TITLE: Dirichlet energy in local coordinates
QUESTION [6 upvotes]: Suppose $f:M\to N$ is a map between Riemannian manifolds. In the literature on Harmonic maps, we often see the Dirichlet energy 
$$E[f]=\int |df|^2 dV.$$
Everywhere I've been reading, in papers and textbooks alike, the text has jumped straight to the equation for $|df|^2$ in local coordinates 
$$g^{\alpha\beta}\frac{\partial f^i}{\partial x^\beta}\frac{\partial f^i}{\partial x^\alpha}$$
where $g_{\alpha\beta}$ is the metric tensor of $M$. 
I know a bit about Riemannian geometry but haven't done any explicit computations with local coordinates before, so I'm struggling to get a derivation for this on my own. To see a derivation would be very instructive to me in how to use local coordinates in future. The sources I've been looking at are 

Variational Methods (Struwe) Chapter 6
A Regularity Theory for Harmonic Maps (Schoen, Uhlenbeck)
Harmonic Maps to Spheres (Solomon)

Summary: Could someone please give a detailed explanation of how to express $|df|^2$ in local coordinates?

REPLY [4 votes]: To add to Jesse's brilliant answer: if $V$ and $W$ are equipped with inner products, one can induce an inner product in ${\rm Hom}(V,W)$ by $\langle T,S\rangle  = {\rm tr}(TS^*)$, where $S^*$ is the adjoint of $S$. This construction goes up to the level of bundles equipped with Riemannian fiber metrics, so it applies for the bundle morphism ${\rm d}f\colon TM\to TN$, and $$|{\rm d}f|^2 = \langle {\rm d}f,{\rm d}f\rangle = {\rm tr}({\rm d}f\,({\rm d}f)^*)$$makes sense. I'll use Jesse's notation for the metrics $\gamma$ and $g$, as well as coordinates $(x^\alpha)$ and $(y^i)$ for $M$ and $N$, respectively. Write ${\rm d}f(\partial_\beta) = (\partial_\beta f^i)\partial_i$ and $({\rm d}f)^*(\partial_j) = A_j^\alpha \partial_\alpha$, for some coefficients $A^\alpha_j$ to be found. The relation $$\gamma(\partial_\beta, ({\rm d}f)^*(\partial_j)) = g({\rm d}f(\partial_\beta),\partial_j)$$reads $$\gamma_{\beta\alpha}A^\alpha_j = (\partial_\beta f^i)g_{ij} \implies A^\alpha_j = \gamma^{\alpha\beta} (\partial_\beta f^i)g_{ij},$$so that we conclude again that $$\langle {\rm d}f,{\rm d}f\rangle = (\partial_\alpha f^i)A^\alpha_i = (\partial_\alpha f^i)\gamma^{\alpha\beta}(\partial_\beta f^j)g_{ji},   $$as wanted.