1

I can prove the following :

There exist a unique map $d:\Omega^k(M)\to \Omega^{k+1}(M)$ such that

  1. $d$ is $\mathbb{R}$-linear
  2. $d^2=0$
  3. For a $0$-differential form (a function) $d$ reduces to the usual differential
  4. $d(\omega\wedge\eta)=(d\omega)\wedge \eta+(-1)^k\omega\wedge d\eta$.

And I can prove that such a map is locally given by $$d(\sum_I a_Idx_{i_1}\wedge \ldots dx_{i_k})=\sum_Ida_I\wedge dx_{i_1}\wedge\ldots\wedge dx_{i_k}$$

where $(X_1,\ldots,\hat X_i,\ldots, X_k)$ means "remove $X_i$ from $(X_1,\ldots,X_k)$". e.g : $(X_1,\hat X_2,X_3)=(X_1,X_3)$

(by the way, I prove the existence part using that formula and the uniqueness part)

Now I want to prove the coordinate-free formula

$$ (d\omega)(X_1,\ldots X_{k+1})=\sum_i(-1)^{i-1}X_i\big( \omega(X_1,\ldots, \hat X_i,\ldots,X_{k+1}) \big)+\sum_{i<j}(-1)^{i+j}\omega\big( [X_i,X_j],X_1,\ldots \hat X_i,\ldots,\hat X_j,\ldots, X_{k+1} \big). \qquad (1) $$

I see two ways to prove that formula.

  1. Prove that it satisfies the 4 conditions.
  2. Prove that it locally reduces to the local expression

This answer goes for options 1. In the hint it gives, it supposes that one already checked that formula (1) is a $(k+1)$-form. It seems to me extremely difficult to prove anti-symmetry (I got into very long unsuccessful computations).

In the same way, here on page 2 it says "simple but messy computation"...

QUESTION: How do I prove formula 1 ? Preferably without going into long and painful computations.

Related:

EDIT. Full proof here. Search for the labels PROPooUOQRooAdvYqx and PROPooTXFRooMtaVrU. Disclamer : I'm the author and the text is not proof-read. Too long to be posted here as answer.

Laurent Claessens
  • 749
  • How do you understand "hat " of Z ? – Adam Wrzesiński Jan 05 '24 at 05:12
  • @Adam Wrzesiński It is a missing element in the list. I edit the question to make it clear. – Laurent Claessens Jan 05 '24 at 05:21
  • 2
    It helps to understand that the bracket terms are there to make this a tensor. After that, you can choose $X_i$ to be coordinate vector fields and then the bracket terms are $0$. – Ted Shifrin Jan 05 '24 at 06:02
  • @Ted Shifrin This seems to be the "Tip" in this question. Can you elaborate ? Showing that it is sufficient to check on the coordinates vectors could be an accepted answer. – Laurent Claessens Jan 05 '24 at 06:21
  • 2
    What I said is in every book on differentiable manifolds. A multilinear map on vector fields defines a tensor iff it is linear over the ring if smooth functions. – Ted Shifrin Jan 05 '24 at 07:02
  • What are those "books" everyone's talking about? These seem to be powerful artifacts hidden in some places called "libraries" that are only accessible to some selected people... Well ... seriously thank for the advise. I will try to show the linearity with respect to the functions. – Laurent Claessens Jan 05 '24 at 08:37
  • If you want to prove a tensor identity, it suffices to fix a frame of vector fields and check that it holds for any combination of the vector fields in the frame. This fact follows directly from the multilinearity of a tensor field. – Deane Jan 06 '24 at 15:08
  • The standard textbooks in differential geometry are by John Lee, Loring Tu. A classic is O’Neill’s Semiriemannian Geometry. There’s a nice French one by Gallot-Hulin-Lafontaine. I would expect this fact to be mentioned in all of these texts but I haven’t checked. – Deane Jan 06 '24 at 15:12

0 Answers0