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Let me try to elaborate a little bit on a not too complicate but not that simple example to see where the symplectic form makes sense. Let us consider a point on the sphere $S^2$, let $$ TS^2 = \{ (x,v) \in S^2 \times {\bf R}^3 \mid x \cdot v = 0 \} $$ Let $$ L : TS^2 - S^2 \to {\bf R} \quad \mbox{with} \quad L(x,v) = \Vert v \Vert $$ be the "length function" as lagrangian. And you look for the variational problem $$ \delta \int L(x(t),\dot x(t))\ dt = \delta \int \Vert \dot x(t) \Vert\ dt = 0. $$ I don't put the limits of the integral on purpose, it would lead to a too long discussion. Since the lagrangian is homogeneous of degree 1 in $v$, we have the Euler identity $$ L(x,v) = \frac{\partial L(x,v)}{\partial v}(v) $$ And the nature of the partial derivative involved above is a map from $TS^2-S^2$ to the cotangent $T^*S^2$ $$ \forall v \in T_xS^2 - \{0\}, \quad \frac{\partial L(x,v)}{\partial v} = \frac{\bar v}{\Vert v \Vert} \in T^*_xS^2 $$ where the bar denotes the transposed. Let's call this map $P$ $$ P : TS^2 - S^2 \to T^*S^2 \quad \mbox{with} \quad P(x,v) = \frac{\partial L(x,v)}{\partial v}. $$ Now let $\lambda = pdx$ the Liouville form on $T^*S^2$, its pullback by $P$, integrated along the curve $\gamma = [t \mapsto (x(t),\dot x(t))]$ is exactly the action $$ \int \Vert \dot x(t) \Vert \ dt = \int_\gamma P^*(\lambda) = \int_{P \circ \gamma} \lambda. $$ Now, let $\tilde \gamma = P \circ \gamma$, this is a path in the image $Y$ of $P$, which is the unit cotangent bundle $$ Y = {\rm Im}(P) = \{ (x, \bar u) \in T^*S^2 \mid \bar u u = 1 \} $$ And the variational condition becomes then $$ \delta \int_{\tilde \gamma} \lambda = \int d\lambda\left(\delta\tilde\gamma(t), \frac{d\tilde \gamma}{dt}\right)\ dt = 0. $$ But $\varpi = d\lambda$ is a 2-form on $Y \simeq US^2 \simeq SO(3)$ which is of odd dimension, actually $3= 2\times 2 -1$. Now, $\varpi$ has a kernel of dimension 1, and $\gamma$ is a solution of the variational problem if and only if $$ \frac{d\tilde \gamma}{dt} \in \ker \varpi_{\tilde \gamma(t)} $$ In this case, the kernel is given explicitly by $$ \frac{dx}{dt} = \alpha u \quad \mbox{and} \quad \frac{du}{dt}= -\alpha x. $$ The quotient space ${\cal S} = Y/\ker\varpi$, the space of solutions of the variational problem, is then equivalent to the sphere $S^2$, thanks to the (SO(3)-moment map) $$ \pi : (x,u) \mapsto x \times u. $$ By construction this space inherits a symplectic form $\omega$ such that $$ \pi^*(\omega) = \varpi. $$ And $({\cal S}, \omega)$ is the space of oriented non parametrized geodesics of the sphere $S^2$ (which by chance is also a sphere $S^2$). Finally what do we get? A space $Y \simeq SO(3)$ made of matrices $y=[x\ u \ x \times u]$, a 1-form $\lambda$, the "action-form" (actually called the "Cartan 1-form"), a characteristic distribution $y \mapsto \ker(d\lambda)$ whose leaves are the pre-images of the point of the sphere $S^2$ by the moment map $\mu : [x\ u \ x \times u] \mapsto x \times u$, and the image of $\mu$ is a symplectic manifolds for the projection $\omega$ of $d\lambda$. Note that in this case $\omega$, proportional to the standard area-form, is closed but not exact.

[[To be continued]]

Let me try to elaborate a little bit on a not too complicate but not that simple example to see where the symplectic form makes sense. Let us consider a point on the sphere $S^2$, let $$ TS^2 = \{ (x,v) \in S^2 \times {\bf R}^3 \mid x \cdot v = 0 \} $$ Let $$ L : TS^2 - S^2 \to {\bf R} \quad \mbox{with} \quad L(x,v) = \Vert v \Vert $$ be the "length function" as lagrangian. And you look for the variational problem $$ \delta \int L(x(t),\dot x(t))\ dt = \delta \int \Vert \dot x(t) \Vert\ dt = 0. $$ I don't put the limits of the integral on purpose, it would lead to a too long discussion. Since the lagrangian is homogeneous of degree 1 in $v$, we have the Euler identity $$ L(x,v) = \frac{\partial L(x,v)}{\partial v}(v) $$ And the nature of the partial derivative involved above is a map from $TS^2-S^2$ to the cotangent $T^*S^2$ $$ \forall v \in T_xS^2 - \{0\}, \quad \frac{\partial L(x,v)}{\partial v} = \frac{\bar v}{\Vert v \Vert} \in T^*_xS^2 $$ where the bar denotes the transposed, that is $\bar v w = v \cdot w$. Let's call this map $P$ $$ P : TS^2 - S^2 \to T^*S^2 \quad \mbox{with} \quad P(x,v) = \left(x,\frac{\partial L(x,v)}{\partial v}\right) = \left(x, \frac{\bar v}{\Vert v \Vert}\right). $$ Now let $\lambda = pdx$ the Liouville form on $T^*S^2$, its pullback by $P$, integrated along the curve $\gamma = [t \mapsto (x(t),\dot x(t))]$ is exactly the action $$ \int \Vert \dot x(t) \Vert \ dt = \int_\gamma P^*(\lambda) = \int_{P \circ \gamma} \lambda. $$ Now, let $\tilde \gamma = P \circ \gamma$, this is a path in the image $Y$ of $P$, which is the unit-cotangent bundle $$ Y = {\rm Im}(P) = \{ (x, \bar u) \in T^*S^2 \mid \bar u u = 1 \} $$ And the variational condition becomes then $$ \delta \int_{\tilde \gamma} \lambda = \int d\lambda\left(\delta\tilde\gamma(t), \frac{d\tilde \gamma}{dt}\right)\ dt = 0. $$ But $\varpi = d\lambda$ is a 2-form on $Y \simeq US^2 \simeq SO(3)$ which is of odd dimension, actually $3= 2\times 2 -1$. Now, $\varpi$ has a kernel of dimension 1, and $\gamma$ is a solution of the variational problem if and only if $$ \frac{d\tilde \gamma}{dt} \in \ker \varpi_{\tilde \gamma(t)} $$ In this case, the kernel is given explicitly by $$ \frac{dx}{dt} = \alpha u \quad \mbox{and} \quad \frac{du}{dt}= -\alpha x. $$ The quotient space ${\cal S} = Y/\ker\varpi$, the space of solutions of the variational problem, is then equivalent to the sphere $S^2$, thanks to the (SO(3)-moment map) $$ \pi : (x,u) \mapsto x \times u. $$ By construction this space inherits a symplectic form $\omega$ such that $$ \pi^*(\omega) = \varpi. $$ And $({\cal S}, \omega)$ is the space of oriented non parametrized geodesics of the sphere $S^2$ (which by chance is also a sphere $S^2$). Finally what do we get? A space $Y \simeq US^2 \simeq SO(3)$ made of couples $(x,u)$ or matrices $y=[x\ u \ x \times u]$, a 1-form $\lambda$, the "action-form" (actually called the "Cartan 1-form"), a characteristic distribution $y \mapsto \ker(d\lambda)$ whose leaves are the pre-images of the point of the sphere $S^2$ by the moment map $\mu : (x,u) \mapsto x \times u$, and the image of $\mu$ is a symplectic manifolds for the projection $\omega$ of $d\lambda$. Note that in this case $\omega$, proportional to the standard area-form, is closed but not exact.

Consider the pullback by $\sigma$ of the $S^1$-principal bundle $\pi : Y \to {\cal S}$, this is a principal bundle on $D^2$, but $D^2$ is contractible, so this fiber bundle is trivial, thus it admits a smooth section, that is a lift $\tilde \sigma : D^2 \to Y$, that is $\pi \circ \tilde \sigma = \sigma$. Now, $$ \int_\sigma \omega = \int_{\pi\circ\tilde\sigma} \omega = \int_{\tilde\sigma} \pi^*(\sigma) = \int_{\tilde\sigma} d\lambda = \int_{\tilde\gamma} \lambda \quad \mbox{with} \quad \tilde\gamma = \partial\tilde\sigma. $$ Let us write $\tilde \gamma(s) = (x_s,\bar u_s) \in Y$, and let us assume that the parameter $s$ runs over $[0,2\pi]$ to describe $\tilde \gamma = \partial \tilde \sigma$, then $$ \int_\sigma \omega = \int_{\tilde\gamma} \lambda = \int_0^{2\pi} \bar u_s \frac{dx_s}{ds} \ ds. $$ And this is the action of the unit vector $s \mapsto u_s$ distribution along the curve $s \mapsto x_s$. And let us remember that the vector $x_s \times u_s$ describes a geodesic of the sphere $S^2$ for all $s$, and $s$ is not the time parameter of this geodesic.

The symplectic area contained in a closed curved, that is the boundary of map of a disc, is the "action along the curve". $$ \int_\sigma \omega = \int_\sigma d\lambda = \int_{\partial \sigma} \lambda = \int_0^{2\pi} \lambda_{\gamma(t)}(\dot \gamma(t)) dt, $$ where $\sigma$ is a smooth map from the disc to $M$, and $\gamma = \partial \sigma$. In all cases, the pullback of the 2-form $\omega$ by $\sigma$ is exact since the disc is contractible, so there exists a primitive $\lambda$, on the disc, and you apply Stokes' theorem.

Let me try to elaborate a little bit on a not too complicate but not that simple example to see where the symplectic form makes sense. Let us consider a point on the sphere $S^2$, let $$ TS^2 = \{ (x,v) \in S^2 \times {\bf R}^3 \mid x \cdot v = 0 \} $$ Let $$ L : TS^2 - S^2 \to {\bf R} \quad \mbox{with} \quad L(x,v) = \Vert v \Vert $$ be the "length function" as lagrangian. And you look for the variational problem $$ \delta \int L(x(t),\dot x(t))\ dt = \delta \int \Vert \dot x(t) \Vert\ dt = 0. $$ I don't put the limits of the integral on purpose, it would lead to a too long discussion. Since the lagrangian is homogeneous of degree 1 in $v$, we have the Euler identity $$ L(x,v) = \frac{\partial L(x,v)}{\partial v}(v) $$ And the nature of the partial derivative involved above is a map from $TS^2-S^2$ to the cotangent $T^*S^2$ $$ \forall v \in T_xS^2 - \{0\}, \quad \frac{\partial L(x,v)}{\partial v} = \frac{\bar v}{\Vert v \Vert} \in T^*_xS^2 $$ where the bar denotes the transposed. Let's call this map $P$ $$ P : TS^2 - S^2 \to T^*S^2 \quad \mbox{with} \quad P(x,v) = \frac{\partial L(x,v)}{\partial v}. $$ Now let $\lambda = pdx$ the Liouville form on $T^*S^2$, its pullback by $P$, integrated along the curve $\gamma = [t \mapsto (x(t),\dot x(t))]$ is exactly the action $$ \int \Vert \dot x(t) \Vert \ dt = \int_\gamma P^*(\lambda) = \int_{P \circ \gamma} \lambda. $$

[[To be continued]]

The symplectic area contained in a closed curved, that is the boundary of map of a disc, is the "action along the curve". $$ \int_\sigma \omega = \int_\sigma d\lambda = \int_{\partial \sigma} \lambda = \int_0^{2\pi} \lambda_{\gamma(t)}(\dot \gamma(t)) dt, $$ where $\sigma$ is a smooth map from the disc to $M$, and $\gamma = \partial \sigma$. In all cases, the pullback of the 2-form $\omega$ by $\sigma$ is exact since the disc is contractible, so there exists a primitive $\lambda$, on the disc, and you apply Stokes' theorem.

[I apologize for the lengthy answer]

Let me try to elaborate a little bit on a not too complicate but not that simple example to see where the symplectic form makes sense. Let us consider a point on the sphere $S^2$, let $$ TS^2 = \{ (x,v) \in S^2 \times {\bf R}^3 \mid x \cdot v = 0 \} $$ Let $$ L : TS^2 - S^2 \to {\bf R} \quad \mbox{with} \quad L(x,v) = \Vert v \Vert $$ be the "length function" as lagrangian. And you look for the variational problem $$ \delta \int L(x(t),\dot x(t))\ dt = \delta \int \Vert \dot x(t) \Vert\ dt = 0. $$ I don't put the limits of the integral on purpose, it would lead to a too long discussion. Since the lagrangian is homogeneous of degree 1 in $v$, we have the Euler identity $$ L(x,v) = \frac{\partial L(x,v)}{\partial v}(v) $$ And the nature of the partial derivative involved above is a map from $TS^2-S^2$ to the cotangent $T^*S^2$ $$ \forall v \in T_xS^2 - \{0\}, \quad \frac{\partial L(x,v)}{\partial v} = \frac{\bar v}{\Vert v \Vert} \in T^*_xS^2 $$ where the bar denotes the transposed. Let's call this map $P$ $$ P : TS^2 - S^2 \to T^*S^2 \quad \mbox{with} \quad P(x,v) = \frac{\partial L(x,v)}{\partial v}. $$ Now let $\lambda = pdx$ the Liouville form on $T^*S^2$, its pullback by $P$, integrated along the curve $\gamma = [t \mapsto (x(t),\dot x(t))]$ is exactly the action $$ \int \Vert \dot x(t) \Vert \ dt = \int_\gamma P^*(\lambda) = \int_{P \circ \gamma} \lambda. $$ Now, let $\tilde \gamma = P \circ \gamma$, this is a path in the image $Y$ of $P$, which is the unit cotangent bundle $$ Y = {\rm Im}(P) = \{ (x, \bar u) \in T^*S^2 \mid \bar u u = 1 \} $$ And the variational condition becomes then $$ \delta \int_{\tilde \gamma} \lambda = \int d\lambda\left(\delta\tilde\gamma(t), \frac{d\tilde \gamma}{dt}\right)\ dt = 0. $$ But $\varpi = d\lambda$ is a 2-form on $Y \simeq US^2 \simeq SO(3)$ which is of odd dimension, actually $3= 2\times 2 -1$. Now, $\varpi$ has a kernel of dimension 1, and $\gamma$ is a solution of the variational problem if and only if $$ \frac{d\tilde \gamma}{dt} \in \ker \varpi_{\tilde \gamma(t)} $$ In this case, the kernel is given explicitly by $$ \frac{dx}{dt} = \alpha u \quad \mbox{and} \quad \frac{du}{dt}= -\alpha x. $$ The quotient space ${\cal S} = Y/\ker\varpi$, the space of solutions of the variational problem, is then equivalent to the sphere $S^2$, thanks to the (SO(3)-moment map) $$ \pi : (x,u) \mapsto x \times u. $$ By construction this space inherits a symplectic form $\omega$ such that $$ \pi^*(\omega) = \varpi. $$ And $({\cal S}, \omega)$ is the space of oriented non parametrized geodesics of the sphere $S^2$ (which by chance is also a sphere $S^2$). Finally what do we get? A space $Y \simeq SO(3)$ made of matrices $y=[x\ u \ x \times u]$, a 1-form $\lambda$, the "action-form" (actually called the "Cartan 1-form"), a characteristic distribution $y \mapsto \ker(d\lambda)$ whose leaves are the pre-images of the point of the sphere $S^2$ by the moment map $\mu : [x\ u \ x \times u] \mapsto x \times u$, and the image of $\mu$ is a symplectic manifolds for the projection $\omega$ of $d\lambda$. Note that in this case $\omega$, proportional to the standard area-form, is closed but not exact.

Now you can ask the same question as previously: "What does mean the area include in a disc $\sigma : D^2 \to {\cal S}$?"

[[To be continued]]

Note 1. that this construction can be applied to any homogeneous lagrangian, and for non-homogeneous lagrangian, first we homogenize them and after we apply this construction.

Bibliography Jean-Marie Souriau, "Structure des Systèmes Dynamiques", Dunod ed., Paris 1970

The symplectic area contained in a closed curved, that is the boundary of map of a disc, is the "action along the curve". $$ \int_\sigma \omega = \int_\sigma d\lambda = \int_{\partial \sigma} \lambda = \int_0^{2\pi} \lambda_{\gamma(t)}(\dot \gamma(t)) dt, $$ where $\sigma$ is a smooth map from the disc to $M$, and $\gamma = \partial \sigma$. In all cases, the pullback of the 2-form $\omega$ by $\sigma$ is exact since the disc is contractible, so there exists a primitive $\lambda$, on the disc, and you apply Stokes' theorem.

The symplectic area contained in a closed curved, that is the boundary of map of a disc, is the "action along the curve". $$ \int_\sigma \omega = \int_\sigma d\lambda = \int_{\partial \sigma} \lambda = \int_0^{2\pi} \lambda_{\gamma(t)}(\dot \gamma(t)) dt, $$ where $\sigma$ is a smooth map from the disc to $M$, and $\gamma = \partial \sigma$. In all cases, the pullback of the 2-form $\omega$ by $\sigma$ is exact since the disc is contractible, so there exists a primitive $\lambda$, on the disc, and you apply Stokes' theorem.

Let me try to elaborate a little bit on a not too complicate but not that simple example to see where the symplectic form makes sense. Let us consider a point on the sphere $S^2$, let $$ TS^2 = \{ (x,v) \in S^2 \times {\bf R}^3 \mid x \cdot v = 0 \} $$ Let $$ L : TS^2 - S^2 \to {\bf R} \quad \mbox{with} \quad L(x,v) = \Vert v \Vert $$ be the "length function" as lagrangian. And you look for the variational problem $$ \delta \int L(x(t),\dot x(t))\ dt = \delta \int \Vert \dot x(t) \Vert\ dt = 0. $$ I don't put the limits of the integral on purpose, it would lead to a too long discussion. Since the lagrangian is homogeneous of degree 1 in $v$, we have the Euler identity $$ L(x,v) = \frac{\partial L(x,v)}{\partial v}(v) $$ And the nature of the partial derivative involved above is a map from $TS^2-S^2$ to the cotangent $T^*S^2$ $$ \forall v \in T_xS^2 - \{0\}, \quad \frac{\partial L(x,v)}{\partial v} = \frac{\bar v}{\Vert v \Vert} \in T^*_xS^2 $$ where the bar denotes the transposed. Let's call this map $P$ $$ P : TS^2 - S^2 \to T^*S^2 \quad \mbox{with} \quad P(x,v) = \frac{\partial L(x,v)}{\partial v}. $$ Now let $\lambda = pdx$ the Liouville form on $T^*S^2$, its pullback by $P$, integrated along the curve $\gamma = [t \mapsto (x(t),\dot x(t))]$ is exactly the action $$ \int \Vert \dot x(t) \Vert \ dt = \int_\gamma P^*(\lambda) = \int_{P \circ \gamma} \lambda. $$

[[To be continued]]