# Dirichlet energy in local coordinates

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?

Let $$f \colon (M,\gamma) \to (N,g)$$ be a map between Riemannian manifolds. Choose local coordinates $$(x^\alpha)$$ on $$N$$ and $$(y^i)$$ on $$N$$. We shall regard $$df$$ as a section of the vector bundle $$T^*M \otimes f^*(TN) \to M$$, so that in our chosen coordinates: $$df = \frac{\partial f^i}{\partial x^\alpha}\,dx^\alpha \otimes \frac{\partial}{\partial y^i}$$ There are several metrics in play here:
• Since $$\gamma$$ is a metric on $$M$$, each $$\gamma|_x$$ is an inner product on $$T_xM$$, which induces an inner product on $$T_x^*M$$. This latter inner product has components $$(\gamma^{\alpha \beta}).$$ Note that $$(\gamma^{\alpha \beta})$$ is the inverse matrix of $$(\gamma_{\alpha \beta})$$.
• Similarly, since $$g$$ is a metric on $$N$$, each $$g|_{f(x)}$$ is an inner product on $$T_{f(x)}N$$, which induces (via the map $$f \colon M \to N$$) an inner product on $$(f^*TN)_x$$. This latter inner product has components $$(g_{ij} \circ f).$$
• Using these two metrics, we get a metric --- which I'll denote $$\langle \cdot, \cdot \rangle$$ --- on $$T^*M \otimes f^*(TN)$$.
Putting this all together, we can now compute: \begin{align*} |df|^2 = \left\langle df, df \right\rangle & = \left\langle \frac{\partial f^i}{\partial x^\alpha}\,dx^\alpha \otimes \frac{\partial}{\partial y^i}, \, \frac{\partial f^j}{\partial x^\beta}\,dx^\beta \otimes \frac{\partial}{\partial y^j} \right\rangle \\ & = \frac{\partial f^i}{\partial x^\alpha}\, \frac{\partial f^j}{\partial x^\beta}\,\left\langle dx^\alpha \otimes \frac{\partial}{\partial y^i}, \,dx^\beta \otimes \frac{\partial}{\partial y^j} \right\rangle \\ & = \frac{\partial f^i}{\partial x^\alpha}\, \frac{\partial f^j}{\partial x^\beta}\,\gamma^{\alpha \beta}\, (g_{ij} \circ f). \end{align*} If the metric $$g$$ on $$N$$ is flat, then we can choose our coordinates $$(y^i)$$ so that $$g_{ij} = \delta_{ij}$$. In that case, our formula reduces to the one you wrote: $$|df|^2 = \frac{\partial f^i}{\partial x^\alpha}\, \frac{\partial f^i}{\partial x^\beta}\,\gamma^{\alpha \beta}$$
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.