Timeline for Tangent bundle of a tensor product bundle
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 4 at 7:59 | comment | added | Keeley Hoek | Under the identification $T(E \otimes F) \cong E \otimes TF$ (in the sense meant above), I believe the connector is actually just $K_{E \otimes F} = \operatorname{id}_E \otimes K_F : E \otimes TF \to E \otimes F$. The reason is that the connection on $E$ is used in the identification $E \otimes TF \cong T(E \otimes F)$ in the first place. I have sorted this out in my answer over here on math.se (at the bottom). | |
| Apr 13, 2022 at 12:55 | vote | accept | Raz Kupferman | ||
| Apr 12, 2022 at 20:01 | history | bounty ended | Raz Kupferman | ||
| Apr 12, 2022 at 8:25 | history | edited | Peter Michor | CC BY-SA 4.0 |
added 82 characters in body
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| Apr 12, 2022 at 7:03 | history | edited | David Roberts♦ | CC BY-SA 4.0 |
small maths fix
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| Apr 12, 2022 at 7:01 | history | edited | Peter Michor | CC BY-SA 4.0 |
answered questions in the comments
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| Apr 12, 2022 at 5:55 | history | edited | Peter Michor | CC BY-SA 4.0 |
added 1156 characters in body
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| Apr 12, 2022 at 4:27 | comment | added | Raz Kupferman | Thanks Peter. Two issues: (1) under this angle, it is no longer transparent that it is a vector bundle over $E\otimes F$. (2) How do horizontal bundles of $TE$ and $TF$ define a horizontal bundle of $T(E\otimes F)$? | |
| Apr 11, 2022 at 19:23 | history | answered | Peter Michor | CC BY-SA 4.0 |