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

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Aug 22, 2021 at 16:40	comment	added	Марина Marina S		Thanks, I also am not sure whether the other map (mod 4) gives an embedding. The reason I asked that is due to eq (4) and the other question as an input mathoverflow.net/q/402173/336737
Aug 22, 2021 at 15:33	comment	added	Robert Bryant		@МаринаMarinaS: I don't really understand your question. When $k\ge 1$, the intersection of the subgroups $\mathrm{Spin}(4k{+}2)$ and the central subgroup $\mathrm{U}(1)\subset\mathrm{U}(2^{2k})$ is the set of $\lambda I$ where $\lambda^4 = 1$, so the first embedding works. However, I don't see how you can get the second map as an embedding. The $\mathrm{U}(1)$ subgroup in $\mathrm{U}(2^{2k})$ has to commute with the subgroup $\mathrm{Spin}(4k{+}2)$, which forces it to be central (i.e., multiples of the identity), and that subgroup intersects $\mathrm{Spin}(4k{+}2)$ in a group of order $4$.
Aug 22, 2021 at 13:14	comment	added	Марина Marina S		"Finally, for 𝑛≥1 there is an embedding Spin(4𝑛+2)↪SU(2^{2𝑛})." I agree with this. I think you agree $$(𝑆𝑝𝑖𝑛(4𝑘+2)×𝑈(1))𝐙/4→𝑈(2^{2𝑘}).$$ But I also ask whether this is true $$(𝑆𝑝𝑖𝑛(4𝑘+2)×𝑈(1))𝐙/2→𝑈(2^{2𝑘})?$$
Aug 21, 2021 at 21:54	comment	added	annie marie cœur		Hi Robert and Marina, I like your post. I find this answer may be partly supporting but also partly contradicting with part of the answer given here: math.stackexchange.com/a/3296484
Aug 21, 2021 at 19:21	comment	added	Марина Marina S		I also have this puzzle mathoverflow.net/q/402173/336737 whether you can shed light on it too?
Aug 21, 2021 at 19:19	comment	added	Марина Marina S		thanks for the refs "F. Reese Harvey's book Spinors and Calibrations" I will read (son)!
Aug 21, 2021 at 19:19	comment	added	Марина Marina S		Thanks, but I also asked extra U(1) factor with appropriate mod structure. Could you illuminate on that too?
Aug 21, 2021 at 19:00	history	edited	Robert Bryant	CC BY-SA 4.0	Fixed a few typos and infelicitous phrasings.
Aug 21, 2021 at 18:44	history	answered	Robert Bryant	CC BY-SA 4.0