Timeline for Tensor contraction and Covariant Derivative
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 22, 2015 at 5:05 | history | edited | Michael Albanese | CC BY-SA 3.0 |
Changed < and > to \langle and \rangle.
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| Dec 17, 2013 at 0:47 | comment | added | Tom Goodwillie | As far as I'm concerned, the motivation is that the equivalent equation $\nabla(Tv)=(\nabla T)v+T(\nabla v)$ is a Leibniz rule. | |
| Dec 16, 2013 at 13:08 | comment | added | Vít Tuček | So what is the intuition or motivation for defining $(\nabla T)(v) = \nabla(Tv) - T(\nabla v)$ for $T\in \mathrm{Hom}(V,W)$? :) It is reminiscent of the definition of action of a group on the same space and thus it seems that the explanation boils down to naturality in the sense of Peter Michor's answer. | |
| Jan 26, 2013 at 17:13 | comment | added | Mariano Suárez-Álvarez | Ah, put that way it is pretty obvious! This gives some sort of explanation for the importance of contraction. (Another, connection-free way to put this «explanation» is that the identity map is obviously an important gadget, and contraction is just the identity up to transposition; if one likes one's tensors to be elements of $V^{\otimes r}\otimes V^{*\otimes s}$, as is classically done, then it is the closest to the identity that one gets) | |
| Jan 26, 2013 at 17:05 | comment | added | Tom Goodwillie | Yes, up to a locally constant factor. Equivalently the identity $V\to V$ is the only map, up to such a factor, that commutes with parallel transport for every connection on $V$. | |
| Jan 26, 2013 at 16:56 | comment | added | Mariano Suárez-Álvarez | (Up to a non-triviality condition, to exclude the zero map, I guess) | |
| Jan 26, 2013 at 16:55 | comment | added | Mariano Suárez-Álvarez | Is contraction the only map $V\otimes V^*\to\mathbb R$ which is parallel wrt every connection on $V$? | |
| Jan 26, 2013 at 16:47 | history | answered | Tom Goodwillie | CC BY-SA 3.0 |