Timeline for Inverse Metric Tensor
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Feb 18, 2015 at 3:12 | comment | added | Robin Ekman | Well, no matter the order of indices you can calculate the contraction in the form a matrix product. It may just be that you need to transpose one of the matrices. We choose to let $g$ act like $g_{\mu\nu} x^\nu$ and so on because then as you say the order is the same as when we do normal matrix multiplication. | |
| Feb 18, 2015 at 3:03 | vote | accept | Myridium | ||
| Feb 18, 2015 at 3:00 | comment | added | Myridium | Something I'm a bit confused about: "The compositions of these maps is a $(1,1) $ tensor.". Am I right in saying this: we desire an inverse map to $ g $ , and if such a thing existed, then its $ C^2,1$ contraction $ g g^{-1} $ (which happens to be expressed in the form of matrix multiplication) must be equal to the identity tensor. So we choose this particular contraction so that we may invoke matrix multiplication? | |
| Feb 18, 2015 at 2:37 | history | answered | Robin Ekman | CC BY-SA 3.0 |