Equal volume iff there is a diffeomorphism

Question

Let M be a compact oriented smooth manifold. Let $w_1$ and $w_2$ be two volume forms. Let integral of both these forms over M be equal i.e vol(M) be equal wrt both forms.

Show that there is a diffeomorphism f from M to M such that $f^*(w_2)=w_1$

Of course if such an f exists then by change of variable formula the volumes shall be equal.

Also it was told in class that apparently this isn't the case for symplectic manifolds and this is a global invariant. Any comments on that?

Answer 1

This is a theorem of Moser:

Assume $\omega_0, \omega_1$ are two volume forms (with the same total mass) on a compact manifold. Then there is a diffeomorphism $\phi$ on $M$ so that $\phi^*\omega_1=\omega_0$.

Proof: Let $$ \omega_s=\omega_0+s(\omega_1-\omega_0). $$ Since $\omega_0$ and $\omega_1$ has the same total mass, they are in the same cohomology class. So there is an $n-1$ form $\eta$ so that $$ \omega_1-\omega_0=d\eta. $$ Observe, (this is most easily seen when writing all the forms in a local coordinate system $(x_1,...,x_n)$, there is a unique vector field $X_s$ so that $$ \iota_{X_s}\omega_s=-\eta. $$

Let $\phi_s$ is the one parameter group of diffeomorphism that is generated by $X_s$.

Compute, at time $s=t$, $$ \begin{aligned} \frac{d}{ds}(\phi_s^*\omega_s)\Big|_{s=t}=&L_{X_t}(\phi_t^*\omega_t)+ \phi_t^*(\omega_1-\omega_0)\\ =&d\iota_{X_t}(\phi_t^*\omega_t)+\iota_{X_t}d(\phi_t^*\omega_t)+ \phi_t^*(\omega_1-\omega_0)\\ =&d\phi_t^*(-\eta)+\phi_t^*(d\eta)\\ =&0. \end{aligned} $$ Here notice, $(\phi_t)_*X_t=X_t$ as a vector field.

Thus $$ \phi_1^*\omega_1=\phi_0^*\omega_0=\omega_0. $$

Answer 2

Let $S^2(1)$ denote the sphere with standard symplectic/volume form $\omega$ with volume 1 and $S^2(r)$ the sphere equipped with the form $r\omega$. Then the manifolds $$S^2(r) \times S^2(R)$$ for $r < R$ are never symplectomorphic unless $(r, R) = (r', R')$. One may check this by seeing that the set of values given by integrating $\omega$ over primitive elements of $H_2(M;\Bbb Z)$ with self-intersection 0 gives an invariant, and the set of values in the case above is $\{-R, -r, r, R\}$.

But the volume of this manifold is $rR$. Thus $S^2(r) \times S^2(1/r)$ gives an uncountable family of symplectic manifolds which are pairwise non-symplectomorphic but do have the same volume.