Let me start with the following tl;dr version of my question

What is a higher-order derivative, in general? How does it relate to the exterior derivative and to differential forms?

Suppose we have a bundle and a section $E \overset{\sigma}{\underset{\pi}{\leftrightarrows}} M$. Assume we have a connection on $E$. (One of the links assumes we also have a connection on $\tau^*M$; assume that also, if you like.) It seems

  • The $k$th order derivative should be a bundle map $\bigoplus_{i=0}^k \otimes^i \tau M \to E$ over M. Perhaps it always will satisfy some condition (such as symmetry) that will allow us to pass to a quotient of $\bigoplus_{i=0}^k \otimes^i \tau M$.
  • The covariant derivative should take us from $k$th order derivatives to $k+1$st order derivatives. It should be multilinear in the entries.

Regarding the last bullet, I could believe that the higher-order derivative might not always be found by an iterated derivative. Like in calculus where we have to recognize that a multidimensional integral isn't the same as an iterated integral, and under nice circumstances, they are equivalent.

One suggestion is that a higher covariant derivative of a section $\sigma: M \to E$ is the prolongation $E \to J^k E$ induced by $\sigma$.

This answer gives an outline of what I'm looking for, but it does so in quite a vague way. This answer seems to be addressing the issue head-on, but I haven't a clue what he's doing with his notation. See also this question about higher covariant derivatives.

This is maybe how an explanation could proceed:

  • A higher-order covariant derivative $\nabla^k s$ of a section of a vector bundle $E \to M$ should be the prolongation $E \to J^k E$of the $k$th jet bundle $J^kE \to E$.
  • Sections of $J^kE \to E$ are elements of $\Gamma \left( \left( \bigoplus\limits_{i=0}^k\operatorname{Sym}^i \tau^* M \right) \otimes E \right)$. So we can think of them like "polynomials of degree $\leq k$ where we plug in tangent vectors and get out elements of $E$." This maps nicely onto Taylor polynomials and $k$th order approximations of a function.
  • The anti-symmetry of the exterior derivative (alternately, the anti-symmetry of the wedge product) kills all but the first-order derivatives. (Becoming a little vague. This is what the first link above says, but does not explain.)

(How can we make the last bullet rigorous? One relevant fact is that if $V$ is a vector space, and we view the symmetric and anti-symmetric tensors of order $k$ as subgroups of $\otimes ^kV$ (really they are quotients), then $\operatorname{Alt}^kV \cap \operatorname{Sym}^k V = 0$ except when $k=1$.)

We'd like to say something like

  • The usual exterior derivative $d: \wedge^k \tau^*M \to \wedge^{k+1} \tau^*M$ or $d: \wedge^k \tau^*M \otimes E \to \wedge^{k+1} \tau^*M \otimes E$ somehow only captures one level at a time(?) of what the higher derivatives do?
  • We also like anti-symmetry for modeling a "volume of a parallelipiped"-type function which is multilinear and vanishes when there are collinear entries (like the determinant). And we like the exterior derivative for being dual to the boundary in Stokes' theorem. Hence we extract this information in differential geometry.

This morning I've been boring through the "Natural Operations..." book by Kolar et al, but boy...it's encyclopedic, and has quite esoteric notation to boot. Kolar claims he's outlined what a higher-order derivative should be in generality, in an article called "On the Absolute Differentiation of Geometric Object Fields" from 1973. As the title suggests, that article is pretty obscure, and some of the math therein is, also, at least to me.

  • $\begingroup$ Too long question!! $\endgroup$ – C.F.G Apr 3 at 13:30
  • 3
    $\begingroup$ @C.F.G I'm open to constructive criticism on my question. But you have fallen far short of that. My question is complicated to explain...not sure how to make it shorter and preserve what it means. $\endgroup$ – Eric Auld Apr 3 at 14:09

I'm just going to focus on how totally anti-symmetrized higher covariant derivatives of $\sigma$ can be expressed in terms of $\sigma$ and $\nabla \sigma$ - no derivatives of $\sigma$ higher than first order. I think that's the main point you're looking for more detail about. (I won't comment on the jet bundles and stuff since I'm mostly unfamiliar with the formalism.)

We know that $\nabla^2 \sigma(X,Y)-\nabla^2 \sigma(Y,X) = \text{Riem}(X,Y) \sigma$. This is as seen in the other question from MathOverflow you've referred to a few times. It is also written in various other styles of notation, such as $d^\nabla(d^\nabla \sigma)(X,Y)$, or in an index-based notation as $$ \nabla_{[i} \nabla_{j]} \sigma^\alpha = \frac{1}{2} R^\alpha{}_{\beta i j} \sigma^\beta. $$ (Square brackets on the lower indices denote anti-symmetrization: e.g. $A_{[ijk]} := \frac{1}{3!} \sum_{\pi} \text{sgn}(\pi) A_{\pi(i)\pi(j)\pi(k)}$, a sum over permutations of $i$, $j$, and $k$. Also, vertical bars can be used to exclude indices from the anti-symmetrization: e.g. $A_{[ij|k|l]} := \frac{1}{3!} \sum_{\pi} \text{sgn}(\pi) A_{\pi(i)\pi(j) k \pi(l)}$.)

The curvature formula takes care of the second covariant derivative - but what about even higher derivatives? At third order, we are looking at $d^\nabla(d^\nabla(d^\nabla \sigma))(X,Y,Z)$, or $\sum_{\pi} \text{sgn}(\pi) \nabla^3 \sigma(\pi(X),\pi(Y),\pi(Z))$, or $\nabla_{[i} \nabla_j \nabla_{k]} \sigma^\alpha$. We can compute it as follows: $$ \begin{align} \nabla_{[i} (\nabla_j \nabla_{k]} \sigma^\alpha) &= \nabla_{[i} ( \frac{1}{2} R^\alpha{}_{|\beta| j k]} \sigma^\beta ) \\ &= \frac{1}{2} (\nabla_{[i} R^\alpha{}_{|\beta| j k]}) \sigma^\beta + \frac{1}{2} R^\alpha{}_{\beta [j k} (\nabla_{i]} \sigma^\beta) \\ &= \frac{1}{2} R^\alpha{}_{\beta [j k} \nabla_{i]} \sigma^\beta. \end{align} $$ (The vanishing of $\nabla_{[i} R^\alpha{}_{|\beta| j k]}$ is the Bianchi identity, although the fact that it vanishes will not really be essential to us.) Translated back to the other notation, this is $$ \nabla^3 \sigma(X,Y,Z) - \nabla^3 \sigma(X,Z,Y) + \nabla^3 \sigma(Y,Z,X) - \nabla^3 \sigma(Y,X,Z) + \nabla^3 \sigma(Z,X,Y) - \nabla^3 \sigma(Z,Y,X) = \text{Riem}(X,Y) \nabla \sigma (Z) + \text{Riem}(Y,Z) \nabla \sigma (X) + \text{Riem}(Z,X) \nabla \sigma (Y). $$

At fourth order, we have: $$ \begin{align} \nabla_{[i} (\nabla_j \nabla_k \nabla_{l]} \sigma^\alpha) &= \nabla_{[i} ( \frac{1}{2} R^\alpha{}_{|\beta| k l} \nabla_{j]} \sigma^\beta ) \\ &= \frac{1}{2} (\nabla_{[i} R^\alpha{}_{|\beta| k l}) (\nabla_{j]} \sigma^\beta ) + \frac{1}{2} R^\alpha{}_{\beta [k l} (\nabla_i \nabla_{j]} \sigma^\beta ) \\ &= 0 + \frac{1}{4} R^\alpha{}_{\beta [k l} R^\beta{}_{|\gamma| i j]} \sigma^\gamma. \end{align} $$ (Aside: The reason I've been preferring index notation is because it makes these expression concise. These notes I found on the internet give the computations without index notation, which you might prefer, although they're very verbose.)

The point is, the expansion of $(d^\nabla)^m \sigma (X_1, ..., X_m)$ or $\nabla_{[i_1} \cdots \nabla_{i_m]} \sigma^\alpha$ contains only various combinations Riemann curvature tensor, $\sigma$, and $\nabla \sigma$. Any time it would have contained an expression that looks like "$\nabla \nabla \sigma$", we replace can it with "$\text{Riem} \, \sigma$" because we've anti-symmetrized over every "$\nabla \nabla$".

The only "miracle" occurs at second order ("$\nabla^2 \sigma(X,Y)-\nabla^2 \sigma(Y,X) = \text{Riem}(X,Y) \sigma$"). Everything else follows from that; there are no "new miracles" at higher orders. That sums up how higher covariant derivatives behave after total anti-symmetrization.

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