4
$\begingroup$

I have the following situation: a smooth compact manifold $M$, without boundary and furnished with a smooth volume $\mu $.

I also have an operator defined on the space of smooth functions with $0$ average w.r.t $\mu$, let's call the space $C^{\infty}_{\mu}(M)$.

This operator is an isomorphism and extends to an isomorphism $H^{s} \rightarrow H^{s}$, for every $s \geq 0$. It also extends to an isometry on $L^{2}$.

My question is the following. Do there exist conditions under which such an operator is realized by pull-back? More formally, if an operator $T$ has the above properties (and eventually some additional ones) does there exist a volume preserving diffeomorphism of $M$, $h \in \mathrm{Diff}^{\infty}_{\mu}(M)$ such that for all $\varphi \in C^{\infty}_{\mu}(M)$ \begin{equation} T.\varphi = \varphi \circ h ^{-1} \end{equation}

I will probably end up giving an explicit construction of $h$ anyway, but an abstract theorem providing its existence would be helpful.

Thanks!

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ It could be $T\phi=2\phi$. Is $T$ an isometry on $H^s$? $\endgroup$
    – Ben McKay
    Commented Sep 19, 2017 at 10:22
  • $\begingroup$ good point. it is certainly an isometry on $L^{2}$. i think it is enough to exclude that case. (i edited the question). $\endgroup$
    – jesus
    Commented Sep 19, 2017 at 10:26

1 Answer 1

Reset to default
4
$\begingroup$

If (and only if) $T$ is also an algebra homomorphism, then it is a pullback by a diffeomorphism. See the introduction to chapter VIII of here.

One might be able to replace algebra homomorphism by order respecting properties of $T$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .