25

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.)

Andrés E. Caicedo
  • 32,193
  • 5
  • 130
  • 233
Toby Bartels
  • 2,654
  • 1
    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 03 '11 at 19:17
  • 3
    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 03 '11 at 22:54
  • 1
    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 03 '11 at 23:29
  • So here's a problem from freshman calculus: Given that $y = x^3 - 3x$, for which values of $x$ does $y$ reach a local maximum or minimum? We compute $\mathrm{d}{y} = (3x^2 - 3) ,\mathrm{d}{x}$, set this to $0$ and solve for $x$ to get two critical points, which we test using the second derivative. Starting from $\mathrm{d}y$ above, we compute $\mathrm{d}^2{y} = 6x ,(\mathrm{d}{x})^2 + (3x^2 - 3) ,\mathrm{d}^2{x}$. Plugging in $x = \pm{1}$, we get $\mathrm{d}^2{y} = \pm{6x^2} ,(\mathrm{d}x)^2$. So $y$ has a local minimum when $x = 1$ and a local maximum when $x = -1$. – Toby Bartels Apr 04 '11 at 00:13
  • 1
    But I'm not sure what exactly I calculated there. At the moment, to be sure that all of my steps are rigorous, I must divide by $\mathrm{d}{x}$ to get $\mathrm{d}{y}/\mathrm{d}{x} = 3x^2 - 3$, then differentiate that to get $\mathrm{d}(\mathrm{d}{y}/\mathrm{d}{x}) = 6x ,\mathrm{d}{x}$. And the test isn't quite whether this is positive or negative; I must divide by $\mathrm{d}{x}$ again first. So this application has lost much of the beauty of the differential-based approach. – Toby Bartels Apr 04 '11 at 00:18
  • 1
    As far as I know, using differentials do not provide any advantage when computing higher derivatives by hand. However, it does work beautifully in designing very simple recursive algorithms for computing higher derivatives of functions in software. Just search for descriptions of "automatic differentiation". It is a lot of fun to implement this using, say, C++ templates. – Deane Yang Apr 04 '11 at 01:43
  • Could you add the definition for cojet? Or give a reference? – Michael Bächtold Apr 04 '11 at 16:07
  • 1
    For jets, you can see Wikipedia http://en.wikipedia.org/wiki/Jet_%28mathematics%29 but the basic idea is that a $k$-jet from $M$ to $N$ at a point $p$ is an equivalence class of smooth functions to $N$ from a neighbourhood of $p$, where two such functions are equivalent if their derivatives up to and including order $k$ are the same. If $N$ is a vector bundle and we restrict to jets that map $p$ to $0$, then these form a vector space. The $k$-cojet space at $p$ is the dual space, and these join to form a vector bundle whose sections I call the degree-$k$ cojet differential forms. – Toby Bartels Apr 05 '11 at 00:28
  • I'm pretty much restricting to the case where $M$ has $1$ dimension and $N$ is $\mathbb{R}$ (the trivial line bundle), but I would like to understand at least higher-dimensional $M$ (which is the usual context for exterior differential forms). So far, I'm working things out for myself, but I'd hate to think that this is new. – Toby Bartels Apr 05 '11 at 00:31
  • 2
    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 05 '11 at 06:23
  • 1
    I don't think Michael is mistaken. His comments seem on target. You might also want to take a look at Griffiths' work on doing calculus of variations with exterior differential systems. – Deane Yang Apr 06 '11 at 15:36
  • Any update on this? – Steven Gubkin Jun 14 '13 at 18:45
  • 2
    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 03:54
  • 1
    @Scott: Please correct me if I am wrong, but I think that the bundle of jets on a manifold with values in a vector space does have a canonical vector bundle structure (given by adding jets). In fact, the transition functions are linear, as pre-composing with a smooth map gives a linear map on jets. It seems that more generally the jets of sections of a vector bundle themselves form a vector bundle. – Ricardo Andrade Jun 17 '13 at 01:17
  • 1
    @Ricardo: Yes, you're correct that if you have a global trivialization of your vector bundle (i.e., taking jets with values in a vector space), you get an additive structure. However, without a fixed trivialization, you don't get linear maps on jets of order greater than one. For example, the transition between the north-pole and south-pole stereographic projections on $S^1$ produces a quadratic term on tangent 2-jet coordinates: if $z = 1/w$, then $d^2z = -\frac{2}{w^3} (dw)^2 - \frac{1}{w^2}d^2w$. – S. Carnahan Jun 18 '13 at 00:23

2 Answers2

6

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

Peter Michor
  • 24,985
4

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 $i \geq 2$.

S. Carnahan
  • 45,116