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Hi. Is there a Haar measure or equivalent on infinite dimensional Lie groups? I've been playing around with $Diff(S^1)$, and at least a direct approach seems quite hopeless. It goes something like this:

Def. element on the group by "Euler coordinates",

$g \doteq \prod\limits_{i=-\infty}^{\infty} e^{\omega^i X_i}$, with $\left[ X_i ,X_j \right] = (j-i)X_{i+j}$.

Now I could define a (left invariant) Maurer-Cartan form as $\Omega_L \doteq g^{-1} dg = X_i \otimes \theta^i$, where $\theta^i = \mathcal L^i_j d\omega^j$. Then the Haar measure is

$d\mu (g) \doteq ||\mathcal L || \bigwedge\limits_i d\omega^i$.

Elements of $\mathcal L$ can be written as

$\mathcal L^i_j = \left( \prod\limits_{n=\infty}^{j+1} \exp(-\omega^n adX_n) \right)^i_j$

Clearly the determinant $||\mathcal L ||$ will be horrible... is there any hope for a manageable explicit expression? I couldn't find any literature on the subject (yet), so I'd appreciate any hints to the right direction.

EDIT: umm and of course the whole question of existence of such a measure should probably addressed...

EDIT 2: I realized that the question setup is a bit misleading: I'm actually looking for a measure on the Virasoro group (with zero central charge), i.e. the Lie group corresponding to the algebra above... maybe the Shavgulidze measure has something to do with it, I don't know...

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I think usually people restrict to fairly "tame" subspaces of infinite-dimensional groups in order to construct an actual measure that's useful. I don't know the literature as well as I'd like but my impression was that the kind of measures you want simply don't exist until you specialize to more modest spaces. – Ryan Budney Dec 6 '11 at 21:56
I think you might need to give up some of the properties of a measure (so there might be some kind of left-invariant form, in a weak sense, but not a measure per se). The reason is that a separated measurable group - roughly speaking, a measure space equipped with compatible group topology - carries a natural topology with respect to which it is a topological group, and that topological group has a "completion" to a locally compact group. (This is in Section 62 of Halmos's book on Measure Theory.) – Yemon Choi Dec 6 '11 at 22:58
IIRC if there was such a thing quantum field theory would be a lot easier. – Steve Huntsman Dec 7 '11 at 3:27
Small correction to my comment: "compatible group topology" should have been "compatible group structure". [The point is that the Weil topology can be defined in terms of the measure algebra structure] – Yemon Choi Dec 7 '11 at 4:26
Doug Pickrell did some work in this direction. You might want to check out If I remember correctly, this paper appeared as part of a book/monograph. – Antun Milas Dec 15 '11 at 20:57
up vote 7 down vote accepted

There is something called the Shavgulidze, or the Malliavin-Shavgulidze measure on Diff of smooth manifolds. You can find a discussion in Differentiable measures and the Malliavin calculus (p. 397, available on google books). It is not quite invariant, but quasi-invariant.

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Thanks for pointng that out! I can't get my hands on the book, and google won't show me some important pages... but at least I have something to google and I've found some references. – H. Arponen Dec 7 '11 at 9:03

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