1
$\begingroup$

Assume that $X$ is a non vanishing vector field on the torus $\mathbb{T}^2$.

We define two linear operators $T,S$ on the space of smooth functions on $\mathbb{T}^2$ as follows:

$T(f)=X.f$
$S(f)= *d(\alpha_{f})$ where $*$ is the Hodge operator and $\alpha_{f}$ is the $1\; \_$ form on the torus with $\alpha_{f}(h)=fX.h$ the later is based on a Riemannian metric on the torus.

Is $D=T^2+S^2$ an elliptic operator?If yes is its index depend on the vector field $X$ and a Riemannain metric on the torus?If the index depends on $X$, are there some dynamical interpretation for the index of this operator?

Improve this question
$\endgroup$
10
  • $\begingroup$ so $h$ is a vector field right ? $\endgroup$
    – Thomas Richard
    Jun 22, 2017 at 8:16
  • $\begingroup$ @ThomasRichard Yes it is a tangent vector. $\endgroup$
    – Ali Taghavi
    Jun 22, 2017 at 8:18
  • 4
    $\begingroup$ Computations show that $Df=|X|_g^2\Delta_g f+\dots$ where the terms in the dots depend on $f$ and $df$. So ellipticity is ok since your vector field is non vanishing, as for the index I have no idea. $\endgroup$
    – Thomas Richard
    Jun 22, 2017 at 8:36
  • 2
    $\begingroup$ What kind of index are you thinking of? The operator described by @ThomasRichard has selfadjoint principal symbol, so the classical definition (dim ker - dim coker) does not give anything interesting. $\endgroup$
    – Sebastian Goette
    Jun 24, 2017 at 9:04
  • 1
    $\begingroup$ In some cases, there are exotic indices, e.g. the parity of dim ker could be an invariant. Please check the book by Lawson-Michelsohn for details. $\endgroup$
    – Sebastian Goette
    Jun 25, 2017 at 7:24

0

Reset to default

Your Answer

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