The metric distance property

As an operator, we can make sense of $[\Dslash, a]$ by conferring on $\sS$ the structure of a Hilbert space: if we write $\det g := \det[g_{ij}]$ for short, then

$$\nu_g := \sqrt{\det g} dx^1 \w dx^2 \wyw dx^n \in \sA^n(M)$$

is the Riemannian volume form (for the given orientation $\eps$ and metric $g$). In the notation, we assume that all local charts are consistent with the given orientation, which just means that $\det[g_{ij}] > 0$ in any local chart. The scalar product on $\sS$ is then given by

$$\braket{\phi}{\psi} := \int_M \pairing{\phi}{\psi} \nu_g \word{for} \phi,\psi \in \sS.$$

On completion in the norm $\Vert\psi\Vert := \sqrt{\braket{\psi}{\psi}}$, we get the Hilbert space $\sH := L^2(M,S)$ of $L^2$-spinors on $M$.

Using the gradient $\grad a := (da)^\3 \in \gX(M)$, we can compute
$$\begin{align*} \Vert[\Dslash, a]\Vert^2 &= \Vert c(da)\Vert^2 = \sup_{x\in M}\Ve... ...t \grad a\bigr\vert _x \bigr\Vert^2 =: \Vert\grad a\Vert _\infty^2. \end{align*}$$
Classically, we compute distances on a (connected) Riemannian manifold by the formula

$$d(x,y) := \inf\set{ \mathrm{length}(\ga) : \ga\:[0,1] \to M;\ \ga(0) = x, \ga(1) = y},$$

with the infimum taken over all piecewise-smooth paths $\ga$ in $M$ from $x$ to $y$. For $a \in \Coo(M)$, we then get
$$\begin{align*} a(y) - a(x) &= a(\ga(1)) - a(\ga(0)) = \int_0^1 \ddt{t} [a(\ga(t)... ...\ga(t)} \bigl( \grad a\bigr\vert _{\ga(t)}, \dot\ga(t) \bigr) dt, \end{align*}$$
and we can estimate this difference by
$$\begin{align*} \vert a(y) - a(x)\vert &\leq \int_0^1 \bigl\vert\grad a\bigr\vert... ...{length}(\ga) \\ &= \Vert[\Dslash, a]\Vert \mathrm{length}(\ga). \end{align*}$$
Thus

$$\sup\set{\vert a(y) - a(x)\vert : a \in C(M), \Vert[\Dslash, a]\Vert \leq 1} \leq \inf_\ga \mathrm{length}(\ga) =: d(x,y).$$ (14)

In this supremum, we can use $a \in C(M)$ not necessarily smooth; $a$ need only be continuous with $\grad a$ ($\nu$-essentially) bounded. Since we have obtained $\vert a(y) - a(x)\vert \leq \Vert\grad a\Vert _\infty d(x,y)$, we see that $a$ need only be Lipschitz on $M$ --with respect to the distance $d$-- with Lipschitz constant $\leq \Vert\grad a\Vert _\infty$. In fact, this is the best general Lipschitz constant: fix $x \in M$, and set $a_x(y) := d(x,y)$. This function lies in $C(M)$, and $\vert a_x(y) - a_x(z)\vert \leq d(y,z)$ by the triangle inequality for $d$. Since $\Vert\grad a_x\Vert _\infty = 1$ by a local geodesic calculation, we see that $a = a_x$ makes the inequality in () sharp:
$$\begin{align} d(x, y) &= \sup\set{ \vert a(y) - a(x)\vert : \Vert\grad a\Vert _\... ...vert a(y) - a(x)\vert : a \in C(M), \Vert[\Dslash, a]\Vert \leq 1}, \end{align}$$
so that $\Dslash$ determines the Riemannian distance $d$, which in turn determines the metric $g$. (The Myers-Steenrod theorem of differential geometry says that $g$ is uniquely determined by its distaqnce function $d$.)