Haar measure on infinite dimensional Lie groups? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T02:41:19Z http://mathoverflow.net/feeds/question/82826 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/82826/haar-measure-on-infinite-dimensional-lie-groups Haar measure on infinite dimensional Lie groups? H. Arponen 2011-12-06T21:39:48Z 2011-12-15T19:49:35Z <p>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:</p> <p>Def. element on the group by "Euler coordinates",</p> <p>$g \doteq \prod\limits_{i=-\infty}^{\infty} e^{\omega^i X_i}$, with $\left[ X_i ,X_j \right] = (j-i)X_{i+j}$.</p> <p>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</p> <p>$d\mu (g) \doteq ||\mathcal L || \bigwedge\limits_i d\omega^i$.</p> <p>Elements of $\mathcal L$ can be written as</p> <p>$\mathcal L^i_j = \left( \prod\limits_{n=\infty}^{j+1} \exp(-\omega^n adX_n) \right)^i_j$</p> <p>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.</p> <p>EDIT: umm and of course the whole question of existence of such a measure should probably addressed...</p> <p>EDIT 2: I realized that the question setup is a bit misleading: I'm actually looking for a measure on the <em>Virasoro group</em> (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...</p> http://mathoverflow.net/questions/82826/haar-measure-on-infinite-dimensional-lie-groups/82834#82834 Answer by Igor Rivin for Haar measure on infinite dimensional Lie groups? Igor Rivin 2011-12-06T23:40:45Z 2011-12-06T23:40:45Z <p>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.</p>