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 $$ \|[D,f]\|_{B(H)} = \|c(df)\|_{B(H)} = \|c(df)^2\|^{1/2}_{B(H)} = \|g^{-1}(df,df)\|^{1/2}_\infty = \|f\|_{d_g}. $$

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    $\begingroup$ 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 $\epsilon$. $\endgroup$ – Nik Weaver Aug 15 '15 at 3:01
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    $\begingroup$ So the answer to question 1 is "yes". $\endgroup$ – Nik Weaver Aug 15 '15 at 3:01
  • $\begingroup$ @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. $\endgroup$ – Branimir Ćaćić Aug 15 '15 at 10:08
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    $\begingroup$ 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 $(\epsilon/2,\epsilon)$. $\endgroup$ – Nik Weaver Aug 15 '15 at 14:05
  • $\begingroup$ 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? $\endgroup$ – Branimir Ćaćić Aug 15 '15 at 15:09

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