4}$ for $n=2k+1$

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Aug 27, 2021 at 13:58	comment	added	Michael Albanese		@wonderich: I do not. That's the first case which isn't covered by my answer.
Aug 25, 2021 at 19:52	comment	added	wonderich		Dear all, Michael and Марина, do any of you know whether $𝑈(5)\subset (𝑆𝑝𝑖𝑛(10)×𝑈(1))/𝐙/4$ is true or false? This is the same $𝐙/4$ as the center of both 𝑆𝑝𝑖𝑛(10) and 𝑈(1). This is your $n=5$ and $k=2$ case. Thanks!
Aug 25, 2021 at 18:31	comment	added	Michael Albanese		In this situation, if you want the diagram to commute, the map $U(n) \to G$ is fixed, it has to be $\varphi\circ\tilde{l}$. So there is an embedding $U(n) \to G$ which makes the diagram commute if and only $\varphi\circ\tilde{l}$ is an embedding. If you don't care about the diagram commuting, then you are just asking if there is an embedding $U(n) \to G$ which is a more general question.
Aug 25, 2021 at 18:31	comment	added	Michael Albanese		If $[(\omega, i)] \in \tilde{l}(U(n))$, then there is $A \in U(n)$ such that $\tilde{l}(A) = [(\omega, i)]$. As my previous comment shows, if such an $A$ were to exist, the map $\varphi\circ\tilde{l}$ would not be injective. It is possible to have an embedding $U(n) \to G$ for $k$ even while the diagram doesn't commute. This is exactly the point I was trying to make by discussing the $k = 0$ case in my answer.
Aug 25, 2021 at 18:03	comment	added	Марина Marina S		(4) can you illuminate more on the final diagram, is it possible to have an embedding for 𝑈(𝑛)→𝐺 for 𝑘 even and $n=2k+1$, while the diagram doesn't commute? Conversely, if the the diagram commutes, do we have the embedding? In short, are there necessary or sufficient conditions between the "diagram commute or not" or "embedding"?
Aug 25, 2021 at 17:41	comment	added	Марина Marina S		(3) In my above, I wrote $T=SU(n)$. Is your notation $𝑙̃ (𝑇)=[(𝜔,𝑖)]$ the same notation of $T$ as mine $T=SU(n)$? thanks! or is that your just some $T \in U(n)$?
Aug 25, 2021 at 17:27	comment	added	Марина Marina S		(2) $A \subset U(n)$?
Aug 25, 2021 at 17:24	comment	added	Марина Marina S		thanks, (1) what is your $A$?
Aug 25, 2021 at 16:53	comment	added	Michael Albanese		If $[(\omega, i)] = \tilde{l}(A)$, then $(\varphi\circ\tilde{l})(A) = \varphi(\tilde{l}(A)) = \varphi([(\omega, i)]) = [(1, 1)]$ which is the identity of $G$, so $A \in \ker(\varphi\circ\tilde{l})$, so $\varphi\circ\tilde{l}$ is not injective.
Aug 25, 2021 at 16:35	comment	added	Марина Marina S		Why not the other way around? "if and only if [(𝜔,𝑖)] $\in$ 𝑙̃ (𝑈(𝑛)) " ?
Aug 25, 2021 at 16:35	comment	added	Марина Marina S		Thanks! I think we have: $$𝜑∘𝑙̃ :𝑈(𝑛)→𝐺$$ $$𝜑:𝑆𝑝𝑖𝑛𝑐(2𝑛)→𝐺$$ $$𝑙̃ :𝑈(𝑛)→𝑆𝑝𝑖𝑛𝑐(2𝑛)$$ to begin with. Could you then clarify this sentence: "Since 𝑙̃ is an embedding (based on the ABS paper), 𝜑∘𝑙̃ is injective if and only if [(𝜔,𝑖)]∉𝑙̃ (𝑈(𝑛))." ? Thanks!
Aug 25, 2021 at 16:27	comment	added	Michael Albanese		Yes, $(\omega, i) \in Spin(2n)\times U(1)$, and $\langle(\omega, i)\rangle$, the subgroup generated by $(\omega, i)$, is isomorphic to $\mathbb{Z}/4$. The image of $(\omega, i)$ under the map $Spin(2n)\times U(1) \to Spin^c(2n)$ is denoted $[(\omega, i)]$. The kernel of the map $\varphi : Spin^c(2n) \to G$ is $\langle[(\omega, i)]\rangle$, which is isomorphic to $\mathbb{Z}/2$. I'd say this notation is pretty standard.
Aug 25, 2021 at 16:17	comment	added	Марина Marina S		May I confirm your notation, is this ⟨(𝜔,𝑖)⟩≅ℤ/4? and the ⟨[(𝜔,𝑖)]⟩ is the kernel of the map 𝜑:𝑆𝑝𝑖𝑛𝑐(2𝑛)→𝐺 ? Are these notations ⟨(𝜔,𝑖)⟩ and ⟨[(𝜔,𝑖)]⟩ standard?
Aug 23, 2021 at 17:59	history	edited	Michael Albanese	CC BY-SA 4.0	added 385 characters in body
Aug 22, 2021 at 21:54	history	edited	Michael Albanese	CC BY-SA 4.0	added 62 characters in body
Aug 22, 2021 at 16:25	history	answered	Michael Albanese	CC BY-SA 4.0

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