When are two symplectic forms "isotopic"?

Question

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.

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:

"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."

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?

Answer 1

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:

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.

Moreover, forms $\omega$ and $\varphi^*(\omega)$ could be joined by a path in a space of symplectic structures.

There is a survey containing the statement of this result and helpful references: http://www.math.sunysb.edu/~dusa/princerev98.pdf

Answer 2

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.

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:

$\bullet$ McMullen and Taubes 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.

$\bullet$ 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 (McDuff). 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 $D$.

$\bullet$ 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).

$\bullet$ 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).

$\bullet$ It's unknown whether two cohomologous symplectic forms on the same 4-manifold, with the same canonical class, are deformation-equivalent (let alone symplectomorphic). Donaldson 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.

Answer 3

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?

Answer 4

In Hofer and Zehnder's book Symplectic Invariants and Hamiltonian Dynamics 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)

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