Existence of isotopy preserving the action

Question

Let $\gamma_1$ and $\gamma_2$ be simple closed curves in $R^4.$ Let $\lambda= x_1 dy_1+ x_2dy_2.$ Suppose that $\int_{\gamma_1} \lambda= \int_{\gamma_2} \lambda.$

I am looking for a reference for the following statement: there exists an isotopy $\{\gamma_t\}$ through simple closed curves such that $\int_{\gamma_t} \lambda$ is constant.

The existence of a smooth isotopy is clear from general position. It seems intuitively obvious that one can just continuously modify the isotopy to ensure that the integral is constant. I am fairly confident that I can write down a proof, but it seems like it would be quite lengthy. I feel like this statement is surely already in the literature (or should follow easily from more general theorems in the literature).

Answer 1

This is not an answer but an idea. If you can show that there is an Lagrangian embedding of $f:S^1\times [0,1]\to\mathbb{R}^4$ such $f|_{S^1\times \{0\}}=\gamma_0$ and $f|_{S^1\times \{1\}}=\gamma_1$, then you are done. Indeed, the Lagrangian (also known as isotropic) embedding is an embedding such that the pullback of the symplectic form $\omega=dx_1\wedge dy_1+dx_2\wedge dy_2$ is zero. Then for $\gamma_t=f|_{S^1\times \{t\}}$ we have that $\int_{\gamma_t}\lambda=\int_{\gamma_0}\lambda$ by the Stokes theorem since $d\lambda=\omega$. Whether such an embedding always exists, I do not know, but there is an extensive literature about Lagrangian embeddings. I am just not an expert in this field.

For example if the curves $\gamma_0$ and $\gamma_1$ are in symplectically orthogonal subspaces (for example $x_1y_1$ and $x_2y_2$), then after a suitable reparametrization of them you can find a Lagrangian homotopy between them (see Lemma 2.2 in [1]). However, it will only be a homotopy and not necessarily isotopy.

[1] D. Allcock, An isoperimetric inequality for the Heisenberg groups. Geom. Funct. Anal. 8 (1998), 219–233.