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Let $(M,\omega = d\alpha)$ be an exact symplectic manifold and $\gamma \in C^\infty(I,M)$ a path in $M$. Given $X_1,X_2 \in \mathfrak{X}(\gamma)$ I would like to compute $$\frac{\partial^2}{\partial\sigma_1\partial\sigma_2}\bigg\vert_{\sigma_1 = \sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}^* \alpha,$$ where $\gamma_{\sigma_1,\sigma_2}$ is a smooth $2$-parameter family of curves such that $\partial_{\sigma_1}\vert_{\sigma_1 = 0} \gamma_{\sigma_1,\sigma_2} = X_1$ and $\partial_{\sigma_2}\vert_{\sigma_2 = 0} \gamma_{\sigma_1,\sigma_2} = X_2$. By the superb answer of @TobiasDiez here we can use the generalised Cartan formula to compute $$\frac{\partial}{\partial \sigma_1}\bigg\vert_{\sigma_1 = 0}\left( \gamma^*_{\sigma_1,0}i_{\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}}(\omega \circ \gamma_{\sigma_1,0}) + d\gamma_{\sigma_1,0}^*i_{\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}}(\alpha \circ \gamma_{\sigma_1,0})\right).$$ Equivalently, $$\frac{\partial}{\partial \sigma_1}\bigg\vert_{\sigma_1 = 0}\left(\omega(\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2},\dot{\gamma}_{\sigma_1,0}) + d(\alpha(\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}))\right).$$ I am not sure how to proceed further, but I think there is a need to choose a (symplectic) connection. If it helps, actually there is an integral in front of this expression and more terms as I want to compute the Hessian of the Rabinowitz action functional

Let $(M,\omega = d\alpha)$ be an exact symplectic manifold and $\gamma \in C^\infty(I,M)$ a path in $M$. Given $X_1,X_2 \in \mathfrak{X}(\gamma)$ I would like to compute $$\frac{\partial^2}{\partial\sigma_1\partial\sigma_2}\bigg\vert_{\sigma_1 = \sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}^* \alpha,$$ where $\gamma_{\sigma_1,\sigma_2}$ is a smooth $2$-parameter family of curves such that $\partial_{\sigma_1}\vert_{\sigma_1 = 0} \gamma_{\sigma_1,\sigma_2} = X_1$ and $\partial_{\sigma_2}\vert_{\sigma_2 = 0} \gamma_{\sigma_1,\sigma_2} = X_2$. By the superb answer of @TobiasDiez here we can use the generalised Cartan formula to compute $$\frac{\partial}{\partial \sigma_1}\bigg\vert_{\sigma_1 = 0}\left( \gamma^*_{\sigma_1,0}i_{\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}}(\omega \circ \gamma_{\sigma_1,0}) + d\gamma_{\sigma_1,0}^*i_{\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}}(\alpha \circ \gamma_{\sigma_1,0})\right).$$ Equivalently, $$\frac{\partial}{\partial \sigma_1}\bigg\vert_{\sigma_1 = 0}\left(\omega(\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2},\dot{\gamma}_{\sigma_1,0}) + d(\alpha(\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}))\right).$$ I am not sure how to proceed further, but I think there is a need to choose a (symplectic) connection. If it helps, actually there is an integral in front of this expression and more terms as I want to compute the Hessian of the Rabinowitz action functional $$\mathrm{Hess}\mathcal{R}^H\vert_{(\gamma,\tau)}\left( (X_1,\tilde{\tau}_1),(X_2,\tilde{\tau}_2)\right) := \frac{\partial^2}{\partial\sigma_1\partial\sigma_2}\bigg\vert_{\sigma_1 = \sigma_2 = 0}\mathcal{R}^H(\gamma_{\sigma_1,\sigma_2},\tau + \sigma_1\tilde{\tau}_1 + \sigma_2\tilde{\tau}_2)$$ with $$\mathcal{R}^H \colon C^\infty(\mathbb{S}^1,M) \times \mathbb{R} \to \mathbb{R}, \qquad \mathcal{R}^H(\gamma,\tau) := \int_0^1 \gamma^*\alpha - \tau\int_0^1 H \circ \gamma.$$

Let $(M,\omega = d\alpha)$ be an exact symplectic manifold and $\gamma \in C^\infty(I,M)$ a path in $M$. Given $X_1,X_2 \in \mathfrak{X}(\gamma)$ I would like to compute $$\frac{\partial^2}{\partial\sigma_1\partial\sigma_2}\bigg\vert_{\sigma_1 = \sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}^* \alpha,$$ where $\gamma_{\sigma_1,\sigma_2}$ is a smooth $2$-parameter family of curves such that $\partial_{\sigma_1}\vert_{\sigma_1 = 0} \gamma_{\sigma_1,\sigma_2} = X_1$ and $\partial_{\sigma_2}\vert_{\sigma_2 = 0} \gamma_{\sigma_1,\sigma_2} = X_2$. By the superb answer of @TobiasDiez here we can use the generalised Cartan formula to compute $$\frac{\partial}{\partial \sigma_1}\bigg\vert_{\sigma_1 = 0}\left( \gamma^*_{\sigma_1,0}i_{\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}}(\omega \circ \gamma_{\sigma_1,0}) + d\gamma_{\sigma_1,0}^*i_{\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}}(\alpha \circ \gamma_{\sigma_1,0})\right).$$ Equivalently, $$\frac{\partial}{\partial \sigma_1}\bigg\vert_{\sigma_1 = 0}\left(\omega(\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2},\dot{\gamma}_{\sigma_1,0}) + d(\alpha(\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}))\right).$$ I am not sure how to proceed further.

Let $(M,\omega = d\alpha)$ be an exact symplectic manifold and $\gamma \in C^\infty(I,M)$ a path in $M$. Given $X_1,X_2 \in \mathfrak{X}(\gamma)$ I would like to compute $$\frac{\partial^2}{\partial\sigma_1\partial\sigma_2}\bigg\vert_{\sigma_1 = \sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}^* \alpha,$$ where $\gamma_{\sigma_1,\sigma_2}$ is a smooth $2$-parameter family of curves such that $\partial_{\sigma_1}\vert_{\sigma_1 = 0} \gamma_{\sigma_1,\sigma_2} = X_1$ and $\partial_{\sigma_2}\vert_{\sigma_2 = 0} \gamma_{\sigma_1,\sigma_2} = X_2$. By the superb answer of @TobiasDiez here we can use the generalised Cartan formula to compute $$\frac{\partial}{\partial \sigma_1}\bigg\vert_{\sigma_1 = 0}\left( \gamma^*_{\sigma_1,0}i_{\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}}(\omega \circ \gamma_{\sigma_1,0}) + d\gamma_{\sigma_1,0}^*i_{\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}}(\alpha \circ \gamma_{\sigma_1,0})\right).$$ Equivalently, $$\frac{\partial}{\partial \sigma_1}\bigg\vert_{\sigma_1 = 0}\left(\omega(\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2},\dot{\gamma}_{\sigma_1,0}) + d(\alpha(\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}))\right).$$ I am not sure how to proceed further, but I think there is a need to choose a (symplectic) connection. If it helps, actually there is an integral in front of this expression and more terms as I want to compute the Hessian of the Rabinowitz action functional

Let $(M,\omega = d\alpha)$ be an exact symplectic manifold and $\gamma \in C^\infty(I,M)$ a path in $M$. Given $X_1,X_2 \in \mathfrak{X}(\gamma)$ I would like to compute $$\frac{\partial^2}{\partial\sigma_1\partial\sigma_2}\bigg\vert_{\sigma_1 = \sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}^* \alpha,$$ where $\gamma_{\sigma_1,\sigma_2}$ is a smooth $2$-parameter family of curves such that $\partial_{\sigma_1}\vert_{\sigma_1 = 0} \gamma_{\sigma_1,\sigma_2} = X_1$ and $\partial_{\sigma_2}\vert_{\sigma_2 = 0} \gamma_{\sigma_1,\sigma_2} = X_2$. By the superb answer of @TobiasDiez here we can use the generalised Cartan formula to compute $$\frac{\partial}{\partial \sigma_1}\bigg\vert_{\sigma_1 = 0}\left( \gamma^*_{\sigma_1,0}i_{\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}}(\omega \circ \gamma_{\sigma_1,0}) + d\gamma_{\sigma_1,0}^*i_{\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,0}}(\alpha \circ \gamma_{\sigma_1,0})\right).$$ Equivalently, $$\frac{\partial}{\partial \sigma_1}\bigg\vert_{\sigma_1 = 0}\left(\omega(\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2},\dot{\gamma}_{\sigma_1,0}) + d(\alpha(\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,0}))\right).$$ I am not sure how to proceed further.

Let $(M,\omega = d\alpha)$ be an exact symplectic manifold and $\gamma \in C^\infty(I,M)$ a path in $M$. Given $X_1,X_2 \in \mathfrak{X}(\gamma)$ I would like to compute $$\frac{\partial^2}{\partial\sigma_1\partial\sigma_2}\bigg\vert_{\sigma_1 = \sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}^* \alpha,$$ where $\gamma_{\sigma_1,\sigma_2}$ is a smooth $2$-parameter family of curves such that $\partial_{\sigma_1}\vert_{\sigma_1 = 0} \gamma_{\sigma_1,\sigma_2} = X_1$ and $\partial_{\sigma_2}\vert_{\sigma_2 = 0} \gamma_{\sigma_1,\sigma_2} = X_2$. By the superb answer of @TobiasDiez here we can use the generalised Cartan formula to compute $$\frac{\partial}{\partial \sigma_1}\bigg\vert_{\sigma_1 = 0}\left( \gamma^*_{\sigma_1,0}i_{\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}}(\omega \circ \gamma_{\sigma_1,0}) + d\gamma_{\sigma_1,0}^*i_{\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}}(\alpha \circ \gamma_{\sigma_1,0})\right).$$ Equivalently, $$\frac{\partial}{\partial \sigma_1}\bigg\vert_{\sigma_1 = 0}\left(\omega(\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2},\dot{\gamma}_{\sigma_1,0}) + d(\alpha(\partial_{\sigma_2}\vert_{\sigma_2 = 0}\gamma_{\sigma_1,\sigma_2}))\right).$$ I am not sure how to proceed further.