MTH-696A: Topics in Geometric Mechanics Assignment 1 (by Dr. Leo Butler)

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Let $\crossproduct : \R^3 \times \R^3 \to \R^3$ be the vector (=cross) product. Define the following operator $\omega$ \begin{equation}\label{eq:omega} \omega_x ( u, v ) := \ip{x}{u \crossproduct v}, \end{equation} for $x \in \R^3$ and $u, v \in T_x \R^3 \equiv \R^3$. Let $e_1, e_2, e_3$ be the standard basis of $\R^3$ and let $x_i$ be coordinates induced by this basis (so $x=x_1 e_1 + x_2 e_2 + x_3 e_3$).
1. Compute $\omega = \sum_{i \lt j} \omega_{ij}(x) \D{x_i} \wedge \D{x_j}$.
  We compute that \begin{align*} \omega_x(e_1, e_2) &= \ip{x}{e_3} = x_3 && \omega_x(e_2, e_3) = \ip{x}{e_1} = x_1 && \omega_x(e_3, e_1) = \ip{x}{e_2} = x_2 \end{align*} \begin{align*} \omega_x &= x_3 \D{x_1} \wedge \D{x_2} + x_1 \D{x_2} \wedge \D{x_3} + x_2 \D{x_3} \wedge \D{x_1} && \text{since $\D{x_i}$ is dual to $e_i$}. \\ \intertext{Thus,} \D{\omega}_x &= \D{x_3} \wedge \D{x_1} \wedge \D{x_2} + \D{x_1} \wedge \D{x_2} \wedge \D{x_3} + \D{x_2} \wedge \D{x_3} \wedge \D{x_1} &&= 3 \D{x_1} \wedge \D{x_2} \wedge \D{x_3}. \end{align*} Alternatively: Let $u, v, w \in T_x \R^3$. Because $\D{\omega}$ is a tensor, it suffices to extend each tangent vector as a ``constant'' vector field on $\R^3$. Then, \begin{align*} \D{\omega}_x(u, v, w) &= \lieder{u}{\omega_x(v, w)} + \lieder{v}{\omega_x(w, u)} + \lieder{w}{\omega_x(u, v)} \\ &\phantom{=} - \omega_x(u, \liebracket{v}{w}) - \omega_x(v, \liebracket{w}{u}) - \omega_x(w, \liebracket{u}{v}) \\ &= \omega_u(v, w) + \omega_v(w, u) + \omega_w(u, v) && \textrm{since $\omega$ is linear in $x$} \\ &= 3 \det \begin{bmatrix} u & v & w \end{bmatrix} = 3 \Omega_x(u, v, w), \end{align*} where $\Omega$ is the ``standard'' volume form on $\R^3$.
2. Compute the exterior derivative of $\omega$, $\D{\omega}$, and show that this equals $\D{x_1} \wedge \D{x_2} \wedge \D{x_3}$, the standard volume element on $\R^3$.
  See above.
Let $\sphere{2} = \set{x \in \R^3 \st \norm{x}=1}$ be the unit sphere. Let $\eta = \omega |_{\sphere{2}}$ be the restriction of the $2$-form $\omega$ defined in (\ref{eq:omega}). Show that $\eta$ is a closed, non-degenerate $2$-form.
Since $\sphere{2}$ is $2$-dimensional, any $3$-form on it is $0$. This shows that $\D{\eta} = \D{\omega}|\sphere{2}$ is zero. To prove non-degeneracy, let $x \in \sphere{2}$ and $v \in T_x\sphere{2}$. Then $w = x \crossproduct v \in T_x\sphere{2}$ and is orthogonal to both $x, v$. Therefore, if $v \neq 0$, then $A=[x\ v\ w]$ has columns that form a basis of $\R^3$ so $\eta_x(v, w) = \det A \neq 0$.