I want to show that given $\alpha,\beta$ two volume forms on a closed manifold such that $\int_M\alpha = \int_M\beta$ how can we show that there exists a diffeomorphism $\phi:M\rightarrow M$ such that $\phi^*\beta = \alpha$.

Attempt: I suspect we have to use Moser's lemma which states that

If $(\omega_t),t\in[0,1]$ is a smooth family of symplectic forms which are cohomologue (i.e. $[\omega_t] = [\omega_0]$ for $t\in[0,1]$). Then there exists a smooth family of diffeomorphisms $(\phi_t),t\in[0,1]$ such that $\phi^*_t\omega_t = \omega_0$ for $t\in[0,1]$.


1 Answer 1


The idea is to build an isotopy $\phi_t : M \to M$, for every $0\leq t \leq 1$, such that $\phi_0^* = Id$ and $\phi = \phi_1$. To do so, we notice that

$$\int_M \alpha = \int_M \beta \implies \exists \ \eta \in \Omega^{(n-1)} (M), \ \beta = \alpha + d \eta $$

Define an $n$-form by $\lambda_t = \alpha + t d\eta$. This is a volume form (why?). Now, since $M$ is compact (you forgot to mention this on your question) an isotopy can be generated by the flow of a time-dependent vector field given by

$$X_t = \left(\frac{d}{dt} \phi_t\right) \circ \phi_t^{-1}$$

It is sufficent to find $\phi_t^* \lambda_t = \alpha$, so

$$\begin{aligned} 0 = \frac{d}{dt} \phi_t^* \lambda_t &= \phi^*_t \left(\mathcal L_{X_t} \lambda_t + \frac{d}{dt}\lambda_t\right)\\&=\phi_t^* (d\iota_{X_t}\lambda_t + \iota_{X_t}\underbrace{d\lambda_t}_{0} + d\eta)\\&=\phi_t^* d\left(\iota_{X_t}\lambda_t + \eta\right) \end{aligned} $$ where you should try to understand each equality.

So the problem boils down to solving

$$\iota_{X_t}\lambda_t + \eta = 0 \tag{1}$$

But since $\lambda_t $ is a volume form the map $\mathfrak X(M) \to \Omega^{(n-1)}(M)$, given by $X \mapsto \iota_{X}\lambda_t$ is an isomorphism for every $t$, so there exists a unique $X_t$ satisfying (1). The flow of $X_t$ defines an isotopy satisfying $\phi^*_t \lambda_t = \alpha$, letting $\phi = \phi_1$ yields the desired isotopy.

  • 1
    $\begingroup$ In your last two equations I believe the $d$ in $d\eta$ should disappear. @Aaron Maroja $\endgroup$
    – Someone
    Commented Apr 7, 2021 at 10:37
  • 1
    $\begingroup$ Thank you, @Something for pointing this out. Already corrected. $\endgroup$ Commented Apr 11, 2021 at 15:49

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