Consider a manifold $L$ included into its tangent bundle $T^\star L$ as the zero section (so $L$ is a Lagrangian in $T^\star L$, hence the notation).
It is standard knowledge that given $\alpha$ a $1$-form on $L$, viewed as a section (hence a map $\alpha : L \to T^\star L$), then $\text{Graph}(\alpha) \subset T^\star N$ is also a Lagrangian submanifold iff $\alpha$ is a closed $1$-form (see Prop. 1.2 of these notes for proof, or most any textbook on symplectic geometry).
However, the result I need is the following:
Prop. Any Hamiltonian deformation of the zero section $L \hookrightarrow T^\star L$ can be written $\text{Graph}(df)$ for $f\in\mathcal{C}^\infty(L)$.
where by Hamiltonian deformation, I mean a deformation by a Hamiltonian isotopy.
Does anyone know where I could find a proof to such a statement? Or to a similar one? I am mainly looking for references (or a proof, if it's so short)
Why I think this is true: though I've never seen it stated in such a straightforward way, a lot of proofs I read seem to use that statement, or one version of it.
The closest I have found is in the discussion after Theorem 1.7 of these notes, which says:
Which seems to confirm that the proposition should indeed be true.
What I need to use this for: Well, I have two exact Lagrangians in a symplectic manifold, and locally look at a neighbourhood which looks like the cotangent bundle of one of them.
I want to say that if my second Lagrangian, $L_2$, is a Hamiltonian deformation of $L_1$, then I can write it as $\text{Graph}(df)$ in $T^\star L_1$. (I know of course that I can always do this locally, since I can write $L_2$ as $\text{Graph}(\alpha)$ where $\alpha$ is closed, and then shrink my neighbourhood to make $\alpha$ exact, which is always possible. However, I'd like to know if I can do this without shrinking my neighbourhood? Which I guess would require a proposition like the one I wrote above).
PS: I don't think that the exactness of my Lagrangians matters so much, though it might. I'm not trying to solve the Lagrangian nearby conjecture or anything, I just want to know whether I can express a Hamiltonian deformation of the zero section as the graph of an exact $1$-form.
Thank you very much,
Have a good day
The answer is no.
The easiest example is the cotangent fiber in $\mathbb{R}^2=T^*\mathbb{R}$. In this case, the cotangent fiber $T^*_0\mathbb{R}$ is Hamiltonian isotopic to the zero section since the rotation by $\pi/2$ is a Hamiltonian map given by the quadratic Hamiltonian function $H(x,y)=c(x^2+y^2)$ (for a suitable $c$ I don't want to compute now). $T^*_0\mathbb{R}$ is not a graph of an exact $1$-form, obviously or by the following discussion.
In fact, Lagrangian $L=\operatorname{graph}(df)$ has the following property: The restriction of the cotangent projection $\pi|_L: L \subset T^*N\rightarrow N$ induces a diffeomorphism. The same is also true for the Legendrian lifting $j^1f=\{(x,df(x),f(x):x\in N\}$, and its front projection $j^0f=\{(x,f(x):x\in N\}$: All natural projections are restricted to a diffeomorphism with $N$. However, it is easy to construct front projections (which are easy to describe since it is in $N\times \mathbb{R}$) of exact Lagrangians Hamiltonian isotopic to zero sections that are not diffeomorphic with $N$.
A typical method to construct such examples is called generating function/family: For any Lagrangian Hamiltonian isotopic to the zero section (assume $N$ is compact), a theorem of Laudenbach–Sikorav tells that there exists a function $S: N\times \mathbb{R}^k\rightarrow \mathbb{R}$ such that a symplectic reduction of $\operatorname{graph}(dS)$ gives you the exact Lagrangian immersion $L \looparrowright T^*N$, and in this case, the front projection of $L$ is described by the Cerf diagram of $S$, i.e. the fiberwise critical values of $S$, which typically cannot induces even bijection with $N$.
For the sentence in Auroux's lecture notes, it means infinitesimal deformation. So, you can assume that $L$ is $C^1$-close to the zero section. In this case, you can prove that $L$ is a graph of an exact $1$-form (Exercise 3.35 of McDuff-Salamon).