# Formula for the distance in noncommutative geometry

Probably the most famous formula in noncommutative geometry is the following formula allowing one to compute distance of two points using the operator theoretic data: $$(1) \ \ d(p,q)=\sup\{|f(p)-f(q)| f \in \mathcal{A}, \|[D\!\!\!/,f]\| \leq 1\}$$ where everything takes place on spin (or spin^c) manifold and $D\!\!\!/$ is the Dirac operator (but at this moment I would not specify what is $\mathcal{A}$). It is defined by the formula $D\!\!\!/ =ic \circ \nabla^S$ where $c$ is the Clifford action and $\nabla^S$ is the spin connection. This definition allows us to prove that, for a function $f$ one has $$(2) \ \ [D\!\!\!/,f]=ic(df)$$ and further it follows from this identity that $\|[D\!\!\!/,f] \|=\|grad(f) \|$. One we have that, we can show that $|f(p)-f(q)| \leq \| grad(f)\| \ell(\gamma)$ where $\gamma$ is piecewise smooth curve joining $p$ and $q$. Taking into account only $f$ such that $\|grad(f)\| \leq 1$ one then obtains $\sup\{|f(p)-f(q)| f \in \mathcal{A}, \|[D\!\!\!/,f]\| \leq 1\} \leq d(p,q)$. In order to get equality one considers the function $d(p,\cdot)$---but the problem is that this function is only continuos but is not smooth. So the first problem is:

Question 1 Is the formula $(1)$ true when one takes supremum only over smooth functions?

And the second thing which I would like to know is the following: as I said to get formula $(1)$ it is enough to know formula $(2)$ so

Question 2 Suppose that we have Riemannian manifold $M$ (not necessarily spin$^c$). Is it always possible to find operator $D$ such that formula (1) holds?

Let $\|\cdot\|_{d_g} : L^\infty(M) \to [0,+\infty]$ be the Lipschitz seminorm induced by the geodesic metric $d_g$ on $M$, defined by $$\forall f \in L^\infty(M), \quad \|f\|_{d_g} := \sup_{x \neq y} \frac{\lvert f(x)-f(y) \rvert}{d_g(x,y)},$$ so that the Lipschitz algebra of the metric space $(M,d_g)$ is $$\operatorname{Lip}(M,d_g) := \{f \in L^\infty(M) \mid \|f\|_{d_g} < +\infty\} = \{f \in C(M) \mid \|f\|_{d_g} < +\infty\},$$ which is a Banach algebra for the norm $\|\cdot\|_{\operatorname{Lip}(M,d_g)}$ defined by $$\forall f \in \operatorname{Lip}(M,d_g), \quad \|f\|_{\operatorname{Lip}(M,d_g)} := \|f\|_\infty + \|f\|_{d_g}.$$ Proving the famous formula, then, boils down to proving the following two claims, where your Question 1 pertains to the first claim, whilst your Question 2 pertains to the second:

Claim 1: For all $x,y \in M$, $$d_g(x,y) = \sup\{\lvert f(x) - f(y) \rvert \mid f \in \operatorname{Lip}(M,d_g), \; \|f\|_{d_g} \leq 1\}.$$

Claim 2: Suppose that $M$ is spin$^\mathbb{C}$ and that $D$ is a spin$^\mathbb{C}$ Dirac operator on a spinor bundle $S \to M$. Then, for any $f \in L^\infty(M)$, $[D,f] \in B(L^2(M,S))$ if and only if $f > \in \operatorname{Lip}(M,d_g)$, in which case $$\|[D,f]\| = \|f\|_{d_g}.$$

Let me first turn to your Question 1. In principle, the algebra you're really taking a supremum over in Claim 1 is $$\{f \in L^\infty(M) \mid df \in L^\infty(M,T^\ast M)\} = \{f \in C(M) \mid df \in L^\infty(M,T^\ast M)\} = \operatorname{Lip}(M,d_g),$$ where, for given $x$, $y \in M$, the supremum, which involves the Lipschitz seminorm $\|\cdot\|_{d_g}$, is attained by $d_g(x,\cdot) \in \operatorname{Lip}(M,d_g)$. Since $C^\infty(M)$ is not dense in $\operatorname{Lip}(M,d_g)$ endowed with the Banach algebra norm induced by $\|\cdot\|_{d_g}$—indeed, the closure, from what I understand, will be the proper closed subalgebra $C^1(M)$ of $C^1$ functions on $M$—I suspect the answer is no, unless there's a result out there that lets you approximate Lipschitz functions by smooth functions in the Lipschitz seminorm alone (whilst necessarily doing violence to the uniform norms). Edit: As Nik Weaver points out in his comment, the answer is indeed yes. Fix $x$, $y \in M$. Let $r = d_g(x,\cdot)$, which is smooth on the complement of $\{x\} \cup \text{cut locus of$x$}$ and has Lipschitz constant $\|r\|_{d_g} = 1$. Then, by mollification on geodesic neighbourhoods of $x$ and of the cut locus of $x$ (cf. the approximation results of Greene and Wu, 1972, 1976, 1979), one can construct for any $\epsilon > 0$ a smooth function $f_\epsilon$ such that $\|r-f_\epsilon\|_\infty < \epsilon/2$ and $\|f_\epsilon\|_{d_g} \leq 1$, so that $$\lvert f_\epsilon(x) - f_\epsilon(y) \rvert = \lvert (r(x)-r(y)) + (r(x)-f_\epsilon(x)) + (r(y) - f_\epsilon(y)) \rvert \geq d_g(x,y) - \epsilon,$$ and hence $$d_g(x,y) - \epsilon \leq \sup\{\lvert f(x) - f(y) \rvert \mid f \in C^\infty(M), \; \|f\|_{d_g} \leq 1 \} \leq d_g(x,y).$$ Note that in context, in Connes's book and in the 1995 paper Noncommutative geometry and reality, $\mathcal{A}$ is either $L^\infty(M)$ or $$\{f \in L^\infty(M) \mid [D,f] \in B(L^2(M,S))\} = \operatorname{Lip}(M,d_g),$$ where $D$ is the spin Dirac operator on the spinor bundle $S \to M$ on the spin manifold $M$.

Let me now turn to your Question 2. The essential point is that Claim 2 holds for any essentially self-adjoint Dirac-type operator on $M$, i.e., a first-order differential operator $D$ on a Hermitian vector bundle $E \to M$ such that $$D^2 = -g^{ij}\partial_i \partial_j + \text{lower order terms};$$ spin$^\mathbb{C}$ and spin Dirac operators are certainly Dirac-type operators, but so is the Hodge–de Rham operator $d+d^\ast$ on $\wedge^\ast T^\ast M$, which only requires an orientation and a Riemannian metric on $M$. The reason is that an essentially self-adjoint Dirac-type operator $D$ always defines a self-adjoint Clifford action $c : T^\ast M \to \operatorname{End}(E)$ on $E$ by $$\forall f \in C^\infty(M), \quad c(df) := i[D,f],$$ so that, in general, if $f \in L^\infty(M)$, then $[D,f] \in B(H)$ if and only if $f \in \operatorname{Lip}(M,d_g)$, with $$\|[D,f]\|_{B(H)} = \|c(df)\|_{B(H)} = \|c(df)^2\|^{1/2}_{B(H)} = \|g^{-1}(df,df)\|^{1/2}_\infty = \|f\|_{d_g}.$$

• You're right that the closure of $C^\infty(M)$ in Lipschitz norm is $C^1(M)$, but that is enough to attain the distance formula. The distance-from-$p$ function $d(p,\cdot)$ is already $C^\infty$ everywhere except at $p$. You can just smooth it out near $p$ and get a function that is $C^\infty$, has Lipschitz number 1, and separates $p$ and any other point $q$ by at least $d(p,q)-\epsilon$, for arbitrary $\epsilon$. – Nik Weaver Aug 15 '15 at 3:01
• So the answer to question 1 is "yes". – Nik Weaver Aug 15 '15 at 3:01
• @NikWeaver Thank you for the correction. I had a silly idée fixe about approximating in the Lipschitz seminorm, so I just didn't see that it was enough to bound it from above. – Branimir Ćaćić Aug 15 '15 at 10:08
• No problem. You don't really need to cite a reference for this construction; just compose the distance function with some $C^\infty$ function $g: [0,\infty) \to [0,\infty)$ satisfying $g(t) = 0$ on $[0,\epsilon/2]$, $g(t) = t - \epsilon$ on $[\epsilon,\infty)$, and something reasonable on $(\epsilon/2,\epsilon)$. – Nik Weaver Aug 15 '15 at 14:05
• This certainly takes care of the problem at the base point $x$, but you're still left, in general, with non-differentiability of the distance function on the cut locus of $x$, no? – Branimir Ćaćić Aug 15 '15 at 15:09