Hamiltonian Vector Field

Hamiltonian Vector Fields

A Riemannian metric on a manifold establishes an isomorphism between the space of tangent vectors and the space of cotangent vectors (i.e. 1-forms). A symplectic structure establishes a similar isomorphism.

Symplectic Duality

Let $(M,\omega )$ be a symplectic manifold. To each vector $v\in T_{p}M$, we associate a 1-form ${\alpha }_v\in T_p^{\ast }M$ by the relation ${\alpha }_v(u):=\omega (u,v),\text{for all }u\in T_{p}M.$Then the correspondence $v\mapsto {\alpha }_v$ is an isomorphism between $T_{p}M$ and $T_p^{\ast }M$. The isomorphism is called the symplectic duality, denoted as $\iota$.

Proof This is straightforward from the construction. $\square$

Consider euclidean space ${\mathbb{R}}^{2n}$ with coordinates $(q_1,\dots ,q_n,p_1,\dots ,p_n)$ and ${\omega }_0=\sum d{q}_j\wedge d{p}_j$. The curve ${\rho }_t=(q(t),p(t))$ is an integral curve for $X_H$ exactly if

$$\{\begin{array}{l}\frac{d{q}_i}{dt}(t)=\frac{\partial H}{\partial p_i}\\ \\ \frac{d{p}_i}{dt}(t)=-\frac{\partial H}{\partial q_i}\end{array}\text{(Hamilton equations)}$$

Let $(M,\omega )$ be a symplectic manifold and let $H:M\to \mathbb{R}$ be a smooth function. Its differential $dH$ is a 1-form. By nondegeneracy, there is a unique vector field $X_H$ on $M$ such that ${\iota }_{X_H}\omega =dH$. Integrate $X_H$. Supposing that $M$ is compact, or at least that $X_H$ is complete, let ${\rho }_t:M\to M,t\in \mathbb{R}$, be the one-parameter family of diffeomorphisms generated by $X_H$:

$$\{\begin{array}{l}{\rho }_0={\text{id}}_M\\ \\ \frac{d{\rho }_t}{dt}\circ {\rho }_t^{-1}=X_H\end{array}$$

Each diffeomorphism ${\rho }_t$ preserves $\omega$, i.e., ${\rho }_t^{\ast }\omega =\omega ,\forall t,$ therefore, every function on (M,ω) gives a family of symplectomorphisms.

Hamiltonian Vector Field

Let $(M,\omega )$ be a symplectic manifold. The Hamiltonian vector field associated with a smooth function $H:M\to \mathbb{R}$, called the Hamiltonian function, is the vector field $X_H\in \mathfrak{X}(M)$ defined by $X_H:={\iota }^{-1}(\text{d}H).$

Equivalent Definition of Hamiltonian Vector Field

Given a smooth function $H:M\to \mathbb{R}$ on a symplectic manifold $(M,\omega )$, the Hamiltonian vector field $X_H$ is the unique vector field such that $\text{d}H=-i_{X_H}\omega =\omega (\cdot ,X_H).$

Proof $X_H$ is the Hamiltonian vector field iff $$X_{H}=\iota^{-1} (\d H) \iff \d H=\iota(X_{H})=\omega(\cdot, X_{H}).$$$\square$

Symplectic Vector Field

A vector field $X$ on $M$ preserving $\omega$ (i.e., such that ${\mathcal{L}}_X\omega =0$) is called a symplectic vector field, or equivalently the flow ${\rho }_t$ of $X$ preserves $\omega$, i.e., ${\rho }_t^{\ast }\omega =\omega ,\forall t.$

\begin{cases} X \text{ is symplectic} & \iff \imath_X\omega \text{ is closed }, \ X \text{ is hamiltonian} & \iff \imath_X\omega \text{ is exact }. \end{cases}

**e.g.** - Suppose $(p_{1},\dots,p_{n},q_{1},\dots,q_{n})$ is a coordinate for $\R^{2n}$, let the standard symplectic form be $\omega=\sum_{i}\d p_{i} \wedge \d q_{i}$. Then the vector field defined by the [[Hamiltonian & Symmetries#^5ef3ed|Hamilton's equations of motion]]: $$\newcommand{\pddf}[2]{\frac{\partial#1}{\partial#2}}\dot p_j = - \frac{\partial H}{\partial q_{j}},\quad \dot q_j = \frac{\partial H}{\partial p_{j}}$$ is the Hamiltonian vector field $X_{H}$ associated with the Hamiltonian function $H\colon \R^{2n}\to\R$. We can show this by checking that the vector field defined by the Hamilton's equations of motion, say $$X(p_{1},\dots,p_{n},q_{1},\dots,q_{n}):=\sum_{i=1}^{n}\left(-\frac{\partial H}{\partial q_{i}}\pddf{}{p_{i}}+\pddf{H}{p_{i}}\pddf{}{q_{i}}\right)$$ satisfies the equation $\d H =-i_{X_{H}}\omega=\omega(\cdot,X).$ Indeed, for any vector field $Y\in\mathfrak{X}(\R^{2n})$, we have $$\newcommand{\L}{\mathcal{L}}\begin{aligned}\omega(Y,X)&=\sum_{i}^{n}(\d p_{i} \wedge \d q_{i})(Y,X) \\ &= \sum_{i=1}^{n} \d p_{i}(Y)\d q_{i}(X) - \d p_{i}(X)\d q_{i}(Y) \\ &= \sum_{i=1}^{n} \d p_{i}(Y) X(q_{i}) - X(p_{i})\d q_{i}(Y) \\ &= \sum_{i=1}^{n} \d p_{i}(Y) \pddf{H}{p_{i}} + \pddf{H}{q_{i}} \d q_{i}(Y) \\ &= \d H(Y). \end{aligned}$$ The gradient vector field of $H$ relative to the euclidean metric is $$ \nabla H := \sum_{i=1}^{n} \left( \frac{\partial H}{\partial q_i} \frac{\partial}{\partial q_i} + \frac{\partial H}{\partial p_i} \frac{\partial}{\partial p_i} \right). $$ If $J$ is the standard (almost) complex structure so that $J(\frac{\partial}{\partial q_i}) = \frac{\partial}{\partial p_i}$ and $J(\frac{\partial}{\partial p_i}) = -\frac{\partial}{\partial q_i}$, we have $JX_H = \nabla H$. $\lozenge$ - Consider the symplectic structure on the torus $\newcommand{\Z}{\mathbb{Z}}T^{2}=\R^{2}/\Z^{2}$ induced by the standard symplectic structure on $\R^{2}$, with coordinates $(p,q)$ and consider the vector field $X=\partial/\partial p$. The the flow of this vector field preserves the symplectic form: $$\begin{aligned}\L_{X}\omega&= \L_{X}(\d p\wedge\d q)\\ &= i_{X}\d\omega + \d (i_{X}(\d p\wedge \d q))\\&= \d((i_{X}\d p)\wedge \d q - \d p \wedge (i_{X}\d q))\\&= \d(\d q)\\&= 0.\end{aligned}$$But this vector field is not a Hamiltonian vector field, because if we assume $X$ is Hamiltonian, then we have $$-\d q=-i_{X}(\d p \wedge\d q)=\d H=\pddf{H}{p}\d p + \pddf{H}{q}\d q,$$it follows that $$\pddf{H}{p}=0,\quad \pddf{H}{q}=-1,$$which means $H(p,q)=-q+C$ for some constant $C\in\R$. Note that $T^{2}$ has no global chart, so $H$ can't be defined smoothly globally. In detail, at some point $z\in T^{2}$, the value of $H$ at $z$ must be the same for all charts that cover $z$. Since $(p,q)\mapsto-q+C$ is injective, these charts must be the same. However, $T^{2}$ has no global chart, so this is impossible. Alternatively, observe that $(p,q)\mapsto-q+C$ is strictly decreasing, so for some appropriate points $z_{1},z_{2}\in T^{2}$, under the chart $(p,q)$ at $z_{1}$ and $(p',q')$ at $z_{2}$, we might have $H(z_{1})<H(z_{2})$ and $H(z_{1})>H(z_{2})$ both hold. This is a contradiction, thus $X$ is not Hamiltonian. $\quad$ ## Conserved Quantities > [!proposition] > > Suppose $X_{H}$ is a Hamiltonian vector field associated with a Hamiltonian function $H\colon M\to\R$ on a symplectic manifold $(M,\omega)$. Then $\L_{X_{H}}\omega=0$ and $\L_{X_{H}}H=0$. ^72585c *Proof* By [[Differential Forms#^04ae6c|Cartan's formula]], we have $$\L_{X_{H}}\omega=\d(i_{X_{H}}\omega)+i_{X_{H}}(\d\omega)=\d(i_{X_{H}}\omega)$$since $\d\omega=0$. By the definition of Hamiltonian vector field, we know that $i_{X_{H}}\omega=-\d H$. Therefore, we have $$\L_{X_{H}}\omega=\d(i_{X_{H}}\omega)=-\d(\d H)=0.$$For the second part, we have $$\L_{X_{H}}H=i_{X_{H}}(\d H)=\d H(X_{H})=\omega(X_{H},X_{H})=0.$$$\square$ > [!remark]+ > > The second equation $\L_{X_{H}}H=0$ means that the Hamiltonian $H$ is conserved along the flow of the Hamiltonian vector field $X_{H}$. In a [[Lagrangian Dynamical System#^466228|natural Lagrangian system]], the Hamiltonian is the total energy of the system, and thus it is conserved. This is a manifestation of the conservation of energy in classical mechanics. > > [!proposition] Symplectic Manifolds Have a Natural Volume Form > > If $(M,ω)$ is a symplectic manifold, then the $n$th wedge of $ω$ is a [[Differential Forms#^923fae|volume form]], and also the flow of $X_{H}$ preserves this volume form. And thus the form $\frac{\omega^n}{n!}$ is called symplectic volume form. ^bec446 *Proof* Clearly $\omega^{n}$ is a $2n$-form on $M$. To see that it is a volume form, we need to check that it is non-degenerate. This follows from the fact that $\omega$ is non-degenerate. To see that the flow of $X_{H}$ preserves this volume form, we show that $\L_{X_{H}}\omega^{n}=0$: $$\begin{aligned}\L_{X_{H}}(\omega^{n})=\d (i_{X_{H}}\omega^{n})+ i_{X_{H}} \d \omega^{n} \end{aligned}.$$Note that $\d\omega^{n}=0$ since $\d\omega=0$. Therefore, we have $$\L_{X_{H}}(\omega^{n})=\d(i_{X_{H}}\omega^{n}).$$Moreover, $$\begin{aligned}\d(i_{X_{H}}\omega^{n})&=\d(i_{X_{H}}(\omega\wedge\omega^{n-1}))\\&=\d((i_{X_{H}}\omega)\wedge \omega^{n-1} + \omega\wedge i_{X_{H}}(\omega^{n-1}))\\&= \d(-\d H\wedge \omega^{n-1}) +\d(\omega\wedge i_{X_{H}}(\omega^{n-1}))\\ &= \d(\omega\wedge i_{X_{H}}(\omega^{n-1}))\\&= \d\omega \wedge i_{X_{H}}(\omega^{n-1}) +\omega\wedge \d(i_{X_{H}}(\omega^{n-1}))\\&=\omega\wedge \d(i_{X_{H}}(\omega^{n-1})), \end{aligned}$$so by induction, we have $\d(i_{X_{H}}\omega^{n})=0$. Thus $\L_{X_{H}}(\omega^{n})=0$. $\square$ > [!proposition] > > There are no compact exact symplectic manifolds. > *Proof* In fact, there is a unique vector field $X:=-\iota^{-1}(\lambda)$, such that $i_{X}\omega=\lambda$. So that $$\L_{X}\omega=\d(i_{X}\omega)+i_{X}\d\omega=\d\lambda=\omega.$$Hence, > [!definition] Action of a Loop > > Given an oriented closed loop $\gamma$ in an exact symplectic manifold $(M, \omega=\d\lambda)$, the action of $\gamma$ is defined as $$A(\gamma):=\int_{\gamma}\lambda.$$ > [!proposition] > > The flow of any Hamiltonian vector field preserves the action. > *Proof* To show this, suppose $h_{t}$ is a flow of a Hamiltonian vector field $X_{H}$, we claim that $$\newcommand{\ddt}{\ddf{}{t}}\ddt A(h_{t}(\gamma))=\ddt\int_{h_{t}(\gamma)}\lambda=\ddt\int_{\gamma}h_{t}^{*}\lambda=\int_{\gamma}\ddt h_{t}^{*}\lambda=\int h_{t}^{*}(\L_{X_{H}}\lambda)$$ >[!proposition] > > For any form $\alpha$, > $$\imath_{[X,Y]}\alpha = \mathcal{L}_X\imath_Y\alpha - \imath_Y\mathcal{L}_X\alpha = [\mathcal{L}_X, \imath_Y]\alpha \;.$$ >[!proposition] > >If $X$ and $Y$ are symplectic vector fields on a symplectic manifold $(M, \omega)$, then $[X, Y]$ is hamiltonian with hamiltonian function $\omega(Y, X)$. *Proof* $\begin{align*} \imath_{[X,Y]}\omega &= \mathcal{L}_X\imath_Y\omega - \imath_Y\mathcal{L}_X\omega \\ &= d\imath_X\imath_Y\omega + \imath_X \underbrace{d\imath_Y\omega}_{0} - \imath_Y \underbrace{d\imath_X\omega}_{0} - \imath_Y\imath_X \underbrace{d\omega}_{0} \\ &= d(\omega(Y, X)) \;. \end{align*}$ >[!corollary] > >Let >$$\begin{align*} \chi(M) &= \{ \text{vector fields on } M \} \\ \chi^{\text{sympl}}(M) &= \{ \text{symplectic vector fields on } M \} \\ \chi^{\text{ham}}(M) &= \{ \text{hamiltonian vector fields on } M \} \;. \end{align*}$$ > The inclusions $(\chi^{\text{ham}}(M), [\cdot, \cdot]) \subseteq (\chi^{\text{sympl}}(M), [\cdot, \cdot]) \subseteq (\chi(M), [\cdot, \cdot])$ are inclusions of Lie algebras. >[!definition] Poisson Bracket > >The Poisson bracket of two functions $f, g \in C^\infty(M; \mathbb{R})$ is $$ \{f, g\} := \omega(X_f, X_g) \;. $$ >[!definition] Poisson Algebra > >A Poisson algebra $(\mathcal{P}, \{\cdot, \cdot\})$ is a commutative associative algebra $\mathcal{P}$ with a Lie bracket $\{\cdot, \cdot\}$ satisfying the Leibniz rule:$$ \{f, gh\} = \{f, g\}h + g\{f, h\} \;. $$ >[!proposition] > > For $(M,\omega)$ a symplectic manifold, $(C^{\infty}(M), \{·, ·\})$ is a Poisson algebra. >[!definition] Hamiltonian System > >A hamiltonian system is a triple $(M, \omega, H)$, where $(M, \omega)$ is a symplectic manifold and $H \in C^{\infty}(M; \mathbb{R})$ is a function, called the hamiltonian function. >[!theorem] > >We have $\{f, H\} = 0$ if and only if f is constant along integral curves of $X_{H}$ . Such $f$ is called integral of motion. >[!definition] Integrable Hamiltonian System > >A hamiltonian system $(M, \omega, H)$ is (completely) integrable if it possesses $n = \frac{1}{2}$ dim $M$ independent integrals of motion, $f_{1} = H, f_{2}, . . . , f_{n}$, which are pairwise in involution with respect to the Poisson bracket, i.e., $\{f_{i} , f_{j}\} = 0, \forall i, j$ >[!lemma] > >Let $(M, \omega, H)$ be an integrable system of dimension $2n$ with integrals of motion $f_1 = H, f_2, \dots, f_n$. Let $c \in \mathbb{R}^n$ be a regular value of $f := (f_1, \dots, f_n)$. The corresponding level set, $f^{-1}(c)$, is a lagrangian submanifold, because it is $n$-dimensional and its tangent bundle is isotropic. >If the hamiltonian vector fields $X_{f_1}, \dots, X_{f_n}$ are complete on the level $f^{-1}(c)$, then the connected components of $f^{-1}(c)$ are homogeneous spaces for $\mathbb{R}^n$, i.e., are of the form $\mathbb{R}^{n-k} \times \mathbb{T}^k$ for some $k$, $0 \le k \le n$, where $\mathbb{T}^k$ is a $k$-dimensional torus. >[!definition] Liouville tori > >Let $(M, \omega, H)$ be an integrable system of dimension $2n$ with integrals of motion $f_1 = H, f_2, \dots, f_n$. Let $c \in \mathbb{R}^n$ be a regular value of $f := (f_1, \dots, f_n)$. >Compact component of $f^{-1}(c)$ would be a torus, called Liouville tori. >[!theorem] Arnold-Liouville > >Let $(M, \omega, H)$ be an integrable system of dimension $2n$ with integrals of motion $f_1 = H, f_2, \dots, f_n$. Let $c \in \mathbb{R}^n$ be a regular value of $f := (f_1, \dots, f_n)$. The corresponding level $f^{-1}(c)$ is a lagrangian submanifold of $M$. >(a) If the flows of $X_{f_1}, \dots, X_{f_n}$ starting at a point $p \in f^{-1}(c)$ are complete, then the connected component of $f^{-1}(c)$ containing $p$ is a homogeneous space for $\mathbb{R}^n$. With respect to this affine structure, that component has coordinates $\varphi_1, \dots, \varphi_n$, known as angle coordinates, in which the flows of the vector fields $X_{f_1}, \dots, X_{f_n}$ are linear. >(b) There are coordinates $\psi_1, \dots, \psi_n$, known as action coordinates, complementary to the angle coordinates such that the $\psi_i$'s are integrals of motion and $\varphi_1, \dots, \varphi_n, \psi_1, \dots, \psi_n$ form a Darboux chart.