Timeline for Tensor Operators
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Sep 10, 2013 at 8:30 | comment | added | joshphysics | @zodiac Notice that $U$ is a representation of the group $G$ on a Hilbert space; it can potentially be a much different beast than $\rho$ which is a representation of $G$ on a finite-dimensional vector space. In particular, it need not be generated by the representation $\rho$ in the way you describe. | |
| Sep 10, 2013 at 7:06 | comment | added | xuanji | In your question I did notice that $U$ and $\rho$ are constrained to both be representations of the same underlying group, but in the standard formulation, the transformation law of the vectors (the $\rho$) completely determines the transformation law of the tensors (the $U$), so I was wondering if there were some inconsistency. Unfortunately I'm not familiar enough with group representations to be sure. | |
| Sep 10, 2013 at 7:01 | comment | added | joshphysics | You've simply described the standard formulation of tensors as multilinear maps that is used both in differential geometry and in algebra; I am well-aware of that stuff. Notice, in fact, that my candidate definition above is written precisely in terms of multilinear maps which is what you're describing here. The notion I want to define here, however, is a bit less straightforward than that I'm afraid. | |
| Sep 10, 2013 at 6:52 | history | edited | xuanji | CC BY-SA 3.0 |
added 5 characters in body
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| Sep 8, 2013 at 20:35 | comment | added | user4552 | This generalizes straightforwardly from rotation to the Lorentz group, and can also be generalized to arbitrary diffeomorphisms. | |
| Sep 8, 2013 at 6:57 | history | answered | xuanji | CC BY-SA 3.0 |