[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*}
[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]
[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 $\Delta_g u_\alpha$.
[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]
[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]
