Timeline for definition of Hessian with respect to connection
Current License: CC BY-SA 3.0
12 events
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| Feb 7, 2012 at 13:52 | comment | added | Pait | Keep in mind that the definition H(f,p)= \nabla_v(df) does depend on the connection at an arbitrary point. This is very useful because you may want a second derivative tensor defined everywhere, not only at critical points, and to accomplish that a connection is needed. At a critical point where df vanishes, the Hessian is indeed independent of the choice of connection. | |
| Feb 7, 2012 at 8:56 | history | edited | agt |
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| Feb 7, 2012 at 8:56 | history | edited | agt |
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| S Feb 4, 2012 at 23:43 | vote | accept | zatilokum | ||
| Feb 4, 2012 at 23:42 | vote | accept | zatilokum | ||
| S Feb 4, 2012 at 23:43 | |||||
| Feb 4, 2012 at 23:42 | vote | accept | zatilokum | ||
| Feb 4, 2012 at 23:42 | |||||
| Feb 4, 2012 at 11:27 | comment | added | Deane Yang | The key point is that the difference of the two Hessians (defined by two different connections) is a linear function of $df$. This follows from the fact that the difference of two connections is a tensor, but you might as well verify this by a direct computation. | |
| Feb 4, 2012 at 10:50 | answer | added | BS. | timeline score: 8 | |
| Feb 4, 2012 at 9:34 | answer | added | Vladimir Dotsenko | timeline score: 15 | |
| Feb 4, 2012 at 2:33 | comment | added | J. GE | Assume $\nabla_i$ are two connections for $i=1,2$. You can calculate $(\nabla_1 -\nabla_2)_XY$ and you will see it's a tensor. | |
| Feb 3, 2012 at 23:47 | comment | added | Deane Yang | Try writing everything out carefully in local co-ordinates. Do a change of co-ordinates and figure out how everything changes. | |
| Feb 3, 2012 at 23:24 | history | asked | zatilokum | CC BY-SA 3.0 |