# Equal volume iff there is a diffeomorphism

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?

• I do not understand your last paragraph. Can you please be more explicit? – Ted Shifrin Jan 31 at 1:51
• Sir,it was told in class that (volume same under different volume forms iff one is a pullback of another) is not true in the symplectic case. I suppose the teacher meant : the volume forms induced from two different symplectic forms (by taking nth power) might give same volume but not be symplectomorphic. – Angry_Math_Person Jan 31 at 14:42

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

• Thank you. The symplectic form version of Moser Theorem was done in class. There we required that the forms be in same cohomology class. I see how you used that. – Angry_Math_Person Jan 30 at 17:40
• Can you elaborate on the last comment in my original post? – Angry_Math_Person Jan 30 at 17:40

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.