MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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