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This is a step of a proof in the book Variational Problems in Geometry by Seiki Nishikawa. I will ignore the background and change some of the statements and notations for simplicity.

Let $(M,g)$ be a smooth Riemannian manifold. Suppose $w:M\to\mathbb R^q$ is an isometric embedding. Let $N$ be a tubular neighborhood of $w(M)$, and $\pi:N\to w(M)$ the canonical projection. Let $\tau(w)$ denote the tension field of $w$. Then since $w$ is an isometric embedding, $\tau(w)$ is orthogonal to $w(M)$ and hence $d\pi(\tau(w))=0$.

What I don't understand is the emphrasized sentence. Why does $w$ being isometric imply $\tau(w)\perp w(M)$?


Edit: The tension field $\tau(w)$ is an $\mathbb R^q$-valued vector field on $M$, or more precisely, a smooth section of the bundle $w^{-1}(T\mathbb R^q)$. It is defined in coordinates ($x^i$ on $M$ and the standard coordinates on $\mathbb R^q$) by $\tau(w)^r=\Delta w^r+g^{ij}\hat\Gamma_{kl}^rw^k_iw_j^l$, where $g$ is the metric on $M$ and $\hat\Gamma_{ij}^r$ are the Christoffel symbols of $\mathbb R^q$. But since w.r.t. the standard coordinates of the Euclidean space all coefficients of its metric are $0$, $\hat\Gamma_{ij}^r=0$ and $\tau(w)=\Delta w$.

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There are a number of ways to see this. One way is to take the covariant derivative of the isometric embedding equation $\partial_iu\cdot\partial_ju = g_{ij}$ and "differentiate by parts". The calculation below is with respect to local coordinates, and $u$ is treated as a $q$-tuple of scalar real-valued functions. Therefore, $\nabla^2_{ij}u = \nabla^2_{ji}u$. \begin{align*} 0 &= \nabla_kg_{ij}\\ &= \nabla_k(\partial_iu\cdot\partial_ju)\\ &= \nabla^2_{ik}u\cdot \partial_ju + \partial_iu\cdot\nabla^2_{jk}u\\ &= \nabla_i(\partial_ku\cdot\partial_ju) + \nabla_j(\partial_iu\cdot\partial_ku) - 2\partial_ku\cdot\nabla^2_{ij}u\\ &= \nabla_ig_{kj} + \nabla_jg_{ik} - 2\partial_ku\cdot\nabla^2_{ij}u\\ &= -2\partial_ku\cdot\nabla^2_{ij}u \end{align*} Since this holds for any $1 \le i, j, k \le \dim M$, it follows that $\nabla^2u(p)$ is normal to $T_pM \subset \mathbb{R}^q$, for every $p \in M$. The tension field is simply $g^{ij}\nabla^2_{ij}u$ and therefore is also normal to $M$.

It's useful to write out the calculation above using Christoffel symbols and the identity $$ g_{kl}\Gamma^l_{ij} = \partial_ku\cdot\partial^2_{ij}u. $$ Here, $\partial^2_{ij}u$ denotes the second partials of $u$ with respect to local coordinates and not its Hessian with respect to $g$.

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