How to compute the index of such operator?

Question

Let $M$ be a compact Riemannian manifold, with $R$ nowhere-vanishing vector field on $M$(whose orbit may be closed/ not closed). $E$ and $F$ are two vector bundle (Edit: which are sub-bundles of $\Omega^*(M, \mathfrak{g})$, namely forms valuded in Lie-algebra $\mathfrak{g}$) over $M$ with the same rank. Denote $\tilde \Gamma(E)$ and $\tilde \Gamma(F)$ the sections of $E$ and $F$ that are invariant under $R$, namely, for instance

$\mathcal{L}_R \tilde \sigma_{E} = 0$.

Let $D$ be some differential operator $D: \tilde \Gamma(E) \to \tilde \Gamma(F)$, which obviously commutes with $\mathcal{L}_R$.

My target $\dim{\rm ker}D - \dim \ker D^*$. What is the corresponding framework to do this? What is the property of $D$ for this quantity to be well defined? Any reference will be appreciated!(I have basic knowledge about usual index-theorem for elliptic complex)

Thanks.

Answer 1

This is a long comment rather than a full answer.

The vector field $R$ generates a one-dimensional foliation on $M$. Therefore, the spaces $\tilde\Gamma(E)$, $\tilde\Gamma(F)$ may be very small, in the worst case $0$, giving index $0$ independent of $D$. In the other extreme, the space $B$ of leaves of $R$ could be a nice Hausdorff space, maybe even a manifold. In this case, depending on the $R$-actions on $E$ and $F$, $\tilde\Gamma(E)$ and $\tilde\Gamma(F)$ could actually be sections of vector bundles on $B$. Then $D$ becomes a differential operator on these bundles, and you will need an ellipticity condition on $D$ to obtain a well-defined index. Note that even small perturbations of $R$ could bring you from one extreme to the other.

For this reason, I suggest that you take a look at index theory for transversally elliptic operators on foliations. I am not an expert here, but maybe Kordyukov's book is a good starting point.

Answer 2

Maybe this helps a little: Locally, each $\tilde \sigma$ is uniquely determined by its values along a slice for the $R$-action; globally this is the orbit space of the $R$-action which might be non-Hausdorff, but it has charts. So $D$ is defined locally there and it has to be elliptic. If $D$ is self-adjoint with respect to any $R$-invariant metric and volume form, its index is zero. If not, you may try to deform the vector field to some canonical form, use homotopy invariance for the index, and try to compute it for the special field.