A review on coisotropic reduction in symplectic, cosymplectic, contact and co-contact Hamiltonian systems

Definition 3.3. Let $(M , \omega)$ be a symplectic manifold. Define the tangent symplectic structure on TM as $\omega_0 = -\mathrm{d}\lambda_0$ where $\lambda_0 = \flat_\omega^*\lambda_M$, and λ_M is the Liouville 1-form on the cotangent bundle.

Recall that λ_M is defined as follows:

$$\begin{equation*} \lambda_M\left(\alpha_x\right)\left(X_{\alpha_x}\right) = \alpha_x\left(d_{\alpha_x}\pi_M \cdot X_{\alpha_x}\right) \end{equation*}$$

where $X_{\alpha_z} \in T_{\alpha_x}(T^*M)$, $\alpha_x \in T^*_xM$, and $\pi_M : T^*M \longrightarrow M$ is the canonical projection. The Liouville 1-form can be also defined as the unique 1-form λ_M on $T^*M$ such that, for every 1-form $\alpha:M \rightarrow T^*M$,

$$\begin{equation*}\alpha^*\left(\lambda_M\right) = \alpha.\end{equation*}$$

In coordinates $(q^i, p_i, \dot{q}^i, \dot{p}_i)$ the tangent symplectic structure is

$$\begin{equation*}\omega_0 = -\mathrm{d}q^i \wedge \mathrm{d}\dot{p}_i - \mathrm{d}\dot{q}^i \wedge \mathrm{d}p_i.\end{equation*}$$

Proposition 3.1. Let $X: M \rightarrow TM$ be a vector field. Then

$$\begin{equation*}X^*\left(\lambda_0\right) = \flat_\omega\left(X\right).\end{equation*}$$

Proof. This is a straight-forward verification. Let $v \in T_qM$, then we have

$$\begin{align*} \langle X^*\lambda_0 ,v \rangle & = \langle \lambda_0, \mathrm{d}_qX \cdot v \rangle = \langle \lambda_M, \mathrm{d}_{X\left(q\right)}\flat_\omega \cdot \mathrm{d}_q X \cdot v \rangle \\ & = \langle \flat_\omega\left(X\right)^* \left(\lambda_M\right), v \rangle = \langle \flat_\omega\left(X\right), v \rangle. \end{align*}$$

 □

Proposition 3.2. Let $X: M \rightarrow TM$ be a vector field. Then X is locally Hamiltonian if and only if X(M) is a Lagrangian submanifold of $(TM, \omega_0)$.

Proof. We only check that X(M) is isotropic using lemma 3.1, since $\operatorname{dim} X(M) = \operatorname{dim} M = \frac{1}{2} \operatorname{dim} TM.$ In fact:

$$\begin{align*} X^*\omega_0 = -X^*\left(\mathrm{d}\theta_0\right) = -\mathrm{d}\left(X^*\theta_0\right) = -\mathrm{d}\left(\flat_\omega\left(X\right)\right), \end{align*}$$

which gives the characterization. □

We can also check this last proposition easily in coordinates. Indeed, let

$$\begin{equation*}X = X^i \frac{\partial }{\partial q^i} + Y_i \frac{\partial }{\partial p_i}.\end{equation*}$$

We have

$$\begin{equation*}-X^* \omega_0 = \frac{\partial Y_i}{\partial q^j} \mathrm{d}q^i \wedge \mathrm{d}q^j + \left( \frac{\partial Y_i}{\partial p_j} + \frac{\partial X^j}{\partial q_i} \right) \mathrm{d}q^i \wedge \mathrm{d}p_j+ \frac{\partial X^i}{\partial p_j} \mathrm{d}p_j \wedge \mathrm{d}p_i,\end{equation*}$$

and thus, X defines a Lagrangian submanifold if and only if

$$\begin{align*} \frac{\partial Y_i}{\partial q^j} - \frac{\partial Y_j}{\partial q^i}& = 0,\\ \frac{\partial Y_i}{\partial p_j} + \frac{\partial X^j}{\partial q_i} & = 0,\\ \frac{\partial X^i}{\partial p_j} - \frac{\partial X^j}{\partial p_i} & = 0. \end{align*}$$

Taking

$$\begin{equation*}\left(G^1, \dots, G^n, G^{n+1}, \dots, G^{2n}\right) : = \left(X^i, -Y_i\right)\end{equation*}$$

and

$$\begin{equation*}\left(x^1, \dots, x^n, x^{n+1}, \dots, x^{2n}\right) : = \left(q^i, p_i\right),\end{equation*}$$

these conditions become

$$\begin{equation*}\frac{\partial G^i}{\partial x^j} = \frac{\partial G^j}{\partial x^i}.\end{equation*}$$

This implies $G^i = \displaystyle \frac{\partial H}{\partial x^i},$ for some local function H. It is clear that locally, we have $X = X_H$.

Now, given a coisotropic submanifold $N \hookrightarrow M$, we define the distribution $(TN)^{\perp_\omega}$ on N as the subbundle of $TM |_N$ consisting of all ω-orthogonal spaces $(T_qN)^{\perp_\omega}$. Note that this distribution is regular and its rank is $\operatorname{dim} M - \operatorname{dim} N$.

Proposition 3.3. Let $(M, \omega)$ be a symplectic manifold and $N \hookrightarrow M$ be a coisotropic submanifold. The distribution $q \mapsto (T_qN)^{\perp_\omega}$ is involutive.

Proof. Let $X,Y$ be vector fields along N with values in $TN^{\perp_\omega}$ and Z be any other vector field tangent to N. Since ω is closed we have

$$\begin{align*} 0 = & \left( \mathrm{d}\omega\right)\left(X,Y,Z\right) = X\left(\omega\left(Y,Z\right)\right) - Y\left(\omega\left(X,Z\right)\right) + Z\omega\left(X,Y\right)\\ & - \omega\left(\left[X,Y\right], Z\right) + \omega\left(\left[X,Z\right], Y\right) - \omega\left(\left[Y,Z\right],X\right) = - \omega\left(\left[X,Y\right],Z\right), \end{align*}$$

since $X,Y$ belong to the orthogonal complement of TN. We conclude that $\omega([X,Y],Z) = 0$ for every field Z tangent to N, that is, $[X,Y] \in (TN)^{\perp_\omega}$. □

Since the distribution is involutive and regular, the Frobenius' Theorem guarantees the existence of a maximal regular foliation $\mathcal{F}$ of N, that is, a decomposition of N into maximal submanifolds tangent to the distribution. In what follows, we suppose that $N / \mathcal{F}$ (the space of all leaves) admits a manifold structure so that the projection

$$\begin{equation*}\pi : N \rightarrow N / \mathcal{F}\end{equation*}$$

is a submersion. The main result is the Weinstein reduction theorem [62]:

Theorem 3.2 (Coisotropic reduction in the symplectic setting). Let $(M , \omega)$ be a symplectic manifold and $N \hookrightarrow M$ be a coisotropic submanifold. If $N / \mathcal{F}$ (the spaces of all leaves under the distribution $q \mapsto (T_qN)^{\perp_\omega}$) admits a manifold structure such that $N \xrightarrow{\pi} N / \mathcal{F}$ is a submersion, there exist an unique 2-form ω_N on $N / \mathcal{F}$ that defines a symplectic manifold structure such that, if $N \xrightarrow{i}M$ is the natural inclusion, then $i^* \omega = \pi ^* \omega_N$. The following diagram summarizes the situation:

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Proof. Uniqueness is guaranteed from the imposed relation since it forces us to define

$$\begin{equation*}\left(\omega_N\right)_{\left[q\right]}\left(\left[u\right],\left[v\right]\right): = \omega\left(u,v\right),\end{equation*}$$

where $[u]: = T\pi(q) \cdot u$. We only need to check that this is a well-defined closed form and that it is non-degenerate.

We begin showing that our definition does not depend on the representative of the vector $[u]$. For this, it is sufficient to observe that $(\omega_N)_{[q]}([u],[v]) = 0$ whenever u is a vector in the distribution.

Furthermore,

$$\begin{equation*}\mathcal{L}_X \omega = \mathrm{d}i_X \omega + i_X \mathrm{d}\omega = 0\end{equation*}$$

for every vector field X in N with values in $(TN^{\perp_\omega})$, and this implies the independence of the point (for every two points in the same leaf of the foliation can be joined by a finite union of flows of such fields).

It is clearly non-degenerate and it is closed, since $\mathrm{d}\pi^* \omega_N = i_*\mathrm{d}\omega = 0$ and π is a submersion. □

Definition 3.4 (Clean intersection). We say that two submanifolds $L, N \hookrightarrow M$ have clean intersection if $L \cap N \hookrightarrow M$ is again a submanifold and $T_q(L \cap N) = T_q L \cap T_qN$, for every $q \in L \cap N$.

Proposition 3.4. Let $L \hookrightarrow M$ be a Lagrangian submanifold and $N \hookrightarrow M$ a coisotropic submanifold. If they have clean intersection and $L_N: = \pi(L \cap N)$ is a submanifold of $N/\mathcal{F}$, L_N is Lagrangian.

Proof. It is sufficient to see that is isotropic and that it has maximal dimension in $N / \mathcal{F}$. It is isotropic since $[u] \in T_q(L _ N)$ implies $\omega_N([u],[v]) = \omega(u,v) = 0$, for every $[v] \in T_q(L _ N)$. Now, since $\operatorname{Ker} \mathrm{d}_q\pi = (T_qN)^{\perp_\omega}$, the kernel-range formula yields

$$\begin{equation} \operatorname{dim} L_N = \operatorname{dim}\left(L \cap N\right) - \operatorname{dim} \left(T_q L \cap \left(T_qN\right)^{\perp_\omega}\right). \end{equation} \tag{ 1 }$$

Furthermore,

$$\begin{equation} \operatorname{dim}\left(L \cap N\right) + \operatorname{dim} \left(T_qL + \left(T_qN\right)^{\perp_\omega}\right) = \operatorname{dim} M, \end{equation} \tag{ 2 }$$

because L is Lagrangian and N coisotropic. Substituting (2) in (1) we obtain

$$\begin{align} \nonumber \operatorname{dim} L_N & = \operatorname{dim} M - \operatorname{dim} \left(T_qL + \left(T_qN\right)^{\perp_\omega}\right)- \operatorname{dim} \left(T_q L \cap \left(T_qN \right)^{\perp_\omega}\right) \nonumber \\ & = \operatorname{dim} M - \operatorname{dim} L - \operatorname{dim} \left(T_qN\right)^{\perp_\omega} = \operatorname{dim} M - \operatorname{dim} L -\left(\operatorname{dim} M - \operatorname{dim} N\right)\nonumber \\ & = \operatorname{dim} N - \operatorname{dim} L = \operatorname{dim} N - \frac{1}{2} \operatorname{dim} \end{align}$$

which is exactly $\frac{1}{2} \operatorname{dim} N/ \mathcal{F}$, as a direct calculation shows. □

As an example of coisotropic reduction, let us take

$$\begin{equation*}M : = \mathbb{R}^{2\left(n +1\right)},\end{equation*}$$

and

$$\begin{equation*}N : = \mathbb{S}^{2n +1}.\end{equation*}$$

Since it has codimension 1, it defines a coisotropic submanifold if we endow M with the natural symplectic form $\omega : = \mathrm{d} q^i \wedge \mathrm{d} p_i.$ It is easy to check that $(TN)^{\perp_\omega}$ is generated by

$$\begin{equation*}\sum_i \left(p_i \frac{\partial }{\partial q^i} - q^i \frac{\partial }{\partial p_i}\right),\end{equation*}$$

where $(q^i, p_i)$ are canonical coordinates in $\mathbb{R}^{2(n +1)}$. Therefore, the leaves of the corresponding foliation are precisely the orbits of the previous vector field in $\mathbb{S}^{2n +1}$. Making the identification

$$\begin{equation*}\mathbb{ R}^{2\left(n +1\right)} = \mathbb{C}^{n +1},\end{equation*}$$

two points $x, y \in \mathbb{S}^{2n +1}$ are in the same orbit if and only if there exist some $\alpha \in \mathbb{C}$ ($|\alpha| = 1$) such that

$$\begin{equation*}\alpha x = y.\end{equation*}$$

This implies that

$$\begin{equation*}\mathbb{S}^{2n +1}/\mathcal{F}\end{equation*}$$

is the complex projective space of complex dimension n, $\mathbb{P}^n \mathbb{C}$. Therefore, we conclude that through coisotropic reduction we can define a natural symplectic structure on $\mathbb{P}^n \mathbb{C},$ for every n.

The above example is taken from exercise 5.3B in [1].

Definition 5.3. Given a cosymplectic manifold $(M, \Omega, \theta)$, we define the symplectic structure on TM as $\Omega_0 : = - \mathrm{d}\lambda_0$, where $\lambda_0 = \flat_{\theta,\Omega}^*\lambda_M$, λ_M being the Liouville 1-form in the cotangent bundle $T^*M$.

There is another expression of Ω₀, namely

$$\begin{equation*}\Omega_0 = -\Omega^c -\theta^c \wedge \theta^v\end{equation*}$$

as one can verify [11]. Here, $\alpha^v, \alpha^c$ denote the complete and vertical lifts of a form α on M to its tangent bundle TM [16]. This implies that in the induced coordinates in TM, $(q^i,p_i, t, \dot{q}^i, \dot{p}_i, \dot{t})$,

$$\begin{equation*}\Omega_0 = -\mathrm{d}q^i \wedge \mathrm{d}\dot{p}_i - \mathrm{d}\dot{q}^i \wedge \mathrm{d}p_i - \mathrm{d} \dot{t} \wedge \mathrm{d}t.\end{equation*}$$

Proposition 5.2. Let $(M, \Omega, \theta)$ be a cosymplectic manifold and $X: M \rightarrow TM$ a vector field. Then X(M) is a Lagrangian submanifold of $(TM, \Omega_0)$ if and only if X is locally a gradient vector field.

Proof. It is easily checked that $X^*\lambda_0 = \flat_{\theta,\Omega}(X)$ (just like in proposition 3.1) and then, X(M) is Lagrangian if and only if

$$\begin{equation*}0 = X^*\Omega_0 = -X^* \mathrm{d}\lambda_0 = - \mathrm{d}\flat_{\theta,\Omega}\left(X\right),\end{equation*}$$

that is, X is locally a gradient vector field. □

We can also check this in coordinates. Indeed, let

$$\begin{equation*}X = X^i \frac{\partial }{\partial q^i} + Y_i \frac{\partial }{\partial p_i} + Z \frac{\partial }{\partial t}\end{equation*}$$

be a vector field on M. $X: M \hookrightarrow TM$ defines a Lagrangian submanifold if and only if $X^*\Omega_0 = 0.$ An easy calculation gives

$$\begin{align} -X^*\Omega_0 = & \left( \frac{\partial X^i}{\partial q^j} + \frac{\partial Y_j}{\partial p_i}\right) \mathrm{d}q^j \wedge \mathrm{d}p_i + \left( \frac{\partial X^i}{\partial t} - \frac{\partial Z}{\partial p_i}\right) \mathrm{d}t \wedge \mathrm{d}p_i + \left( \frac{\partial Y_i}{\partial t} + \frac{\partial Z}{\partial q^i}\right) \mathrm{d}q^i \wedge \mathrm{d}t\nonumber\\ & \frac{\partial X^i}{\partial p_j} \mathrm{d}p_j \wedge \mathrm{d}p_i + \frac{\partial Y_i}{\partial q^j} \mathrm{d}q^j \wedge \mathrm{d}q^i. \end{align}$$

Therefore, X defines a Lagrangian submanifold of $(TM, \Omega_0)$ if and only if

$$\begin{align} \frac{\partial X^i}{\partial q^j} + \frac{\partial Y_j}{\partial p_i} = 0, \end{align} \tag{ 3 }$$

$$\begin{align}\frac{\partial X^i}{\partial t} - \frac{\partial Z}{\partial p_i} = 0, \end{align} \tag{ 4 }$$

$$\begin{align}\frac{\partial Y_i}{\partial t} + \frac{\partial Z}{\partial q^i} = 0, \end{align} \tag{ 5 }$$

$$\begin{align}\frac{\partial X^i}{\partial p_j} - \frac{\partial X^j}{\partial p_i} = 0, \end{align} \tag{ 6 }$$

$$\begin{align}\frac{\partial Y_i}{\partial q^j} - \frac{\partial Y_j}{\partial q^i} = 0. \end{align} \tag{ 7 }$$

The equations above can be summarized by taking

$$\begin{equation*}\left(G^1, \dots, G^n, G^{n+1} , \dots, G^{2n}, G^{2n +1}\right) : = \left(X^i, - Y_i, Z\right),\end{equation*}$$

$$\begin{equation*}\left(x^1, \dots, x^n, x^{n+1}, \dots, x^{2n}, x^{2n +1}\right) : = \left(q^i, p_i, t\right),\end{equation*}$$

since they translate to

$$\begin{equation*}\frac{\partial G^i}{\partial x^j} = \frac{\partial G^j}{\partial x_i}.\end{equation*}$$

We conclude that $G^i = \displaystyle \frac{\partial H}{\partial x^i}$, for some local function H, that is, locally, $X = \operatorname{grad} H$.

In general, the Hamiltonian and evolution vector field do not define Lagrangian submanifolds in $(TM,\Omega_0)$. However, modifying the form we can achieve this. First, let us study the Hamiltonian vector field X_H . We have

$$\begin{equation*}X_H^* \Omega_0 = \left(\operatorname{grad} H - \mathcal{R}\left(H\right) \mathcal{R}\right) ^* \Omega_0 = - \mathrm{d}\left(\mathcal{R}\left(H\right) \theta\right) = - \mathrm{d}\left(\mathcal{R}\left(H\right)\right) \wedge \theta.\end{equation*}$$

The form defined as

$$\begin{equation*}\Omega_H : = \Omega_0 + \left( \mathrm{d}\mathcal{R}\left(H\right) \wedge \theta \right)^v \end{equation*}$$

is a symplectic form and has the local expression

$$\begin{equation*} \Omega_H = - \mathrm{d}q \wedge \mathrm{d}\dot{p}_i - \mathrm{d}\dot{q}^i \wedge \mathrm{d}p_i - \mathrm{d}\dot{t} \wedge \mathrm{d}t + \mathrm{d}\left( \frac{\partial H}{\partial t} \right) \wedge \mathrm{d}t. \end{equation*}$$

Also,

$$\begin{equation*}X_H^* \Omega_H = - \mathrm{d}\mathcal{R}\left(H\right) \wedge \theta + \mathrm{d}\left(\mathcal{R}\left(H\right)\right) \wedge \theta = 0.\end{equation*}$$

We have proved that X_H defines a Lagrangian submanifold of $(TM, \Omega_H).$ Furthermore, since

$$\begin{equation*}\mathcal{R} ^* \Omega_0 = 0,\end{equation*}$$

it follows that the evolution vector field $\mathcal{E}_H$ also defines a Lagrangian submanifold of $(TM , \Omega_H)$.

This also gives a way of interpreting both vector fields as Lagrangian submanifolds of a the cosymplectic submanifold $(TM \times \mathbb{R}, \Omega_H, \mathrm{d}s),$ taking the coordinate in $\mathbb{R}$ to be constant.

We can interpret the orthogonal complement defined by the Poisson structure using the cosymplectic structure. We note that $\Omega|_\mathcal{H}$ defined as Ω restricted to the distribution $\mathcal{H}$ induces a symplectic vector space in each $\mathcal{H}_q$ and thus we have a symplectic vector bunde $\mathcal{H} \rightarrow M$. If $\Delta_q \subseteq \mathcal{H}_q$, we have the $\Omega|_\mathcal{H}$-orthogonal complement

$$\begin{equation*}\left(\Delta_q\right)^{\perp_{\Omega|_\mathcal{H}}} = \left\{v \in \mathcal{H} \,\, | \,\, \Omega\left(v,w\right) = 0,\,\, \forall w \in \Delta_q\right\}.\end{equation*}$$

Proposition 5.3. Let $\Delta_q \subseteq T_qM$. Then $\Delta_q^{\perp_\Lambda} = (\Delta_q\cap \mathcal{H})^{\perp_{\Omega|_{\mathcal{H}}}}.$

Proof. Let $v \in \Delta_q^{\perp_\Lambda}$, that is, $v = \sharp_\Lambda(\alpha)$ with $\alpha \in \Delta_q^0$. This implies that v is horizontal. We only need to check that $\Omega(v,w) = 0$ for every $w \in \Delta_q \cap \mathcal{H}_q$. Indeed, since $\theta(w) = 0$,

$$\begin{align*} \Omega\left(\sharp_\Lambda\alpha, w\right) & = - \Omega\left(w, \sharp_\Lambda\alpha\right) - \theta\left(w\right)\theta\left(\sharp_\Lambda \alpha\right) = -\left(\flat_{\theta, \Omega}w\right)\left(\sharp_\Lambda \alpha\right) = - \Lambda\left(\alpha, \flat_{\theta, \Omega} w\right) \\ & = - \Omega\left(\sharp_{\theta, \Omega} \alpha, w\right) = - \Omega\left(\sharp_{\theta, \Omega} \alpha, w\right) - \theta\left(\sharp_{\theta, \Omega}\alpha\right)\theta\left( w\right) = - \alpha\left(w\right) = 0. \end{align*}$$

Now we compare dimensions. We distinguish two cases, if $\theta \in \Delta_q^0$, we have

$$\begin{equation*}\operatorname{dim} \Delta_q^{\perp_\Lambda} = \operatorname{dim} \Delta_q^0 - 1 = 2n - \operatorname{dim} \Delta_q\end{equation*}$$

which is exactly $\operatorname{dim}(\Delta_q \cap \mathcal{H}_q)^{\perp_{\Omega|_\mathcal{H}}}$, for $\Delta_q \subseteq \mathcal{H}_q$ and $(\mathcal{H}_q, \Omega |_\mathcal{H})$ is symplectic. Now, if $\theta \not \in \Delta_q^0$, then

$$\begin{equation*}\operatorname{dim} \Delta_q^{\perp_\Lambda} = 2n +1 - \operatorname{dim} \Delta_q\end{equation*}$$

and, since $\Delta_q \not \subseteq \mathcal{H}_q$, we have $\operatorname{dim} (\Delta_q \cap \mathcal{H}_q) = \operatorname{dim} \Delta_q - 1$ which implies that $\operatorname{dim} (\Delta_q \cap \mathcal{H}_q)^{\perp_{\Omega|_\mathcal{H}}} = 2n + 1 - \operatorname{dim} \Delta.$ □

This last proposition clarifies the situation. The Λ-orthogonal of a subspace Δ is just the symplectic orthogonal of the intersection with the symplectic leaf. This means that coisotropic reduction in cosymplectic geometry will be performed in each leaf of the characteristic distribution $\mathcal{H}$. Also, because the Λ-orthogonal complement is just the symplectic complement of the intersection with $\mathcal{H}$, we have the following properties:

• i)  
  $(\Delta_1 \cap \Delta_2)^{\perp_\Lambda} = \Delta_1^{\perp_\Lambda} + \Delta_2^{\perp_\Lambda}.$
• ii)  
  $(\Delta_1 + \Delta_2)^{\perp_\Lambda} = \Delta_1^{\perp_\Lambda} \cap \Delta_2^{\perp_\Lambda}.$
• iii)  
  $(\Delta^{\perp_\Lambda})^{\perp_\Lambda} = \Delta \cap \mathcal{H}.$

It will also be important to distinguish submanifolds $N \hookrightarrow M$ acording to the position relative to the distributions $\mathcal{H},\mathcal{V}$.

Definition 5.4 (Horizontal, non-horizontal and vertical submanifolds). Let $i: N \hookrightarrow M$ be a submanifold. N will be called a:

• i)  
  Horizontal submanifold if $T_qN \subseteq \mathcal{H}_q$ for every $q \in N$;
• ii)  
  Non-horizontal submanifold if $T_qN \not \subseteq \mathcal{H}_q$ for every $q \in N$;
• iii)  
  Vertical submanifold if the Reeb vector field is tangent to N, that is, $\mathcal{R}(q) \in T_qN$ for every $q \in N$.

Remark 5.1. Note that if $N\hookrightarrow M$ is a vertical submanifold, then N is non-horizontal.

Lagrangian submanifolds are characterized as follows:

Lemma 5.1. Let $L\hookrightarrow M$ be a Lagrangian submanifold and $q \in L$. Then

• i)  
  If $T_qL \subseteq \mathcal{H}_q$, $\operatorname{dim} T_qL^{\perp_\Lambda} = \operatorname{dim} M - \operatorname{dim} L -1$.
• ii)  
  If $T_qL \not \subseteq \mathcal{H}_q$, $\operatorname{dim} T_qL^{\perp_\Lambda} = \operatorname{dim} M - \operatorname{dim} L$.

and, in either case, $\operatorname{dim} T_qL^{\perp_\Lambda} = n$, where $\operatorname{dim} M = 2n +1$.

Proof. 

• i)  
  Since $\theta \in T_qL^0$ we have

  $$\begin{equation*}\operatorname{dim} T_qL^{\perp_ \Lambda} = \operatorname{dim} \sharp_\Lambda\left(T_qL^0\right) = \operatorname{dim} M - \operatorname{dim} L - \operatorname{dim} \left(\operatorname{Ker} \sharp_\Lambda \cap T_qL\right) = \operatorname{dim} M - \operatorname{dim} L - 1.\end{equation*}$$
• ii)  
  It follows from the previous calculation using that $\theta \not \in T_qL^0$ because

  $$\begin{equation*}\operatorname{dim}\left(\operatorname{Ker}\sharp_\Lambda \cap T_qL\right) = 0.\end{equation*}$$

The proof of the equality $\operatorname{dim} T_qL^{\perp_\Lambda} = n$ is straightforward using that $T_qL \cap \mathcal{H}_q$ is a Lagrangian subspace of $(\mathcal{H}_q, \Omega|_\mathcal{H})$. □

Lemma 5.1 guarantees that either $\operatorname{dim} L = n$, in which case L is horizontal, or $\operatorname{dim} L = n+1$, in which case L is non-horizontal. We have the following useful characterization of Lagrangian submanifolds:

Lemma 5.2. Let $L \hookrightarrow M$ be a submanifold. We have

• i)  
  If $\operatorname{dim} L = n$, then L is Lagrangian if and only if $i^* \theta = 0, i^* \Omega = 0$.
• ii)  
  If L is non-horizontal and $\operatorname{dim} L = n+1$, then L is Lagrangian if and only if $i^*\Omega = 0$.

Proof. Both assertions are proved by a comparison of dimensions. □

Proposition 5.4. Let $i:N \hookrightarrow M$ be a coisotropic submanifold. Then the distribution $(TN)^{\perp_\Lambda}$ is involutive.

Proof. We start proving that $\mathcal{H}$ is an involutive distribution. Let $X,Y$ be vector fields tangent to $\mathcal{H}$. Since θ is closed we have

$$\begin{align*} 0 & = \left(\mathrm{d}\theta\right)\left(X,Y\right) = X\left(\theta\left(Y\right)\right) - Y\left(\theta\left(X\right)\right) - \theta\left(\left[X,Y\right]\right) = - \theta\left(\left[X,Y\right]\right), \end{align*}$$

that is, $[X,Y]$ is tangent to $\mathcal{H}.$

Denote $\Omega_0: = i^* \Omega$. Let $X,Y$ be vector fields in N tangent to $(TN)^{\perp_\Lambda}$. Using proposition 5.3, $[X,Y] \in (TN)^{\perp_\Lambda}$ if and only if $[X,Y] \in (TN \cap \mathcal{H})^{\perp_{\Omega |_\mathcal{H}}}$. In order to see this, we take an arbitrary vector field Z on N tangent to $\mathcal{H}$ and check that $\Omega_0([X,Y],Z) = 0$. Because Ω is closed, we have

$$\begin{align*} 0 & = i^*\left( \mathrm{d}\Omega\right) = \left( \mathrm{d}\Omega_0\right)\left(X,Y,Z\right) = X\left(\Omega_0\left(Y,Z\right)\right) - Y\left(\Omega_0\left(X,Z\right)\right) + Z\left( \Omega_0\left(X,Y\right)\right) \\ & \quad - \Omega_0\left(\left[X,Y\right],Z\right) + \Omega_0\left(\left[X,Z\right],Y\right) - \Omega_0\left(\left[Y,Z\right],X\right) = - \Omega_0\left(\left[X,Y\right],Z\right)\end{align*}$$

where we have used that $X,Y,Z,[Y,Z], [X,Z]$ are horizontal (since $\mathcal{H}$ is involutive) and that $X,Y \in (TN \cap \mathcal{H})^{\perp_{\Omega |_\mathcal{H}}}$. □

We shall now study coisotropic reduction of a vertical submanifold $N \hookrightarrow M$. Let $q \in N$. We have $\operatorname{dim} (T_qN)^0 = \operatorname{dim} M - \operatorname{dim} N$. Since N is vertical, $\theta \not \in (T_qN)^0$ and we have

$$\begin{equation*}\operatorname{dim} \left(T_qN\right)^{\perp_\Lambda} = \operatorname{dim} \sharp_\Lambda\left(T_qN\right)^0 = \operatorname{dim} M - \operatorname{dim} N - \operatorname{dim} \left(\operatorname{Ker} \sharp_\Lambda \cap \left(T_qN\right)^0\right) = \operatorname{dim} M - \operatorname{dim} N.\end{equation*}$$

In particular, $(TN)^{\perp_\Lambda}$ is a regular distribution.

Theorem 5.1 (Vertical coisotropic reduction in the cosymplectic setting). Let $(M, \Omega, \theta)$ be a cosymplectic manifold and $i:N \hookrightarrow M$ be an coisotropic vertical submanifold. Denote by $\mathcal{F}$ the maximal foliation of the involutive regular distribution $(TN)^{\perp_\Lambda}$. If the space of all leaves $N/ \mathcal{F}$ admits a manifold structure such that the projection $\pi : N \rightarrow N/ \mathcal{F}$ is a submersion, then there exist unique θ_N , ω_N such that

$$\begin{align*} i^* \omega & = \pi^* \omega_N,\\ i^* \theta & = \pi^* \theta_N, \end{align*}$$

and they define a cosymplectic structre on $N / \mathcal{F}.$ The following diagram summarizes the situation:

Zoom In Zoom Out Reset image size

Proof. Uniqueness is clear from the imposed relation. Denote $\Omega_0: = i^* \Omega$, $\theta_0 : = i^*\theta$. We only need to verify that the following forms are closed, well defined and define a cosymplectic structure:

$$\begin{align*} \Omega_N\left(\left[u\right],\left[v\right]\right) : = \Omega_0\left(u,v\right),\\ \theta_N\left(\left[u\right]\right): = \theta_0\left(u\right), \end{align*}$$

where $[u]: = T\pi(q) \cdot u \in T_{[q]}N/ \mathcal{F}$. If they were well defined, it is clear that they are smooth and closed since $\pi^* \mathrm{d} \theta_N = \mathrm{d} \theta_0 = 0$, $\pi^*\Omega_N = \mathrm{d} \Omega_0 = 0$ and π is a submersion.

Let us first check that these definitions do not depend on the chosen representatives of the vectors. It suffices to observe that for vectors in the distribution, say $v \in (T_qN)^{\perp_\Lambda}$, we have $i_v \Omega_0 = 0$ and $i_v \theta_0 = 0$. This easily follows from proposition 5.3 using that the horizontal proyection of every vector $u \in T_qN$ is tangent to N (here we use the condition $\mathcal{R}(q) \in T_qN$). To see the independence of the point in the leaf chosen, it is enough to observe that

$$\begin{equation*}\mathcal{L}_X \Omega_0 = 0; \,\,\, \mathcal{L}_X \theta_0 = 0 \end{equation*}$$

for every vector field on N tangent to the distribution $(T_qN)^{\perp_\Lambda}$ (since every two points in the same leaf of the foliation can be joined by a finite union of flows of such fields). Indeed, we have

$$\begin{align*} \mathcal{L}_X\Omega_0 & = i_X \mathrm{d}\Omega_0 + \mathrm{d}i_X \Omega_0 = 0,\\ \mathcal{L}_X \theta_0 & = i_X \mathrm{d}\theta_0 + \mathrm{d}i_X\theta_0 = 0. \end{align*}$$

Now we check that they define a cosymplectic structure. Assuming $k = \operatorname{dim} N$ and $2n +1 = \operatorname{dim} M$, from the remark above we have

$$\begin{equation*}\operatorname{dim} N/\mathcal{F} = \operatorname{dim} N - \operatorname{rank}\, \left(TN\right)^{\perp_\Lambda} = 2k - 2n -1 = 2\left(k - n -1 \right) + 1\end{equation*}$$

and hence, $(N/\mathcal{F},\Omega_N,\theta_N)$ is a cosymplectic manifold if and only if

$$\begin{equation*}\theta_N \wedge \Omega_N^{k-n-1}\neq 0,\end{equation*}$$

which is equivalent to $\theta_0 \wedge \Omega_0^{k-n-1} \neq 0$, because π is a submersion. For every point $q \in N$, TqN can be decomposed in

$$\begin{equation*}TqN = TqN_\mathcal{H} \oplus \mathcal{V}\end{equation*}$$

where $T_qN_\mathcal{H} = (T_qN) \cap \mathcal{H}_q$. It is easy to see that $(T_qN)_\mathcal{H}$ is a coisotropic subspace of $(\mathcal{H}_q, \Omega|_\mathcal{H})$. This implies (using symplectic reduction) that there are $\operatorname{dim}(T_qN)_\mathcal{H} - \operatorname{dim} (T_qN)_\mathcal{H}^{\perp_{\widetilde \Omega}} = k - 1 - (2n +1 -k) = 2k - 2n - 2$ horizontal vectors, say $u_1, \dots, u_{2k-2n-2}$ such that $\Omega_0^{k-n-1}(u_1, \dots, u_{2k-2n-2})\neq 0$. Taking the last vector to be $\mathcal{R}(q)$, it is clear that

$$\begin{equation*}\left(\theta_0 \wedge \Omega_0^{k-n-1}\right)\left(\mathcal{R}\left(q\right), u_1, \dots, u_{2k-2n-2}\right) \neq 0.\end{equation*}$$

 □

5.3.1. Projection of Lagrangian submanifolds.

Now we will prove that Lagrangian submanifolds $L \hookrightarrow M$ project to Lagrangian submanifolds in $N/\mathcal{F}.$

Proposition 5.5 (Projection of horizontal Lagrangian submanifolds is Lagrangian). Under the hypotheses of theorem 5.1, let $L \hookrightarrow M$ be an horizontal Lagrangian submanifold such that L and N have clean intersection. If $L_N: = \pi(L \cap N)$ is a submanifold of $N/ \mathcal{F}$, then L_N is Lagrangian.

Proof. Let $\mathcal{H}_N$ be the horizontal distribution in $N/\mathcal{F}$. It is clear that L_N is horizontal, because $T_{[q]}L_N = T\pi(q)(T_qL \cap T_qN)$. Using proposition 5.3 we have

$$\begin{equation*}T_{\left[q\right]}L_N^{\perp_{\Lambda_N}} = \left(T_{\left[q\right]}L_N \cap \mathcal{H}_ N\right)^{\perp_{\Omega_N|_{\mathcal{H}_N}}} = T_{\left[q\right]}L_N^{\perp_{\Omega_N|_{\mathcal{H}_N}}}.\end{equation*}$$

We will check that

$$\begin{equation*}T_{\left[q\right]}L_N \subseteq T_{\left[q\right]}L_N^{\perp_{\Lambda_N}}\end{equation*}$$

and prove that $\operatorname{dim} L_N = \operatorname{dim} N- n - 1$, which with lemma 5.2 together with the calculation of the dimension of $N / \mathcal{F}$ done in theorem 5.1, yields the result.

Let $[v], [w] \in T_{[q]}L_N$. Then

$$\begin{equation*}\Omega_N\left(\left[v\right],\left[w\right]\right) = \Omega\left(v,w\right) = 0,\end{equation*}$$

since L is Lagrangian. Because $[w]$ is arbitrary, this last calculation implies that $[v] \in T_{[q]}L_N^{\perp_{\Omega_N|_{\mathcal{H}_N}}} \subseteq T_{[q]}L_N^{\perp_{\Lambda_N}}$. Now,

$$\begin{equation} \operatorname{dim} L_N = \operatorname{dim} \left(L \cap N\right) - \operatorname{dim} \left(T_qL \cap \left(T_qN\right)^{\perp_\Lambda}\right). \end{equation} \tag{ 8 }$$

Furthermore, since L is Lagrangian and horizontal, $(T_qL \cap (T_qN)^{\perp_\Lambda})^{\perp_\Lambda} = T_qL \cap \mathcal{H}_q + (T_qN^{\perp_\Lambda})^{\perp_\Lambda} = T_qL + (T_qN^{\perp_\Lambda})^{\perp_\Lambda}$ and thus (using that $T_qL \cap (T_qN^{\perp_\Lambda})^{\perp_\Lambda}$ is necessarily horizontal),

$$\begin{equation*} \operatorname{dim} \left(T_qL \cap \left(T_qN\right)^{\perp_\Lambda}\right) = \operatorname{dim} M - \operatorname{dim}\left(T_qL + \left(T_qN^{\perp_\Lambda}\right)^{\perp_\Lambda}\right) - 1. \end{equation*}$$

Since $\operatorname{dim}(T_qL\cap \mathcal{H}_q + (T_qN^{\perp_\Lambda})^{\perp_\Lambda}) = \operatorname{dim}(T_qL\cap \mathcal{H}_q + T_qN) - 1$ (which comes from the fact that $\operatorname{dim} (T_qN^{\perp_\Lambda})^{\perp_\Lambda} = \operatorname{dim} T_qN - 1$) we have

$$\begin{equation} \operatorname{dim}\left(T_qL \cap \mathcal{H}_q + T_qN^{\perp_\Lambda}\right) = \operatorname{dim} M - \operatorname{dim}\left(T_qL \cap T_qN^{\perp_\Lambda}\right). \end{equation} \tag{ 9 }$$

Substituting (9) in (8) and using $\operatorname{dim} L = n$, we conclude

$$\begin{align*} \operatorname{dim} L_N & = \operatorname{dim} \left(L \cap N\right) - \left( \operatorname{dim} M - \operatorname{dim}\left(T_qL + T_qN\right)\right) \\ & = \operatorname{dim} \left(L \cap N\right) - \operatorname{dim} M + \operatorname{dim} L + \operatorname{dim} N - \operatorname{dim}\left(L \cap N\right)\\ & = -2n -1 + n + \operatorname{dim} N = \operatorname{dim} N - n - 1. \end{align*}$$

 □

Proposition 5.6 (Projection of non-horizontal Lagrangian submanifold is Lagrangian). Under the hypotheses of theorem 5.1, let $L \hookrightarrow M$ be a non-horizontal Lagragian submanifold. If L and N have clean intersection and $L_N: = \pi(L \cap N) \hookrightarrow N/\mathcal{F}$ is a submanifold, then L_N is Lagrangian.

Proof. The proof follows the same lines as that of proposition 5.5. That L_N is isotropic follows easily from proposition 5.3. However, in order to calculate $\operatorname{dim} L_N$, we need to distinguish whether $L \cap N$ is horizontal or not.

• i)  
  If $L\cap N$ is horizontal, we need to check that $\operatorname{dim} L_N = \operatorname{dim} N - n - 1$, since L_N is horizontal. Because $(T_qL \cap T_qN^{\perp_\Lambda})^{\perp_\Lambda} = T_qL \cap \mathcal{H}_q + (T_qN^{\perp_\Lambda})^{\perp_\Lambda}$, we have

  $$\begin{align*} \operatorname{dim}\left(T_qL\cap \mathcal{H}_q + \left(T_qN^{\perp_\Lambda}\right)^{\perp_\Lambda}\right) & = \operatorname{dim} M - \operatorname{dim} \left(T_qL \cap T_qN^{\perp_\Lambda}\right) - 1. \end{align*}$$

  It is easy to check that $\operatorname{dim}(T_qL\cap \mathcal{H}_q + (T_qN^{\perp_\Lambda})^{\perp_\Lambda}) = \operatorname{dim}(T_qL\cap \mathcal{H}_q + T_qN) - 1$ and thus,

  $$\begin{equation*}\operatorname{dim}\left(T_qL \cap T_qN^{\perp_\Lambda}\right) = \operatorname{dim} M - \operatorname{dim} \left(T_qL\cap \mathcal{H}_q + T_qN\right).\end{equation*}$$

  We conclude that

  $$\begin{align*} \operatorname{dim} L_N & = \operatorname{dim}\left(L \cap N\right) - \operatorname{dim} \left(T_qL \cap T_qN^{\perp_\Lambda}\right)\\ & = \operatorname{dim} \left(L \cap N\right) - \left(\operatorname{dim} M - \operatorname{dim}\left(T_qL \cap \mathcal{H}_q +T_qN\right) \right) \\ & = \operatorname{dim}\left(L \cap N\right) - \left(\operatorname{dim} M - \operatorname{dim}\left(T_qL \cap \mathcal{H}_q\right) - \operatorname{dim} N+ \operatorname{dim}\left(T_qL \cap \mathcal{H}_q \cap T_qN\right)\right)\\ & = \operatorname{dim}\left(L \cap N\right) - \operatorname{dim} M + \operatorname{dim} L - 1 + \operatorname{dim} N - \operatorname{dim}\left(T_qL \cap T_qN\right)\\ & = \operatorname{dim} N - \operatorname{dim} M + \operatorname{dim} L - 1 = \operatorname{dim} N - 2n - 1 + n + 1 - 1 = \operatorname{dim} N - n -1, \end{align*}$$

  where we have used that $T_qL \cap \mathcal{H}_q \cap T_qN = T_qL \cap T_qN$, since $N \cap L$ is horizontal, and $\operatorname{dim} T_qL \cap \mathcal{H} _q = \operatorname{dim} L - 1$, because $T_qL \not \subseteq \mathcal{H}_q$.
• ii)  
  If $L\cap N$ is not horizontal, we need to check that $\operatorname{dim} L_N = \operatorname{dim} N - n$. This follows from the same calculation done in i), using that

  $$\begin{equation*}\operatorname{dim}\left(T_qL \cap \mathcal{H}_q \cap T_qN\right) = \operatorname{dim}\left(T_qL \cap T_qN\right) -1.\end{equation*}$$

 □

We will restrict the study to horizontal coisotropic submanifolds $N\hookrightarrow M$, that is, manifolds satisfying $T_qN \subseteq \mathcal{H}_q$ for every $q\in N$. Note that in this case the distribution $(TN)^{\perp_\Lambda}$ is also regular, since

$$\begin{equation*}\operatorname{dim} \left(T_qN\right)^{\perp_\Lambda} = \operatorname{dim} M - \operatorname{dim} N - \operatorname{dim} \left(\operatorname{Ker} \sharp_\Lambda\cap \left(T_qN\right)^0\right) = \operatorname{dim} M - \operatorname{dim} N - 1.\end{equation*}$$

Theorem 5.2 (Horizontal coisotropic reduction in the cosymplectic setting). Let $(M, \Omega, \theta)$ be a cosymplectic manifold and $i: N \hookrightarrow M$ be an horizontal coisotropic submanifold. Denote by $\mathcal{F}$ the space of leaves determined by the regular and involutive distribution $(TN)^{\perp_\Lambda}.$ If $N/ \mathcal{F}$ admits a manifold structure such that $\pi : N \rightarrow N/ \mathcal{F}$ is a submersion, then there exists a unique 2- form $\Omega_N$ in $N/\mathcal{F}$ such that

$$\begin{equation*}\pi^* \Omega_N = i^* \Omega\end{equation*}$$

and $(N/\mathcal{F}, \Omega_N)$ is a symplectic manifold.

Proof. Since N is horizontal and the horizontal distribution is integrable, N will be contained in an unique symplectic leaf and thus, we are performing symplectic reduction. The proof is just repeating what has been done in theorem 3.2. □

We can generalize this process to arbitrary submanifolds. Let $N \hookrightarrow M$ be a coisotropic submanifold. Since in general we cannot guarantee the well-definedness of the 2-form in the quotient, we will reduce the intersection of N with each one of the symplectic leaves. It is clear that $TN\cap \mathcal{H}$ is an involutive distribution, since TN and $\mathcal{H}$ are. If this distribution was regular, for every $q \in N$ there would exist an unique maximal leaf of the distribution, say S_q . Notice that $S_q \hookrightarrow M$ is an horizontal submanifold. We can perform coisotropic reduction on each of these submanifolds.

5.4.1. Projection of Lagrangian submanifolds

Proposition 5.7. Let $L \hookrightarrow M$ be a Lagrangian submanifold. If L and N have clean intersection and $L_N: = \pi(L\cap N)$ is a submanifold, then L_N is Lagrangian.

Proof. Let $q \in N \cap L$, we have to prove that

$$\begin{equation*}T_{\left[q\right]} L_N = \mathrm{d}_q\pi \cdot \left( T_q L \cap T_q N\right)\end{equation*}$$

is a Lagrangian subspace of $T_{[q]} N/\mathcal{F} = T_{[q]}N / \left( T_{q} N\right)^{\perp_{\Lambda}}.$ Since N is horizontal, $T_q N \subseteq \mathcal{H}_q$. Now, from proposition 5.3, we know that

$$\begin{equation*}\left(T_qN\right) ^{\perp_{\Omega|_{\mathcal{H}}}} = \left(T_q N\right)^{\perp_\Lambda} \subseteq T_qN,\end{equation*}$$

that is, T_q N is a coisotropic subspace of $\mathcal{H}_q$, with its natural symplectic structure. A similar argument shows that T_q L is a Lagrangian subspace of $\mathcal{H}_q.$ Now, from symplectic reduction (linear symplectic reduction) we conclude that

$$\begin{equation*} T_{\left[q\right]} L_N = \mathrm{d}_q\pi \cdot \left( T_q L \cap T_q N\right)\end{equation*}$$

is a Lagrangian submanifold of $T_qN/\left( T_{q} N\right)^{\perp_{\Lambda}} = T_{[q]} N/ \mathcal{F}$, with the symplectic structure induced by $\Omega_q|_{\mathcal{H}}$, which coincides sith the symplectic structure induced by $\Omega_q$, as a quick check shows. □

Definition 7.2 (Hamiltonian vector field). Let $H \in \mathcal{C}^\infty(M)$. Define the Hamiltonian vector field of H as

$$\begin{equation*}X_H : = \sharp_ \Lambda\left(\mathrm{d}H\right) - H \mathcal{R} = \sharp_\eta\left(\mathrm{d}H\right) - \left(\mathcal{R}\left(H\right) + H\right)\mathcal{R}.\end{equation*}$$

Locally, it has the local expression

$$\begin{equation*}X_H = \frac{\partial H}{\partial p_i} \frac{\partial }{\partial q^i} - \left(\frac{\partial H}{\partial q^i} + p_i\frac{\partial H}{\partial z} \right) \frac{\partial }{\partial p_i} + \left( p_i \frac{\partial H}{\partial p_i} - H\right) \frac{\partial }{\partial z}.\end{equation*}$$

The dynamics corresponding to a Hamiltonian vector field X_H are determined by

$$\begin{align*} \frac{\mathrm{d} q^i}{\mathrm{d} t} & = \frac{\partial H}{\partial p_i},\\ \frac{\mathrm{d} p_i}{\mathrm{d} t} & = - \frac{\partial H}{\partial q^i} - p_i \frac{\partial H}{\partial z},\\ \frac{\mathrm{d} z}{\mathrm{d} t} & = p_i \frac{\partial H}{\partial p_i} - H. \end{align*}$$

For instance, if we take as Hamiltonian

$$\begin{equation*}H : = \frac{p^2}{2m} + \frac{\kappa^2 m q^2}{2} + \gamma z,\end{equation*}$$

the equations determined by X_H are

$$\begin{align*} \frac{\mathrm{d} q}{\mathrm{d} t} & = \frac{p}{m},\\ \frac{\mathrm{d} p}{\mathrm{d} t} & = - \kappa^2 m q - \gamma p,\\ \frac{\mathrm{d} z}{\mathrm{d} t} & = \frac{p^2}{2m} - \frac{\kappa^2 m q^2}{2} - \gamma z, \end{align*}$$

which are precisely the equations for the damped harmonic oscillator.

We can define a symplectic structure in TM taking $\Omega_0 : = \flat_\eta^* \Omega_M$, where $\Omega_M$ is the canonical symplectic structure in $T^*M$. In local coordinates $(q^i, p_i, z, \dot{q}^i, \dot{p}_i, \dot{z})$, it has the expression

$$\begin{align*} \Omega_0 = \,& p_ip_j \mathrm{d}q^i \wedge \mathrm{d}\dot{q}^j + \left(\left(1 + \delta_i ^j\right)p_i \dot{q}^j - \delta_i^j\dot{z}\right) \mathrm{d}q^i \wedge \mathrm{d}p_j - \mathrm{d}q^i \wedge \mathrm{d}\dot{p}_i - p_i \mathrm{d}q^i \wedge \mathrm{d}\dot{z}\\ &+ \mathrm{d}p_i \wedge \mathrm{d}\dot{q}^i + \mathrm{d}z \wedge \mathrm{d}\dot{z} - p_i \mathrm{d}z \wedge \mathrm{d}\dot{q}^i - \dot{q}^i \mathrm{d}z \wedge \mathrm{d}p_i \\ = \,& \mathrm{d}q^i \wedge \left(p_ip_i \mathrm{d}q^j + \left(1 + \delta_i^j\right)p_i \dot{q}^j \mathrm{d}p_j - \dot{z} \mathrm{d}p_i - \mathrm{d}\dot{p}_i - p_i \mathrm{d}\dot{z}\right) \\ & + \mathrm{d}p_i \wedge \mathrm{d}\dot{q}^i + \mathrm{d}z \wedge \left( \mathrm{d}\dot{z} - p_i\mathrm{d}\dot{q}^i - \dot{q}^i \mathrm{d}p_i\right). \end{align*}$$

Definition 7.3 (Gradient vector field). Given a Hamiltonian H on M, define the gradient vector field of H as

$$\begin{equation*}\operatorname{grad} H : = \sharp_\eta\left(dH\right).\end{equation*}$$

Locally, it is given by

$$\begin{equation*}\operatorname{grad} H = \frac{\partial H}{\partial p_i} \frac{\partial }{\partial q^i} - \left(\frac{\partial H}{\partial q^i} + p_i\frac{\partial H}{\partial z} \right) \frac{\partial }{\partial p_i} + \left( p_i \frac{\partial H}{\partial p_i} + \frac{\partial H}{\partial z}\right) \frac{\partial }{\partial z}.\end{equation*}$$

We have the following relation between both vector fields

$$\begin{equation*}X_H = \operatorname{grad} H - \left(\mathcal{R}\left(H\right) + H \right) \mathcal{R}.\end{equation*}$$

Just like in the previous sections, a vector field $X: M \rightarrow TM$ is locally a gradient vector field if and only if it defines a Lagrangian submanifold in $(TM , \Omega_0)$. The proof is straight-forward, checking that

$$\begin{equation*}X^*\Omega_0 = - \mathrm{d}X^{\flat_\eta}.\end{equation*}$$

We can also interpret the Hamiltonian vector field X_H as a Lagrangian submanifold of TM, but we need to modify slightly the symplectic form. It is easy to verify that

$$\begin{equation*}X_H^*\Omega_0 = \operatorname{d}\left(\mathcal{R}\left(H\right) \eta + H\eta\right),\end{equation*}$$

therefore, taking

$$\begin{equation*} \Omega_H: = \Omega_0 - \operatorname{d}\left(\mathcal{R}\left(H\right) \eta + H\eta\right)^v,\end{equation*}$$

we have

$$\begin{equation*}X_H^*\Omega_H = 0.\end{equation*}$$

It is clear that $\Omega_H$ is a symplectic form. We have proved:

Proposition 7.1. The Hamiltonian vector field $X_H: M \rightarrow TM$ defines a Lagrangian submanifold of the symplectic manifold $(TM, \Omega_H)$.

Now we study evolution vector fields, an important vector field in the application of contact geometry to thermodynamics.

Definition 7.4. Given a Hamiltonian H, we define the evolution vector field as

$$\begin{equation*}\mathcal{E}_H : = X_H + H \mathcal{R}.\end{equation*}$$

Locally, the evolution vector field is written

$$\begin{equation*}\mathcal{E}_H = \frac{\partial H}{\partial p_i} \frac{\partial }{\partial q^i} - \left(\frac{\partial H}{\partial q^i} + p_i\frac{\partial H}{\partial z} \right) \frac{\partial }{\partial p_i} + p_i \frac{\partial H}{\partial p_i}\frac{\partial }{\partial z}.\end{equation*}$$

Let us see how we can modify the symplectic form Ω₀ in such a way that $\mathcal{E}_H$ defines a Lagrangian submanifold. We have

$$\begin{equation*}\mathcal{E}_H^*\Omega_0 = X_H^* \Omega_0 + \left(H \mathcal{R}\right) ^* \Omega_0 = \operatorname{d}\left(\mathcal{R}\left(H\right) \eta + H\eta\right) - \operatorname{d}\left(H \eta\right) = \operatorname{d}\left( \mathcal{R}\left(H\right) \eta\right),\end{equation*}$$

and thus, $\mathcal{E}_H$ defines a Lagrangian submanifold of $(TM , \widetilde \Omega_H),$ where

$$\begin{equation*}\widetilde \Omega_H = \Omega_0 - \operatorname{d}\left(\mathcal{R}\left(H\right) \eta\right).\end{equation*}$$

We can also interpret Hamiltonian and evolution vector fields as Legendrian submanifolds of a certain contact structure defined on $TM \times \mathbb{R}.$

Definition 7.5. Let $(M, \eta)$ be a contact manifold. Define the contact form on $TM \times \mathbb{R}$ as

$$\begin{equation*}\hat{\eta} : = \eta^c + t \eta^v,\end{equation*}$$

where η^c , η^v are the complete and vertical lifts [16]. It is easily checked that $\hat{\eta}$ defines a contact structure [15].

In local coordinates it has the expression:

$$\begin{equation*}\hat{\eta} = \mathrm{d}\dot{z} - \dot{p}_i \mathrm{d}q^i - p_i \mathrm{d}\dot{q}^i + t \mathrm{d}z - tp_i \mathrm{d}q^i.\end{equation*}$$

We have the following [15]:

Proposition 7.2. Let $X_H: M \rightarrow TM$ be a Hamiltonian vector field. Then, the submanifold defined by the immersion

$$\begin{equation*}i: M \hookrightarrow TM \times \mathbb{R};\,\,\,\,p \mapsto \left(X_H\left(p\right), \mathcal{R}\left(H\right)\right)\end{equation*}$$

is a Legendrian submanifold of $(TM\times \mathbb{R}, \hat{\eta})$.

Proof. Using the properties of complete and vertical lifts we have

$$\begin{equation*}\left(X_H \times \mathcal{R}\left(H\right)\right) ^* \hat{\eta} = \mathcal{L}_{X_H}\eta + \mathcal{R}\left(H\right) \eta.\end{equation*}$$

Using lemma 7.2 it will be sufficient to see that $\mathcal{L}_{X_H} \eta = - \mathcal{R}(H).$ This is a straight-forward verification since

$$\begin{equation*}\mathcal{L}_{X_H} \eta = \mathrm{d}i_{X_H} \eta + i_{X_H} \mathrm{d}\eta = - \mathrm{d}H + \mathrm{d}H - \mathcal{R}\left(H\right) \eta. \end{equation*}$$

 □

A thermodynamical system is characterized by the following variables

$$\begin{equation*}\left(H, T, S, P_i, V^i\right),\end{equation*}$$

corresponding to energy, temperature, entropy, and the generalized pressures and volumes, respectively. We will denote $q^i = V^i$, following the notation used so far. The energy of the system is a function

$$\begin{align*} H : & \; T^\ast Q \times \mathbb{R} \rightarrow \mathbb{R}; \,\,\left(q^i, p_i, S\right) \mapsto H\left(q^i,p_i, S\right). \end{align*}$$

The first law of thermodynamics can be written as

$$\begin{equation*}\mathrm{d} H = \delta \mathcal{Q} - \delta \mathcal{W},\end{equation*}$$

where $\delta \mathcal{Q}$ and $\delta \mathcal{W}$ are one forms representing the heat and work, respectively. We will assume that these forms have local expressions

$$\begin{equation*}\delta \mathcal{Q} = T \mathrm{d}S, \,\, \delta \mathcal{W} = P_i \mathrm{d}q^i,\end{equation*}$$

for some functions $P_i, T$ that physically represent the conjugate variables to qⁱ (pressure, if qⁱ represented volume), and S (temperature, if S was the entropy), respectively. Then, the first law of thermodynamics reads

$$\begin{equation*}\mathrm{d} H = T\mathrm{d} S - P_i \mathrm{d}q^i.\end{equation*}$$

If we were studying an isolated system, energy would be conserved, that is

$$\begin{equation*}0 = T\mathrm{d} S - P_i \mathrm{d}q^i\end{equation*}$$

or, dividing by T,

$$\begin{equation*}0 = \mathrm{d} S - \frac{P_i}{T} \mathrm{d} q^i.\end{equation*}$$

Identifying $p_i = P_i/T$, after a change of variables if necessary, we obtain

$$\begin{equation*}0 = \mathrm{d}S - p_i \mathrm{d}q^i.\end{equation*}$$

Defining

$$\begin{equation*}\eta : = \mathrm{d} S - p_i \mathrm{d}q^i,\end{equation*}$$

the canonical contact form in $T^\ast Q \times \mathbb{R},$ we conclude that isolated processes take place in Legendrian submanifolds of $T^\ast Q \times \mathbb{R},$ motivating their study.

It turns out that the integral curves of the evolution vector field $\mathcal{E}_H$ can be interpreted as an isolated system

Proposition 7.3. The integral curves of $\mathcal{E}_H$ satisfy

$$\begin{equation*}\frac{\mathrm{d} H}{\mathrm{d} t} = 0.\end{equation*}$$

Furthermore, locally

$$\begin{equation*}\frac{\mathrm{d} S}{\mathrm{d} t} = p_i \mathrm{d}q ^i = \frac{P_i}{T} \mathrm{d} q^i.\end{equation*}$$

Proof. Indeed, by definition we know that

$$\begin{equation*}i_{\operatorname{grad} H} \mathrm{d} \eta + \eta\left(H\right) \eta = \mathrm{d} H.\end{equation*}$$

Therefore,

$$\begin{align*} \frac{\mathrm{d} H}{\mathrm{d} t} & = \mathrm{d} H \cdot \mathcal{E}_H = \mathrm{d} H \cdot X_H + H \mathcal{R}\left(H\right) = \mathrm{d} H \cdot \operatorname{grad} H - \left(\mathcal{R}\left(H\right) + H\right) \mathcal{R}\left(H\right) + H \mathcal{R}\left(H\right)\\ & = \left(i_{\operatorname{grad} H} \mathrm{d} \eta + \eta\left(H\right) \eta\right)\cdot \operatorname{grad} H - \left(\mathcal{R}\left(H\right)\right)^2 = 0. \end{align*}$$

Now, the last part is a consequence of the equality

$$\begin{equation*}\mathrm{d} H = \mathrm{d} S - p_i \mathrm{d}q^i.\end{equation*}$$

 □

For more details on the subject, we refer to [4, 47, 48, 51, 52], (see also two modern approaches [9, 10]).

Coisotropic reduction in contact manifolds has been developed in [15] (see also [41, 57]).

The following definition will result useful. Given a subspace $\Delta_q \subseteq T_qM$, we define the $\mathrm{d}\eta$-orthogonal complement as

$$\begin{equation*}\Delta_q^{\perp_{\mathrm{d}\eta}}: = \left\{v \in T_qM,\,|\,\, \mathrm{d}\eta\left(v,w\right) = 0\,\, \forall w \in \Delta_q\right\}.\end{equation*}$$

Proposition 7.4. Let $\Delta_q \subseteq T_qM$ be a subspace. Then

$$\begin{equation*}\Delta_q^{\perp_{d\eta}} \cap \mathcal{H}_q \subseteq \Delta_q^{\perp_\Lambda}.\end{equation*}$$

Furthermore, if $\mathcal{R}(q) \in \Delta_q$ or $\Delta_q \subseteq \mathcal{H}_q$, the equality holds.

Proof. Let $v \in \Delta_q^{\perp_{d\eta}}\cap \mathcal{H}_q$ and take $\alpha : = i_v \mathrm{d} \eta$. It is clear that $\alpha \in \Delta_q^0$. We will prove that $\sharp_\Lambda(-\alpha) = v$. Indeed, for $\beta \in T^*_qM$, since $b_\eta(v) = \alpha$ (as a direct calculation shows), we have

$$\begin{align*} \langle \beta, \sharp_\Lambda\left(\alpha\right) \rangle & = \Lambda\left(\alpha, \beta\right) = \Omega\left(\sharp_\eta\left(\alpha\right), \sharp_\eta\left(\beta\right)\right) = \Omega\left(v, \sharp_\eta\left(\beta\right)\right) \\ & = -\Omega\left( \sharp_\eta\left(\beta\right),v\right) - \eta\left(\sharp_\Lambda\left(\beta\right)\right)\eta\left(v\right) = - \langle \flat_\eta\left(\sharp_\eta\left(\beta\right)\right), v\rangle = -\langle \beta, v \rangle, \end{align*}$$

that is, $v = \sharp_\Lambda(-\alpha)$.

Now, if $\mathcal{R}(q) \in \Delta_q$, we compare dimensions. Since $\operatorname{Ker} \sharp_\Lambda = \langle \eta\rangle$ and $\eta \not \in \Delta_q^0$, we have

$$\begin{equation*}\operatorname{dim} \Delta_q^{\perp_\Lambda} = \operatorname{dim} \Delta_q^0 = \operatorname{dim} M - \operatorname{dim} \Delta_q.\end{equation*}$$

Furthermore, $\Delta^{\perp_{\mathrm{d}\eta}}\cap \mathcal{H}_q$ has the same dimension, since

$$\begin{align*} \Delta_q^{\perp_{\mathrm{d}\eta}} \cap \mathcal{H}_q & = \left(\Delta_q \cap \mathcal{H}_q \oplus \mathcal{V}_q\right)^{\perp_{\mathrm{d}\eta}} \cap \mathcal{H}_q = \left(\Delta_q \cap \mathcal{H}_q\right)^{\perp_{\mathrm{d}\eta}} \cap \mathcal{V}_q^{\perp_{\mathrm{d}\eta}} \cap \mathcal{H}_q\\ & = \left(\Delta_q \cap \mathcal{H}_q\right)^{\perp_{\mathrm{d}\eta}} \cap \mathcal{H}_q. \end{align*}$$

This latter is just the symplectic complement in $(\mathcal{H}_q, \mathrm{d}\eta |_{\mathcal{H}})$ and hence,

$$\begin{equation*}\operatorname{dim} \left(\Delta_q^{\perp_{\mathrm{d}\eta}}\cap \mathcal{H}_q\right) = \operatorname{dim} \mathcal{H}_q - \operatorname{dim} \left(\Delta_q \cap \mathcal{H}_q\right) = \operatorname{dim} M - 1 - \left(\operatorname{dim} \Delta_q - 1\right).\end{equation*}$$

Now, if $\Delta_q \subseteq \mathcal{H}_q$, $\eta \in \Delta_q^0$ and, thus,

$$\begin{equation*}\operatorname{dim} \Delta_q^{\perp_{\mathrm{d}\eta}} = \operatorname{dim} M - \operatorname{dim} \Delta_q - 1.\end{equation*}$$

Since $\Delta_q^{\perp_{\mathrm{d}\eta}} \cap \mathcal{H}_q$ is just the symplectic complement of $\Delta_q$ we have

$$\begin{equation*}\operatorname{dim} \left(\Delta_q^{\perp_{\mathrm{d}\eta}} \cap \mathcal{H}_q\right) = \operatorname{dim} \mathcal{H}_q - \operatorname{dim} \Delta_q = \operatorname{dim} M - 1 - \operatorname{dim} \Delta_q.\end{equation*}$$

 □

This proposition allows us to characterize Legendrian submanifolds:

Lemma 7.1. If $L\hookrightarrow M$ is a Legendrian submanifold, then L is horizontal and $\operatorname{dim} L = n$ (where $\operatorname{dim} M = 2n+1$). Furthermore, if L is horizontal and isotropic (or coisotropic) with $\operatorname{dim} L = n$, L is Legendrian.

Proof. Since $\sharp_\Lambda$ takes values in $\mathcal{H}$, it is clear that every Legendrian submanifold is horizontal. Since L is horizontal,

$$\begin{equation*}\operatorname{dim} T_qL^{\perp_\Lambda} = \operatorname{dim} M - \operatorname{dim} L - 1.\end{equation*}$$

From the previous equation and using that $T_qL^{\perp_\Lambda} = T_qL$, we deduce that $\operatorname{dim} L = \operatorname{dim} M - \operatorname{dim} L - 1$. This implies $\operatorname{dim} L = n$. The last property is easily seen via a direct comparison of dimensions. □

We will also need characterization of isotropic submanifolds in contact geometry:

Lemma 7.2. A submanifold $N \hookrightarrow M$ is isotropic if and only if $i^* \eta = 0$.

Proof. Necessity is clear, since $\sharp_\Lambda$ takes values in $\mathcal{H}$. Now suppose that N is horizontal. We have $i^*\eta = 0$ and thus, $i^* \mathrm{d}\eta = 0$. This implies that $T_qN \subseteq T_qN^{\perp_{\mathrm{d}\eta}} \cap \mathcal{H}_q \subseteq T_qN^{\perp_\Lambda},$ using proposition 7.4. □

Proposition 7.5. Let $i:N \hookrightarrow M$ be a coisotropic submanifold such that $\mathcal{R}(q) \in T_qN$ for every $q\in N$ or $T_qN \subseteq \mathcal{H}_q$ for every $q \in N$. Define $\eta_0: = i^*\eta$. Then

$$\begin{equation*}T_qN^{\perp_\Lambda} = \operatorname{Ker}{\mathrm{d}\eta_0} \cap \operatorname{Ker} \eta_0.\end{equation*}$$

Proof. Let $q\in N$. Proposition 7.4 implies that

$$\begin{equation*}T_qN^{\perp_\Lambda} = T_qN^{\perp_\eta} \cap \mathcal{H}_q.\end{equation*}$$

But, since T_q N is coisotropic, then it is just $\operatorname{Ker} \mathrm{d}\eta_0 \cap \operatorname{Ker} \eta_0.$ □

We then have the following result:

Proposition 7.6. Let $i: N \hookrightarrow M$ be a coisotropic subamnifold such that $\mathcal{R}(q) \in T_qN$ for every $q \in N$ or $T_qN \subseteq \mathcal{H}_q$ for every $q \in N$. Then, the distribution $TN^{\perp_\Lambda}$ defined by $q \mapsto T_qN^{\perp_\Lambda}$ is involutive.

Proof. Denote $\eta_0: = i^*\eta$ and let $X, Y$ be vector fields along N taking values in $TN^{\perp_\Lambda}$. Proposition 7.5 implies that

$$\begin{equation*}i_X \mathrm{d}\eta_0 = i_Y \mathrm{d}\eta_0 = 0;\,\,\,\, i_X \eta_0 = i_Y \eta_0 = 0.\end{equation*}$$

It suffices to check that

$$\begin{equation*}i_{\left[X,Y\right]} \mathrm{d}\eta_0 = 0;\,\,\,\, i_{\left[X,Y\right]}\eta_0 = 0.\end{equation*}$$

Indeed, taking Z an arbitrary vector field in N, we have

$$\begin{align} 0 & = \operatorname{d}^2\eta_0\left(X,Y,Z\right) = X\left( \mathrm{d}\eta_0\left(Y,Z\right)\right) - Y\left( \mathrm{d}\eta_0\left(X,Z\right)\right) + Z\left( \mathrm{d}\eta_0\left(X,Y\right)\right)\nonumber\\ & \quad - \mathrm{d}\eta_0\left(\left[X,Y\right],Z\right) + \mathrm{d}\eta_0\left(\left[X,Z\right],Y\right) - \mathrm{d}\eta_0\left(\left[Y,Z\right], X\right) = -\mathrm{d}\eta_0\left(\left[X,Y\right],Z\right), \end{align}$$

where we have used that $X,Y \in \operatorname{Ker} \mathrm{d} \eta_0.$ In a similar way we obtain

$$\begin{align*} 0 = \mathrm{d}\eta_0\left(X,Y\right) = X\left(\eta_0\left(Y\right)\right) - Y\left(\eta_0\left(X\right)\right) - \eta_0\left(\left[X,Y\right]\right) = -\eta_0\left(\left[X,Y\right]\right), \end{align*}$$

that is, $[X,Y] \in \operatorname{Ker} \mathrm{d}\eta_0 \cap \operatorname{Ker} \eta_0 = TN^{\perp_\Lambda}.$ □

We will restrict the study to vertical submanifolds, that is, submanifolds satisfying $\mathcal{R}(q) \in T_qN$, for every $q \in N$. Notice that if N is a coisotropic vertical submanifold, the distribution $(TN)^{\perp_\Lambda}$ is regular of rank

$$\begin{equation*}\operatorname{rank} \, \left(TN\right)^{\perp_\Lambda} = \operatorname{dim} M - \operatorname{dim} N.\end{equation*}$$

Theorem 7.1 (Vertical coisotropic reduction in the contact setting). Let $(M, \eta)$ be a contact manifold and $i :N \hookrightarrow M$ be a coisotropic submanifold such that $\mathcal{R}(q) \in T_qN$ for every $q \in N$. If the space of all leaves $N /\mathcal{F}$ admits a manifold structure such that the projection $\pi: N \rightarrow N/\mathcal{F}$ is a submersion, then there exists a unique 1-form η_N in $N/\mathcal{F}$ such that $\pi^*\eta_N = i_*\eta$ and $(N/\mathcal{F},\eta_N)$ is a contact manifold.

Proof. Denote $\eta_0: = i^*\eta$. Uniqueness is clear from the imposed relation since it forces us to define

$$\begin{equation*}\eta_N\left(\left[u\right]\right): = \eta_0\left(u\right).\end{equation*}$$

It only remains to check well-definedness and that it defines a contact manifold. That this definition does not depend on the chosen representative vector is clear since a vector tangent to the distribution is necessarily in the kernel of η. Furthermore, if X is a vector field tangent to the distribution $TN^{\perp_\Lambda}$, proposition 7.5 implies

$$\begin{equation*}\mathcal{L}_X \eta_0 = 0 = \mathrm{d}i_X \eta_0 + i_X\mathrm{d}\eta_0 = 0,\end{equation*}$$

since $X \in \operatorname{Ker} \mathrm{d}\eta_0 \cap \operatorname{Ker} \eta_0$.

To check that it is a contact manifold, we calculate the dimension of $N /\mathcal{F}$. We have that $\operatorname{rank} \, (T_qN)^{\perp_\Lambda} = \operatorname{dim} M - \operatorname{dim} N$. We conclude, taking $k: = \operatorname{dim} N$, that

$$\begin{equation*}\operatorname{dim} N/\mathcal{F} = 2 \operatorname{dim} N - \operatorname{dim} M = 2\left(k-n-1\right) + 1\end{equation*}$$

and therefore, $(N/\mathcal{F}, \eta_N)$ is a contact manifold if and only if

$$\begin{equation*}\eta_N \wedge \left( \mathrm{d}\eta_N\right)^{k-n-1}\neq 0.\end{equation*}$$

Since π is a submersion, this is equivalent to $\eta_0 \wedge ( \mathrm{d}\eta_0)^{k-n-1}\neq0$. This is straightforward using proposition 7.5. □

7.4.1. Projection of Legendrian submanifolds.

Now we check that the image of a Legendrian submanifold $L \hookrightarrow M$ under the projection $\pi: N \rightarrow N/\mathcal{F}$ is again a Legendrian submanifold.

Proposition 7.7. Let $L \hookrightarrow M$ be a Legendrian submanifold such that L and N have clean intersection. If $L_N: = \pi(L \cap N)$ is a submanifold of $N/ \mathcal{F}$, then L_N is Legendrian.

Proof. It suffices to check that L_N is horizontal, isotropic and $\operatorname{dim} L_N = \operatorname{dim} N - n - 1$ using lemma 7.2. Since L is horizontal, L_N is horizontal and thus, L_N is isotropic.

Comparing dimensions, we have

$$\begin{equation} \operatorname{dim} T_{\left[q\right]}L_N = \operatorname{dim} T_qL\cap T_qN - \operatorname{dim} T_qL \cap T_qN^{\perp_\Lambda}. \end{equation} \tag{ 10 }$$

Now, since $(T_qL \cap T_qN^{\perp_\Lambda})^{\perp_\Lambda} = T_qL + (T_qN^{\perp_\Lambda})^{\perp_\Lambda}$ and $(T_qL \cap T_qN)^{\perp_\Lambda}$ is horizontal, we have

$$\begin{equation} \operatorname{dim}\left( T_qL + \left(T_qN^{\perp_\Lambda}\right)^{\perp_\Lambda}\right) = \operatorname{dim} M - \operatorname{dim}\left(T_qL \cap T_qN^{\perp_\Lambda}\right) - 1. \end{equation} \tag{ 11 }$$

Using $\operatorname{dim}(T_qL \cap T_qN^{\perp_\Lambda}) = \operatorname{dim}(T_qL + T_qN) - 1$ and substituting (11)) in (10), we obtain

$$\begin{align*} \operatorname{dim} L_N& = \operatorname{dim} L \cap N - \left( \operatorname{dim} M - \operatorname{dim}\left(T_qL + T_qN \right)\right) \\ & = \operatorname{dim} L \cap N - \operatorname{dim} M + \operatorname{dim} N + \operatorname{dim} L - \operatorname{dim} L \cap N \\ & = \operatorname{dim} N - 2n - 1 + n = \operatorname{dim} N - n -1. \end{align*}$$

 □

We will restrict the study to horizontal coisotropic submanifolds $N \rightarrow M$, that is, manifolds satisfying $T_qN \subseteq \mathcal{H}_q$ for every $q\in N$.

Remark 7.2. Notice that in this case reduction is trivial, since the only coisotropic horizontal submanifolds of a contact manifold are those that are Legendrian. This would imply

$$\begin{equation*}\operatorname{dim} N/\mathcal{F} = 0,\end{equation*}$$

making the resulting manifold trivial.

Given an arbitrary coisotropic submanifold $N \hookrightarrow M$, we cannot guarantee the well-definedness of the 2-form in the quotient $N/\mathcal{F}$ (actually, in the contact setting, we cannot even guarantee the integrability of $TN^{\perp_\Lambda}$) so this time (referring to horizontal reduction in cosymplectic geometry) we cannot obtain a foliation of N in symplectic leaves, since $TN \cap \mathcal{H}|N$ is not integrable in the general setting.

Remark 7.3. The triviality of this case makes the projection of Lagrangian submanifolds trivial.

In order to distinguish the different dynamics discussed in this paper and their physical nature, it is useful to recall their Lagrangian formulation.

Assume that $L : TQ \longrightarrow \mathbb{R}$ is a Lagrangian function, where Q is a n-dimensional configuration manifold. Then, $L = L(q^i, \dot{q}^i)$, where $(q^i)$ are coordinates in Q and $(q^i, \dot{q}^i)$ are the induced bundle coordinates in TQ (positions and velocities). We will assume that L is regular, that is, the Hessian matrix

$$\begin{equation*} \left( \frac{\partial^2 L}{\partial \dot{q}^i \partial \dot{q}^j} \right) \end{equation*}$$

is regular. Using the canonical endomorphism S on TQ locally defined by

$$\begin{equation*} S = \mathrm{d} q^i \otimes \frac{\partial}{\partial \dot{q}^i}. \end{equation*}$$

One can construct a 1-form λ_L defined by

$$\begin{equation*} \lambda_L = S^* \left(\mathrm{d}L\right) \end{equation*}$$

and the 2-form

$$\begin{equation*} \omega_L = - \mathrm{d}\lambda_L. \end{equation*}$$

Then, ω_L is symplectic if and only if L is regular.

In that case, we have the corresponding vector bundle isomorphism

$$\begin{eqnarray*} &&\flat_{\omega_L} : T\left(TQ\right) \longrightarrow T^*\left(TQ\right)\\ &&\flat_{\omega_L} \left(v\right) = i_v \, \omega_L \end{eqnarray*}$$

and the Hamiltonian vector field

$$\begin{equation*} \xi_L = X_{E_L}, \end{equation*}$$

defined by

$$\begin{equation*} \flat_{\omega_L}\left(\xi_L\right) = \mathrm{d}E_L, \end{equation*}$$

where $E_L = \Delta(L) -L$ is the energy, and $\Delta = \dot{q}^i \, \frac{\partial}{\partial \dot{q}^i}$ is the Liouville vector field on TQ.

The vector field ξ_L (the Euler–Lagrange vector field) is a second order differential equation, that is, its integral curves are just the tangent lifts of its projections to Q. These projections are called the solutions of ξ_L , and satisfy the usual Euler–Lagrange equations

$$\begin{equation} \frac{\mathrm{d}}{\mathrm{d}t} \left(\frac{\partial L}{\partial \dot{q}^i}\right) - \frac{\partial L}{\partial q^i} = 0. \end{equation} \tag{ 12 }$$

Next, we recall here the geometric formalism for time-dependent Lagrangian systems. In this case, we also have a regular Lagrangian $L : TQ \times \mathbb{R} \longrightarrow \mathbb{R}$, and we consider the cosymplectic structure given by the pair $(\Omega_L, \mathrm{d}z)$ (in this case z represents time), where

$$\begin{equation*} \Omega_L = - \mathrm{d} \lambda_L. \end{equation*}$$

It is esay to check that L is regular if and only if

$$\begin{equation*} \mathrm{d}z \wedge \Omega_L^n \not = 0. \end{equation*}$$

In that case, we have a cosymplectic structure and, defining $(W^{ij})$ to be the inverse matrix of

$$\begin{equation*}\left(W_{ij}\right) = \left( \frac{\partial^2 L}{\partial \dot{q}^i \partial \dot{q}^j} \right).\end{equation*}$$

The Reeb vector field of the cosymplectic manifold $(TQ, \Omega_L, \mathrm{d}z)$ is locally given by

$$\begin{equation*} \mathcal R = \frac{\partial}{\partial z} - W^{ij} \frac{\partial^2 L}{\partial \dot{q}^j \partial z} \, \frac{\partial}{\partial \dot{q}^i}. \end{equation*}$$

Recall that we had the vector bundle isomorphism

$$\begin{eqnarray*} &&{\flat}_{\mathrm{d}z, \Omega_L} : T\left(TQ \times \mathbb{R}\right) \longrightarrow T^*\left(TQ \times \mathbb{R}\right)\\ &&{\flat}_{\mathrm{d}z, \Omega_L} \left(v\right) = i_v \, \Omega_L + \mathrm{d}z \left(v\right) \, \mathrm{d}z \end{eqnarray*}$$

which gives the vector fields defined in section 5. In particular, we have the evolution vector field

$$\begin{equation*}\mathcal{E}_{E_L} = X_{E_L} + \mathcal{R},\end{equation*}$$

where E_L is the energy of the system,

$$\begin{equation*}E_L = \Delta\left(L\right) - L = \dot{q}^i \frac{\partial L}{\partial q^i} - L.\end{equation*}$$

Now, if $(q^i(t), \dot{q}^i(t), z(t))$ is an integral curve of ${\mathcal E}_L$, then its projection to Q satisfies the usual Euler–Lagrange equations

$$\begin{equation} \frac{\mathrm{d}}{\mathrm{d}t} \left(\frac{\partial L}{\partial \dot{q}^i}\right) - \frac{\partial L}{\partial q^i} = 0, \end{equation} \tag{ 13 }$$

since $z = t + constant$.

Finally, let $L : TQ \times \mathbb{R} \longrightarrow \mathbb{R}$ be a Lagrangian function, $L = L(q^i, \dot{q}^i, z)$, where z is a global coordinate on $\mathbb{R}$, representing action.

We will assume that L is regular, that is, the Hessian matrix

$$\begin{equation*} \left( \frac{\partial^2 L}{\partial \dot{q}^i \partial \dot{q}^j} \right) \end{equation*}$$

is regular. So, we construct a 1-form λ_L defined by

$$\begin{equation*} \lambda_L = S^* \left(\mathrm{d}L\right) \end{equation*}$$

where now S and $S^*$ are the natural extension of S.

Now, the 1-form

$$\begin{equation*} \eta_L = \mathrm{d}z - \frac{\partial L}{\partial \dot{q}^i} \, dq^i \end{equation*}$$

is a contact form on $TQ \times \mathbb{R}$ if and only if L is regular, and then

$$\begin{equation*} \eta_L \wedge \left(\mathrm{d}\eta_L\right)^n \not = 0. \end{equation*}$$

The corresponding Reeb vector field is

$$\begin{equation*} {\mathcal R} = \frac{\partial}{\partial z} - W^{ij} \frac{\partial^2 L}{\partial \dot{q}^j \partial z} \, \frac{\partial}{\partial \dot{q}^i}. \end{equation*}$$

The energy of the system is defined as in the precedent cases by

$$\begin{equation*} E_L = \Delta \left(L\right) - L \end{equation*}$$

where $\Delta = \dot{q}^i \, \frac{\partial}{\partial \dot{q}^i}$ is the Liouville vector field on TQ extended in the usual way to $TQ \times \mathbb{R}$. The vector bundle isomorphism of contact manifolds defined in section 7 is given by

$$\begin{equation*} {\flat_{\eta_L}} : T\left(TQ \times \mathbb{R}\right) \longrightarrow T^* \left(TQ \times \mathbb{R}\right);\,\,\, {\flat_{\eta_L}} \left(v\right) = i_v \left(\mathrm{d}\eta_L\right) + \left(i_v \eta_L\right) \cdot \eta_L. \end{equation*}$$

We shall denote its inverse by ${\sharp_{\eta_L}} = ({\flat_{\eta_L})}^{-1}$.

Denote by ξ_L the unique vector field defined by the equation

$$\begin{equation} {\flat_{\eta_L}} \left({\xi}_L\right) = \mathrm{d}E_L - \left(\mathcal R\left(E_L\right) + E_L\right) \, \eta_L. \end{equation} \tag{ 14 }$$

Note that this is precisely the Hamiltonian vector field of E_L defined in section 7. A direct computation from equation (14) shows that, if $(q^i(t), \dot{q}^i(t), z(t))$ is an integral curve of $\bar{\xi}_L$, we obtain

$$\begin{equation*} {\ddot{q}}^i \, \frac{\partial}{\partial \dot{q}^i}\left(\frac{\partial L}{\partial \dot{q}^j}\right) + \dot{q}^i \, \frac{\partial}{\partial q^i}\left(\frac{\partial L}{\partial \dot{q}^j}\right) + \dot{z} \frac{\partial}{\partial z}\left(\frac{\partial L}{\partial \dot{q}^i}\right) - \frac{\partial L}{\partial q^i} = \frac{\partial L}{\partial \dot{q}^i} \frac{\partial L}{\partial z} \end{equation*}$$

with an additional equation $\dot{z} = L$. These equations correspond to the generalized Euler–Lagrange equations considered by G Herglotz in 1930.

$$\begin{equation} \frac{\mathrm{d}}{\mathrm{d}t} \left(\frac{\partial L}{\partial \dot{q}^i}\right) - \frac{\partial L}{\partial q^i} = \frac{\partial L}{\partial \dot{q}^i} \frac{\partial L}{\partial z}. \end{equation} \tag{ 15 }$$

Notice that Herglotz equations depend on the action, so this type of Lagrangians are called in physics dependent on the action.

In order to give additional differences between the usual Hamilton principle and Herglotz principle, it is interesting to recall briefly the last one and how it is a natural generalization of the former one.

Let $L:TQ \times \mathbb{R} \to \mathbb{R}$ be a Lagrangian function.

Fix $q_1,q_2 \in Q$ and an interval $[a,b] \subset \mathbb{R}$. We denote by $\Omega(q_1,q_2, [a,b]) \subseteq({\cal C}^\infty([a,b]\to Q))$ the space of smooth curves ξ such that $\xi(a) = q_1$ and $\xi(b) = q_2$. This space has the structure of an infinite dimensional smooth manifold whose tangent space at ξ is given by the set of vector fields over ξ that vanish at the endpoints, that is,

$$\begin{equation} \begin{aligned} T_\xi \Omega\left(q_1,q_2, \left[a,b\right]\right) = & \left\{ v_\xi \in {\cal C}^\infty\left(\left[a,b\right] \to TQ\right) \mid \tau_Q \circ v_\xi = \xi, \,v_\xi\left(a\right) = 0, \, v_\xi\left(b\right) = 0 \right\}. \end{aligned} \end{equation} \tag{ 16 }$$

We will consider the following maps. Fix an initial action $c \in \mathbb{R}$. Let

$$\begin{equation} \mathcal{Z}:\Omega\left(q_1,q_2, \left[a,b\right]\right) \to {\cal C}^\infty \left(\left[a,b\right] \to \mathbb{R}\right) \end{equation} \tag{ 17 }$$

be the operator that assigns to each curve ξ the curve $\mathcal{Z}(\xi)$ that solves the following ODE:

$$\begin{equation} \frac{\mathrm{d} \mathcal{Z}\left(\xi\right)\left(t\right)}{\mathrm{d}t} = L\left(\xi\left(t\right), \dot \xi\left(t\right), \mathcal{Z}\left(\xi\right)\left(t\right)\right), \quad \mathcal{Z}\left(\xi\right)\left(a\right) = c. \end{equation} \tag{ 18 }$$

Now we define the action functional as the map which assigns to each curve the solution to the previous ODE evaluated at the endpoint:

$$\begin{equation} \begin{aligned} \mathcal{A}: \Omega\left(q_1,q_2, \left[a,b\right]\right) &\to \mathbb{R},\\ \xi &\mapsto \mathcal{Z}\left(\xi\right)\left(b\right), \end{aligned} \end{equation} \tag{ 19 }$$

that is, $\mathcal{A} = ev_b \circ \mathcal{Z}$, where $ev_b: \zeta \mapsto \zeta(b)$ is the evaluation map at b. We have

Theorem 8.1. Let $L: TQ \times \mathbb{R} \to \mathbb{R}$ be a Lagrangian function and let $\xi\in \Omega(q_1,q_2, [a,b])$ be a curve in Q. Then, $(\xi,\dot\xi, \mathcal{Z}(\xi))$ satisfies the Herglotz's equations if and only if ξ is a critical point of $\mathcal{A}$.

Remark 8.1. This theorem generalizes Hamilton's Variational Principle. In the case that the Lagrangian is independent of the $\mathbb{R}$ coordinate (i.e. $L(q^i,\dot{q}^i,z) = \hat{L}(q^i, \dot{q}^i)$) the contact Lagrange equations reduce to the usual Euler–Lagrange equations. In this situation, we can integrate the ODE of (19) and we get

$$\begin{equation} \mathcal{A}\left(\xi\right) = \int_a^b \hat L\left(\xi\left(t\right),\dot\xi\left(t\right)\right)\mathrm{d} t + \frac{c}{b-a}, \end{equation} \tag{ 20 }$$

that is, the usual Euler–Lagrange action up to a constant.

Given a Lagrangian function $L : TQ \times \mathbb{R} \longrightarrow \mathbb{R}$ we can define the Legendre transformation

$$\begin{equation*} \mathbb{F}L : TQ \times \mathbb{R} \longrightarrow T^*Q \times \mathbb{R} \end{equation*}$$

given by

$$\begin{equation*} \mathbb{F}L \left(q^i, \dot{q}^i, z\right) = \left(q^i, \hat{p}_i, z\right) \end{equation*}$$

where

$$\begin{equation*} \hat{p}_ i = \frac{\partial L}{\partial \dot{q}^i}. \end{equation*}$$

A direct computation shows that

$$\begin{equation*} \mathbb{F}L ^* \lambda_Q = \lambda_L, \end{equation*}$$

where λ_Q is the canonical Liouville form on $T^*Q$, here extended to the product manifold $T^*Q \times \mathbb{R}$. If the Lagrangian does not depend on time, then the Legendre transformation reduces to a bundle morphism

$$\begin{equation*} \mathbb{F}L : TQ \longrightarrow T^*Q \end{equation*}$$

and then the pull-back of the canonical symplectic form ω_Q on $T^*Q$ is just ω_L . In addition, the energy E_L corresponds via the Legendre transformation to the Hamiltonian energy H such that

$$\begin{equation*} \left(\mathbb{F}L\right)^*\left(H\right) = E_L. \end{equation*}$$

Since the corresponding geometric structures (symplectic, cosymplectic and contact) on both sides are preserved by the Legendre transformation, one concludes that corresponding dynamics are $\mathbb{F}L$-related (see [14] for the details).

Just like in previous sections, define the gradient vector field of certain Hamiltonian $H\in \mathcal{C}^\infty(M)$ as

$$\begin{equation*} \operatorname{grad} H = \sharp_{\theta, \eta}\left(\mathrm{d}H\right).\end{equation*}$$

Locally, the gradient vector field is expressed:

$$\begin{equation*}\operatorname{grad} H = \frac{\partial H}{\partial p_i} \frac{\partial }{\partial q^i} - \left(\frac{\partial H}{\partial q^i} + p_i\frac{\partial H}{\partial z} \right) \frac{\partial }{\partial p_i} + \left( p_i \frac{\partial H}{\partial p_i} + \frac{\partial H}{\partial z}\right) \frac{\partial }{\partial z} + \frac{\partial H}{\partial t} \frac{\partial }{\partial t}.\end{equation*}$$

We can define a symplectic structure in TM taking

$$\begin{equation*}\Omega_0 : = \flat_{\theta, \eta} ^* \Omega_M,\end{equation*}$$

where $\Omega_M$ is the canonical symplectic form on the cotangent bundle. In the induced coordinates $(q^i, p_i, z, \dot{q}^i, \dot{p}_i, \dot{z}),$ Ω₀ takes the form

$$\begin{align*} \Omega_0 = \,& \mathrm{d}q^i \wedge \left(p_ip_i \mathrm{d}q^j + \left(1 + \delta_i^j\right)p_i \dot{q}^j \mathrm{d}p_j - \dot{z} \mathrm{d}p_i - \mathrm{d}\dot{p}_i - p_i \mathrm{d}\dot{z}\right) \\ & + \mathrm{d}p_i \wedge \mathrm{d}\dot{q}^i + \mathrm{d}z \wedge \left( \mathrm{d}\dot{z} - p_i\mathrm{d}\dot{q}^i - \dot{q}^i \mathrm{d}p_i\right) + \mathrm{d}t \wedge \mathrm{d}\dot{t}. \end{align*}$$

It is easy to verify that a vector field $X: M \rightarrow TM$ is a locally gradient vector field if and only if it defines a Lagrangian submanifold in $(TM , \Omega_0).$

Definition 9.2 (Hamiltonian vector field). Given a Hamiltonian H on M, define its Hamiltonian vector field as

$$\begin{equation*}X_H: = \sharp_{\theta, \eta}\left(\mathrm{d}H\right) - \left(\mathcal{R}_z\left(H\right) + H\right) \mathcal{R}_z + \left(1 - \mathcal{R}_t\left(H\right)\right) \mathcal{R}_t.\end{equation*}$$

The Hamiltonian vector field has the local expression

$$\begin{equation*}X_H = \frac{\partial H}{\partial p_i} \frac{\partial }{\partial q^i} - \left(\frac{\partial H}{\partial q^i} + p_i\frac{\partial H}{\partial z} \right) \frac{\partial }{\partial p_i} + \left( p_i \frac{\partial H}{\partial p_i} - H\right) \frac{\partial }{\partial z} + \frac{\partial }{\partial t}.\end{equation*}$$

Integral curves of this vector field satisfy

$$\begin{align*} \frac{\mathrm{d} q^i}{\mathrm{d} \lambda} & = \frac{\partial H}{\partial p_i},\\ \frac{\mathrm{d} p_i}{\mathrm{d} \lambda} & = - \frac{\partial H}{\partial q^i} - p_i \frac{\partial H}{\partial z},\\ \frac{\mathrm{d} z}{\mathrm{d} \lambda} & = p_i \frac{\partial H}{\partial p_i} - H,\\ \frac{\mathrm{d} t}{\mathrm{d} \lambda} & = 1, \end{align*}$$

which are equivalent to the time-dependent version of the contact equations:

$$\begin{align*} \frac{\mathrm{d} q^i}{\mathrm{d} t} & = \frac{\partial H}{\partial p_i},\\ \frac{\mathrm{d} p_i}{\mathrm{d} t} & = - \frac{\partial H}{\partial q^i} - p_i \frac{\partial H}{\partial z},\\ \frac{\mathrm{d} z}{\mathrm{d} t} & = p_i \frac{\partial H}{\partial p_i} - H. \end{align*}$$

In general, X_H does not define a Lagrangian submanifold of $(TM, \Omega_0)$; but, just like in the cosymplectic and contact scenario, we can achieve this by modifying the symplectic form. Indeed, since

$$\begin{equation*}X_H^* \Omega_0 = - \mathrm{d}X_H^{\flat_{\theta, \eta}} = \mathrm{d}\left(\left(\mathcal{R}_z\left(H\right) + H\right) \eta\right) - \mathrm{d}\left(\left(1 - \mathcal{R}_t\left(H\right)\right) \theta \right),\end{equation*}$$

defining

$$\begin{equation*} \Omega_H : = \Omega_0 - \operatorname{d}\left(\left(\mathcal{R}_z\left(H\right) + H\right) \eta\right) + \operatorname{d}\left(\left(1 - \mathcal{R}_t\left(H\right)\right) \theta\right),\end{equation*}$$

we have that X_H defines a Lagrangian submanifold of $(TM, \Omega_H).$

Now we interpret the Hamiltonian vector field X_H as a Legendrian submanifold of $TM \times \mathbb{R} \times \mathbb{R}$ with the cocontact structure given by the forms

$$\begin{equation*}\widetilde \eta : = \eta^c + s \eta^v + \theta^c + e \theta^v; \,\,\, \widetilde \theta = \theta^c,\end{equation*}$$

where (s, e) are the parameters in $\mathbb{R} \times \mathbb{R}$. In local coordinates $(q^i, p_i, z, t, \dot{q}^i, \dot{p}_i, \dot{z}, \dot{t}, s,e)$, these forms have the expression:

$$\begin{align*} \widetilde \eta & = \mathrm{d}\dot{z} - \dot{p}_i \mathrm{d}q^{i} - p_i \mathrm{d}\dot{q}^i + s \mathrm{d}z - sp_i \mathrm{d}q^i + \mathrm{d}\dot{t} + e \mathrm{d}t, \\ \widetilde \theta & = \mathrm{d}\dot{t}. \end{align*}$$

It is easy to see that these forms define a cocontact structure. Now, given a vector field $X: M \rightarrow TM$ and two functions $f, g$ on M, define

$$\begin{equation*}X \times f \times g: M \rightarrow TM \times \mathbb{R} \times \mathbb{R}; \,\,\, p \mapsto \left(X\left(p\right), f\left(p\right), g\left(p\right)\right).\end{equation*}$$

Applying the properties of complete and vertical lifts, namely $X^*\alpha^c = \mathcal{L}_X(\alpha)$, we have

$$\begin{equation*}\left(X \times f \times g\right)^* \widetilde \eta = \mathcal{L}_X \eta + f \eta + \mathcal{L}_X \theta + g \theta\end{equation*}$$

and

$$\begin{equation*}\left(X \times f \times g\right)^*\widetilde \theta = \mathrm{d}\theta\left(X\right).\end{equation*}$$

Proposition 9.1. Let H be a Hamiltonian on M. Then $X_H \times \mathcal{R}_z(H) \times 0$ defines a Legendrian submanifold of $(TM \times \mathbb{R} \times \mathbb{R}, \widetilde \theta, \widetilde \eta).$

Proof. Using the observation above and lemma 9.1, it is sufficient to observe that

$$\begin{equation*}\mathcal{L}_{X_H}\eta = - \mathcal{R}_z\left(H\right) \eta, \,\, \mathcal{L}_{X_H}\theta = 0.\end{equation*}$$

 □

A cocontact manifold is also a Jacobi manifold defining

$$\begin{equation*}\Lambda\left(\alpha, \beta\right) : = -\mathrm{d}\eta\left(\sharp\left(\alpha\right), \sharp\left(\beta\right)\right), \,\, E = -\mathcal{R}_z,\end{equation*}$$

and thus, we have the Λ -orthogonal and the corresponding definitions of isotropic, coisotropic or Legendrian submanifolds and distributions.

Notice that $\mathcal{H}_t$ is an integrable distribution and that each leaf of its foliation inherits a contact structure. Indeed, $\mathcal{H}_t$ is the characteristic distribution defined by the Jacobi structure, $\mathcal{S}$.

Now we give a symplectic interpretation of the Λ-orthogonal. Notice that the restriction of $\mathrm{d}\eta$ to $\mathcal{H}_{tz}$ defines a symplectic structure on the distribution. Denote by $\perp_{\mathrm{d}\eta|}$ its symplectic orthogonal. The Λ-orthogonal is just the symplectic orthogonal of the intersection with $\mathcal{H}_{tz}$.

Proposition 9.2. Given a distribution Δ on a cocontact manifold $(M, \eta, \theta)$,

$$\begin{equation*}\Delta^{\perp_\Lambda} = \left(\Delta \cap \mathcal{H}_{tz}\right)^{\perp_{\mathrm{d}\eta |}}.\end{equation*}$$

Proof. We check one inclusion and compare dimensions:

Let $\alpha \in \Delta^0_q$ and $u \in \Delta^0_q \cap (\mathcal{H}_{tz})_q.$ We will see that $d\eta_q(u,\sharp_\Lambda(\alpha)) = 0.$ Indeed,

$$\begin{align*} \mathrm{d}\eta_q\left(u, \sharp_\Lambda\left(\alpha\right)\right) & = \mathrm{d}\eta_q\left(u, \sharp_\Lambda\left(\alpha\right)\right) + \theta_q\left(u\right) \theta_q\left(\sharp_\Lambda\left(\alpha\right)\right) + \eta_q\left(u\right) \eta_q\left(\sharp_\Lambda\left(\alpha\right)\right) \\ & = \langle \flat\left(u\right), \sharp_\Lambda\left(\alpha\right) \rangle = \Lambda_q\left(\alpha, \flat\left(u\right)\right) = - \mathrm{d}\eta_q\left(\sharp\left(\alpha\right), \sharp \left(\flat\left(u\right)\right)\right)\\ & = - \mathrm{d}\eta_q\left(\sharp\left(\alpha\right), u\right) = - \mathrm{d}\eta_q\left(\sharp\left(\alpha\right),u\right) - \theta_q\left(\sharp\left(\alpha\right)\right)\theta_q(u) - \eta_q(\sharp(\alpha))\eta_q(u)\\ & = -\langle \flat(\sharp(\alpha)), u\rangle = - \alpha(u) = 0, \end{align*}$$

that is, $\Delta^{\perp_\Lambda} \subseteq (\Delta \cap (\mathcal{H}_{tz})_q)^{\perp_{\mathrm{d}\eta|}}.$

Now we compare both dimensions. Let $k: = \operatorname{dim} \Delta, r_q : = \operatorname{dim}(\Delta_q^0 \cap\langle \theta_q, \eta_q\rangle).$ Since

$$\begin{equation*}\Delta_q^0 \cap \langle \theta_q, \eta_q\rangle = \left(\Delta_q \oplus \left(\mathcal{H}_{tz}\right)_q\right)^0,\end{equation*}$$

we have

$$\begin{align*} r_q & = \operatorname{dim} \left(\Delta_q^0 \cap \langle \theta_q, \eta_q\rangle\right) = \operatorname{dim}\left(\Delta_q \oplus \left(\mathcal{H}_{tz}\right)_q\right)^0 = 2n +2 - \operatorname{dim} \left(\Delta_q \oplus \left(\mathcal{H}_{tz}\right)_q\right)\\ & = 2n + 2 - \left(\operatorname{dim} \Delta_q + \operatorname{dim} \left(\mathcal{H}_{tz}\right)_q - \operatorname{dim}\left(\Delta_q \cap \left(\mathcal{H}_{tz}\right)_q\right) \right)\\ & = 2n + 2 - k - 2n + \operatorname{dim} \left(\Delta_q \cap \left(\mathcal{H}_{tz}\right)_q\right)\\ & = 2 + \operatorname{dim} \left(\Delta_q \cap \left(\mathcal{H}_{tz}\right)_q\right) - k, \end{align*}$$

which implies that

$$\begin{equation*}\operatorname{dim} \left(\Delta_q \cap \left(\mathcal{H}_{tz}\right)_q\right) = k + r_q - 2.\end{equation*}$$

It only remains to observe that

$$\begin{equation*}\operatorname{dim} \Delta^{\perp_\Lambda} = 2n + 2 - k - r_q = \operatorname{dim} \Delta^0 - \operatorname{dim} \left(\Delta^0 \cap \operatorname{Ker} \sharp_\Lambda\right)\end{equation*}$$

and that

$$\begin{equation*}\operatorname{dim}\left(\Delta \cap \left(\mathcal{H}_{tz}\right)_q\right)^{\perp_{\mathrm{d}\eta |}} = 2n + 2 - k - r_q = 2n - \operatorname{dim}\left(\Delta \cap \left(\mathcal{H}_{tz}\right)_q\right). \end{equation*}$$

 □

Now we can give a characterization of Legendrian submanifolds:

Lemma 9.1.  $i:L \rightarrow M$ is a Legendrian submanifold ($(TL)^{\perp_\Lambda} = TL$) if and only if $\operatorname{dim} L = n$ and

$$\begin{equation*}i^*\theta = 0, i^*\eta = 0.\end{equation*}$$

Proof. Necessity is clear, since L is necessarily horizontal. Sufficiency follows from $i^* \mathrm{d}\eta = 0$ which, together with $\operatorname{dim} L = n$, implies that T_q L is a Lagrangian submanifold of $(\mathcal{H}_{tz})_q$, for every $q\in L$. Using proposition 9.2, we have the equivalence. □

This allows us to express the Λ-orthogonal complement of a coisotropic distribution in a more convenient way:

Corollary 9.1. Let Δ be a coisotropic distribution on M ($\Delta^{\perp_\Lambda}\subseteq \Delta$), then

$$\begin{equation*}\Delta^{\perp_\Lambda} = \operatorname{Ker} \mathrm{d}\eta_0|_{\Delta \cap \mathcal{H}_{tz}} \cap \operatorname{Ker} \eta_0 \cap \operatorname{Ker} \theta_0,\end{equation*}$$

where η₀ and θ₀ are the restrictions of η and θ to Δ, respectively.

Corollary 9.2. Let $i: N \hookrightarrow M$ be a coisotropic submanifold of $(M,\theta, \eta).$ Then the distribution $TN^{\perp_\Lambda}$ is involutive.

Proof. Denote $\eta_0 : = i^*\eta$, $\theta_0 : = i^*\theta$ and let $X,Y$ be vector fields on N tangent to the distribution $TN^{\perp_\Lambda}$ and Z be an arbitrary tz-horizontal vector field on N. Using corollary 9.2, we only need to check that $[X,Y] \in \operatorname{Ker} \mathrm{d}\eta_0|_{\Delta \cap \mathcal{H}_{tz}} \cap \operatorname{Ker} \eta_0 \cap \operatorname{Ker} \theta_0.$ Indeed, expanding the expressions $0 = \operatorname{d}^2 \eta_0(X,Y,Z)$, $0 = \mathrm{d}\eta_0(X,Y)$, $0 = \mathrm{d}\theta_0(X,Y)$; we obtain

$$\begin{align*} 0 & = - \mathrm{d}\eta_0\left(\left[X,Y\right], Z\right), \\ 0 & = - \eta_0\left(\left[X,Y\right]\right), \\ 0 & = - \theta_0\left(\left[X,Y\right]\right). \end{align*}$$

 □

Now, given a coisotropic submanifold $N \hookrightarrow M$, since $TN^{\perp_\Lambda}$ is involutive, it provides a maximal foliation, $\mathcal{F}$. We assume that $N/\mathcal{F}$ inherits a manifold structure such that the canonical projection $\pi: N \rightarrow N/\mathcal{F}$ is a submersion.

Just like in the previous cases, for the well-definedness and non-degeneracy of the forms in the quotient, we need to restrict the coisotropic submanifolds we are studying. Consequently, we will say that a submanifold $N \hookrightarrow M$ is:

• i)  
  t-vertical (resp. z-vertical) if $\mathcal{V}_t\subseteq TN$ (resp. $\mathcal{V}_z\subseteq TN$).
• ii)  
  tz-vertical, if it is both t-vertical and z-vertical.
• iii)  
  t-horizontal (resp. z-horizontal) if $\mathcal{H}_t \subseteq TN$ (resp. $\mathcal{H}_z \subseteq TN$).
• iv)  
  tz-horizontal if it is both t-horizontal and z-horizontal, that is, if $TN \subseteq \mathcal{H}_{tz}$.

Let $i: N \hookrightarrow M$ be a tz-vertical submanifold. It is easy to check that under these conditions

$$\begin{equation*}TN^{\perp_\Lambda} = \operatorname{Ker} \mathrm{d}\eta_0 \cap \operatorname{Ker} \eta_0 \cap \operatorname{Ker} \theta_0,\end{equation*}$$

and that $(TN)^{\perp_\Lambda}$ is a regular distribution of rank

$$\begin{equation*}\operatorname{rank}\, \left(TN\right)^{\perp_\Lambda} = \operatorname{dim} M - \operatorname{dim} N. \end{equation*}$$

Theorem 9.1 (tz-vertical coisotropic reduction). Let $i: N \hookrightarrow M$ be a tz-vertical submanifold of a cocontact manifold $(M, \theta, \eta).$ Denote by $\mathcal{F}$ the maximal foliation induced by the integrable distribution $TN^{\perp_\Lambda}$ on N. If $N/\mathcal{F}$ admits a manifold structure such that the canonical projection $\pi: N \rightarrow N/\mathcal{F}$ defines a submersion, then there exists unique forms $\theta_N, \eta_N$ on $N/\mathcal{F}$ such that $(N/\mathcal{F}, \theta_N, \eta_N)$ defines a cocontact structure and

$$\begin{equation*}i^*\theta = \pi^*\theta_N, \,\, i^*\eta_N = \pi^*\theta_N.\end{equation*}$$

Before proving the theorem, let us calculate the dimension of the quotient. Let $k + 2: = \operatorname{dim} N$. We have

$$\begin{equation*}\operatorname{rank}\, TN^{\perp_\Lambda} = 2n + 2 - \left(k + 2\right) = 2n - k\end{equation*}$$

and, therefore,

$$\begin{equation*}\operatorname{dim} N/\mathcal{F} = 2\left(k - n\right) + 2.\end{equation*}$$

Proof. Uniqueness is clear, since π is a submersion. We only need to check the well-definedness taking

$$\begin{equation*}\theta_N\left(\left[u\right]\right) : = \theta_0\left(u\right), \,\, \eta_N\left(\left[u\right]\right): = \eta_0\left(u\right).\end{equation*}$$

Independence of the vector is clear, using proposition 9.2. For independence on the point, let X be a vector field on N tangent to the distribution. It is easy to check that

$$\begin{equation*}\mathcal{L}_X \eta_0 = 0 , \mathcal{L}_X \theta_0 = 0;\end{equation*}$$

and thus, well-definedness follows. For non-degeneracy, it is enough to proof that

$$\begin{equation*}\theta_0 \wedge \eta_0 \wedge \left(\mathrm{d}\eta_0\right)^{k-n} \neq 0.\end{equation*}$$

This follows easily from $TN^{\perp_\Lambda} = \operatorname{Ker} \mathrm{d}\eta_0 \cap \operatorname{Ker} \eta_0 \cap \operatorname{Ker} \theta_0.$ □

9.3.1. Projection of Legendrian submanifolds

Proposition 9.3. Let $L \hookrightarrow M$ be a Legendrian submanifold and $i: L \hookrightarrow M$ be a tz-vertical coisotropic submanifold. If L and N have clean intersection and $L_N = \pi(L \cap N)$ is a submanifold in $N/\mathcal{F}$, L_N is Legendrian.

Proof. Using lemma 9.1, L_N is clearly isotropic. Now we need to check that $\operatorname{dim} L_N = k - n$, given that $\operatorname{dim} N/\mathcal{F} = 2(k-n) + 2.$ We have

$$\begin{equation} \operatorname{dim} \pi\left(L \cap N\right) = \operatorname{dim}\left(L \cap N\right) - \operatorname{rank}\, \left(TL \cap \left(TN\right)^{\perp_\Lambda}\right). \end{equation} \tag{ 21 }$$

Furthermore, since $(TL \cap (TN)^{\perp_\Lambda}) = TL + (TN^{\perp_\Lambda})^{\perp_\Lambda}$ and $TL \cap (TN)^{\perp_\Lambda}$ is tz-horizontal,

$$\begin{equation} \operatorname{rank} \,\left(TL + \left(TN^{\perp_\Lambda}\right)^{\perp_\Lambda}\right) = 2n - \operatorname{rank}\, \left(TL \cap TN^{\perp_\Lambda}\right). \end{equation} \tag{ 22 }$$

Now, using the Grassman formula:

$$\begin{align} \operatorname{rank} \,\left(TL + \left(TN^{\perp_\Lambda}\right)^{\perp_\Lambda}\right) & = \operatorname{dim} L + \operatorname{rank} \, \left(TN^{\perp_\Lambda}\right)^{\perp_\Lambda} - \operatorname{rank} \, \left(TL \cap \left(TN^{\perp_\Lambda}\right)^{\perp_\Lambda}\right) \end{align} \tag{ 23 }$$

$$\begin{align} & = \operatorname{dim} L + \left(\operatorname{dim} N - 2\right) - \operatorname{dim}\left(L \cap N\right) \end{align} \tag{ 24 }$$

$$\begin{align} & = n + k - \operatorname{dim}\left(L \cap N\right). \end{align} \tag{ 25 }$$

From (22) and (25) we obtain

$$\begin{equation} \operatorname{rank} \,\left(TL \cap TN^{\perp_\Lambda}\right) = n - k + \operatorname{dim}\left(L \cap N\right). \end{equation} \tag{ 26 }$$

Substituting in (21) yields $\operatorname{dim} \pi(L \cap N) = k - n.$ □

Suppose $i: N \hookrightarrow M$ is a t-vertical and z-horizontal coisotropic submanifold. This time we have

$$\begin{equation*}\left(TN\right)^{\perp_\Lambda} = \operatorname{Ker} \theta_0,\end{equation*}$$

since η₀ = 0 implies $d\eta _0 = 0$. We conclude that $(TN)^{\perp_\Lambda} = TN \cap \mathcal{H}_{tz}$, which implies that

$$\begin{equation*}\operatorname{dim} N/\mathcal{F} = 1.\end{equation*}$$

This means that reduction is trivial, leaving the trivial cosymplectic submanifold of dimension 1:

Theorem 9.2. Let $i: N \hookrightarrow M$ be a t-vertical, z-horizontal coisotropic submanifold of a coconatct manifold $(M , \theta, \eta)$. Denote by $\mathcal{F}$ the maximal foliation defined by the distribution $(TN)^{\perp_\Lambda}$. If $N/\mathcal{F}$ has a manifold structure such that the canonical projection $\pi: N \rightarrow N/\mathcal{F}$ defines a submersion, then $N/\mathcal{F}$ is one-dimensional and there exists and unique volume form θ_N on $N/\mathcal{F}$ such that

$$\begin{equation*}i^* \theta = \pi^* \theta_N.\end{equation*}$$

Remark 9.1. Given the triviality of reduction in the t-vertical and z-horizontal case, projection of Legendrian submanifolds in M will always result in 0-dimensional Lagrangian submanifolds in $N/\mathcal{F}$.

Let $i: N \rightarrow M$ be a z-vertical and t-horizontal coisotropic submanifold. It is easy to check that this time we have the equality

$$\begin{equation*}TN = \operatorname{Ker} \mathrm{d}\eta_0 \cap \operatorname{Ker} \eta_0.\end{equation*}$$

Since $\mathcal{H}_t$ is integrable, we have that coisotropic reduction of N is actually happening in one of the leaves of the foliation that inhertits a contact structure from the cocontact structure. We conclude, from theorem 7.1:

Theorem 9.3. Let $i: N \rightarrow M$ be a z-vertical and t-horizontal coisotropic submanifold of a cocontact manifold $(M, \theta, \eta)$. Denote by $\mathcal{F}$ the maximal foliation on N defined by the distribution $TN^{\perp_\Lambda}$. If $N/\mathcal{F}$ has a manifold structure such that the canonical projection $\pi: N \rightarrow N/\mathcal{F}$ defines a submersion, then there exists an unique form η_N such that $(N/\mathcal{F}, \eta_N)$ is a contact manifold and

$$\begin{equation*}i^*\eta = \pi^* \eta_N.\end{equation*}$$

9.5.1. Projection of Legendrian submanifolds

Proposition 9.4. Let $L \hookrightarrow M$ be a Legendrian submanifold. If L and N have clean intersection and $L_N = \pi(L \cap N)$ is a submanifold in $N/\mathcal{F}$, L_N is Legendrian in $(N/\mathcal{F}, \theta_N)$.

Proof. It is clearly horizontal and, therefore, using lemma 7.2, it is isotropic. Now, supposing $k = \operatorname{dim} N$, we only need to check that

$$\begin{equation*}\operatorname{dim} L_N = k - n,\end{equation*}$$

since $\operatorname{dim} N/\mathcal{F} = 2(k - n) + 1.$ This is straight-forward, following the same steps given in proposition 9.3. □

Let $i: N \hookrightarrow M$ be a tz-horizontal coisotropic submanifold. Since η₀ = 0, $\mathrm{d}\eta_0 = 0$ and θ₀ = 0, we have

$$\begin{equation*}\left(TN\right)^{\perp_\Lambda} = TN,\end{equation*}$$

wich implies that

$$\begin{equation*}\operatorname{dim} N/\mathcal{F} = 0,\end{equation*}$$

leaving a trivial symplectic manifold, having as many points as path components of N. This means that if $N/\mathcal{F}$ admits a manifold structure, it will be a symplectic manifold.

Remark 9.2. Proceeding as in the previous cases we immediately obtain that the projection of Legendrian submanifolds is trivial.

We can define a symplectic structure on TM taking

$$\begin{equation*}\Omega_0 : = \flat_{\lambda, \omega} ^* \Omega_M,\end{equation*}$$

where $\Omega_M$ is the canonical symplectic form on $T^*M$. In Darboux coordinates it has the expression:

$$\begin{align*} \Omega_0 & = \mathrm{d}q^j \wedge \mathrm{d}\left(\dot{q}^ia_ia_j + \dot{p}_ib^ia_j+ \dot{z}a_j - \dot{p}_j \right) + \mathrm{d}p_j \wedge \mathrm{d}\left( \dot{q}^ia_ib^j + \dot{p_i}b^ib^j + \dot{z}b^j + \dot{q}^j \right) \\ & \quad + \mathrm{d}z \wedge \mathrm{d}\left( \dot{q}^ia_i + \dot{p}_ib^i + \dot{z}\right). \end{align*}$$

Definition 10.2 (Gradient vector field). Given a Hamiltonian $H \in \mathcal{C}^ \infty(M),$ define the gradient vector field of H as

$$\begin{equation*}\operatorname{grad} H: = \sharp_{\lambda, \omega}\left(\mathrm{d}H\right).\end{equation*}$$

In Darboux coordinates, the gradient vector field is written

$$\begin{equation*}\operatorname{grad} H = \left( \frac{\partial H}{\partial p_i} - b^i \frac{\partial H}{\partial z} \right) \frac{\partial }{\partial q^i} + \left( - \frac{\partial H}{\partial q^i} + a_i \frac{\partial H}{\partial z}\right) \frac{\partial }{\partial p_i} + \left( b^i\frac{\partial H}{\partial q^i} - a_i\frac{\partial H}{\partial p_i} + \frac{\partial H}{\partial z}\right) \frac{\partial }{\partial z}.\end{equation*}$$

It is easily checked that $X: M \rightarrow TM$ is locally a gradient vector field if and only if X(M) is a Lagrangian submanifold of $(TM, \Omega_0)$. Indeed, we have the equality

$$\begin{equation*}X^*\Omega_0 = - \mathrm{d}\flat_{\lambda, \omega}\left(X\right).\end{equation*}$$

When Λ comes from a Jacobi bracket on M, that is, when

$$\begin{equation*}\mathrm{d}\lambda = f \omega, \,\,\, \left[f \mathcal{R}, \Lambda\right] = 0,\end{equation*}$$

for some function f on M, we have the Hamiltonian vector field of the Jacobi structure $(\Lambda, f \mathcal{R})$:

$$\begin{equation*}X_H = \sharp_\Lambda\left(\mathrm{d}H \right) + f H \mathcal{R} = -\operatorname{grad} H + \left(\mathcal{R}\left(H\right) + fH \right)\mathcal{R} .\end{equation*}$$

In Darboux coordinates it has the expression:

$$\begin{equation*}X_H = \left( - \frac{\partial H}{\partial p_i}+ b^i \frac{\partial H}{\partial q^i}\right) \frac{\partial }{\partial q^i} + \left( \frac{\partial H}{\partial q^i} - a_i \frac{\partial H}{\partial z}\right) \frac{\partial }{\partial p_i} + \left( a_i \frac{\partial H}{\partial p_i} - b^i \frac{\partial H}{\partial q^i} + f H\right) \frac{\partial}{\partial z} {z}.\end{equation*}$$

Let us interpret the Hamiltonian vector field as a Lagrangian submanifold of TM, with some symplectic form. First, observe that

$$\begin{align*} X_H ^* \Omega_0 = - \mathrm{d}\left(\flat_{\lambda, \omega}\left(\mathrm{d}H\right)\right) = - \mathrm{d}\left( \mathcal{R}\left(H\right) \lambda + fH \lambda\right). \end{align*}$$

Therefore, defining the symplectic form

$$\begin{equation*}\Omega_H : = \Omega_0 + \mathrm{d}\left(\mathcal{R}\left(H\right) \lambda + fH \lambda\right)^v,\end{equation*}$$

we have that X_H defines a Lagrangian submanifold of $(TM, \Omega_H).$

Theorem 10.1 (Vertical coisotropic reduction in stable Hamiltonian structures). Let $i: N \hookrightarrow M$ be a vertical coisotropic submanifold such that $(TN)^{\perp_\Lambda}$ defines an integrable distribution. Let $\mathcal{F}$ be the set of leaves and suppose that $N/\mathcal{F}$ admits a manifold structure such that the canonical projection $\pi: N \rightarrow N/\mathcal{F}$ defines a submersion. If $i^* \mathrm{d}\lambda = 0$ in $TN \cap \mathcal{H}$, then $N/\mathcal{F}$ admits an unique stable Hamiltonian system structure $(\omega_N, \lambda_N)$ such that $\pi^*\omega_N = i^*\omega$ and $\pi^*\lambda_N = i^*\lambda$. The following diagram summarizes the situation:

Zoom In Zoom Out Reset image size

Proof. The proof is similar to the the proof of theorem 5.1. Asking $i^*d\lambda = 0$ is necessary to guarantee the well-definedness of λ_N in the quotient using

$$\begin{equation*}\mathcal{L}_X \lambda_0 = i_X\mathrm{d}\lambda-0 + \mathrm{d}i_X\lambda_0 = 0,\end{equation*}$$

where $\lambda_0 = i^*\lambda$. It would only remain to check that

$$\begin{equation*}\operatorname{Ker} \,\,\omega_N \subseteq \operatorname{Ker} \mathrm{d}\lambda_N.\end{equation*}$$

Indeed, since $\operatorname{Ker} \omega_N = \langle \mathcal{\mathcal{R}_N} \rangle$, and $\mathcal{R}_N = \pi_* \mathcal{R}$, it follows from

$$\begin{equation*}\pi^*\left(i_{\mathcal{R}_N}\mathrm{d}\lambda_N\right) = i_\mathcal{R} \mathrm{d}\lambda = 0.\end{equation*}$$

 □

10.2.1. Projection of Lagrangian submanifolds.

Theorem 10.2 (Horizontal coisotropic reduction in stable Hamiltonian structures). Let $i: N \rightarrow M$ be a coisotropic horizontal submanifold such that $(TN)^{\perp_\Lambda}$ defines an integrable distribution. Let $\mathcal{F}$ be the set of leaves of the foliation and suppose that $N/\mathcal{F}$ admits a manifold structure such that the canonical projection $\pi: N \rightarrow N/\mathcal{F}$ defines a submersion. Then $N/\mathcal{F}$ admits and unique symplectic structure ω_N such that $\pi^*\omega_N = i^*\omega$. The following diagram summarizes the situation:

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Proof. The proof is similar to the the proof of theorem 3.2. □

10.3.1. Projection of Lagrangian submanifolds