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I'm trying to understand the Weil representation and hope there are some experts around who can set me straight. Let $F$ be a non-Archimedean local field (I don't mind assuming that the characteristic of $F$ is not $2$ if it simplifies things; also the situation over $\mathbb{R}$ is similar but obviously I need a proof which does not rely on any covering space theory.) and fix a nontrivial continuous additive character $\psi : F \to \mathbb{C}^{\times}$. Then by the Stone-von Neumann theorem there is a unique up to isomorphism irreducible smooth representation of the Heisenberg group with central character $\psi$, and one sees the existence of a projective representation of the symplectic group in this space by the uniqueness part of that theorem.

So my question is: why does this not lift to an ordinary representation? All the articles I have read refer this claim to Weil's original paper, which is quite long, and my French is not so good. I think one can see this from the fact that the metaplectic group (I'm talking about the two-sheeted cover) is not a trivial extension of the symplectic group by using the fact that the latter group is perfect. So if one constructs the metaplectic group via Maslov cocycles, one needs to show a certain cocycle is not a coboundary. But how?

Thanks for the help. I hope my question is clear enough.

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  • $\begingroup$ 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. $\endgroup$
    – David E Speyer
    Commented Jul 6, 2012 at 16:58
  • $\begingroup$ @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 $\mathbb{Z}/4\mathbb{Z}$? $\endgroup$
    – Justin Campbell
    Commented Jul 6, 2012 at 22:57
  • $\begingroup$ 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 $4$. $\endgroup$
    – David E Speyer
    Commented Jul 6, 2012 at 23:35
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    $\begingroup$ 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 $\endgroup$
    – Peter Woit
    Commented Jul 8, 2012 at 18:04
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    $\begingroup$ @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. $\endgroup$
    – Justin Campbell
    Commented Jul 9, 2012 at 17:05

4 Answers 4

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Take a look at Proposition 5.8 in Rao, On some explicit formulas in the theory of Weil representation, Pacific Journal of Mathematics 157 (1993), 335-371. This is actually a paper that dates back to 1978. Basically, it comes down to the Hilbert symbol being non-trivial in the p-adic case.

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  • $\begingroup$ Hello David, nice to see you at MO! $\endgroup$
    – Victor Protsak
    Commented May 8, 2011 at 22:30
  • $\begingroup$ Thanks. Hope all is well with you. Math keeps me grounded as a Dean. $\endgroup$
    – mander
    Commented May 9, 2011 at 1:11
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    $\begingroup$ I was aware of this paper and honestly find it more unreadable than Weil's original one, despite the fact that it is written in English. This may be unrealistic, but I was hoping for something with fewer "explicit formulas." To be fair, I should not have said "All the articles I have read refer this claim to Weil's original paper," since this is not quite true... $\endgroup$
    – Justin Campbell
    Commented May 9, 2011 at 3:45
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Have a look at

Lion & Vergne, The Weil representation, Maslov index, and Theta series.

I haven't got it with me but I think I remember that it contains what you want.

Also there is

Teruji Thomas, The trace of the Weil representation (on arXiv).

This paper is quite explicit but not "as explicit" as Rao (I know what you mean). It's got constructions of the metaplectic groups, and again I think I remember that what you're asking is an easy consequence of these.

Some random comments now. The Maslov index gives you an extension of $Sp_{2n}(k)$ by the Witt group $W(k)$. For $k=\mathbb{R}$ the connected component of 1 in this extension is the simply-connected extension of $Sp_{2n}(\mathbb{R})$ (this fact is in Lion-Vergne), so it's certainly non-trivial. The metaplectic group is a quotient of this. Note that for $n=1$ you get an extension of $SL_2(\mathbb{R})$, and the inverse image of $SL_2(\mathbb{Z})$ is the braid group $B_3$, proving again that the extension is non-trivial (see Kassel & Turaev, Braid groups, appendix, and Milnor's book on algebraic K-theory).

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    $\begingroup$ The first chapter of the cited book is available, along with a lot of material about the Maslov index, on this webpage: maths.ed.ac.uk/~aar/maslov.htm (whose legality is debatable, but whose usefulness certainly isn't). $\endgroup$
    – Maxime Bourrigan
    Commented May 9, 2011 at 8:26
  • $\begingroup$ Sorry to un-accept the answer after over a year, but I still can't figure out what's going on, even with all of these references and a lot of hard work. The answer certainly depends on some nontrivial arithmetic: at the very least the fact that the Hilbert symbol is nontrivial on non-Archimedean local fields. $\endgroup$
    – Justin Campbell
    Commented Jul 6, 2012 at 4:37
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Hi All, The topic is old but in case some of you may be interested in the following additional comments, I may recommend to read Hans Reiter: Lecture Notes in Mathematics 1382 "Metaplectic Groups and Segal Algebras". Basically, it is a translation in english of Weil seminal paper (Acta Math 111), but placed in a slightly generalized context. So it may help if you want to understand the way Weil introduced the metaplectic group.

To answer the original post. The projective representation lifts to an ordinary representation of the metaplectic group "by construction", because the metaplectic group is actually a central extension of the symplectic group and the cocycle used in this central extensionm is actually the 2-cocycle of the projective representation. But of course is you define the metaplectic group through assuming the existence of a 2-fold cover this may not be direct, but it is not the way followed by Weil in his paper. ps: I also recommend Ranga Rao's paper which is still a good reference on this subject...

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I would like the credit to go to Peter Woit for suggesting Section I.6 of Stephen Kudla's "Notes on the Local Theta Correspondence," which contains a nice proof, but he posted this only as a comment and the bounty ends today.

The text of Woit's comment is

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

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    $\begingroup$ Just in case there are curious cynics out there: by accepting his own answer, Justin did not recover the bounty (see mathoverflow.net/users/3544?tab=reputationhistory#sort-top). It's too bad Peter didn't post his comment as an answer. @Justin: consider editing Peter's comment verbatim into this answer. $\endgroup$
    – Anton Geraschenko
    Commented Jul 12, 2012 at 19:33
  • $\begingroup$ I have edited following @AntonGeraschenko's suggestion. $\endgroup$
    – LSpice
    Commented Jun 21, 2018 at 21:56

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