# Is there a convenient differential calculus for cojets?

I understand the basics of exterior differential geometry and how to do calculus with exterior differential forms. I know how to use this to justify the notation dy/dx as a literal ratio of the differentials dy and dx (by treating x and y as scalar-valued functions on a 1-dimensional manifold and introducing division formally). I would like to extend this to second derivatives. Ideally, this would justify the notation d2y/dx2 as a literal ratio.

I can't do this with the exterior differential, since both d2y and dx ∧ dx are zero in the exterior calculus. It occurs to me that this would work if, instead of exterior differential forms (sections of the exterior bundle), I used sections of the cojet bundle (cojet differential forms). In particular, while degree-2 exterior forms may be written in local coordinates as linear combinations of dxi ∧ dxj for i < j (so on a 1-dimensional manifold the only exterior 2-form is zero), degree-2 cojet forms may be written in local coordinates as linear combinations of d2x and dxi · dxj for i ≤ j (so on a 1-dimensional manifold the cojet 2-forms at a given point form a 2-dimensional space).

I know some places to read about cojets (and more so about jets) theoretically, but I don't know where to learn about practical calculations in a cojet calculus analogous to the exterior calculus. In particular, I don't know any reference that introduces the concept of the degree-2 differential operator d2, much less one that gives and proves its basic properties. I've even had to make up the notation ‘d2’ (although you can see where I got it) and the term ‘cojet differential form’. I can work some things out for myself, but I'd rather have the confidence of seeing what others have done and subjected to peer review.

(Incidentally, I don't think that it is quite possible to justify d2y/dx2; the correct formula is d2y/dx2 − (dy/dx)(d2x/dx2); we cannot let d2x/dx2 vanish and retain the simplicity of the algebraic rules. It would be better to write ∂2y/∂x2; the point is that this is the coefficient on dx2 in an expansion of d2y, just as ∂y/∂xi is the coefficient of dy on xi when y is a function on a higher-dimensional space. The coefficient of d2y on d2x, which would be ∂2y/∂2x, is simply dy/dx again.)

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Could you provide an example of a calculation that you would like to be able to do or do more easily using such a calculus? –  Deane Yang Apr 3 '11 at 19:17
If you want to see a modern approach to the formal theory of PDE's (i.e., a cohomological approach to the Cartan-Kahler theorem, which was developed originally using exterior differential systems), look at the work of Hubert Goldschmidt, which builds on work by Spencer, Quillen, Guillemin, and Guillemin-Sternberg. See, for example, Chapters IX and X of the book "Exterior differential systems" by Bryant, Chern, Goldschmidt, and Griffiths. –  Deane Yang Apr 3 '11 at 22:54
I didn't want to get into the context, in case people started discussing that instead. But there is a place to discuss that, in this old thread. I want to understand the theory behind a differentials-based approach to teaching freshman calculus, as advocated by Dray and Manogue (pdf). For first derivatives, I know how to make everything that they write formally correct. But what about higher derivatives? –  Toby Bartels Apr 3 '11 at 23:29
Maybe I'm mistaken but if $\Gamma(N)$ is the module of sections of a vector bundle over $M$ then $\Gamma(J^K(N)^*$ is "the same" as the module of linear differential operators from $\Gamma(N)$ to $C^\infty(R)$. I.e. differential operators associating a function to a section. There is a differential d on these objects (if you allow the order k to change) called the Spencer complex (see for example the articles of Spencer on overdetermined system or the book cited by Dean or books by Vinogradov and Lychagin). But i don't know if this d is the one you are looking for. –  Michael Bächtold Apr 5 '11 at 6:23
The $k$-jet bundle isn't a vector bundle for $k \geq 2$. The fibers are vector spaces, but the transitions are not linear maps. In particular, it cannot be dual to a co-jet vector bundle. –  S. Carnahan Jun 16 '13 at 3:54

Maybe the following links help: Gerd Kainz, Peter W. Michor: Natural transformations in differential geometry. Czechoslovak Math. J. 37 (1987), 584-607, accessible as scanned paper under: http://www.mat.univie.ac.at/~michor/nat-transf.pdf. A slightly more extended version is in chapter 8 of: Ivan Kolár, Jan Slovák, Peter W. Michor: Natural operations in differential geometry. Springer-Verlag, Berlin, Heidelberg, New York, (1993), vi+434 pp. which is accessible via http://www.mat.univie.ac.at/~michor/kmsbookh.pdf

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As far as I can tell, your example computation in the comments is a computation in the Hasse-Schmidt ring of a polynomial algebra. Given a commutative ring $A$ and $A$-algebras $f:A \to B$ and $A \to R$, an order $k$ Hasse-Schmidt differential from $B$ to $R$ is a $k+1$-tuple $(D_0,\ldots,D_k)$ of $A$-module maps from $B$ to $R$ satisfying:

1. $D_i(f(a)) = 0$ for all $i \geq 1$ and all $a \in A$.
2. $D_i(b_1 \cdot b_2) = \sum_{j=0}^i D_j(b_1) D_{i-j}(b_2)$.

We write $Der^k_A(B,R)$ for the set of order $k$ differentials from $B$ to $R$. There is a Hasse-Schmidt algebra $HS^k_{B/A}$ with universal $k$-derivation that represents the functor $Der^k_A(B,-)$, and its relative spectrum over $\operatorname{Spec} B$ is the relative $k$th jet space of $B/A$. For example, $HS^0_{B/A} = B$, and $HS^1_{B/A} = Sym_B^*(\Omega_{B/A})$ yields the tangent bundle. You can find this information in Vojta's EGA-style exposition.

Concretely, here is your example: Let $A$ be a ring such as $\mathbb{R}$, and let $B = A[x]$. It is not hard to show that $HS^k_{B/A} \cong B[x^{(1)},x^{(2)},\ldots,x^{(k)}]$, with canonical maps $y \mapsto y^{(i)}$. In terms of ordinary differentials, we have $x^{(i)} = \frac{1}{i!}d^ix$, and in particular, if we were to write the higher Leibniz rule with differentials, we would need some $\binom{i}{j}$ factors. At any rate, repeated use of the Leibniz relation yields $d^2(x^3-3x) = (3x^2-3)d^2x + 6x(dx)^2$.

If you want to differentiate something twice, you use the fact that for any $y$, $dy$ is equal to $d^0 y' dx$ for some $y' \in B$, then apply the quotient rule: $d\left(\frac{dy}{dx}\right) = \frac{dx\cdot d^2 y - dy \cdot d^2 x}{(dx)^2}$. In particular, you find that $d\left(\frac{dy}{dx}\right)/dx \neq \frac{d^2 y}{(dx)^2}$, because $\frac{d^2x}{(dx)^2} \neq 0$ in this ring. If you really want the somewhat misleading notation $\frac{d^ky}{dx^k}$ to literally denote a $k$th derivative, you have to mod out by the ideal generated by $d^ix$ for $i \geq 2$.

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Thanks, that looks promising! –  Toby Bartels Jun 18 '13 at 18:01