Timeline for metaplectic group does not split
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Jul 12, 2012 at 18:57 | vote | accept | Justin Campbell | ||
| Jul 12, 2012 at 18:57 | answer | added | Justin Campbell | timeline score: 2 | |
| Jul 9, 2012 at 17:05 | comment | added | Justin Campbell | @Peter Woit: Excellent! If you post this as an answer I'll accept it. By the way, I've been reading your notes on Lie groups and their representations, which I am enjoying immensely. | |
| Jul 8, 2012 at 18:04 | comment | added | Peter Woit | There's a relatively straight-forward argument in Section I.6 of Stephen Kudla's "Notes on the Local Theta Correspondence", available at www.math.toronto.edu/~skudla/castle.pdf This may just be a restatement of the Rao argument. As mentioned elsewhere, it comes down to invoking non-triviality of the Hilbert symbol | |
| Jul 7, 2012 at 17:31 | history | edited | Justin Campbell | CC BY-SA 3.0 |
added 29 characters in body; edited tags
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| Jul 7, 2012 at 17:16 | comment | added | Justin Campbell | @David Speyer: It is certainly confusing. I think that everything is worked out over $\mathbb{R}$ in the paper of Lion and Vergne mentioned below. There we can apply covering space theory to the problem, but this doesn't help in the non-Archimedean case... | |
| Jul 7, 2012 at 16:43 | comment | added | David E Speyer | The more I look at this, the more confused I get. You probably shouldn't assume that I got the $Sp_2(\mathbb{R})$ case right either. | |
| Jul 6, 2012 at 23:35 | comment | added | David E Speyer | I thought that there should also be a direct way to do it from the description in terms of Fourier transforms, but I'm not seeing it right now. | |
| Jul 6, 2012 at 23:35 | comment | added | David E Speyer |
The reason I knew it was the following: Look at the nonsplit torus $\left( \begin{smallmatrix} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta \right)$ in $SL_2(\mathbb{R})$. If you look at the action of the corresponding Lie algebra on the metaplectic rep, the weights are in $\mathbb{Z}+1/2$. (I'm using the presentation in Section 2 of Woit's notes math.columbia.edu/~woit/notes21.pdf ). So the preimage in the metaplectic group is the nontrivial double cover of this one and, in particular, the preimages of $\theta =0$ and $\pi$ form a cyclic group of order $4$.
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| Jul 6, 2012 at 22:57 | comment | added | Justin Campbell | @David Speyer: I haven't thought about it in precisely this way before. How does one see that the preimage of $\pm 1$ over $\mathbb{R}$ is $\mathbb{Z}/4\mathbb{Z}$? | |
| Jul 6, 2012 at 16:58 | comment | added | David E Speyer | Would it work to just compute the preimage in the metaplectic group of $\pm \mathrm{Id}$? For $Mp(2, \mathbb{R})$, it is $\mathbb{Z}/4$. If we had a splitting of $0 \to \mathbb{Z}/2 \to Mp(2, \mathbb{R}) \to Sp(2, \mathbb{R}) \to 0$, it would also split $0 \to \mathbb{Z}/2 \to \mathbb{Z}/4 \to \mathbb{Z}/2 \to 0$, a contradiction. I don't know how to compute the preimage of $\pm \mathrm{Id}$ in the general case, though. | |
| Dec 1, 2011 at 12:36 | answer | added | Eric Chopin | timeline score: 2 | |
| May 9, 2011 at 16:00 | vote | accept | Justin Campbell | ||
| Jul 6, 2012 at 4:34 | |||||
| May 9, 2011 at 7:40 | answer | added | Pierre | timeline score: 5 | |
| May 8, 2011 at 22:13 | answer | added | mander | timeline score: 5 | |
| May 8, 2011 at 21:44 | history | asked | Justin Campbell | CC BY-SA 3.0 |