# How does this orthogonality follow from the map being an isometry?

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

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 $$g$$.