**ds**to vanish on that kernel. As was already discussed here, the knowledge of D gives back the metric. Moreover the noncommutative integral, in the form of the Dixmier trace, gives back the volume form. Thus the integral of a function

**f**in dimension n is simply given by

where the "cut" integral is the Dixmier trace ie the functional that assigns to an infinitesimal of order one the coefficient of the logarithmic divergency in the series that gives the sum of its eigenvalues.

I will not try to justify the heuristic definition of the line element any further. It is more interesting to put it to the test, to

*question it,*and I will discuss an example of an issue which left me perplex for quite sometime but has a pretty resolution.

and one extends the above formula giving D^2 for a product of two spaces and forms the following sum:

We make no commutativity hypothesis and even drop the self-adjointness of D_mu which is not needed. We want a formula for the inverse of the square of D ie for:

in terms of the inverse matrix:

which plays a role similar to the g\mu\nu of Riemannian geometry, and of the operators

where the notation with z stresses the fact that we do not even assume self-adjointness of the various D_\mu.

It sounds totally hopeless since one needs a formula for the inverse of a sum of noncommuting operators. Fortunately it turns out that there is a beautiful simple formula that does the job in full generality. It is reminiscent of the definition of distances as an infimum. It is given by:

The infimum is taken over all decompositions of the given vector as a sum:

Note that this formula suffices to determine the operator ds^2 completely, since it gives the value of the corresponding positive quadratic form on any vector in Hilbert space. The proof of the formula is not difficult and can be done by applying the technique of Lagrange multipliers to take care of the above constraint on the free vectors \xi^\mu.