}_1 = \mathrm{d}_2.

    • CommentRowNumber7.
    • CommentAuthorMike Shulman
    • CommentTimeApr 3rd 2014

    On the other hand, it might be that because of the equality of mixed partials, d 1d 2x\mathrm{d}_1\mathrm{d}_2 x and d 2d 1x\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.)

    • CommentRowNumber8.
    • CommentAuthorMike Shulman
    • CommentTimeApr 4th 2014

    I think what I want is the cofree space with two commuting vector fields cogenerated by XX; that should ensure that d 1d 2=d 2d 1\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 XX has two vector fields vv and ww, then I think that vv and ww commute iff the two composites XvTXTwT 2XX \xrightarrow{v} T X \xrightarrow{T w} T^2 X and XwTXTvT 2XswapT 2XX \xrightarrow{w} T X \xrightarrow{T v} T^2 X \xrightarrow{swap} T^2 X are equal. These maps define two natural transformations J ×2T 2J^{\times 2} \to T^2 over the identity, hence two maps of comonads J ×2C(T 2)J^{\times 2} \to C(T^2), where C(T 2)C(T^2) is the cofree comonad generated by T 2T^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 ×2J^{\times 2} and C(T 2)C(T^2) dually to here, or by mixing these equalizers into the sequential-limit construction of J ×2J^{\times 2}.