I try to find in a direct calculation the first variation of the symplectic action functional.
The action functional sends any smooth path $\gamma$ in $T^*M$ to $$\mathcal{A}_H(\gamma)=\int_0^1 (\lambda(\dot{\gamma}(t))+H_t(\gamma(t)))dt$$
where $\lambda$ is the Liuoville form on the cotangent bundle ($\lambda=pdq$ such that the symplectic form is $\omega=-d\lambda$). Suppose $\delta \gamma$ is a vector field along the path $\gamma$.
The variation of $\mathcal{A}_H$ at $\gamma$ in direction $\delta \gamma$ is
$$d\mathcal{A}_H(\gamma)\delta \gamma=\int_0^1\Big (-\omega(\dot{\gamma}(t),\delta \gamma(t))+dH_t(\delta\gamma(t))\Big )dt+ \lambda(\delta\gamma(1))-\lambda(\delta\gamma(0))$$
To see this, let $\tilde{\gamma}(s,t)$ be a variation of $\gamma$, i.e. $\gamma(0,t)=\gamma(t)$ and $\left.\frac{\partial}{\partial s}\right \vert _{s=0} \tilde{\gamma}(s,t)=\delta \gamma(t)$.
We calculate now $\left.\frac{\partial}{\partial s}\right\vert_{s=0}\mathcal{A}_H(\tilde{\gamma})$ which is the same as $d\mathcal{A}_H(\gamma)\delta \gamma$. The derivative of the first term in the integral is
\begin{align*} \left.\frac{\partial}{\partial s}\right\vert_{s=0}\lambda_{\tilde{\gamma}(s,t)}\bigg(\frac{\partial}{\partial t}\tilde{\gamma}(s,t)\bigg)&\color{red}=d\lambda_{\gamma(t)}\bigg(\dot{\gamma}(t),\delta \gamma(t)+\frac{\partial}{\partial t}\delta\gamma(t)\bigg)\\&\color{red}=d\lambda_{\gamma(t)}\Big(\dot{\gamma}(t),\delta \gamma(t)\Big)+\frac{\partial}{\partial t}\lambda_{\gamma(t)}(\delta\gamma(t)) \end{align*} whereas the second \begin{align*} \left.\frac{\partial}{\partial s}\right\vert_{s=0}H_t(\tilde{\gamma}(s,t))&=(dH_t)_{\gamma(t)}\bigg (\left.\frac{\partial}{\partial s}\right\vert_{s=0}\tilde{\gamma}(s,t)\bigg)=(dH_t)_{\gamma(t)}(\delta\gamma(t))\\ \end{align*}
By the fundamental theorem of integration for the latter term of the first formula we get the formula for the variation.
I know my calculations should be correct but can anyone explain precisely what's going on where I marked the equalities red? Is it even correct what I am doing or do I need to introduce a connection? If there is a nice solution via Lie derivative I would be happy as well. I elaborated the formula by some kind of chain rule but only got a result because I know what the result should be.
I have another suggestion:
Write $\gamma_s(t)=\widetilde{\gamma}(s,t)$.
Then using the Lie derivative $L_{\delta\gamma}$
\begin{align*} \left.\frac{\partial}{\partial s}\right\vert_{s=0}\int_0^1\lambda\Big(\frac{\partial}{\partial t}\gamma(s,t)\Big)dt &= \int_0^1\left.\frac{\partial}{\partial s}\right\vert_{s=0}\gamma_s^*\lambda=\int_0^1\gamma_0^*(L_{\delta\gamma}\lambda)=\int_0^1\gamma^*(\iota_{\delta\gamma}d\lambda+d\iota_{\delta\gamma}\lambda)\\ &=\int_0^1 -\omega(\delta\gamma,\dot{\gamma})+d\lambda(\delta\gamma)= \int_0^1 -\omega(\delta\gamma,\dot{\gamma})dt +\lambda(\delta\gamma(1))-\lambda(\delta\gamma(0)) \end{align*}
Does this make sense?