# A corollary of Moser's theorem

In my symplectic geometry class we gave the following result :

Let $$M$$ be a closed symplectic manifold, $$Q\subset M$$ a closed submanifold and $$\omega_t$$ a family of symplectic forms such that $$[\frac{d}{dt}\omega_t]=0$$ and $$(\frac{d}{dt}\omega_t)|_Q=0$$ for any $$t\in [0,1]$$. Then there exists an isotopy $$\phi_t :M\rightarrow M$$ such that $$(\phi_t)_Q=id$$ and $$\phi_t^*(\omega_t)=\omega_0$$ for any $$t\in [0,1]$$.

Now there's a step on the proof where we realized there is a mistake and so far no one was able to fix it. I have tried to look for this result in symplectic geometry books but I had no sucess so I am wondering if anyone knows if this result is actually true and how one can prove it , or if there is a counterexample ?

Any help is appreciated. Thanks in advance.

If the cohomology class $$[\frac{d\omega_t}{dt}]$$ is merely assumed to lie in the (absolute) cohomology $$H^2_{dR}(M)$$, then the result is false. If it is assumed to lie in the relative cohomology $$H^2_{dR}(M,Q)$$, then the result is true under the following formulation (to be explained below):

Let $$M$$ be a closed symplectic manifold, $$i_Q : Q \subset M$$ a closed submanifold and $$\omega_t$$ a family of symplectic forms such that $$\left[ \frac{d \omega_t}{dt} \right] = 0$$ in $$H^2(M,Q)$$ and $$i_Q^*\frac{d\omega_t}{dt} = 0$$ for any $$t \in [0,1]$$. Then there exists an isotopy $$\phi_t : M \to M$$ such that $$\phi_t|_Q = id$$ and $$\phi_t^*\omega_t = \omega_0$$ for every $$t \in [0,1]$$.

Absolute case: For $$[\frac{d\omega_t}{dt}] = 0$$ in $$H^2_{dR}(M)$$ , one can produce counter-examples to the statement by taking $$Q$$ to be a contractible embedded circle in a closed surface $$M$$. Indeed, pick an embedded smooth disc $$D$$ bounded by $$Q$$ and define $$\omega_t$$ by changing the symplectic form $$\omega_0$$ away from a neighborhood of the circle in such a way that the area of $$D$$ disc increases with $$t$$ while keeping the area of the whole surface fixed. Then any isotopy $$\phi_t$$ produced from the family $$(\omega_t)_{t \in [0,1]}$$ by Moser's trick is such that the $$\omega_0$$-area of $$D_t := \phi_t^{-1}(D)$$ increases with $$t$$, which implies $$\phi_t|_Q \neq id$$.

Relative case: (Preliminary remark: Let $$i_Q : Q \subset M$$ denote the inclusion map. Given a differential form $$\alpha$$ on $$M$$, we are going to consider two types of 'restrictions' to $$Q$$, (i) the pullback form $$i_Q^*\alpha$$ on $$Q$$ and (ii) the restriction $$\alpha|_{Q}$$ of $$\alpha$$ to $$Q$$, by which I mean the form induced on the pullback bundle $$i_Q^*TM = T_{Q}M$$. The pullback $$i_Q^*\alpha$$ only captures the value of $$\alpha|_Q$$ on the subbundle $$(i_Q)_*TQ \subset T_Q M$$.)

The desired property $$\phi_t|_Q = id$$ amounts to $$X_t|_Q = 0$$, where $$X_t$$ is the vector field produced by Moser's trick via the equation $$\lambda_t = \iota_{X_t}\omega_t$$ where $$d\lambda_t = \frac{d \omega_t}{dt}$$. Hence, for $$\phi_t$$ to have the desired property, we need to be able to find for every $$t \in [0,1]$$ a primitive 1-form $$\lambda_t$$ for $$\frac{d \omega_t}{dt}$$ whose restriction $$\lambda_t|_Q$$ to $$Q$$ vanishes. This is only possible if $$\frac{d \omega_t}{dt}$$ is 'well-behaved relative to $$Q$$', in a sense we now explore.

The counter-examples given in the absolute case suggest that for the statement to hold, one should further require that all members of the family $$(\omega_t)_{t \in [0,1]}$$ give the same area to every 2-cycle with boundary in $$Q$$. To satisfy this requirement, it suffices to ask that for every $$t \in [0,1]$$, the exact two-form $$\frac{d\omega_t}{dt}$$ admits a primitive 1-form $$\lambda_{t}$$ whose pullback $$i_Q^*\lambda_{t}$$ to $$Q$$ is exact. (Note: The condition $$i_Q^*\frac{d\omega_t}{dt} = 0$$ already implies that $$i_Q^*\lambda_{t}$$ is closed.)

This last condition can be expressed in relative cohomological terms as defined in Bott's & Tu's Differential forms in algebraic topology. Namely, we define the de Rham complex of $$M$$ relative to $$Q$$ as the cone of $$i_Q$$, i.e. as the complex $$\Omega^*(M, Q) := \Omega^*(M) \oplus \Omega^{*-1}(Q)$$ with differential given by $$d(\alpha, \eta) = (d\alpha, i_Q^*\alpha - d\eta)$$. Then the condition stated in the previous paragraph is precisely that for every $$t \in [0,1]$$, the relative 2-form $$\left( \frac{d \omega_t}{dt}, 0 \right) \in \Omega^2(M, Q)$$ should be exact; In other words, the relative cohomology class $$\left[ \frac{d \omega_t}{dt} \right] := \left[ \left( \frac{d \omega_t}{dt}, 0 \right) \right] \in H^2(M, Q)$$ should be required to vanish.

Under this assumption, we can find for every $$t \in [0,1]$$ a primitive 1-form $$\lambda_t$$ to $$\frac{d\omega_t}{dt}$$ whose pullback to $$Q$$ is exact, say $$i_Q^*\lambda_t = df_t$$ with $$f_t : Q \to \mathbb{R}$$. Now, by Whitney's extension theorem, there exists $$F_t : M \to \mathbb{R}$$ such that $$i_Q^* F_t = f_t$$ and $$dF_t|_Q = \lambda_t|_Q$$. Let's define $$\lambda'_t := \lambda_t - dF_t$$; This is another primitive 1-form for $$\frac{d\omega_t}{dt}$$ which satisfies $$\lambda'_t|_Q = 0$$. Since all the above choices can be made to depend smoothly on $$t$$, we can apply Moser's trick with this family $$\lambda'_t$$ to obtain the desired isotopy.