Timeline for Exterior square of $\operatorname{Sp}(4,\mathbb{C})$ is isomorphic to $\operatorname{SO}(5,\mathbb{C})$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Aug 23, 2022 at 15:52 | vote | accept | mhahthhh | ||
| Aug 22, 2022 at 20:04 | history | edited | LSpice | CC BY-SA 4.0 |
Capitalise title (forgot to do it in my previous edit)
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| Aug 22, 2022 at 18:20 | answer | added | David E Speyer | timeline score: 7 | |
| Aug 22, 2022 at 17:36 | review | Close votes | |||
| Aug 28, 2022 at 3:01 | |||||
| Aug 22, 2022 at 16:32 | history | edited | LSpice | CC BY-SA 4.0 |
TeX
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| Aug 22, 2022 at 16:29 | comment | added | LSpice | … Then you try to show this using an isomorphism of ${\bigwedge}^2\mathbb C^4$ with $\mathbb C^5$ that intertwines the actions, but of course there isn't one, because ${\bigwedge}^2\mathbb C^4$ is 6D and $\mathbb C^5$ is 5D. Instead one way to proceed is to consider the 5D representation ${\bigwedge}^2(\mathbb C^4)^*/\mathbb C\omega$, where $\omega$ is the symplectic form—which is preserved by $G$ by definition! | |
| Aug 22, 2022 at 16:28 | comment | added | LSpice | As written, your equation doesn't make sense; one cannot take the exterior square of a group, and I think you don't mean to take the exterior square of the Lie algebra. I suspect what you mean is that $G = \operatorname{Sp}(4, \mathbb C)$ is isomorphic to $\operatorname{Spin}(5, \mathbb C)$—or you could say that it has a mapping onto $H = \operatorname{SO}(5, \mathbb C)$ that annihilates the centre. ($G$ and $H$ are not isomorphic.) … | |
| Aug 22, 2022 at 16:23 | history | asked | mhahthhh | CC BY-SA 4.0 |