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.



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> 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 $d\pi(\tau(w))=0$.
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What I don't understand is the emphrasized sentence. Why does $w$ being isometric imply $\tau(w)\perp w(M)$?




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**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 $\tau(w)=\Delta w$.



How does this orthogonality follow from the map being an isometry?
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 $g$.


