22
$\begingroup$

Let $\lambda = (\lambda_1, \ldots, \lambda_r)$ and $\mu = (\mu_1, \ldots, \mu_r)$ be partitions such that $\mu_j = \lambda_j +1$ for one index $j$ and $\mu_i = \lambda_i$ for all other $i$. Then there is a natural transformation $\alpha_{\mu/\lambda}: \mathbb{S}_{\lambda}(V) \otimes V \to \mathbb{S}_{\mu}(V)$, where $\mathbb{S}_{\kappa}$ denotes the $\kappa$-Schur functor; $\alpha_{\mu/\lambda}$ is unique up to scaling.

For a smooth manifold $X$, let's define a $\mu/\lambda$-differential operator to be a map $\delta$ from sections of $\mathbb{S}_{\lambda} T^{\ast} X$ to sections of $\mathbb{S}_{\mu} T^{\ast} X$ such that, for any smooth function $f$ and section $v$, we have the Leibniz rule $$\delta(f v) = \alpha_{\mu/\lambda}(df \otimes v) + f \delta(v).$$

Let's define a natural $\mu/\lambda$-differential operator to be a choice $\delta_X$ of a $\mu/\lambda$-differential operator on each manifold $X$ such that, if $\phi: X \to Y$ is a smooth map, then $\phi^{\ast} \circ \delta_Y = \delta_X \circ \phi^{\ast}$.

Are the only natural differential operators scalar multiples of the exterior derivative $d$ with $\mu= 1^{k+1}$ and $\lambda = 1^k$?

Motivation: Just curiosity. I've been trying to make $d$ sound natural this term, and one thing that I've said a lot is that it is the only thing that commutes with pullback, so I'm curious if this formalization of that this is true.

Improve this question
$\endgroup$
11
  • 9
    $\begingroup$ Natural Operations in Differential Geometry is dedicated to answering questions of this form: emis.de/monographs/KSM/kmsbookh.pdf $\endgroup$
    – Qiaochu Yuan
    Mar 22, 2015 at 19:13
  • $\begingroup$ I don't know the answer to the question, but I think it is reasonable to call Lie derivatives and contractions natural differential operators. They are defined in a coordinate independent way and they satisfy the graded Leibniz rule when they act on p-forms (Lie derivatives are degree 0 and contraction is degree -1) $\endgroup$
    – Daniel Barter
    Mar 22, 2015 at 19:38
  • 4
    $\begingroup$ @Daniel: the Lie derivative and contractions are natural with respect to diffeomorphisms, but they are not natural with respect to arbitrary smooth maps (since vector fields can't be transported along arbitrary smooth maps). $\endgroup$
    – Qiaochu Yuan
    Mar 22, 2015 at 19:44
  • $\begingroup$ @David: I haven't thought very hard about this, but can't you do things like consider an extension of $d$ to symmetric powers of $1$-forms? $\endgroup$
    – Qiaochu Yuan
    Mar 23, 2015 at 3:11
  • 3
    $\begingroup$ @QiaochuYuan: There is no natural differential operator even on $S^2(T^*M)$ with the properties that David has assumed. However, there is a nonlinear differential operator on $S^2_n(T^*M)$, the bundle of nondegenerate quadratic forms on $M$ that is natural for diffeomorphisms, namely the operator that assigns to each nondegenerate quadratic tensor its Levi-Civita connection, which is a section of the natural affine bundle whose sections are the torsion-free connections on $M$. $\endgroup$
    – Robert Bryant
    Mar 23, 2015 at 9:00

2 Answers 2

Reset to default
18
$\begingroup$

I think your question, the way it is stated, makes one want to classify unary and binary (depending how far you generalise the question as written) invariant differential operators on tensor fields. This has been done for unary operators by an awful lot of people, and the statement indeed is that $d$ is the only operator of that sort. More interestingly, there exists a full classification of binary invariant operators, this was done by Grozman around 1980, and is documented in http://arxiv.org/abs/math/0509562 .

Improve this answer
$\endgroup$
2
  • 3
    $\begingroup$ Following links from your paper, the answer to the specific question I asked is Theorem 5.7 in "Natural vector bundles and natural differential operators" by Terng jstor.org/stable/2373910, and appears to have been proved around the same time by several Russian authors whom I haven't read yet. Thanks for the reference! I definitely appreciate your point about how the problem becomes different with only diffeomorphisms in the picture -- and much harder. $\endgroup$
    – David E Speyer
    Mar 25, 2015 at 16:03
  • 1
    $\begingroup$ Invariance under any smooth map, as in Speyer's question, is considered by Freed-Hopkins (ArXiv:1301.5959), where they apply the uniqueness of $d$ to obtain the uniqueness of the Chern-Weil construction. This is re-written, with the classical Diff-invariance language, in the paper by Navarro-Sancho mentioned by Khavkine. $\endgroup$
    – José Navarro
    Nov 13, 2015 at 12:48
10
$\begingroup$

Only a partial answer. The recent preprint arXiv:1412.0840 by Navarro and Sancho addresses precisely this question, but restricted only maps from forms to forms (say $1^p$ to $1^q$, in your notation). Here's the relevant part from the abstract:

We prove that the only natural operations between differential forms are those obtained using linear combinations, the exterior product and the exterior differential.

I'm not sure about maps between covariant tensor corresponding to arbitrary Young diagrams.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ I was not aware of this paper, thanks! Interestingly enough, this result that you quote from their abstract was also obtained in the master thesis of my student Lucas Mason-Brown in the beginning of 2014 :-) $\endgroup$
    – Vladimir Dotsenko
    Mar 25, 2015 at 11:40
  • 4
    $\begingroup$ That statement, as is naively quoted, may already be found in Palais' paper from 1959. Subsequents refinements, as in KMS-book or the paper by Navarro-Sancho, have to do with what you understand by "natural operation" (v.gr., in this last paper, the authors consider morphisms of sheaves covariant with respect to arbitrary smooth maps). $\endgroup$
    – José Navarro
    Nov 13, 2015 at 12:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.