Almost Complex Structures: 'Tame' versus 'Compatible' - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T00:55:58Z http://mathoverflow.net/feeds/question/106694 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106694/almost-complex-structures-tame-versus-compatible Almost Complex Structures: 'Tame' versus 'Compatible' Chris Gerig 2012-09-08T23:02:13Z 2012-09-11T22:59:51Z <p>Consider a symplectic manifold $(M,\omega)$ and the space of almost complex structures (a.c.s. for short). These are $J:TM\to TM$ with $J^2=-\text{id}$. An a.c.s. $J$ is <strong>$\omega$-tame</strong> when $\omega(v,Jv)>0$, and $J$ is <strong>$\omega$-compatible</strong> when it is $\omega$-tame and $\omega(J\cdot,J\cdot)=\omega(\cdot,\cdot)$. The set of either such $J$ forms a contractible subspace. Note that in either scenario we can form a Riemannian metric $g_J$ by twisting $\omega$ in some way with $J$.</p> <p>In practice I sometimes see $J$ being tame, and other times (more often) see it being compatible, but I am unsure of the restrictions that each impose. <strong>Is there any context in which it is necessary/helpful to use one condition over the other?</strong> This need not be related to pseudo-holomorphic curves in Floer theory. <strong>Do certain results fail when I relax "compatible" to "tame"?</strong></p> <p>A fundamental difference I see is with the Levi-Civita connection $\nabla$ associated to the metric $g_J$. Here a compatible or tame $J$ is not necessarily parallel ($\nabla J\ne 0$) under $\nabla$. It <em>is</em> preserved under the modified connection $\tilde{\nabla}_XY=\nabla_XY-\frac{1}{2}J(\nabla_XJ)Y-\frac{1}{4}(\nabla_{JY}J+J\nabla_YJ)X$, and its torsion is a multiple of the Nijunheis tensor. Now when $J$ is compatible then this connection preserves the metric, but not when $J$ is simply tame. I don't really know if this affects its use.<br> Even more different (quoted from McDuff-Salamon's big textbook) is that compatible $J$ minimize the energy of a J-holomorphic curve in its homology class, but not necessarily for tame $J$.</p> http://mathoverflow.net/questions/106694/almost-complex-structures-tame-versus-compatible/106708#106708 Answer by YangMills for Almost Complex Structures: 'Tame' versus 'Compatible' YangMills 2012-09-09T04:46:06Z 2012-09-09T04:46:06Z <p>If you assume that $M$ is compact and of dimension $4$ then Donaldson has conjectured that if $J$ is $\omega$-tame then $J$ is $\omega'$-compatible for some symplectic form $\omega'$, see Question 2 <a href="http://arxiv.org/abs/math/0607083" rel="nofollow">here.</a> By classical results of Gromov and Taubes this is known to hold if $M=\mathbb{CP}^2$. It is also not hard <a href="http://arxiv.org/abs/0708.2520" rel="nofollow">to verify this</a> when $J$ is integrable (so $(M,J)$ is a compact complex surface), using the classification of surfaces. It is <a href="http://arxiv.org/abs/1203.4331" rel="nofollow">also known</a> to hold when $(M,J)$ is homogeneous.</p> <p>In this same paper Donaldson proposed a way to attack this question by suitably extending the Calabi-Yau theorem to the symplectic case. Such an extension is still conjectural, but see <a href="http://arxiv.org/abs/0901.1501" rel="nofollow">this survey</a> for some work that has been done in this direction. </p> <p>On the other hand Taubes has developed a different way of attacking Donaldson's question, and <a href="http://projecteuclid.org/euclid.jsg/1309546043" rel="nofollow">has succeeded</a> in the case when $b^+(M)=1$ and $J$ is suitably generic.</p> <p>In higher dimensions ($6$ or more) the analogous statement is false.</p> http://mathoverflow.net/questions/106694/almost-complex-structures-tame-versus-compatible/106726#106726 Answer by Sam Lisi for Almost Complex Structures: 'Tame' versus 'Compatible' Sam Lisi 2012-09-09T12:31:33Z 2012-09-09T12:31:33Z <p>I think YangMills's answer is fantastic. I will go in a slightly different direction, and address the "usefulness" of each definition part of the question.</p> <p>A very useful feature of tame acs is that tameness is an open condition. This means that it is more straightforward to talk about generic perturbations, e.g. so that somewhere injective curves are transverse. The whole discussion can be carried through with compatible, but it becomes more involved (most recently, I've seen this discussed in some detail in Massot-Niederkruger-Wendl on filling questions for higher dimensional contact manifolds.) The compatible case works, if I am not mistaken, because the space of compatible $J$ is a Banach manifold.</p> <p>A useful feature of a compatible almost complex structure $J$ is that for a $J$-holomorphic curve, $u^* \omega = |du|^2_J d \operatorname{vol}$. If you have a tamed almost complex structure, you obtain an inequality that is good enough for compactness, but the proofs become a little more involved.</p> <p>To the best of my knowledge, there is nothing that anyone has proved for $J$-holomorphic curves for compatible $J$ that is believed to be false for tame $J$. Many results, however, are only proved for compatible $J$ because it makes life easier. I personally would love to see an example of a result that wasn't overly technical for which the difference mattered. </p>