[proofplan] We expand the definition of the [tension field](/page/Tension%20Field) $\tau(u)$ as the contraction of the two covariant domain indices of $\nabla du$ against the inverse metric $g^{-1}$ in the coordinate frames on the domain and target. The covariant derivative of $du$ has three contributions: ordinary second derivatives of the coordinate functions $u_\alpha$, correction terms from the Levi-Civita connection of $M$, and quadratic correction terms from the Levi-Civita connection of $N$. The first two contributions combine exactly into the coordinate formula for the [Laplace-Beltrami operator](/page/Laplace-Beltrami%20Operator) $\Delta_g u_\alpha$, while the target connection terms remain as the quadratic Christoffel expression. [/proofplan]

[step:Expand $du$ in the chosen coordinate frames] Let $(U,x)$ be the chosen domain coordinate patch on $M$, with coordinate map $x: U \to x(U) \subseteq \mathbb{R}^m$, and let $(V,y)$ be the chosen target coordinate patch on $N$, with coordinate map $y: V \to y(V) \subseteq \mathbb{R}^n$, such that $u(U) \subset V$. Write $x=(x_1,\dots,x_m)$ and $y=(y_1,\dots,y_n)$. On $U$, for each $i=1,\dots,m$, let $\partial_{x_i}: U \to TU$ denote the $i$-th coordinate vector field. Along $u$, for each $\alpha=1,\dots,n$, let $\partial_{y_\alpha}\big|_u: U \to u^*TN$ denote the pulled-back $\alpha$-th target coordinate vector field. Since $u_\alpha = y_\alpha \circ u$, the differential of $u$ satisfies \begin{align*} du(\partial_{x_i}) = \sum_{\alpha=1}^n \partial_{x_i}u_\alpha\, \partial_{y_\alpha}\big|_u. \end{align*} [/step]

[step:Compute the covariant derivative of $du$]Let $T^*M$ denote the cotangent bundle of $M$, and let $u^*TN$ denote the pullback bundle over $U$ whose fibre at $p \in U$ is $T_{u(p)}N$. Thus $T^*M \otimes u^*TN$ denotes the vector bundle over $U$ whose sections are $u^*TN$-valued one-forms on $U$; the differential $du$ is such a section. Let $\nabla^M$ denote the Levi-Civita connection of $g$, let $\nabla^N$ denote the Levi-Civita connection of $h$, let $\nabla^u$ denote the [pullback connection](/page/Pullback%20Connection) on $u^*TN$ induced by $\nabla^N$, and let $\nabla$ denote the induced connection on $T^*M \otimes u^*TN$. By definition, \begin{align*} (\nabla du)(\partial_{x_i},\partial_{x_j}) = \nabla^u_{\partial_{x_i}}\bigl(du(\partial_{x_j})\bigr) - du\bigl(\nabla^M_{\partial_{x_i}}\partial_{x_j}\bigr), \end{align*} Using \begin{align*} \nabla^M_{\partial_{x_i}}\partial_{x_j} = \sum_{k=1}^m \Gamma^k_{ij}\,\partial_{x_k} \end{align*} and \begin{align*} \nabla^u_{\partial_{x_i}}\bigl(\partial_{y_\beta}\big|_u\bigr) = \sum_{\gamma=1}^n \sum_{\alpha=1}^n \partial_{x_i}u_\gamma\, \Gamma^\alpha_{\gamma\beta}(u)\, \partial_{y_\alpha}\big|_u, \end{align*} the product rule gives \begin{align*} (\nabla du)(\partial_{x_i},\partial_{x_j}) = \sum_{\alpha=1}^n \left( \partial_{x_i}\partial_{x_j}u_\alpha - \sum_{k=1}^m \Gamma^k_{ij}\,\partial_{x_k}u_\alpha + \sum_{\beta,\gamma=1}^n \Gamma^\alpha_{\beta\gamma}(u)\, \partial_{x_i}u_\beta\,\partial_{x_j}u_\gamma \right) \partial_{y_\alpha}\big|_u. \end{align*}[/step]

[guided]Let $T^*M$ denote the cotangent bundle of $M$, and let $u^*TN$ denote the pullback bundle over $U$ whose fibre at $p \in U$ is $T_{u(p)}N$. Let $\nabla^u$ denote the [pullback connection](/page/Pullback%20Connection) on $u^*TN$ induced by $\nabla^N$. The object $du$ is a section of $T^*M \otimes u^*TN$, so its covariant derivative must differentiate both the target-valued part and correct for the variation of the input vector field on $M$. Thus, for coordinate vector fields, the definition is \begin{align*} (\nabla du)(\partial_{x_i},\partial_{x_j}) = \nabla^u_{\partial_{x_i}}\bigl(du(\partial_{x_j})\bigr) - du\bigl(\nabla^M_{\partial_{x_i}}\partial_{x_j}\bigr). \end{align*} We first compute the pullback-connection term. From the previous step, \begin{align*} du(\partial_{x_j}) = \sum_{\beta=1}^n \partial_{x_j}u_\beta\,\partial_{y_\beta}\big|_u. \end{align*} Applying the product rule for $\nabla^u_{\partial_{x_i}}$ gives \begin{align*} \nabla^u_{\partial_{x_i}}\bigl(du(\partial_{x_j})\bigr) = \sum_{\beta=1}^n \partial_{x_i}\partial_{x_j}u_\beta\,\partial_{y_\beta}\big|_u + \sum_{\beta=1}^n \partial_{x_j}u_\beta\, \nabla^u_{\partial_{x_i}}\bigl(\partial_{y_\beta}\big|_u\bigr). \end{align*} The pullback connection differentiates the target coordinate frame along the velocity $du(\partial_{x_i})$. Since \begin{align*} du(\partial_{x_i}) = \sum_{\gamma=1}^n \partial_{x_i}u_\gamma\,\partial_{y_\gamma}\big|_u \end{align*} and \begin{align*} \nabla^N_{\partial_{y_\gamma}}\partial_{y_\beta} = \sum_{\alpha=1}^n \Gamma^\alpha_{\gamma\beta}\,\partial_{y_\alpha}, \end{align*} we get \begin{align*} \nabla^u_{\partial_{x_i}}\bigl(\partial_{y_\beta}\big|_u\bigr) = \sum_{\gamma=1}^n\sum_{\alpha=1}^n \partial_{x_i}u_\gamma\,\Gamma^\alpha_{\gamma\beta}(u)\, \partial_{y_\alpha}\big|_u. \end{align*} Therefore \begin{align*} \nabla^u_{\partial_{x_i}}\bigl(du(\partial_{x_j})\bigr) = \sum_{\alpha=1}^n \left( \partial_{x_i}\partial_{x_j}u_\alpha + \sum_{\beta,\gamma=1}^n \Gamma^\alpha_{\gamma\beta}(u)\, \partial_{x_i}u_\gamma\,\partial_{x_j}u_\beta \right) \partial_{y_\alpha}\big|_u. \end{align*} Now we subtract the domain-connection correction. Since \begin{align*} \nabla^M_{\partial_{x_i}}\partial_{x_j} = \sum_{k=1}^m \Gamma^k_{ij}\,\partial_{x_k}, \end{align*} linearity of $du$ gives \begin{align*} du\bigl(\nabla^M_{\partial_{x_i}}\partial_{x_j}\bigr) = \sum_{k=1}^m \Gamma^k_{ij}\,du(\partial_{x_k}) = \sum_{\alpha=1}^n \left( \sum_{k=1}^m \Gamma^k_{ij}\,\partial_{x_k}u_\alpha \right) \partial_{y_\alpha}\big|_u. \end{align*} Subtracting this expression yields \begin{align*} (\nabla du)(\partial_{x_i},\partial_{x_j}) = \sum_{\alpha=1}^n \left( \partial_{x_i}\partial_{x_j}u_\alpha - \sum_{k=1}^m \Gamma^k_{ij}\,\partial_{x_k}u_\alpha + \sum_{\beta,\gamma=1}^n \Gamma^\alpha_{\gamma\beta}(u)\, \partial_{x_i}u_\gamma\,\partial_{x_j}u_\beta \right) \partial_{y_\alpha}\big|_u. \end{align*} Since the Levi-Civita connection of $h$ is torsion-free, $\Gamma^\alpha_{\gamma\beta}=\Gamma^\alpha_{\beta\gamma}$. Relabelling the dummy indices gives the displayed form \begin{align*} (\nabla du)(\partial_{x_i},\partial_{x_j}) = \sum_{\alpha=1}^n \left( \partial_{x_i}\partial_{x_j}u_\alpha - \sum_{k=1}^m \Gamma^k_{ij}\,\partial_{x_k}u_\alpha + \sum_{\beta,\gamma=1}^n \Gamma^\alpha_{\beta\gamma}(u)\, \partial_{x_i}u_\beta\,\partial_{x_j}u_\gamma \right) \partial_{y_\alpha}\big|_u. \end{align*}[/guided]

[step:Trace the covariant Hessian with the inverse metric] By definition, the tension field is obtained by contracting the two covariant domain slots of $\nabla du$ with the inverse metric. In the coordinate frame, this contraction is \begin{align*} \tau(u) = \sum_{i,j=1}^m g^{ij}\, (\nabla du)(\partial_{x_i},\partial_{x_j}). \end{align*} Substituting the formula from the previous step gives \begin{align*} \tau(u) = \sum_{\alpha=1}^n \left[ \sum_{i,j=1}^m g^{ij} \left( \partial_{x_i}\partial_{x_j}u_\alpha - \sum_{k=1}^m \Gamma^k_{ij}\,\partial_{x_k}u_\alpha \right) + \sum_{i,j=1}^m\sum_{\beta,\gamma=1}^n g^{ij}\Gamma^\alpha_{\beta\gamma}(u) \partial_{x_i}u_\beta\partial_{x_j}u_\gamma \right] \partial_{y_\alpha}\big|_u. \end{align*} Thus the component $\tau(u)_\alpha$ equals the bracketed expression. [/step]

[step:Identify the domain contribution with the Laplace-Beltrami operator]Let $G: U \to (0,\infty)$ be the smooth function defined by $G(p)=\det(g_{ij}(p))$, where $(g_{ij}(p))$ is the coordinate matrix of $g$ at $p \in U$. For every smooth map $f: U \to \mathbb{R}$, the coordinate formula for the scalar Laplace-Beltrami operator satisfies \begin{align*} \Delta_g f = \sum_{i,j=1}^m g^{ij} \left( \partial_{x_i}\partial_{x_j}f - \sum_{k=1}^m \Gamma^k_{ij}\,\partial_{x_k}f \right). \end{align*} Indeed, expanding the divergence form gives \begin{align*} \Delta_g f = \frac{1}{\sqrt{G}} \sum_{i,j=1}^m \partial_{x_i} \left( \sqrt{G}\,g^{ij}\,\partial_{x_j}f \right). \end{align*} Applying the product rule to each summand then yields \begin{align*} \Delta_g f = \sum_{i,j=1}^m g^{ij}\partial_{x_i}\partial_{x_j}f + \sum_{j=1}^m \left[ \frac{1}{\sqrt{G}} \sum_{i=1}^m \partial_{x_i}(\sqrt{G}\,g^{ij}) \right]\partial_{x_j}f. \end{align*} We now verify the needed coordinate identity for the Levi-Civita connection. We use the derivative-of-inverse identity for a smooth invertible matrix-valued function, derived by differentiating $A^{-1}A=I$, and apply it to the metric matrix. For coordinate indices $r,s,\ell \in \{1,\dots,m\}$, differentiating $\sum_{a=1}^m g^{ra}g_{as}=\delta^r_s$ with respect to $x_\ell$ gives \begin{align*} \partial_{x_\ell}g^{rs} = -\sum_{a,b=1}^m g^{ra}g^{sb}\partial_{x_\ell}g_{ab}. \end{align*} Setting $r=i$, $s=j$, and $\ell=i$ inside the subsequent sum over $i$ gives the contracted identity needed below. The Jacobi determinant formula, derived by differentiating $\det A$ along a smooth matrix path and written as $\partial(\det A)=(\det A)\operatorname{tr}(A^{-1}\partial A)$, combined with the ordinary chain rule for the square root, gives \begin{align*} \partial_{x_i}\sqrt{G} = \frac{1}{2}\sqrt{G}\sum_{a,b=1}^m g^{ab}\partial_{x_i}g_{ab}. \end{align*} Hence \begin{align*} \frac{1}{\sqrt{G}} \sum_{i=1}^m \partial_{x_i}(\sqrt{G}\,g^{ij}) = \sum_{i=1}^m \partial_{x_i}g^{ij} + \frac{1}{2}\sum_{i,a,b=1}^m g^{ij}g^{ab}\partial_{x_i}g_{ab}. \end{align*} Using the Christoffel formula \begin{align*} \Gamma^j_{ik} = \frac{1}{2}\sum_{b=1}^m g^{jb} \left( \partial_{x_i}g_{kb} + \partial_{x_k}g_{ib} - \partial_{x_b}g_{ik} \right), \end{align*} contracting with the symmetric tensor $g^{ik}$, and relabelling dummy indices gives \begin{align*} \sum_{i,k=1}^m g^{ik}\Gamma^j_{ik} = \sum_{i,k,b=1}^m g^{ik}g^{jb}\partial_{x_i}g_{kb} - \frac{1}{2}\sum_{i,k,b=1}^m g^{ik}g^{jb}\partial_{x_b}g_{ik}. \end{align*} After relabelling dummy indices in the determinant term, this is exactly \begin{align*} \frac{1}{\sqrt{G}} \sum_{i=1}^m \partial_{x_i}(\sqrt{G}\,g^{ij}) = - \sum_{i,k=1}^m g^{ik}\Gamma^j_{ik}. \end{align*} Substituting this identity therefore gives \begin{align*} \Delta_g f = \sum_{i,j=1}^m g^{ij}\partial_{x_i}\partial_{x_j}f - \sum_{i,k,j=1}^m g^{ik}\Gamma^j_{ik}\partial_{x_j}f. \end{align*} Relabelling the dummy indices $k$ and $j$ yields \begin{align*} \Delta_g f = \sum_{i,j=1}^m g^{ij} \left( \partial_{x_i}\partial_{x_j}f - \sum_{k=1}^m \Gamma^k_{ij}\partial_{x_k}f \right). \end{align*} Applying this identity to $f=u_\alpha$ identifies the first part of $\tau(u)_\alpha$ with $\Delta_g u_\alpha$.[/step]

[guided]The purpose of this step is to justify that the two domain terms obtained from tracing $\nabla du$ are exactly the scalar Laplace-Beltrami operator applied to a coordinate component. Let $G: U \to (0,\infty)$ be the smooth function $G(p)=\det(g_{ij}(p))$, where $(g_{ij}(p))$ is the coordinate matrix of $g$ at $p \in U$. For a smooth map $f: U \to \mathbb{R}$, the divergence-form definition gives \begin{align*} \Delta_g f = \frac{1}{\sqrt{G}} \sum_{i,j=1}^m \partial_{x_i} \left( \sqrt{G}\,g^{ij}\,\partial_{x_j}f \right). \end{align*} Applying the product rule to each summand, we obtain \begin{align*} \Delta_g f = \sum_{i,j=1}^m g^{ij}\partial_{x_i}\partial_{x_j}f + \sum_{j=1}^m \left[ \frac{1}{\sqrt{G}} \sum_{i=1}^m \partial_{x_i}(\sqrt{G}\,g^{ij}) \right]\partial_{x_j}f. \end{align*} It remains to identify the coefficient of $\partial_{x_j}f$. We use two matrix identities, but we record why they apply here. The derivative-of-inverse identity follows by differentiating $A^{-1}A=I$, and the Jacobi determinant formula follows by differentiating $\det A$ along a smooth matrix path. Applied to the metric matrix, the inverse identity says that, for coordinate indices $r,s,\ell \in \{1,\dots,m\}$, \begin{align*} \partial_{x_\ell}g^{rs} = -\sum_{a,b=1}^m g^{ra}g^{sb}\partial_{x_\ell}g_{ab}. \end{align*} In the coefficient of $\partial_{x_j}f$, we use this with $r=i$, $s=j$, and $\ell=i$ under the sum over $i$. The Jacobi determinant formula is $\partial(\det A)=(\det A)\operatorname{tr}(A^{-1}\partial A)$; followed by the ordinary chain rule for the square root, it gives \begin{align*} \partial_{x_i}\sqrt{G} = \frac{1}{2}\sqrt{G}\sum_{a,b=1}^m g^{ab}\partial_{x_i}g_{ab}. \end{align*} Substituting these identities into the product-rule coefficient yields \begin{align*} \frac{1}{\sqrt{G}} \sum_{i=1}^m \partial_{x_i}(\sqrt{G}\,g^{ij}) = \sum_{i=1}^m \partial_{x_i}g^{ij} + \frac{1}{2}\sum_{i,a,b=1}^m g^{ij}g^{ab}\partial_{x_i}g_{ab}. \end{align*} Using the Christoffel formula for the Levi-Civita connection of $g$ and contracting with the symmetric tensor $g^{ik}$ gives \begin{align*} \sum_{i,k=1}^m g^{ik}\Gamma^j_{ik} = \sum_{i,k,b=1}^m g^{ik}g^{jb}\partial_{x_i}g_{kb} - \frac{1}{2}\sum_{i,k,b=1}^m g^{ik}g^{jb}\partial_{x_b}g_{ik}. \end{align*} After relabelling dummy indices, the preceding two displayed formulas give \begin{align*} \frac{1}{\sqrt{G}} \sum_{i=1}^m \partial_{x_i}(\sqrt{G}\,g^{ij}) = - \sum_{i,k=1}^m g^{ik}\Gamma^j_{ik}. \end{align*} Therefore \begin{align*} \Delta_g f = \sum_{i,j=1}^m g^{ij}\partial_{x_i}\partial_{x_j}f - \sum_{i,k,j=1}^m g^{ik}\Gamma^j_{ik}\partial_{x_j}f. \end{align*} Relabelling the dummy indices $k$ and $j$ gives \begin{align*} \Delta_g f = \sum_{i,j=1}^m g^{ij} \left( \partial_{x_i}\partial_{x_j}f - \sum_{k=1}^m \Gamma^k_{ij}\partial_{x_k}f \right). \end{align*} Taking $f=u_\alpha$ proves that the traced domain-connection contribution is precisely $\Delta_g u_\alpha$.[/guided]

[step:Combine the scalar Laplacian and target connection terms] From the trace computation, \begin{align*} \tau(u)_\alpha = \sum_{i,j=1}^m g^{ij} \left( \partial_{x_i}\partial_{x_j}u_\alpha - \sum_{k=1}^m \Gamma^k_{ij}\,\partial_{x_k}u_\alpha \right) + \sum_{i,j=1}^m\sum_{\beta,\gamma=1}^n g^{ij}\Gamma^\alpha_{\beta\gamma}(u) \partial_{x_i}u_\beta\partial_{x_j}u_\gamma. \end{align*} The previous step identifies the first summation with $\Delta_g u_\alpha$. Hence, for every $\alpha=1,\dots,n$, \begin{align*} \tau(u)_\alpha = \Delta_g u_\alpha + \sum_{i,j=1}^m\sum_{\beta,\gamma=1}^n g^{ij}\Gamma^\alpha_{\beta\gamma}(u) \partial_{x_i}u_\beta\partial_{x_j}u_\gamma. \end{align*} This is exactly the asserted coordinate formula for the tension field. [/step]