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I'm interested in an infinite dim'l Heisenberg group associated to the vector space $V = L\mathbb{C}/\mathbb{C}$ = {$f \colon S^1 \to \mathbb{C}$|$f$ smooth}/(const. maps). The group is $\mathbb{C}^\times \times V$ with group law

$(z,f)(z',g) = (zz' e^{\pi i (f,g)}, f+g)$

where $(f,g) = \int fdg$ is a symplectic form.


There's a pairing $e(f,g) = e^{2\pi i (f,g)}$. The isotropic subspaces $W \subset V$ are the ones s.t. $e = 1$ on $W \times W$. General theory says for such $W$ you can construct a representation $F(W)$. Whenever you have a Lagrangian (= maximal isotropic) subspace $L \subset V$ you get up to equivalence a unique irreducible representation where $\mathbb{C}^\times$ acts by scalars.

(actually I've heard this for groups that are extensions by $U(1)$ and then then there is a unique unitary representation; but I'm guessing it works with $\mathbb{C}^\times$ too by just removing unitary)

One way to describe this representation is as continuous maps $\phi\colon V \to \mathbb{C}$ that satisfy $\phi(v + l) = e^{\pi i (v,\ l)}\phi(v)$ and $\int_{V/L} |\phi|^2 dk < \infty$ where $dk$ is a Haar measure on $V/L$.


For $V = L\mathbb{C}/\mathbb{C}$, $z = e^{i\theta}$ it seems that $z^k$ for $k \in \mathbb{Z} - 0$ forms a basis. Also $L^\pm =$ the vector spaces spanned by the positive/negative powers of $z$ are Lagrangian. `Span' here doesn't mean finite linear combinations but linear combinations where the coefficients form maybe absolutely convergent series?

My question is what is an example of a $\phi \in F(L^+)$?

I think all such $\phi$ should be described as follows. Let $p_\pm \colon V \to L^\pm$ be the projections. Then $\phi(v) = e^{i\pi (p_-(v),\ p_+(v))}\overline{\phi}(p_i(V))$ where $\overline{\phi} \in L^2(L_-;dk)$

Among my difficulties with answering this question is that $L_-$ is still a really big space and I don't know what a Haar measure would be on this a space.

I should say, the answer to this question wont really help me in any research per se; I ask it because morally I feel better talking about $F(L)$ if I could write down at least one of its elements.

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You want the integral to be finite? –  S. Carnahan Sep 23 '10 at 14:20
Have you read "Loop groups" by Pressley and Segal? They address some of the analytic issues there. –  Victor Protsak Sep 23 '10 at 15:44
+1 for providing background. –  Theo Johnson-Freyd Sep 23 '10 at 16:21
@Scott thanks for pointing that out; I fixed it. @Victor I've skimmed through that book; there's a lot of information in there maybe I'll take another look thanks. –  solbap Sep 23 '10 at 20:33
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up vote 4 down vote accepted

I'll give a description on the level of the polynomial Lie algebra, and then wave my hands about integrating and completing. As Victor Protsak mentioned in the comments, you can find a more precise treatment in section 9.5 of Pressley and Segal. There, the unitary representation arises from a choice of complex structure.

The Lie algebra of the Heisenberg group is (topologically) spanned by operators $\{ x_k \}_{k \neq 0}$ and $c$, where $x_k$ describes the tangent vector corresponding to the basis vector $z^k$ in the group, and $c$ exponentiates to the central torus. The element $c$ is central in the Lie algebra, and the other generators obey the commutation relation $[x_j, x_k] = j\delta_{j,-k} c$ (although you may need a factor of 1/2 with your choice of normalization). The Lie algebra of $L^+$ is topologically spanned by $x_k$ for $k$ positive, and the corresponding statement holds for $L^-$ with $i$ negative.

The Fock representation is some completion of $\mathbb{C}[x_{-1},x_{-2},\dots,]$, i.e., ``finite energy'' elements are just finite sums of monomials in the generators of $L^-$. The action is given by the following rules:

  1. When $k$ is negative, $x_k$ acts by multiplication.
  2. When $k$ is positive, $x_k$ acts by $k\frac{\partial}{\partial x_{-k}}$.
  3. $c$ acts by the identity.

Exponentiating will give you a description of the action of group elements of the form $z^k$, and finite sums thereof, on finite energy elements of the representation. Modulo normalization, this will yield essentially the formula you gave in the background section. However, I don't think a Haar measure on $V/L$ exists.

Here is an explicit element: there is a distinguished vacuum vector $1$ (sometimes written $\Omega$ or $|0\rangle$), which is the unit element of the polynomial ring above. It is annihilated by all $x_i$ for $i$ positive, so all $z^k$ act by identity on it when $k>0$. The action of $z^k$ for $k<0$ is by exponentiating $x_k$ in the completion. Elements of the central torus act by ordinary scaling.

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