The Jacobian of the nearest point map is an orthogonal projection map.

Question

I'm reading a book about harmonic maps and at some point they said something I did not understand. Let $(N, h)$ be a compact Rimannian manifold. By the isometric embedding theorem by Nash, the exists $L \ge 1$ such that $N$ is isometrically embedded into an Euclidean space $\mathbb R^L$. Moreover, there exists $\delta = \delta(N) > 0$ such that the nearest point projection map $\Pi_N:N_\delta \to N$ is smooth, where $N_\delta$ consists on the points $y \in \mathbb R^L$ satisfying $d(y, N) = \inf_{z \in N}|z - y| < \delta$, and $\Pi_N(y) \in N$ is such that $|y - \Pi_N(y)| = d(y, N)$ for $y \in N_\delta$. Now they claim that $$P(y) = \nabla \Pi_N(y): \mathbb R^L \to T_yN$$ for $y \in N$, is an orthogonal projection map. I get that $\nabla \Pi_N(y)$ (which is the jacobian of $\Pi$ at $y$) goes from $T_y \mathbb R^L\cong \mathbb R^L$ to $T_y N$, but why is it supposed to be an orthogonal projection map ?

Answer 1

Let $\Xi$ denote the normal bundle of $N$, and let $\Pi(y)=x$ for $y\in N_\delta$. Note that $\Pi^{-1}(x)=\Xi_x\cap B(x,\delta)$ is the set of normal vectors at $x$ of length $<\delta$; indeed, $\Pi(y)=x$ if and only if $y=x+\nu(x)$ for some $\nu(x)\in\Xi_x$ of length $<\delta$.

Now, consider $x_0$ and $y_0$ fixed, with $\|y_0-x_0\|=r_0$ and $\Pi(y_0)=x_0$. We claim that there is a section $\sigma$ of the unit normal bundle so that $y_0=x_0+r_0\sigma(x_0)$ and $\text{im}(d\sigma(x_0)) = T_{x_0}N\subset\Bbb R^L$. (We say $\sigma$ is parallel at $x_0$; this is an easy argument analogous to the existence of a frame of the tangent bundle whose Christoffel symbols vanish at a given point.) We can think of $\Sigma$, the image of $r_0\sigma$, as a natural (local) parallel copy of $N$. Now we observe that $$T_{y_0}\Bbb R^L = T_{y_0}\Sigma \oplus \Xi_{x_0}.$$ Of course, $d\Pi_{y_0}$ maps $\Xi_{x_0}$ to $0$; however, $d\Pi_{y_0}$, mapping $T_{y_0}\Sigma$ to $T_{y_0}\Sigma=T_{x_0}N$, is rarely the identity map. So I question the author's use of the terminology "orthogonal projection map." The simplest example shows what's going on in general: If $N$ is the unit circle and $y_0=r_0x_0$ with $r_0>1$, then $d\Pi_{y_0}$ maps the tangent vector $v_0$ to the circle of radius $r_0$ to the tangent vector $v_0/r$ of the circle of radius $1$.