Let $(M^n, g), (N, h)$ be closed Riemannian manifolds. The Dirichlet energy density of a map $u \in C^2(M,N)$ is defined to be \[ e(u) := (g \otimes u^*h)(\mathrm{d}u, \mathrm{d}u). \] In this expression, $\mathrm{d}u \in \Gamma(T^*M \otimes u^* TN)$ is the pushforward of $u$; it is a section of a vector bundle over $M$, where the value at the point $p$ is the pushforward $T_p M \to T_{u(p)} N$, under the canonical identification of maps $T_p M \to T_{u(p)} N$ with elements of $(T_p M)^* \otimes T_{u(p)} N = (T_p M)^* \otimes (u^* TN)_p$.

The Dirichlet energy of $u$ is the total Dirichlet energy density: \[ E(u) := \frac{1}{2} \int_M e(u) \mathrm{vol}_M. \] In other words, \[ E(u) = \frac{1}{2} |\mathrm{d}u|^2_{L^2} \] is half the squared length of $\mathrm{d}u$ in the $L^2$-norm on sections of $T^* M \otimes u^* TN$.

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 \[ \left(\frac{\partial U}{\partial t}\right)_p = \left. \frac{\mathrm{d}}{\mathrm{d}t} u_t(p)\right\rvert_{t=0} \in T_{u(p)} N. \]

The Euler-Lagrange equations for the Dirichlet energy functional are derived from \begin{align*} \left. \frac{\mathrm{d}}{\mathrm{d}t} E(u_t)\right\rvert_{t=0} &= \left(\nabla_{\frac{\mathrm{d}}{\mathrm{d}t}} \mathrm{d}u_t, \mathrm{d}u_0 \right)_{L^2} \\ &= \left(\mathrm{d}_{u^* TN} \frac{\partial U}{\partial t}, \mathrm{d}u \right)_{L^2} \\ &= \left( \frac{\partial U}{\partial t}, \mathrm{d}^*_{u^* TN} (\mathrm{d} u) \right)_{L^2}, \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 Weitzenböck identity for why these two expressions are equal.

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*}

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.