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--- dirichlet_energy [2022/08/31 14:39] – [Euler-Lagrange equations] spencer
+++ dirichlet_energy [2022/09/05 05:44] (current) – [Euler-Lagrange equations] spencer
@@ Line 13: / Line 13: @@
 ===== Euler-Lagrange equations =====
+These computations are done in Eells--Lemaire (2.4).
 Consider a family of maps $u_t \in C^2(M,N)$ such that $u_0 = u$.
 Let $U : \mathbb{R} \times M \to N$ be the map defined by $U(t,x) = u_t(x)$.
+Let $\nabla_{\mathrm{d}/\mathrm{d}t}$ be the covariant derivative, and then compute on a vector field $X$ on $M$:
+\begin{align*}
+(\nabla_{\mathrm{d}/\mathrm{d} t} \mathrm{d} u_t)X &= \nabla_{\mathrm{d}/\mathrm{d}t}(\mathrm{d}u_t \cdot X) - \mathrm{d}u_t \cdot \nabla_{\mathrm{d}/\mathrm{d} t} X \\
+&= \nabla_{\mathrm{d}/\mathrm{d}t}(\mathrm{d}U \cdot X) - 0\\
+&= \nabla_X (\mathrm{d}U \cdot \mathrm{d}/\mathrm{d} t) + \mathrm{d}U\left[ \frac{\mathrm{d}}{\mathrm{d}t}, X\right] \\
+&= \nabla_X \frac{\partial U}{\partial t} + 0.
+\end{align*}
 The family of maps generates a section $\frac{\partial U}{\partial t}$ of the bundle $u^*TN$ over $M$ by \[ \left(\frac{\partial U}{\partial t}\right)_p = \left. \frac{\mathrm{d}}{\mathrm{d}t} u_t(p)\right\rvert_{t=0} \in T_{u(p)} N. \]
-Write $\partial_t U$ for this section.
-The bundle $u^*TN$ has a natural connection $\nabla$, which is the pullback by $u$ of the Levi-Civita connection on $TN$. Thus $\nabla (\partial_t U) \in \Gamma(T^* M \otimes u^* TN)$.
-Since on ($u^*TN$-valued) $0$-forms the connection agrees with the exterior covariant derivative it induces, one may also write $\mathrm{d} (\partial_t U)$ for this section of $T^* M \otimes u^* TN$, or $\mathrm{d}_{u^* TN} (\partial_t U)$ to emphasize the bundle over with respect to which the exterior derivative is being taken.
-For fixed $p \in M$, pullback by the constant map $\psi_p : \mathbb{R} \to M$ defined by $\psi_p(t) = p$ allows one to pull back the bundle $T^* M  \otimes u^* TN$ over $M$ to one on $\mathbb{R}$; then the covariant derivative $\nabla_{\frac{\mathrm{d}}{\mathrm{d}t}} \psi_p^* \mathrm{d}u_t$ evaluated at $t=0$ is an element of $\psi_p^*(T^*M \otimes u^*TN)(0) = T_p^*M \otimes T_{u(p)} N$.
-On the other hand, $\psi_p^* \mathrm{d} u_t = \mathrm{d} u_t(p)$. **TODO**
 The Euler-Lagrange equations for the Dirichlet energy functional are derived from
@@ Line 32: / Line 34: @@
 \end{align*}
 and so it is necessary and sufficient for $u$ to be a critical point of the functional $E$ that $\mathrm{d}^*_{u^* TN} (\mathrm{d} u) = 0$, or more compactly that $\mathrm{d}^* (\mathrm{d}u) = 0$.
+A map that is a critical point of the functional $E$ is called //harmonic//.
+Write $\tau(u) := -\mathrm{d}^* (\mathrm{d}u) = \mathrm{tr} \nabla \mathrm{d}u$ for the //torsion field of $u$//.
+See the page [[weitzenboeck_identity|Weitzenböck identity]] for why these two expressions are equal.
+===== Properties of the Dirichlet energy =====
+==== Invariance under pullbacks ====
+Let $\Phi : M \to M$ be a $C^2$-diffeomorphism, and consider the Dirichlet energy of the pullback map $\Phi^* u$ with respect to the pullback metric $\Phi^* g$.
+From the change of variables formula for integrals,
+\begin{align*}
+E(u,g) &= \int_M (g \otimes u^* h)(\mathrm{d}u, \mathrm{d}u) \mathrm{vol}(g)\\
+ &= \int_M (\Phi^* g \otimes \Phi^* u^* h)(\Phi^* \mathrm{d}u, \Phi^* \mathrm{d}u) \mathrm{vol}(\Phi^* g) \\
+&= E(\Phi^* u, \Phi^* g)
+\end{align*}
+==== Invariance under conformal change in dimension 2 ====
+When the source manifold $M$ is a surface (that is, $n = 2$), then the Dirichlet energy is invariant under conformal change.
+Suppose $\widetilde{g} = e^{2\phi} g$ is conformally equivalent to the metric $g$ on $M$.
+We compute the adjoint $\mathrm{d}^*$ of the exterior covariant derivative in the $L^2$ norm induced by the pair of metrics $(g,h)$ as well as the adjoint $\widetilde{\mathrm{d}}^*$ induced by the pair of metrics $(\widetilde{g}, h)$.
+Let $\sigma,\tau \in \Gamma(T^*M \otimes u^*TN)$.
+Using [[conformal_change_formulae]],
+\begin{align*}
+(\sigma, \tau)_{L^2(g)} &= \int_M (g \otimes u^*h)(\sigma, \tau) \mathrm{vol}(g) \\
+&= \int_M e^{-2\phi} (g \otimes u^*h)(\sigma, \tau) e^{2\phi} \mathrm{vol}(g) \\
+&= \int_M (\widetilde{g} \otimes u^*h)(\sigma, \tau) \mathrm{vol}(\widetilde{g}) \\
+&= (\sigma, \tau)_{L^2(\widetilde{g})}.
+\end{align*}
+So the $L^2(g)$ and $L^2(\widetilde{g})$ norms are identical, and thus
+\[ E(u,g) = \frac{1}{2} |\mathrm{d}u|^2_{L^2(g)} = \frac{1}{2} |\mathrm{d}u|^2_{L^2(\widetilde{g})} = E(u,\widetilde{g}). \]
+Moreover, the proof reveals what breaks when $n \neq 2$: one is left to integrate a factor of $e^{(n-2)\phi}$ in order to change the norm from $L^2(g)$ to $L^2(\widetilde{g})$.
+To pull this factor out of the integral for all $\sigma, \tau$ requires that $e^{(n-2)\phi}$ is constant; that is, that either $n = 2$ or $\phi$ is a constant.