Computing a differential on the loop space of a Symplectic manifold

Question

Suppose we have a symplectic manifold $(M,\omega)$ such that $\omega$ is exact ,i.e, $\omega=d\alpha$. Now let $L_0$ and $L_1$ be two lagrangian submanifolds and consider $\Omega(L_0,L_1)$ to be the set of smooth paths $\gamma:[0,1]\rightarrow M$ such that $\gamma(0)\in L_0$ and $\gamma(1)\in L_1$. Now the "tangent space" at a point of this space is just the space of vector fields $\xi(t)$ along that curve, such that $\xi(i)\in T_{\gamma(i)}L_i$ for $i=0,1$.

Define $A:\Omega(L_0,L_1)\rightarrow \mathbb{R}$ as $A(\gamma)=\int_{0}^{1}\gamma^*\alpha$, in local coordinate we have $A(\gamma)=\int_{0}^{1}\alpha(\frac{\partial \gamma}{\partial t})dt$.

Now given a family of smooth almost-complex structures $J_t$ we have a canocanil family of riemannian metrics $g_t$ and can define the inner product in $\Omega(L_0,L_1)$ as $\langle v_1, v_2\rangle=\int_{0}^{1}g_t(v_1(t),v_2(t))dt$.

Now with this I would like to compute the gradient of $A$. My attempt is the following :

$dA_{\gamma}(\xi)=\frac{\partial }{\partial s}|_{s=0}A(c(s))$, where $c(s):(-\epsilon,\epsilon)\rightarrow \Omega(L_0,L_1)$ such that $c(0)=\gamma$ and $\frac{\partial }{\partial s}|_{s=0}c(s,t)=\xi(t)$, and this will be equal to $\int_{0}^{1} \frac{\partial }{\partial s}|_{s=0} \alpha(\frac{c(s,t)}{\partial t})dt$ , and now from here I don't know what to do. I belive the answer is supposed to be that $grad A(\gamma)=J_t\frac{\partial \gamma}{\partial t}$, but I don't know how to get there.

Any help is appreciated. Thanks in advance.

Answer 1

First, recall the following formula:

Proposition. Let $V$ be a smooth manifold, $(\lambda_t)_t$ be a smooth path of $p$-forms of $V$ and $X_t$ be a time-dependent vector field on $V$ whose local flow is $(\phi_t)_t$, then the following holds true: $$\frac{\operatorname{d}}{\operatorname{d}t}{{\phi_t}^*\lambda_t}_{\vert t=t_0}={\phi_{t_0}}^*\left(\frac{\operatorname{d}}{\operatorname{d}t}{\lambda_t}_{\vert t=t_0}+L_{X_{t_0}}\lambda_t\right).$$

Proof. This follows easily writing the infinetisimal rate of change of $t\mapsto {\phi_t}^*\lambda_t$ at $t_0$. $\Box$

Then, writing $c_s(t)=c(s,t)$ and using Cartan's magic formula for the Lie derivative of time-dependent vector fields (It suffices to show that the formula holds for functions and is well-behaved with respect to exterior derivative and wedge product. A full proof can be found in Lemma B.2. of An introduction to contact topology by H. Geiges) gives: $$\frac{\partial}{\partial s}{c_s}^*\alpha_{\vert s=0}=\gamma^*L_\xi\alpha=\gamma^*\left(\iota_\xi\operatorname{d}\alpha+\operatorname{d}\iota_{\xi}\alpha\right).$$ Using that $\omega=\operatorname{d}\alpha$, $\xi$ is a section of $\gamma^*TM$ and that the exterior derivative and pullback commute, yields: $$\frac{\partial}{\partial s}{{c_s}^*\alpha}_{\vert s=0}=\omega(\xi,\dot{\gamma})+\operatorname{d}\iota_\xi\alpha.$$ At last, recalling that $g(\cdot,\cdot)=\omega(\cdot,J\cdot)$ and $J^2=-1$, $\xi(0)\in T_{\gamma(0)}L_0$ and $\xi(1)\in T_{\gamma(1)}L_1$ gives: $$T_\gamma A(\xi)=\int_0^1g(\xi,-J\dot{\gamma})+\alpha(\xi(1))-\alpha(\xi(0)).$$ I am not able to conclude as $\alpha(\xi(1))-\alpha(\xi(0))$ may not vanish...