Timeline for When does the shape operator commute with a derivative?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Nov 8, 2023 at 17:51 | vote | accept | Bumblebee | ||
| S Sep 2, 2023 at 3:45 | history | suggested | Ali Taghavi |
I add two tags
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| Sep 1, 2023 at 20:45 | review | Suggested edits | |||
| S Sep 2, 2023 at 3:45 | |||||
| Aug 8, 2023 at 5:05 | history | edited | LSpice | CC BY-SA 4.0 |
Typo
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| Aug 8, 2023 at 0:15 | history | became hot network question | |||
| Aug 7, 2023 at 23:29 | answer | added | Robert Bryant | timeline score: 18 | |
| Aug 7, 2023 at 14:20 | comment | added | Mikhail Katz | As Shifrin mentioned there in a comment, the key is to look at the eigenvectors. Note that the shape operator is selfadjoint and therefore diagonalizable. Your commutation condition implies that $d\phi$ sends eigenvectors to eigenvectors, and moreover the corresponding eigenvalues must be equal. | |
| Aug 7, 2023 at 14:07 | history | edited | Bumblebee | CC BY-SA 4.0 |
added 3 characters in body
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| Aug 7, 2023 at 13:51 | comment | added | Bumblebee | @DanielAsimov: Noted, and edited accordingly. | |
| Aug 7, 2023 at 13:48 | history | edited | Bumblebee | CC BY-SA 4.0 |
deleted 2 characters in body; edited title
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| Aug 7, 2023 at 13:47 | comment | added | Daniel Asimov | I would use the word "derivative" rather than "differential" for what you mean. And denote it with a capital D, not a lowercase one, which has a different meaning. | |
| Aug 7, 2023 at 13:42 | history | asked | Bumblebee | CC BY-SA 4.0 |