Timeline for Definition of a spin group
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Oct 28, 2023 at 20:00 | comment | added | LSpice | @Kevin, re, but, as @Eric points out, the set of products of elements of the empty set is the singleton set containing the identity (because $\prod_{x \in \emptyset} x = 1$, which I was sure was an answer to Interesting examples of vacuous / void entities but doesn't seem to be), not the empty set. | |
| Oct 28, 2023 at 18:47 | answer | added | S. Carnahan♦ | timeline score: 5 | |
| Aug 5, 2022 at 13:15 | history | edited | Eric | CC BY-SA 4.0 |
Revert incorrect edit
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| Aug 5, 2022 at 10:58 | history | edited | YCor | CC BY-SA 4.0 |
formatting, added tag
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| Aug 4, 2022 at 17:22 | comment | added | Kevin | There aren't any elements with norm 1, so as is literally written $Pin(V)$ is empty. If you assume it at least contains 1, you get the trivial group. That's silly! | |
| Aug 4, 2022 at 16:59 | history | edited | Eric | CC BY-SA 4.0 |
added 71 characters in body
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| Aug 4, 2022 at 16:58 | comment | added | Eric | I don't follow; why do the later definitions preclude a group structure? I assume the empty product ($1$) of no elements of $V$ is still considered an element of the spin group. | |
| Aug 4, 2022 at 15:50 | comment | added | Kevin | Strange, it seems like the latter definitions must be assuming positive definiteness. Otherwise it doesn't even define a group (no identity), let alone a double cover of $SO$! | |
| Aug 4, 2022 at 15:40 | history | edited | Eric | CC BY-SA 4.0 |
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| S Aug 4, 2022 at 13:20 | review | First questions | |||
| Aug 4, 2022 at 13:24 | |||||
| S Aug 4, 2022 at 13:20 | history | asked | Eric | CC BY-SA 4.0 |