1
$\begingroup$

Assume that $X$ is a vector field on a $n$ dimensional manifold $M$.Let $0\leq i,j \leq n$.

1.Is there a linear isomorphism $T:\Omega^i(M) \to \Omega^j(M)$ with $L_X T=T L_X$?

2.Is there a linear isomorphism $T:\Omega^i(M)/Z^i(M) \to \Omega^j(M)/Z^j(M)$ with $TL_X=L_XT$?

By $Z^i(M)$ we mean the space of closed differential $i$_form.

For a motivation please see the following post:

Fredholm index vs. Limit cycle theory

$\endgroup$
  • 1
    $\begingroup$ The spaces you consider need not be isomorphic, so in general, this answer is no. $\endgroup$ – Matthias Ludewig May 21 '17 at 15:24
  • $\begingroup$ @MatthiasLudewig What is a precise example of this non isomorphicity? $\endgroup$ – Ali Taghavi May 21 '17 at 16:34
1
$\begingroup$

For appropriate choices of $i$, $j$ (e.g. $i+j \neq n$), $\Omega^i(M)$ and $\Omega^j(M)$ have different ranks as $C^\infty(M)$ module, so at least they cannot be isomorphic as modules. Maybe they could be isomorphic as topological vector spaces, but I don't see why, and in any case, this would not be very natural.

If you choose $i=j$, then of course $T = \mathrm{id}$ works.

More generally, setting $T= \psi_t^*$ for any $t \in \mathbb{R}$ does the job, where $\psi_t$ is the flow of $X$, and $\psi_t^*$ denotes pullback via the flow.

$\endgroup$
  • $\begingroup$ @MittihasLudewig (+1) and thank you for your answers to my questions. I think that it is natural to compare $\Omega^{i}(M)$ and $\Omega^{j}(M)$ as two vector spaces not two modules, since $L_{X}$ is not a $C^{\infty}(M))$-linear operator. $\endgroup$ – Ali Taghavi May 22 '17 at 11:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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