Timeline for Problem arising in metrizability of connections: Simultaneously skewsymmetrizing matrices
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Jan 29, 2020 at 21:44 | vote | accept | Mike Cocos | ||
| Dec 1, 2019 at 23:24 | comment | added | Mike Cocos | I hope I'm not becoming annoying but there are more things related to the local metrizability. On an affinely flat manifold one can deform the flat connection $\nabla$ into any globally metric connection $D$ by the standard formula $$\nabla(t)=(1-t)\nabla+tD$$ If one can show that $\nabla(t)$ is locally metric for all $t$ then the Euler class of the manifold equals the Euler class of the flat connection and therefore is zero. This is the main reason I got interested in locally metric connections. | |
| Dec 1, 2019 at 23:15 | comment | added | Mike Cocos | A possible obstruction for the global metrizability of a connection in any vector bundle is obtained by calculating the Euler class of the connection. The Euler class of any locally metric connection is well defined via the Pfaffian of its curvature matrix. If its Euler class doesn't equal the Euler characteristic of the manifold then the connection cannot be globally metric. I believe that calculating the Euler class of the connection is much simpler than calculating its holonomy group. | |
| Dec 1, 2019 at 23:06 | comment | added | Mike Cocos | I am not sure what you mean by getting $U$ and $V$ to be unimodular? I am only interested in the local problem. If $U$ and $V$ skew-symmetrize the curvature matrix and since they are non singular just dividing by their determinant will make them unimodular. I am not even pretending to be trying to solve the problem on the whole manifold. All I am interested in is a practical way( using local frames) to determine if a connection is LOCALLY metric. I believe that if there are more than $ {n-1 \choose 2}$ linearly independent matrices among the $S_{ij}$ then algorithm works | |
| Dec 1, 2019 at 9:53 | comment | added | Robert Bryant | @MikeCocos: Actually, you need more than ${n-1}\choose2$ linearly independent matrices in order for this argument to work. However, your argument still doesn't address how you are going to get the matrices $U$ and $V$ to be unimodular. In other words, you still have to do an integration to determine the volume form of the metric. It can't be determined algebraically (and it may not be globally possible), as I pointed out in my example. | |
| Nov 30, 2019 at 22:05 | comment | added | Mike Cocos | Thank you Professor Bryant! Always to the point. So it turns out that if we take a a local frame $\sigma=(\sigma_1, \sigma_2,...,\sigma_m )$ and calculate the curvature matrix $\Omega$ with respect to this frame and then look at the matrices $S_{ij}$(with real smooth functions as entries) defined by the equation $$ \Omega=\sum_{i<j} (\sigma^i \wedge \sigma^j) S_{ij} $$ and if there are ${n-1\choose 2}$ linearly independent among them then the algorithm works. In particular since for $n=2$ there's only one matrix in that family the algorithm should work as well. | |
| Nov 29, 2019 at 12:25 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Added a remark about 'generic' generation of so(n).
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| Nov 29, 2019 at 10:10 | history | answered | Robert Bryant | CC BY-SA 4.0 |