Haar measure on infinite dimensional Lie groups?

2011-12-06T21:39:48Z

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...

Answer by Igor Rivin for Haar measure on infinite dimensional Lie groups?

2011-12-06T23:40:45Z

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.

Haar measure on infinite dimensional Lie groups? - MathOverflow

Haar measure on infinite dimensional Lie groups?

Answer by Igor Rivin for Haar measure on infinite dimensional Lie groups?