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What is a precise example of the following situation:

A compact Riemanian manifold $M$ admits two vector field $X,Y$ such that the the operator $$Z\mapsto R(X,Y)Z$$ Would be an elliptic operator and the index of operator can be computed precisely?

In this question $R$ is the curvature tensor associated to the Riemannian metric.

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    $\begingroup$ This is a skew-symmetric endomorphism of the tangent bundle $TM$. As such it is elliptic if and only if it is a bundle isomorphism in which case the index is zero. Note that if $X,Y$ are collinear at a point the endomorphism is zero at that point. In particular it shows that for this operator to be invertible it is necessary that you can find two vector fields which are linearly independent at each point of $M$. This constrains the topology of $M$. For example, the Euler characteristic has to be zero. $\endgroup$
    – Liviu Nicolaescu
    Commented Apr 4, 2018 at 14:29
  • $\begingroup$ @DeaneYang yes. Before that I ask this question I did not pay attention to the fact that the curvature is tensorial in all its 3 variables.thank you for comments of you and Liviu. $\endgroup$
    – Ali Taghavi
    Commented Apr 4, 2018 at 16:33
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    $\begingroup$ Liviu’s answer can also be said as follows: the differential operator is zeroth order and therefore just a matrix-valued function of the manifold. It is elliptic if and only if the matrix is invertible. $\endgroup$
    – Deane Yang
    Commented Apr 4, 2018 at 16:57

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