Timeline for Finite subgroup of $\operatorname{Sp}(2n,K)$
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 15, 2023 at 10:16 | comment | added | user488802 | Sorry, why a unique symplectic form on V implies a unique conjugacy class? | |
| Sep 15, 2023 at 2:36 | comment | added | LSpice | @JoshuaMundinger, re, ah, right, thanks. | |
| Sep 15, 2023 at 2:02 | comment | added | Joshua Mundinger | @LSpice The question is a little ambiguous. If we take $Q/Z(G)$ in the statement to mean $Q/Q \cap Z(G)$, then $Q/Q \cap Z(G)$ being abelian means $-1 \in Q$ maps into $Z(G)$. | |
| Sep 15, 2023 at 0:11 | comment | added | LSpice | @JoshuaMundinger, re, that was my question: why must $Z(Q)$ map to $Z(G)$? Certainly it must for an irreducible representation, but I don't see it in general. | |
| Sep 14, 2023 at 23:53 | comment | added | Joshua Mundinger | @user488802 the tensor product of two vector spaces $V^* \otimes \rho$ with alternating bilinear forms carries a symmetric bilinear form: the tensor product of the forms. | |
| Sep 14, 2023 at 23:49 | comment | added | Joshua Mundinger | @LSpice $Z(G) = \{\pm 1\}$ so an injective homorphism sending $Z(Q)$ into $Z(G)$ must send $-1$ to $-1$. | |
| Sep 14, 2023 at 22:50 | history | edited | Kenta Suzuki | CC BY-SA 4.0 |
added 249 characters in body
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| Sep 14, 2023 at 21:54 | comment | added | LSpice | I don't understand—why couldn't an Abelian group like $Q/Z(G)$ have an element $-1$ acting trivially on some components? (Clearly not on all, since the homomorphism is injective.) | |
| Sep 14, 2023 at 21:53 | history | edited | LSpice | CC BY-SA 4.0 |
`\operatorname`
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| Sep 14, 2023 at 20:59 | history | answered | Kenta Suzuki | CC BY-SA 4.0 |