Timeline for Ellipticity of certain differential operator associated to a pair of vector field via curvature tensor
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 4, 2018 at 16:57 | comment | added | Deane Yang | 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. | |
| Apr 4, 2018 at 16:33 | comment | added | Ali Taghavi | @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. | |
| Apr 4, 2018 at 14:29 | comment | added | Liviu Nicolaescu | 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. | |
| Apr 4, 2018 at 8:27 | history | asked | Ali Taghavi | CC BY-SA 3.0 |