4}\to U(2^{2k})$

Current License: CC BY-SA 4.0

10 events

when toggle format	what		by	license	comment
Aug 22, 2021 at 16:37	history	edited	Марина Marina S	CC BY-SA 4.0	added 8 characters in body
Aug 22, 2021 at 0:48	history	edited	Марина Marina S	CC BY-SA 4.0	added 4 characters in body
Aug 21, 2021 at 18:44	answer	added	Robert Bryant		timeline score: 2
Aug 21, 2021 at 18:31	comment	added	Марина Marina S		YES, "A representation 𝜌 of Spin(2𝑛,ℂ) is called spinoral if it does not descend to the orthogonal group SO(2𝑛,ℂ) " --> so here I meant that this injective group homomorphism is pinned down by the spinor or spinoral representation of Spin(2𝑛,ℂ), no the vector representation of SO(2𝑛,ℂ).
Aug 21, 2021 at 18:18	comment	added	LSpice		I emphasise "descend to $\operatorname{SO}(2n, \mathbb C)$" because that's where you made the claim: "[$\rho$] does not descend to the orthogonal group $\operatorname{SO}(2n, \mathbb C)$ (equivalently, $\rho$ is injective)". But I would be equally sceptical of the real version of the claim; I would definitely believe it on the level of central characters, but it's not clear to me on the level of the entire representation.
Aug 21, 2021 at 17:56	comment	added	Марина Marina S		I am not sure why you emphasize "descend to SO(2𝑛,ℂ)". Here I focus on to Spin(2𝑛)=Spin(2𝑛;ℝ). I put a Note p.s. on Spin(2𝑛,ℂ) just for the convenience of sufficient studying complex representations
Aug 21, 2021 at 17:41	comment	added	LSpice		Why is $\rho$ injective if and only if it does not descend to $\operatorname{SO}(2n, \mathbb C)$? Do you mean to require only the central character of $\rho$ to be injective?
Aug 21, 2021 at 17:41	history	edited	LSpice	CC BY-SA 4.0	`\eqref` and `\DeclareMathOperator`
Aug 21, 2021 at 17:13	history	edited	Марина Marina S	CC BY-SA 4.0	edited title
Aug 21, 2021 at 17:05	history	asked	Марина Marina S	CC BY-SA 4.0