1
$\begingroup$

I was wondering whether the following commuting diagram property was enough to characterize the exterior derivative (although I'm suspicious. Is this answered somewhere in the literature? my question came about after posting a comment in What is the exterior derivative intuitively?)

Let $f$ be any smooth map $f:M\to N$ where $M$ and $N$ are smooth manifold and consider the map $f^*: A(N)\to A(M)$ between the spaces of smooth sections of the exterior form bundles $\Omega(N^*)$ and $\Omega(M^*)$, $A(N)$ and $A(M)$, associated to the manifolds $N$ and $M$. Then the exterior derivative is the unique map $d$ such that the following diagram commutes (and its properties follow from this characterization):

$\require{AMScd}$ \begin{CD} A(N) @>d>> A(N)\\ @V f^* V V @VV f^* V\\ A(M)@>>d> A(M) \end{CD}

Thanks.

$\endgroup$
6
  • 5
    $\begingroup$ I'm pretty sure that $d = 0$ makes your diagram commute...? $\endgroup$ Nov 19, 2020 at 18:25
  • 4
    $\begingroup$ This (and linked paper) contains what you're after I believe: math.stackexchange.com/questions/918949/… $\endgroup$ Nov 19, 2020 at 18:49
  • 3
    $\begingroup$ The paper @PaulReynolds references: Palais - Natural operations on differential forms. $\endgroup$
    – LSpice
    Nov 19, 2020 at 19:05
  • 1
    $\begingroup$ The identity also makes your diagram commute! I think you further want to require that $d$ is a derivation of degree $1$, and even then I think you can at best hope to recover $d$ up to a scalar. $\endgroup$ Nov 19, 2020 at 20:40
  • 2
    $\begingroup$ @PaulReynolds Definitely what I was thinking about, thank you! $\endgroup$ Nov 19, 2020 at 22:54

0

Your Answer

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

Browse other questions tagged or ask your own question.