Timeline for Diagonalization of a matrix of differential operators
Current License: CC BY-SA 3.0
6 events
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| Sep 22, 2011 at 8:27 | vote | accept | Alexander Vais | ||
| Sep 2, 2011 at 8:24 | comment | added | Alexander Vais | Here is the other question. It can be summarized as: Is the 1-form Bochner Laplacian conjugate to the double of some scalar operator? The De Rham Laplacian is. Unfortunately I could not find any reference discussing this issue. | |
| Sep 1, 2011 at 15:00 | comment | added | Robert Bryant | Thanks for the correction. Actually, my mistake was the typo of putting the minus sign in the wrong place in the formula for the explicit solution $A$. I've fixed this now. By the way, I should say that the point is that your operator is conjugate to the double of a scalar operator on the nontrivial line bundle over the circle. What is this `other question' that you refer to? | |
| Sep 1, 2011 at 14:56 | history | edited | Robert Bryant | CC BY-SA 3.0 |
fixed typos and grammar
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| Sep 1, 2011 at 14:40 | comment | added | Alexander Vais |
This is a nice solution (minor correction: should be $A^{-1} M A$ since $AMA^{-1}=-\partial^2 + 2J \partial -\frac{3}{4}$). Indeed, your $m$ has the correct eigenvalues if antiperiodic boundary conditions are applied. So by going to the twisted bundle, the problem simplifies as we can reduce the rank from $2$ to $1$. Still this is not quite the solution I was hoping for, because I cannot see how to generalize it. My original motivation was my other question where I was wondering if there exists a "rank 1" version of the Bochner Laplacian (defined on the rank 2 bundle $T^*M$) on a surface.
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| Aug 31, 2011 at 12:15 | history | answered | Robert Bryant | CC BY-SA 3.0 |