most recent 30 from http://mathoverflow.net 2013-05-22T07:02:01Z http://mathoverflow.net/feeds/question/3929 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3929/infinite-dimensional-newlander-nirenberg-theorem Spinorbundle 2009-11-03T12:37:38Z 2009-11-12T21:33:22Z

<p>The Newlander-Nirenberg theorem states that an almost complex structure is integrable if and only if the Nijenhuis tensor vanishes. I heard that this statement is not true in infinite dimensions, since for example the Loop space of a Riemannian 3-manifold is counterexample. (In fact, I think NN fails for Fréchet manifolds in general(?)) So my question is: </p> <p>Is the Newlander Nirenberg theorem valid for Banach- or Hilbertmanifolds? If not, is it possible to weaken the statement (or some conditions) such that it remains true for some class of infinite dimensional manifolds?</p> <p>EDIT: Added the tag "open-problem", since NN for Hilbertmanifolds seems to be an open problem.</p>

http://mathoverflow.net/questions/3929/infinite-dimensional-newlander-nirenberg-theorem/4501#4501 Marco Gualtieri 2009-11-07T07:00:41Z 2009-11-07T07:00:41Z

<p>As far as I understand, in a paper of Petyi [On the ∂-equation in a Banach space. Bull. Soc. Math. France 128 (2000), no. 3, 391–406.] it is shown that the NN theorem does not hold for Banach manifolds in general. However, as you may know, the NN theorem has an easy proof when the almost complex structure is assumed to be real analytic. In this case, the NN theorem is an easy consequence of the Frobenius theorem, which is true for Banach manifolds (reference? Smale?). There is a paper by Daniel Beltita [<a href="http://arxiv.org/abs/math/0407395" rel="nofollow">http://arxiv.org/abs/math/0407395</a>] which confirms that NN is true in this real-analytic case. It would be interesting to see exactly where the NN proof breaks down for Banach manifolds, and exactly what one should assume to allow it to go through. But I am not aware of any such detailed work. </p>