Lipschitz Seminorms Associated to a Metric and Seminorms from Dirac Operators

Question

I am learning to work with spectral triples and given a spectral triple $(A, H, D)$ with representation $\pi$, a seminorm can be defined on $A$ via $L_D := \|[ D, \pi(f) ]\|_{B(H)}$. View $A$ as $C(X)$ for some compact Hausdorff space $X$. With respect to this seminorm the spectral distance on $X$ can be defined by:

$$d_X(x,y) := \sup \{ |f(x) - f(y)| : L_D (f) \leq 1 \}.$$

Also, the Lipschitz seminorm associated to $d_X$ is given by:

$$L_{d_X}(f) := \sup \left\{ \frac{|f(x) - f(y)|}{d_X(x,y)}\ \middle | \ x,y \in X, x \neq y \right\}.$$

My question is: "Does $L_D$ always coincide with $L_{d_X}$ or does this depend on particular properties of the Dirac operator $D$?" I would begin by trying to show that $L_{d_X} (f) \leq L_D(f)$ for arbitrary $f \in A$ and vice versa, but I don't see how such arguments can be made without knowing more properties about the Dirac operator $D$.

Answer 1

Whats true is that $L_{d_X}(f) ≤ L_{D}(f)$ for all $f\in A$ and if $A$ is commutative also $L_{d_X}(f)=0\implies L_{D}(f)=0$. I would guess that the inequality $L_{D}(f)≤L_{d_X}(f)$ is true if $A$ is commutative, but have not been able to prove it. However if $A$ is not commutative the object $X$ can be really dumb and it should be easy to find counter-examples to the second inequality. Now these considerations should clearly depend on the definition of a spectral triple, which I am not familiar with. I will write down what I think you are working with:

A spectral triple is a triple $(A, H, D)$ where $H$ is a Hilbert space, $D$ is a densely define self-adjoint operator on $H$ and $A$ is a (here apparently commutative and unital) $C^*$ algebra together with a representation $\pi: A\to B(H)$.

First we check $L_D(f)=0$ $\iff$ $L_{d_X}(f)=0$, here the direction $\impliedby$ makes use of $A$ being commutative.

Suppose $L_D(f)=0$, then you get for all $x,y$ you get that either $f(x)=f(y)$ or $d_X(x,y)=\infty$, hence $\frac{|f(x)-f(y)|}{d_X(x,y)}=0$ always and then $L_{d_X}(f)=0$. Now if $L_{d_X}(f)=0$ then whenever $f(x)\neq f(y)$ there must be some $g\in A$ with $L_D(g)=0$ and $g(x)\neq g(y)$. The $C^*$ sub-algebra of $A$ generated by all such $g$ then separate the points that $f$ separates and as such $f$ is a limit of such $g$ by Stone-Weierstraß. In particular you will get $[f, D]=0$ and $L_D(f)=0$.

Next the inequality $L_{d_X}≤ L_D$ follows from elementary considerations:

Let $f\in A$ with $L_D(f)\neq0$ and by rescaling assume $L_D(f)=1$. Then $d_X(x,y) ≥ |f(x)-f(y)|$ for any $x,y$ by the definition of $d_X$. Hence $\frac{|f(x)-f(y)|}{d_X(x,y)}≤ 1$ for all $x,y$ and $L_{d_X}(f) ≤ L_D(f)$.