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 $U(n)$.


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.

  • $\begingroup$ Thanks for the answer. It may take me some time to digest it. $\endgroup$ – Eric O. Korman Apr 22 '12 at 15:46

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).

  • $\begingroup$ Thanks for the response. It turns out that the reps do not descend to $U(n)$ in general but do to $SU(n)$. $\endgroup$ – Eric O. Korman Apr 22 '12 at 15:49

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


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