When are two symplectic forms "isotopic"? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-24T16:12:05Zhttp://mathoverflow.net/feeds/question/34237http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/34237/when-are-two-symplectic-forms-isotopicWhen are two symplectic forms "isotopic"?Dick Palais2010-08-02T10:21:05Z2011-03-05T20:37:20Z
<p>I've been skulking around MathOverflow for about a month, reading questions and answers and comments, and I guess it's about time I asked a question myself, so here is one has interested me for a long time.</p>
<p>Suppose $M$ is a compact even dimensional smooth manifold with two symplectic forms $\omega_0$ and $\omega_1$ When are they "isotopic", i.e., when does there exist a 1-parameter family of diffeos $\phi_t$ of $M$, starting from the identity, such that $\phi_1^*(\omega_0) = \omega_1$? Of course a necessary condition is that $\omega_0$ and $\omega_1$ should define the same 2-dimensional cohomology classes. Is this also sufficient? One can ask the same question for volume forms. I asked Juergen Moser about this twenty-five years ago, and he came back with an elegant proof of sufficiency for the volume element case a few months later in a well-known paper in TAMS. He remarks in that paper as follows:</p>
<p>"The statement concerning 2-forms was also suggested by R. Palais. Unfortunately
it seems very difficult to decide when two 2-forms which are closed,
belong to the same cohomology class and are nondegenerate can be deformed
homotopically into each other within the class of these differential forms."</p>
<p>So my question is, what if any progress has been made on this question. Poking around here and in Google hasn't turned up anything. Does anyone know if there are any progress?</p>
http://mathoverflow.net/questions/34237/when-are-two-symplectic-forms-isotopic/34239#34239Answer by Pietro Majer for When are two symplectic forms "isotopic"?Pietro Majer2010-08-02T10:39:11Z2010-08-02T10:51:10Z<p>In Hofer and Zehnder's book <em>Symplectic Invariants and Hamiltonian Dynamics</em> they've a proof of Darboux theorem that uses a deformation argument to pass from a symplectic form $\omega_0$ to a symplectic $\omega_1$ in the case of $R^{2n}$; a one parameter family $\phi_t$ is found solving an ODE. Here it is, at page 10 (I'm not sure if this helps for the case of a compact manifold M too)</p>
<p><a href="http://books.google.it/books?id=lKDFmGLU54sC&printsec=frontcover&dq=symplectic+invariants+and+hamiltonian+dynamics&source=bl&ots=RrcKPkxMY4&sig=BY1fMnWbOBjC2lmG2lRMkL_3iME&hl=it&ei=VZ9WTO-iLIWWsQamu_zhAQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBoQ6AEwAA#v=onepage&q&f=false" rel="nofollow">http://books.google.it/books?id=lKDFmGLU54sC&printsec=frontcover&dq=symplectic+invariants+and+hamiltonian+dynamics&source=bl&ots=RrcKPkxMY4&sig=BY1fMnWbOBjC2lmG2lRMkL_3iME&hl=it&ei=VZ9WTO-iLIWWsQamu_zhAQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBoQ6AEwAA#v=onepage&q&f=false</a></p>
http://mathoverflow.net/questions/34237/when-are-two-symplectic-forms-isotopic/34249#34249Answer by Petya for When are two symplectic forms "isotopic"?Petya2010-08-02T12:07:52Z2010-08-02T12:27:02Z<p>It is known, for a long time now, that there exist examples of symplectic forms in the same cohomology class which are non-isotopic. I do not remember if there exists such example in the dimension $4$, but in dimension $6$ there are different examples. Here is an example constructed by Dusa McDuff:</p>
<p>Let $X$ be a product $S^2\times S^2\times T^2$ ($T^2$ is a torus $(\mathbb R/2\pi\mathbb Z)^2$ with angle coordinates $(\psi,\gamma)$) and $\omega$ is a sum $\omega_1\oplus\omega_2\oplus\omega_3$ of area forms on factors. We suppose that total areas of the first and of the second factor coincides. Consider the map $\varphi \colon X \to X$, where $\varphi (x,y,\psi,\gamma) = (x, T_{x,\psi}(y),\psi,\gamma)$, where $T_{x,\psi}$ is the rotation around $x$ on the angle $\psi$. Then forms $\omega$ and $\varphi^*(\omega)$ define the same cohomology class and non-isotopic.</p>
<p>Moreover, forms $\omega$ and $\varphi^*(\omega)$ could be joined by a path in a space of symplectic structures.</p>
<p>There is a survey containing the statement of this result and helpful references:
<a href="http://www.math.sunysb.edu/~dusa/princerev98.pdf" rel="nofollow">http://www.math.sunysb.edu/~dusa/princerev98.pdf</a></p>
http://mathoverflow.net/questions/34237/when-are-two-symplectic-forms-isotopic/34267#34267Answer by Tim Perutz for When are two symplectic forms "isotopic"?Tim Perutz2010-08-02T15:09:40Z2010-08-02T15:09:40Z<p>There is a cheap way to find cohomologous but non-isotopic (in fact, non-deformation equivalent) symplectic forms: start with a symplectic manifold and pull back the symplectic form via a diffeomorphism that alters the Chern classes of a taming almost complex structure. </p>
<p>As this makes clear, there is a cluster of interrelated questions about isotopy, deformation and diffeomorphism for (closed) symplectic manifolds. (N.B.: A deformation is a path in the space of symplectic forms; by Moser's lemma, such a path is an isotopy if it fixes the symplectic class). In four dimensions:</p>
<p>$\bullet$ <a href="http://www.math.harvard.edu/~ctm/papers/home/text/papers/sw/sw.pdf" rel="nofollow">McMullen and Taubes</a> found a 4-manifold $X$ with two symplectic forms $\omega_0$ and $\omega_1$, not equivalent under the relation generated by deformation and diffeomorphism. They prove using Seiberg-Witten theory that the first Chern classes of these structures are in different orbits of the diffeomorphism group.</p>
<p>$\bullet$ On symplectic 4-manifolds $X$ with "enough" holomorphic curves (e.g. those diffeomorphic to a blow-up of a rational or ruled complex surface), any deformation of symplectic structures in the same cohomology class can be homotoped rel endpoints to an isotopy (<a href="http://www.math.sunysb.edu/~dusa/defjnrev.pdf" rel="nofollow">McDuff</a>). The modification uses "inflation", a topologically trivial symplectic sum of $X$ (along a symplectic divisor $D$) with a ruled surface. Algebraic geometers would call this a deformation to the normal cone of $D$.</p>
<p>$\bullet$ It's not known whether cohomologous, deformation-equivalent symplectic forms on a 4-manifold can ever fail to be isotopic (contrast the 6-dimensional picture painted in Petya's answer).</p>
<p>$\bullet$ On $\mathbb{CP}^2$, it's natural to guess that every symplectic form deforms to plus or minus the standard one. From this we could deduce that $\pi_0 Diff(\mathbb{CP}^2)=\mathbb{Z}/2$. This is an open problem (it doesn't follow from the remarkable results about $\mathbb{CP}^2$ found by Gromov and by Taubes).</p>
<p>$\bullet$ It's unknown whether two cohomologous symplectic forms on the same 4-manifold, with the same canonical class, are deformation-equivalent (let alone symplectomorphic). <a href="http://arxiv.org/abs/math/0607083" rel="nofollow">Donaldson</a> sketched an intriguing programme
to work towards positive results on this problem. It involves the development of a continuity method for an elliptic equation related to that which appears in Yau's proof of the Calabi conjecture; I understand that Weinkove has made some progress on this. </p>
http://mathoverflow.net/questions/34237/when-are-two-symplectic-forms-isotopic/50651#50651Answer by Patrick I-Z for When are two symplectic forms "isotopic"?Patrick I-Z2010-12-29T13:00:58Z2010-12-29T13:00:58Z<p>Just a reformulation of Dick's question: How to describe the orbits of the identity component of the group of diffeomorphisms of a compact manifold, acting naturally on the subspace of cohomology classes of its symplectic forms? </p>