I am working on an exercise for my differential geometry class and would like to know if my solution to a problem is correct. The problem is as follows:

A symplectic manifold $S$ is - by definition - endowed with a closed non-degenerate 2-form $\omega$. The map

$$T(S) \longrightarrow T^{\ast}(S) : \, v \longmapsto -i_v \, \omega$$

is an isomorphism (by non-degeneracy of $\omega$), so it has an inverse $$I: T^{\ast}(S) \longrightarrow T(S) \quad .$$

This can be used to assign to any function $H \in C^{\infty}(S)$ ("Hamiltonian") the vector field $I dH$ ("Hamiltonian vector field"). Let $S$ be described locally by $2n$ coordinates

$$\lbrace x^j \rbrace_{j=1,...,2n} \enspace = \enspace \lbrace p_i, q^i \rbrace_{i=1,...,n} \qquad \text{with} \quad \omega \enspace = \enspace \sum_{i=1}^n dp_i \land dq^i \quad .$$

Show that the flow equation determined by the Hamiltonian vector field corresponds to Hamilton's equations, i.e.

$$\frac{dx^j}{dt} \, \frac{\partial}{\partial x^j} \enspace = \enspace I \, dH \qquad \Longleftrightarrow \qquad \frac{dp_i}{dt} \enspace = \enspace - \frac{\partial H}{\partial q^i} \quad , \qquad \frac{dq^i}{dt} \enspace = \enspace \frac{\partial H}{\partial p_i} \quad .$$

My solution is provided as an answer. Please let me know if it is correct and if not, where my errors are.


The equation

$$\frac{dx^j}{dt} \, \frac{\partial}{\partial x^j} \enspace = \enspace I \, dH$$

can be rewritten as

\begin{align*} \frac{dx^j}{dt} \, \frac{\partial}{\partial x^j}\enspace &= \enspace I \, dH \\ &= \enspace I \Big( \frac{\partial H}{\partial x^j} \, dx^j \Big) \\ &= \enspace I \Big( \frac{\partial H}{\partial q^i} \, dq^i + \frac{\partial H}{\partial p_i} \, dp_i \Big) \quad . \end{align*}

Now consider the LHS of this equation. We find

\begin{align*} \text{LHS} \enspace &= \enspace \frac{dx^j}{dt} \, \frac{\partial}{\partial x^j} \\ &= \enspace \frac{dq^i}{dt} \, \frac{\partial}{\partial q^i} + \frac{dp_i}{dt} \, \frac{\partial}{\partial p_i} \\ &= \enspace \dot{q}^i \, \frac{\partial}{\partial q^i} + \dot{p}_i \, \frac{\partial}{\partial p_i} \end{align*}

Multiplying both the LHS and the RHS with $dq^k$ yields for the LHS:

\begin{align*} dq^k \cdot \text{LHS} \enspace &= \enspace \dot{q}^i \, dq^k \cdot \frac{\partial}{\partial q^i} + \dot{p}_i \, dq^k \cdot \frac{\partial}{\partial p_i} \\ &= \enspace \dot{q}^i \, \delta^k_i + \dot{p}_i \cdot 0 \\ &= \enspace \dot{q}^k \end{align*}

Since $I$ is an isomoprhism, the RHS becomes under this multiplication:

\begin{align*} dq^k \cdot \text{RHS} \enspace &= \enspace dq^k \cdot I \Big( \frac{\partial H}{\partial q^i} \, dq^i + \frac{\partial H}{\partial p_i} \, dp_i \Big) \\ &= \enspace dq^k \cdot I \Big( dq^i \Big) \, \frac{\partial H}{\partial q^i} + dq^k \cdot I \Big( dp_i \Big) \, \frac{\partial H}{\partial p_i} \end{align*}

In order to get (one of) the desired Hamilton equations, we need to show that $dq^k \cdot I \Big( dq^i \Big)$ equals $0$ and that $dq^k \cdot I \Big( dp_i \Big)$ is equal to $\delta^k_i$. This is done in the following way: Let $\partial_i \equiv \frac{\partial}{\partial q^i}$. Then

\begin{align*} I^{-1} \big( \partial_i \big) \enspace &= \enspace - i_{\partial_i} \omega \\ &= \enspace - i_{\partial_i} \sum_{\ell = 1}^n dp_{\ell} \land dq^{\ell} \\ &= \enspace - \sum_{\ell = 1}^n \big( i_{\partial_i} dp_{\ell} \big) \land dq^{\ell} - dp_{\ell} \land \big( i_{\partial_i} dq^{\ell} \big) \\ &= \enspace - \sum_{\ell = 1}^n 0 \land dq^{\ell} - dp_{\ell} \land \delta^{\ell}_i \\ &= \enspace dp_{i} \land 1 \\ &= \enspace dp_{i} \end{align*}

We therefore know that $I^{-1} \Big( \frac{\partial}{\partial q^i} \Big) = dp_i$, meaning that $I(dp_i) = \frac{\partial}{\partial q^i}$. Analogously we get $I(dq^i) = - \frac{\partial}{\partial p_i}$. Inserting this in the above equation for the RHS we get:

\begin{align*} dq^k \cdot \text{RHS} \enspace &= \enspace dq^k \cdot I \Big( \frac{\partial H}{\partial q^i} \, dq^i + \frac{\partial H}{\partial p_i} \, dp_i \Big) \\ &= \enspace - dq^k \cdot \frac{\partial}{\partial p_i} \, \frac{\partial H}{\partial q^i} + dq^k \cdot \frac{\partial}{\partial q^i} \, \frac{\partial H}{\partial p_i} \\ &= \enspace \delta^k_i \, \frac{\partial H}{\partial p_i} \\ &= \enspace \frac{\partial H}{\partial p_k} \quad . \end{align*}

Putting all together we find

\begin{align*} dq^k \cdot \text{LHS} \enspace = \enspace \underbrace{\dot{q}^k \enspace = \enspace \frac{\partial H}{\partial p_k}}_{\text{Hamilton equation of motion}} \enspace = \enspace dq^k \cdot \text{RHS} \end{align*}

In a similar approach we get the other Hamilton equation of motion.


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