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Let $\Theta$ be a closed Riemannian manifold with Riemannian metric $\langle \cdot, \cdot \rangle_{x} \colon T_x \Omega \times T_x \Omega \to \mathbb R_{\ge 0}$ and $\Omega := \Theta \times \mathbb R_{\ge 0}$. Let $\xi \colon \Theta \to \mathbb R$ be differentiable and $h \colon \mathbb R_{\ge 0} \to \mathbb R_{\ge 0}$, $r \mapsto r^2$. What is the gradient of $\Phi \colon \Omega \to \mathbb R$, $(\theta, r) \mapsto \xi(\theta) h(r)$? The metric on $\Omega$ is $$ \label{eq:omega} \tag{$\star$} \langle (r_1, \partial \theta_1), (r_2, \partial \theta_2) \rangle_{(r, \theta)} := \frac{1}{\alpha} r_1 r_2 + \frac{r^2}{\beta} \langle \partial \theta_1, \partial \theta_2 \rangle_{\theta} $$ for $x = (r, \theta) \in \Omega$ and $(r_1, \partial \theta_1), (r_2, \partial \theta_2) \in T_x \Omega \cong \mathbb R \times T_{\theta} \Theta$.

My work. Let us first identify the derivative and then identify the gradient via $d \Phi(X) = \langle \text{grad}(\Phi), X \rangle$, where $X$ is a vector field.

The derivative at each point $x \in \Omega$ is the linear function $d \Phi_x \colon T_x \Omega \to T_{\Phi(x)} \mathbb R \cong \mathbb R$ defined by $d \Phi_x(v)(f) := v(f \circ \Phi)$ for $p \in T_x \Omega$ and $f \in C^{\infty}(\mathbb R)$, where $T_x \Omega$ is the tangent space, the space of linear functions on $C^{\infty}(M)$ that satisfy the Leibniz rule at $x$ (in the notation of John M. Lee's Introduction to Smooth Manifolds). The chain rule is $ d(G \circ F)_p = d G_{F(p)} \circ d F_p. $

We can write $\Phi(x) = (\xi \circ \pi_{\Theta}) \cdot (h \circ \pi_{\mathbb R_{\ge 0}})$ for $x \in \Omega$, where $\pi$ are the projections onto one of the factors of $\Omega$, as a product of functions defined on $\Omega$.

Using the chain and product rule, we now should have for $x \in \Omega$ $$ d \Phi_x = d \xi_{\pi_{\Theta}(x)} \circ d (\pi_{\Theta})_{x} \cdot (h \circ \pi_{\mathbb R_{\ge 0}})(x) + (\xi \circ \pi_{\Theta})(x) \cdot d h_{\pi_{\mathbb R_{\ge 0}}(x)} \circ d (\pi_{\mathbb R_{\ge 0}})_{x}. $$ We have $d (\pi_{\Theta})_x \colon T_{x} \Omega \to T_{\pi(x)} \Theta = T_{\theta} \Theta$ for $x = (\theta, r) \in \Omega$ and $T_{x} \Omega = T_{\theta} \Theta \times T_{r} \mathbb R_{\ge 0} \cong T_{\theta} \Theta \times \mathbb R$. It should be the case (but I can't prove it) that hence $d (\pi_{\Theta})_x = \pi_{T_{\pi_{\Theta}(x)} \Theta}$, so that $$ d \Phi_x = d \xi_{\theta} \circ \pi_{T_{\theta} \Theta} \cdot r^2 + \xi(\theta) \cdot d h_{r} \circ \pi_{\mathbb R}(x) = d \xi_{\theta} \circ \pi_{T_{\theta} \Theta} \cdot r^2 + \xi(\theta) \cdot 2 r, $$ that is, $d \Phi_x$ acts on $v = (v_{\theta}, v_r) \in T_{x} \Omega$ as $$ v \mapsto r^2 \cdot d_{\theta} \xi \circ v_{\theta} + \xi(\theta) \cdot 2 v_{r}. $$ Hence we can write $$ d \Phi_x = \begin{pmatrix} r^2 \cdot d \xi_{\theta} \\ 2 r \xi(\theta) \end{pmatrix} $$ or is that not correct? I would rather expect something like $\nabla \Phi(\theta, r) = (\nabla \xi(\theta), 2 r)^{T}$.

ViktorStein
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1 Answers1

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Let us first compute the differential of $\psi \colon \Omega \to \mathbb R$, $\text{d} \Phi \colon \Omega \to T^* \Omega$, given by $x \mapsto \text{d}\Phi_x$, where $\text{d}\Phi_x(v) := v \Phi$ for $v \in T_x \Omega$.

Due to the chain rule $\text{d}(G \circ F)_p = \text{d}G_{F(p)} \circ \text{d} F_x$ and the product rule $\text{d}(F \cdot G)_p = F(p) \cdot \text{d}G_p + \text{d}F_p \cdot G(p)$ we have for $x = (\theta, r) \in \Omega$ $$ \text{d}\psi_x = (\xi \circ \pi_{\Theta})(x) \cdot \text{d} h_r \circ \text{d}[\pi_{\mathbb R_{\ge 0}}]_x + h(r) \cdot \text{d}\xi_{\theta} \circ \text{d}[\pi_{\Theta}]_x. $$ Now we use the fact that the differential of the projection onto $\Theta$ is the projection onto $T \Theta$ (and analogously for the second factor): $d[\pi_{\Theta}]_{x} = \pi_{T_{\theta} \Theta}$. Furthermore, since $h$ is a map between vector spaces, the differential coincides with the Fréchet differential: $d h_r$ is the linear map $t \mapsto h'(r) \cdot t = 2 r t$. Hence $$ \text{d}\psi_x = \xi(\theta) \cdot 2 r \cdot \pi_{T_r \mathbb R_{\ge 0}} + r^2 \cdot \text{d}\xi_{\theta} \circ \pi_{T_{\theta} \Theta}, $$ so that for $w := (w_{\theta}, w_r) \in T_{\theta} \Theta \times T_r \mathbb R_{\ge 0} \cong T_{x} \Omega$ we have $$ \text{d} \psi_x(w) = \xi(\theta) \cdot 2 r w_r + r^2 \cdot \text{d}\xi_{\theta}(w_{\theta}) = \xi(\theta) \cdot 2 r w_r + r^2 \cdot \langle \nabla \xi(\theta), w_{\theta} \rangle_{\theta}, $$ where in the last step we used the defining relation for the gradient: $\langle \nabla \xi(\theta), w_{\theta} \rangle_{\theta} = d \xi_{\theta}(w_\theta)$.

We can now identify the gradient: for $w := (w_{\theta}, w_r) \in T_{\theta} \Theta \times T_r \mathbb R_{\ge 0} \cong T_{x} \Omega$ we have \begin{align*} \text{d} \psi_x(w) & \overset{!}{=} \langle \nabla \psi(x), w \rangle_{x} = \left\langle \begin{pmatrix} \nabla_{\theta} \psi(x) \\ \nabla_r \psi(x) \end{pmatrix}, \begin{pmatrix} w_{\theta} \\ w_r \end{pmatrix} \right\rangle_{x} \\ & \overset{(\star)}{=} \frac{1}{\alpha} \nabla_r \psi(x) w_r + \frac{r^2}{\beta} \langle \nabla_{\theta} \psi(x), w_{\theta} \rangle_{\theta}, \end{align*} so that $$ \nabla \psi(x) = \begin{pmatrix} \nabla_{\theta} \psi(x) \\ \nabla_r \psi(x) \end{pmatrix} = \begin{pmatrix} \beta \nabla \xi(\theta) \\ 2 \alpha r \xi(\theta) \end{pmatrix}. $$

ViktorStein
  • 5,024