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Let $M$ be the configuration space of the planar double pendulum. See figure 1.
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The configuration space of the planar double pendulum is the set of pairs of angles $(x, y) \mod 2 \pi$. What is the standard name for this surface?
The torus, $\Torus{2}$.
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Determine the kinetic energy of the system depicted. The bobs of masses $m$ and $M$ move on the inflexible rods of lengths $l$ and $L$ and the suspension point $P$ is fixed.
Let $z = l \exp(ix)$ and $w = L \exp(iy)$ so the position of the body of mass $m$ is $z$ and that of mass $M$ is $Z=z+w$. The velocities are $\dot{z}=z i \dot{x}$ and $\dot{Z} = z i \dot{x} + w i \dot{y}$. We get that $\norm{\dot{z}}=l\norm{\dot{x}}$ and $\norm{\dot{Z}}^2 = l^2 \norm{\dot{x}}^2 + 2 l L \cos(x-y) \dot{x}\dot{y} + L^2 \norm{\dot{y}}^2$. The kinetic energy is $\frac{1}{2} m \dot{z}^2 + \frac{1}{2} M \norm{\dot{Z}}^2.$
Planar double pendulum.
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The configuration space of the planar double pendulum is the set of pairs of angles $(x, y) \mod 2 \pi$. What is the standard name for this surface?
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Let $M$ be an $n$-manifold and let $g$ be a Riemannian metric on $M$.
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Suppose that $x_i$ and $y_j$ are smooth coordinate systems on $M$, and $x = \phi(y)$ for a smooth diffeomorphism $\phi$. Show that
\[ \D{x_1} \wedge \cdots \wedge \D{x_n} = \det \Phi \cdot \D{y_1} \wedge \cdots \wedge \D{y_n}, \] where
\[ \Phi = \begin{bmatrix} \displaystyle\didi{\phi_i}{y_j} \end{bmatrix} \] is the Jacobian matrix of the coordinate transformation.
We have that $\D{x_i} = \sum_{j=1}^n \didi{\phi_i}{y_j} \D{y_j}.$ Then $\D{x_1} \wedge \cdots \wedge \D{x_n} = \sum_{j_1,\ldots, j_n} \left( \Phi^1_{j_1} \D{y_{j_1}} \right) \wedge \cdots \wedge \left( \Phi^1_{j_n} \D{y_{j_n}} \right) = \det(\Phi) \D{y_1} \wedge \cdots \wedge \D{y_n}.$
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In a local system of coordinates $x_i$ on $M$, one can define an $n$-form
\[ \Omega = \sqrt{\det \mathbf{g}} \D{x_1} \wedge \cdots \wedge \D{x_n}, \] where
\[ \mathbf{g} = [ g_{ij} ] \]
is the matrix of $g = \sum_{i, j=1}^n g_{ij} \D{x_i} \cdot \D{x_j}$ in the local coordinates. Use the first part to show that if $M$ is orientable, then $\Omega$ is a tensor field on $M$. What goes wrong if $M$ is not orientable?
Let $\Omega'$ (resp. $\mathbf{g}' = [ g'_{ij} ]$) denote $\Omega$ (resp. $\mathbf{g} = [ g_{ij} ]$) in the $y$ coordinates. We want to show that $\phi^* \Omega' = \Omega$, i.e. that the formula for $\Omega$ holds in all coordinate systems. Indeed, we compute that $g = \sum_{ij} g_{ij} \D{x_i} \symmetricproduct \D{x_j} = \sum_{i, j, k, l} g_{ij} \Phi^i_k \Phi^j_l \D{y_k} \symmetricproduct \D{y_l} = \sum_{k, l} g'_{kl} \D{y_k} \symmetricproduct \D{y_l}$. This shows that $\mathbf{g}' = \Phi' \mathbf{g} \Phi$. Then $\Omega = \sqrt{\det\mathbf{g}} \det(\Phi) \D{y_1} \wedge \cdots \wedge \D{y_n} = \sqrt{\det \mathbf{g}'} \D{y_1} \wedge \cdots \wedge \D{y_n} = \Omega'$. We have used the positivity of $\det \Phi$ for the last step.
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The $n$-form defined above is non-degenerate. Prove this.
Since $g$ is a metric, $\mathbf{g}$ is positive definite so its determinant is always positive.
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The $n$-form $\Omega$ is called a Riemannian volume form. Compute the Riemannian $2$-form for the sphere in $\R^3$ of radius $r$. [Hint: it is easiest to use spherical coordinates on $\R^3$.]
Let $(\theta, \phi)$ be spherical coordinates, so that $(x, y, z) = f(\theta, \phi) = r(\cos \theta \sin \phi, \sin \theta \sin \phi, \cos \phi)$ for $0 \lt \phi \lt \pi$ and $0 \lt \theta \lt 2 \pi$. The vector fields $X = \D{f} \cdot \didi{}{\theta}$ and $Y = \D{f} \cdot \didi{}{\phi}$ determine the Riemannian metric by $g = \ip{X}{X} \D{\theta}^2 + 2 \ip{X}{Y} \D{\theta} \symmetricproduct \D{\phi} + \ip{Y}{Y} \D{\phi}^2$. We compute that $X = r(-\sin \theta \sin \phi, \cos \theta \sin \phi, 0)$ and $Y = r(\cos \theta \cos \phi, \sin \theta \cos \phi, -\sin \phi)$. This gives \begin{align} \mathbf{g} &= \begin{bmatrix} r^2 \sin^2 \phi & 0 \\ 0 & r^2 \end{bmatrix} && \Omega = r^2 \sin \phi \D{\theta} \wedge \D{\phi}. \end{align}
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Suppose that $x_i$ and $y_j$ are smooth coordinate systems on $M$, and $x = \phi(y)$ for a smooth diffeomorphism $\phi$. Show that
\[ \D{x_1} \wedge \cdots \wedge \D{x_n} = \det \Phi \cdot \D{y_1} \wedge \cdots \wedge \D{y_n}, \] where
\[ \Phi = \begin{bmatrix} \displaystyle\didi{\phi_i}{y_j} \end{bmatrix} \] is the Jacobian matrix of the coordinate transformation.
