We know that a cotangent bundle $T^\star M$ has a canonical symplectic form and $M$ is a natural Lagrangian submanifold of it. A well known result is that any submanifold $X=\{(p,f(p)): p\in M\}$, where $f$ is a closed one form is Lagrangian. Denote by $[f]$ the de Rham cohomology class of $f$. Assume that we flow $X$ in a Hamiltonian direction to $Y$, then $Y$ will be a Lagrangian submanifold of $T^\star M$. My question is can we write it as $Y={(p,g(p)): p\in M}$ for some closed one form $g$.? If so, do we have $ [g]=[f]?$ Thanks in advance!



On the de Rham cohomology of 1-forms in cotangent bundle.
As has been noted in Peter Michor's answer, a Hamiltonian isotopy can certainly move the graph of a closed one-form to a Lagrangian submanifold that is not the graph of a one-form.


However, in the special case that the starting and ending submanifolds are both graphs of closed one-forms, the de Rham cohomology classes will in fact be equal. (It is important that we are assuming here that the isotopy is Hamiltonian and not just symplectic.) This can be seen as follows:


Let $\lambda=\sum p\_idq\_i$ be the canonical one-form $T^{\ast}M$, so the symplectic form is $d\lambda$. In general, if $\iota:L\to T^{\ast}M$ is a Lagrangian embedding, then the Lagrangian condition amounts to the statement that $\iota^{\ast}\lambda$ is closed. Consequently, there is a well-defined de Rham 
cohomology class $[\iota^{\ast}\lambda]\in H^1(L;\mathbb{R})$, generally called the **Liouville class**.
In the special case that $L=M$ and the Lagrangian embedding is a closed one-form $\sigma:M\to T^{\ast}M$
(viewed as a section), the definition of the canonical one-form $\lambda$ is such that $\sigma^{\ast}\lambda=\sigma$.
In particular, the Liouville class of a closed one-form is just the cohomology class of 
the one-form.


I claim now that the Liouville class is invariant under Hamiltonian isotopies of the Lagrangian submanifold.
In other words, if $\{\phi\_t\}$ is a Hamiltonian isotopy, obtained as the flow of the Hamiltonian vector
field $X\_H$ of a function $H:T^{\ast}M\to \mathbb{R}$ (i.e. $i\_{X\_H}d\lambda=-dH$), and if $\iota$ is some Lagrangian embedding, I
claim that the Liouville class of $\phi\_t\circ \iota$ is independent of $t$. (For ease of notation I'll assume $H$
is time-independent, but the time-dependent case is a straightforward modification.) 
Indeed this follows fairly quickly from Cartan's magic formula, as follows: 
$$
\frac{d}{dt}(\phi\_{t}\circ\iota)^{\ast}\lambda=\iota^{\ast}\left(\frac{d}{dt}\phi\_{t}^{\ast}\lambda \right)
=\iota^{\ast}\phi\_{t}^{\ast}\mathcal{L}\_{X\_H}\lambda=\iota^{\ast}\phi\_{t}^{\ast}(di\_{X\_H}\lambda+i\_{X\_H}d\lambda)
$$ $$=\iota^{\ast}\phi\_{t}^{\ast}d\left(i\_{X\_H}\lambda-H\right) 
$$


which is exact; hence the cohomology class of $(\phi\_{t}\circ\iota)^{\ast}\lambda$ is indeed independent of $t$. In particular if $\iota$
is equal to the section of $T^{\*}M$ given by a closed one-form $\sigma$ and if $\phi\_1\circ\iota$ is equal to the section given by $\tau$ then
looking at their respective Liouville classes shows that $\sigma$ and $\tau$ are cohomologous.


