Timeline for The normalizer of SU(n) in U(m)?
Current License: CC BY-SA 4.0
26 events
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| Feb 25, 2021 at 17:53 | comment | added | wonderich | Sorry, now I think that the normalizer of this SU(5) inside Spin(10) at least contains a SU(5) $\times \mathbb{Z}/4$ where $\mathbb{Z}/4$ is a center of Spin(10). | |
| Feb 25, 2021 at 14:22 | comment | added | Will Sawin | @wonderich Please do not ask me the same question twice. I am not sure what the answer is. | |
| Feb 25, 2021 at 2:01 | comment | added | wonderich | @Will Sawin: the normalizer of this SU(5) inside U(16) is U(5) × U(1)$^2$. Suppose we ask differently: the normalizer of this SU(5) inside Spin(10) [instead of U(16)], do we get U(5)? | |
| Feb 23, 2021 at 21:28 | comment | added | wonderich | @Will, thanks so much! the normalizer of this SU(5) inside U(16) is U(5) $\times$ U(1)$^2$. Suppose we ask differently: the normalizer of this SU(5) inside Spin(10) [instead of U(16)], do we get U(5)? or do we get extra how many U(1) factors still? (I just wanted to make sure I did not miss your logic.) What precisely it is? | |
| Feb 23, 2021 at 20:58 | comment | added | Will Sawin | @wonderich Maybe this helps? | |
| Feb 23, 2021 at 20:58 | history | edited | Will Sawin | CC BY-SA 4.0 |
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| Feb 23, 2021 at 20:18 | comment | added | wonderich | (I hope you can correct me on your logic --- I am sorry to possibly mix up your logic in that paragraph.) | |
| Feb 23, 2021 at 20:17 | history | edited | wonderich | CC BY-SA 4.0 |
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| Feb 23, 2021 at 20:00 | history | edited | wonderich | CC BY-SA 4.0 |
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| Feb 23, 2021 at 19:59 | comment | added | wonderich | Excuse me, could we break your whole-sentence paragraph into two sentences to make it clear about the logic? This is rather long to follow... "The kernel of this natural map, for any πΊβπ», is πΊπΆπ»(πΊ), where πΆπ»(πΊ) is the centralizer of πΊ in π», since every element of the kernel acts by conjugation as an inner automorphism, i.e. is an element of πΊ times something that acts trivially by conjugation, i.e. is an element of πΊ times something in its centralized. So the normalizer is πΊπΆπ»(πΊ)." | |
| Feb 23, 2021 at 19:18 | history | edited | Will Sawin | CC BY-SA 4.0 |
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| Feb 23, 2021 at 19:17 | comment | added | Will Sawin | @wonderich Sorry, by $Z(G)$ I meant the centralizer, and by $G Z(G)$ the set of all products of an element of $G$ with an element of the centralizer of $G$. | |
| Feb 23, 2021 at 19:16 | comment | added | wonderich | Can I make sure the way you defined $GZ(G)$? The $Z(G)$ is usually the center but now the centralizer. And $GZ(G)$ is? | |
| Feb 22, 2021 at 22:50 | vote | accept | wonderich | ||
| Feb 22, 2021 at 22:50 | comment | added | wonderich | Thanks for your answer, I accepted your wonderful answer. I ask another one which is more subtle mathoverflow.net/questions/384694, maybe it is obvious to you. | |
| Feb 21, 2021 at 22:35 | comment | added | wonderich | @Will Sawin, thanks so much. I have a naive question, suppose we take a special unitary subgroup of $SU(5)$ which contains both the $U(2)=U(1)\times SU(2)/\mathbb{Z}/2$ and $U(3)=U(1)\times SU(3)/\mathbb{Z}/3$ but mod out the shared $\mathbb{Z}/2$ and $\mathbb{Z}/3$ subgroups between two of $U(1)$. Define this special unitary group as $S(U(2)\times U(3))$ which is the subgroup of $SU(5)$. What would be the normalizer of this $S(U(2)\times U(3))$ within $U(16)$? | |
| Feb 21, 2021 at 20:11 | comment | added | Will Sawin | @wonderich Yes, because the centralizer consists of matrices that act as scalars on each irreducible representation (Schur's lemma), and thus is $U(1) \times U(1) \times U(1)$ (and because of the outer automorphism argument). | |
| Feb 21, 2021 at 20:09 | comment | added | wonderich | I think you are correct. Thanks -- and from your answer, it also means that there are no additional discrete sectors like $\mathbb{Z}/N$ outside the U(5)×U(1)×U(1)? Thank you so much! (I will accept the answer in 1 week if all minor issues are clear to me) | |
| Feb 19, 2021 at 14:00 | comment | added | Will Sawin | @wonderich Do you disbelieve the argument in my previous comment? | |
| Feb 19, 2021 at 4:32 | comment | added | wonderich | @Will Sawin, I was wondering whether the π(5) and two π(1) share a common $\mathbb{Z}/2$, so we need to mod out $\mathbb{Z}/2$ twice...? | |
| Feb 19, 2021 at 2:28 | history | edited | LSpice | CC BY-SA 4.0 |
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| Feb 19, 2021 at 2:22 | comment | added | Will Sawin | @LSpice I mean $\overline{\bf 5} \oplus {\bf 10} \oplus {\bf 1}$. | |
| Feb 19, 2021 at 2:19 | comment | added | LSpice | What is "this representation" in "an isomorphism betwee this representation and the complex conjugate"? | |
| Feb 19, 2021 at 1:35 | comment | added | Will Sawin | @wonderich Yes, since every nontrivial element of $U(5)$ acts nontrivially on the five-dimensional representation, but every element of $U(1) \times U(1)$ acts trivially on that representation. | |
| Feb 19, 2021 at 1:33 | comment | added | wonderich | Thanks very much for the quick answer! +1. It is possible, are you sure that there are no shared finite normal subgroup between each of $U(5)$ and two $U(1)$? | |
| Feb 19, 2021 at 1:26 | history | answered | Will Sawin | CC BY-SA 4.0 |