Timeline for definition of Hessian with respect to connection
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Feb 5, 2012 at 18:26 | comment | added | Vladimir Dotsenko | Morally, think of a tensor as of a multilinear function (well, a linear operator, to be precise). But you've to be a bit more careful, since you are working with manifolds, and things depend on the point where you evaluate stuff etc. A safe way to deal with this sort of stuff is to talk about sections of vector bundles. E.g., in our case, $A$ is a $C^\infty(M)$-linear operator from $\Gamma(TM)\otimes_{C^\infty(M)}\Gamma(T^*M)$ to $\Gamma(T^*M)$, where $\Gamma(E)$ denotes global sections of a bundle $E$ (e.g. $\Gamma(TM)$ would be vector fields). But your point-wise intuition is correct, yes. | |
| Feb 5, 2012 at 13:47 | comment | added | zatilokum | My last question is about $A$, what does it mean being a "tensor"? What is the difference being a tensor as real valued multilinear function on $V$ . For example, is domain of $A$ $T_pM \times T_p^\ast M$?. If so, it is not the same as real valued multilinear function. | |
| Feb 5, 2012 at 10:06 | comment | added | Vladimir Dotsenko | @zatilokum: basically, when you calculate the Hessian by the formula $\nabla_v(df)$ at the point $p$, you cannot yet say "since $df=0$ at $p$, we get zero, since $\nabla$ depends on more than mere behaviour at point $p$. (Since it is not a tensor.) However, the difference is a tensor (a linear construction from tangent/cotangent vectors <i>at each point</i>!), and so you are able to conclude $A(X,df)=A(X,0)=0$. This is actually a very good example to fully understand what the condition of being a tensor that differential geometers are so keen to check is good for! | |
| Feb 4, 2012 at 23:43 | vote | accept | zatilokum | ||
| Feb 4, 2012 at 23:42 | vote | accept | zatilokum | ||
| Feb 4, 2012 at 23:42 | |||||
| Feb 4, 2012 at 22:33 | comment | added | zatilokum | I know for any connection $\nabla$, $\nabla_X(0)= 0$. How $A(X, 0)=O$ give an information for independency of connection? | |
| Feb 4, 2012 at 15:23 | comment | added | Deane Yang | I agree with your latest comment, too! I was unable to appreciate properly the light and fresh air without being locked in the Christoffel prison first. | |
| Feb 4, 2012 at 13:47 | comment | added | Vladimir Dotsenko | Fair enough. Different people surely learn in different ways! I was scared to death by Christoffel symbols, and this explanation was like what one must feel enjoying the light and fresh air after having been locked in a basement for a while, or something of the sort. | |
| Feb 4, 2012 at 11:24 | comment | added | Deane Yang | I agree that in the long run your approach is the right and much clearer way to do things. And if you can understand it when you see it for the first time, then that's even better. But, alas, I didn't understand these formulas and calculations when I first saw them and only did after I unwound the formulas into local co-ordinates and watched the Christoffel symbols appear and disappear. | |
| Feb 4, 2012 at 9:34 | history | answered | Vladimir Dotsenko | CC BY-SA 3.0 |