6
$\begingroup$

Consider the metaplectic representation of $Mp(n)$ on $L^2(\mathbb R^n)$. We can view $U(n)$ as a subgroup of $Sp(n)$ and so inside $Mp(n)$ is a double cover $\tilde U(n)$ of $U(n)$. The restriction of the metaplectic representation to $\tilde U(n)$ commutes with the Hamiltonian of the harmonic oscillator: $H = \sum_i (x_i^2 - \frac{\partial^2}{\partial x_i^2})$ and so decomposes as a direct sum of finite dimensional representations of $\tilde U(n)$ on the eigenspaces of $H$.

I am looking for a reference that discusses these representations of $\tilde U(n)$. Specific things I would like to know are if each of these representations is irreducible and if any of them descend to representations of $U(n)$.

$\endgroup$

3 Answers 3

2
$\begingroup$

I believe the double cover you are looking for is just $\tilde{U}(n) = \tilde{U}(1) \times SU(n)$, where $$\tilde{U}(1) = \lbrace\exp(\phi T) : 0 \leq \phi < 4\pi\rbrace$$ is the double cover of $U(1)$ and $T$ is a formal generator of its Lie algebra. This cover factors through the usual cover $U(1)\times SU(n) \to U(n)$ as $$(\exp(\phi T),U) \mapsto (e^{i\phi/2},U) \mapsto e^{i\phi/2} U.$$ So the problem reduces to classifying the $\tilde{U}(1)$ representations and the $SU(n)$ ones, and the answer is straightforward. First decompose
$$L^2(\mathbb{R}^n) \simeq \bigoplus_{N=0}^\infty V_N,$$ where $V_N$ is the eigenspace of $H$ with eigenvalue $2N+1$. Then $\tilde{U}(1)\times SU(n)$ should act irreducibly on each $V_N$ as $$(\exp(\phi T),U) \mapsto e^{iN\phi/2} \cdot \mathrm{Sym}^N U.$$

It is easy to see this. The $\tilde{U}(1)$ action is the same as with the usual $L^2(\mathbb{R})$ case. You can easily derive the $SU(n)$ representations using the Fock space isomorphism of $L^2(\mathbb{R}^n)$ with a Hilbert space of analytic functions: $$L^2(\mathbb{R}^n) \simeq L^2_\mathrm{hol}\(\mathbb{C}^n,\pi^{-n}e^{-||z||^2/2}dz).$$ Under this isomorphism, $V_N$ maps to the space of homogeneous polynomials of degree $N$, and $SU(n)$ acts on polynomials in the usual way as $(Uf)(z) = f(Uz)$.

This is probably all contained in Bargmann's classic paper.

$\endgroup$
1
  • $\begingroup$ Thanks for the answer. It may take me some time to digest it. $\endgroup$ Apr 22, 2012 at 15:46
1
$\begingroup$

As a place-holder answer: while, unfortunately I do not know a good reference offhand, I believe (based on a memory of having done the computation at least twice myself) that direct computation shows that the restriction to meta-U(n) does descend to U(n).

Conceivably various expository papers of Steve Kudla include either-or-both discussion of the various pairs inside metaplectic groups (as your meta-version of U(1)xU(n)) and the sort of splitting property of interest.

Edit: Oops! Indeed, as commented and amplified in another answer, the descent is just to SU(n).

$\endgroup$
1
  • $\begingroup$ Thanks for the response. It turns out that the reps do not descend to $U(n)$ in general but do to $SU(n)$. $\endgroup$ Apr 22, 2012 at 15:49
0
$\begingroup$

I just found that Borel and Wallach's * Continuous cohomology, discrete subgroups, and representations of reductive groups* does a detailed analysis of this representation (which they call the oscillatory representation). It's on google books: http://books.google.com/books?id=_EZY9LhAxosC&lpg=PA257&ots=NIQffrLQCA&dq=borel%20wallach%20cohomology%20lie&pg=PA151#v=onepage&q&f=false

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.