Timeline for A lie Subgroup of SO(4n)
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jan 11, 2013 at 12:54 | comment | added | Robert Bryant | It's because $\mathrm{Sp}(n)\times\mathrm{Sp}(1)$ is the product of two (almost) simple groups, so there is no nontrivial homomorphism of this product into $\mathrm{T}^1$. This implies that the embedding $$\mathrm{Sp}(n)\times\mathrm{Sp}(1)\longrightarrow \mathrm{SO}(4n)\times \mathrm{T}^1$$ must actually go into $\mathrm{SO}(4n)\times \lbrace e\rbrace$. | |
| Jan 11, 2013 at 11:10 | comment | added | user30435 | I'm sorry for this probably easy question: How does an embedding of $Sp(n)\times Sp(1)$ as a subgroup of $SO(4n)\times T^1$ yield being a subgroup of $SO(4n)$? | |
| Jan 3, 2013 at 17:52 | comment | added | Robert Bryant | @Nerd-Math: Oh, sorry. I didn't read the question carefully. The answer is still 'no' because such an inclusion would cause an embedding of $\mathrm{Sp}(n)\times\mathrm{Sp}(1)$ as a subgroup of $\mathrm{SO}(4n)\times T^1$, and this is not possible, since it would have to be a subgroup of $\mathrm{SO}(4n)$, which is clearly impossible. | |
| Jan 3, 2013 at 16:55 | comment | added | Nrd-Math | Pardon me if I am wrong. Your argument implies $Sp(n)\times Sp(1).T^1$ is not a Lie subgroup of $SO(4n)$. Isn't it? whereas the question is about $SO(4n)\times T^1$. | |
| Jan 3, 2013 at 15:52 | vote | accept | Nrd-Math | ||
| Jan 3, 2013 at 12:45 | comment | added | Robert Bryant | @Nerd-Math: Look up Schur's Lemma; it's basic. Maybe I don't understand what your $\mathrm{Sp}(1).T^1$ means, but I assume that you mean a subgroup of $\mathrm{SO}(4n)$ that commutes with the elements of $\mathrm{Sp}(n)$ and that contains both a copy of $\mathrm{Sp}(1)$ and a separate torus $T^1$ (i.e., a circle). However, the ring of $\mathbb{R}$-linear maps of $\mathbb{H}^n$ to itself that commute with all the elements of $\mathrm{Sp}(n)$ (i.e., its commuting ring) is the linear maps that come from multiplying by elements of $\mathbb{H}$ on the right, so the extra $T^1$ is not possible. | |
| Jan 3, 2013 at 11:05 | comment | added | Nrd-Math | Thanks a lot for your helpful comments. I guess there is a misunderstanding in the second question. There, I mean $Sp(n)\times (Sp(1).T^1)$ not $(Sp(n)\times (Sp(1)).T^1$. Could you introduce me a good reference for what you,ve used in your comments, for example "commuting ring of an irreducible representation", please. | |
| Jan 2, 2013 at 13:23 | history | edited | Robert Bryant | CC BY-SA 3.0 |
added a sentence of explanation
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| Jan 2, 2013 at 13:17 | history | answered | Robert Bryant | CC BY-SA 3.0 |