# Connes’ Distance Function

In classical geometry, the distance between two points $p$ and $q$ is given by the length of any shortest path from $p$ to $q$. However, this definition is not valid in quantum mechanics, where the concept of path between two points is not well defined.  The idea introduced by Alain Connes in noncommutative geometry consists of defining a spectral distance $d ( p , q )$ from values taken by operator observables rather than from classical coordinates. In this way, the concept of geometrical point is not used, which allows the spectral distance to be applied to both classical and quantum spaces.

DISTANCE ON THE REAL LINE

In order to simplify the calculations, we choose an example of usual distance between two points $p$ and $q$ on the real line

$\boxed{d ( p , q ) = | p - q |}$

INFIMUM FORMULATION

In Riemannian geometry, the distance $d ( p , q )$ between two points $p$ and $q$ on a manifold $M$ can be formulated as an infimum

$\boxed{d ( p , q ) = \inf \{ L ( \gamma ) : \gamma \textrm{ path joining } p \textrm{ to } q \}}$

where the length $L ( \gamma )$ of a path $\gamma$ joining $p$ to $q$ is computed from the metric $g_{\mu \nu}$

$\displaystyle L ( \gamma ) = \int_{p}^{q} \sqrt{g_{\mu \nu} dx^{\mu} dx^{\nu}}$

In the case of the real line with unit metric

$d ( p , q ) = \displaystyle L ( [ p , q ] ) = | p - q |$

SUPREMUM FORMULATION

The distance on the real line can also be expressed as a supremum taken over continuously derivable functions on $\mathbb{R}$

$\boxed{d ( p , q ) = \sup_{f \in C^{1} ( \mathbb{R} )} \{ | f ( p ) - f ( q ) | : | f ' ( x ) | \leq 1 \textrm{ for all } x \in \mathbb{R} \}}$

At first sight, it can be strange to define a distance as a supremum whereas the usual definition uses an infimum. However, with this original approach, the primary objects are observables $f$ rather than points $p$ and $q$, which is the inverse of the usual point of view. The advantage of this method is that it does not use the concept of path and thus can be generalized to quantum spaces, as explained further.

Proof

Let us consider the trivial function

$\displaystyle h : x \mapsto x$

We want to show that it corresponds to the supremum, that is to say that $d ( p , q ) = | h ( p ) - h ( q ) | = | p - q |$. To show this, we first note that $h \in C^1 ( \mathbb{R} )$ and $| h ' ( x ) | = 1 \leq 1$ for all $x \in \mathbb{R}$, thus $h$ is in the set defining the supremum. Suppose that $h$ is not the supremum. Then, there must exist $f \in C^1 ( \mathbb{R} )$ such that $| f ' ( x ) | \leq 1$ for all $x \in \mathbb{R}$ and $| f ( p ) - f ( q ) | > | h ( p ) - h ( q ) |$. This leads to

$\displaystyle | f ( p ) - f ( q ) | > | p - q |$

By the mean value theorem, there exists $c \in [ p , q ]$ such that

$\displaystyle | f ' ( c ) | = \frac{| f ( p ) - f ( q ) |}{| p - q |}$

From the two previous equations we deduce

$| f ' ( c ) | > 1$

which is in contradiction with $| f ' ( x ) | \leq 1$ for all $x \in \mathbb{R}$

OPERATOR FRAMEWORK

Although the above supremum formulation considers observables as primary objects, the expression is still incompatible with the operator framework of quantum theory. The approach suggested by noncommutative geometry is to replace points by states and  functions by operators, which is exactly the way we do calculations in quantum mechanics.

First, a state $\omega_p$ is associated to each point $p$ as follows

$\displaystyle \omega_p ( f ) = f ( p )$

Thus, a point is considered as an evaluator for observables, which reverses the classical point of view. This is called a state in accordance with the mathematical definition of positive linear functional on an algebra, which is in the above case $C^1 ( \mathbb{R} ) \rightarrow \mathbb{R}$. Note that complex numbers are preferred in quantum mechanics but they are not used here for simplification.

Next, the subexpression $| f ' ( x ) | \leq 1$ for all $x \in \mathbb{R}$ must be rewritten to be compatible with operator theory. In fact, each function $f$ can be considered as an operator acting by pointwise multiplication on the space of square integrable functions $L^2 ( \mathbb{R} )$. That is to say $( f \psi ) ( x ) = f ( x ) \psi ( x )$ for any vector $\psi \in L^2 ( \mathbb{R} )$. Then the above subexpression becomes equivalent to

$\displaystyle \left \Vert \left [ \frac{d}{dx} , f \right ] \right \Vert \leq 1$

where $[ , ]$ denotes the commutator and $\Vert . \Vert$ the operator norm.

Proof

Indeed, let us develop the expression for $\psi \in L^2 ( \mathbb{R} )$

$\displaystyle \left [ \frac{d}{dx} , f \right ] \psi = \frac{d}{dx} f \psi - f \frac{d}{dx} \psi = f ' \psi$

Then we compute the operator norm

$\displaystyle \left \Vert \left [ \frac{d}{dx} , f \right ] \right \Vert = \Vert f ' \Vert = \sup_{x \in \mathbb{R}} | f ' ( x ) |$

Therefore $\left \Vert \left [ \frac{d}{dx} , f \right ] \right \Vert \leq 1$ is equivalent to $\sup_{x \in \mathbb{R}} | f ' ( x ) | \leq 1$ which in turn is equivalent to $| f ' ( x ) | \leq 1$ for all $x \in \mathbb{R}$

Finally the supremum expression can be rewritten in the operator framework as

$\boxed{d ( p , q ) = \sup_{f \in C^1 ( \mathbb{R} )} \left \{ | \omega_p ( f ) - \omega_q ( f ) | : \left \Vert \left [ \frac{d}{dx} , f \right ] \right \Vert \leq 1 \right \}}$

CONNES SPECTRAL TRIPLE

Intuitively, a spectral triple encodes geometrical information in an algebraic way.

Spectral triple : a spectral triple $( A , \mathcal{H} , D )$ is composed of a unital C*-algebra $A$ acting on a Hilbert space $\mathcal{H}$ and a self-adjoint operator $D$ with compact resolvent such that $[ D , a ]$ is bounded for all $a \in A$.

In this context, if $\sigma$ and $\sigma '$ are two states, their distance is defined by

$\boxed{d ( \sigma , \sigma ' ) = \sup_{a \in A} \{ | \sigma ( a ) - \sigma ' ( a ) | : \Vert [ D , a ] \Vert \leq 1 \}}$

EXAMPLE : RIEMANNIAN MANIFOLD

Let $M$ be a compact Riemannian manifold. Although being a classical space, it is a good example to understand how noncommutative geometry works in the commutative case. A spectral triple $( A , \mathcal{H} , D )$ is defined as follows.

We consider $A = C ( M )$ the algebra of continuous functions over $M$. Then, we choose $\mathcal{H}$ to be the Hilbert space of square integrable differentiable forms, the elements of $A$ acting by pointwise multiplication on $\mathcal{H}$. Finally, a first order differential operator $D$ on $\mathcal{H}$ must be provided. A good choice is the Dirac operator, that is the square root of the Laplace operator

$\displaystyle D^2 = \Delta$

It is well known that the Laplace-de Rham operator satisfies

$\displaystyle \Delta = d \delta + \delta d$

where $d$ is the exterior derivative and $\delta$ the exterior coderivative (obtained from $d$ and the Hodge star). Since $d^2 = \delta^2 = 0$ we put

$\displaystyle D = d + \delta$

The distance between two points $p$ and $q$ considered as pure states $\omega_p$ and $\omega_q$ is thus

$\boxed{d ( \omega_p , \omega_q ) = \sup_{f \in A} \{ | \omega_p ( f ) - \omega_q ( f ) | : \Vert [ D , f ] \Vert \leq 1 \}}$

This is a generalization of the previous formula on the real line. Moreover, although the metric $g_{\mu \nu}$ is not explicitly present in this expression, it is recovered by the distance. In other terms, $D$ encodes the metric of the Riemannian structure.