Monday, August 13, 2007

Harmonic mean

This "post" is mainly an attempt to see if one can manage to use formulas in the blog and discuss some real stuff somehow. The formulas should be really visible, a bit like with transparencies. So as a pretext, I'll start by discussing an issue related to the basic Ansatz:

which gives the operator theoretic line element in terms of the Dirac operator in the general framework of "metric" noncommutative geometry. The kernel of the operator D is finite dimensional and one takes 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.

The point is to understand what happens when one takes the product of two noncommutative geometries. One gets the following relation for the squares of the corresponding Dirac operators:

where we abuse notations by removing the tensor product by the identity operator that normally goes with each of the operators D_j. Now this relation is quite different from the simple Pythagorean relation of the classical line elements whose square simply add up and it thus raises the question of reconciling the above Ansatz with the simple formula of addition of the squares of the Dirac operators. More generally, one can consider a bunch of NC spaces with Dirac operators D_mu and combine them as follows: One starts with a positive matrix of operators in Hilbert space:

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.


Christophe de Dinechin said...

I am curious how you generated the equations. They look good. Did you use MathML or anything like that? Or did you embed images generated using some other tool?

AC said...

Dear Christophe

What I did was to embed small images inside the text, first I wrote a pdf file and then I extracted small portions of the pdf using adobe professional and saving them as jpeg.

Anonymous said...

there are easier ways to insert TeX into web pages; for example
just typing the web-link

[img alt="\int_{x=1}^{10} f(x) dx" src="\LARGE\!\int_{x=1}^{10} f(x) dx.gif" align=center border=0/]

will generate the corresponding formula


(you may skip alt= and align= border= tags)

Christophe de Dinechin said...

Dear ac,

I see you did it the hard way. I was hoping there was some simpler way.

Dear Anonymous,

Thanks a lot for the tip about TeXify. I tried it (see, it "works", but it does not look as good as on this blog, in particular because there is no anti-aliasing.

I could go for MathML, but so many browsers don't support it, so it's almost useless as a communication medium.

The folks at Wikipedia have found a good way to do it, but you can't reuse a Wikipedia formula anywhere, as all you get is a cached picture with a cryptic name encoding.

Oh well, thanks for the tips.

Unknown said...

Dear ac,

in some
products of geometries one is able to recover Pythagorean relation for the line elements. This is the case for instance in the 2-sheet model obtained as the product of a manifold M (with Dirac DE) by an internal geometry consisting in two copies of the complex numbers. The distance extracted from the total Dirac operator D corresponds to a line element ds whose square is the sum of the square of the line element dsE of the manifold with the square dsI of the line element of the internal geometry

Pythagorean relation

It seems that the explanations lies in the internal Dirac operator DI that has a square proportional to the identity

square of DI

Indeed under this condition the square of D can be seen in two ways:

-the sum of the squares of DE and DI

D^2 as a sum of squares

-the square of the Dirac operator of a manifold Mx[0,1], with the chirality as the extra Dirac matrix

D^2 as the square of an extended D_E

We assume the operator "derivative in the extra dimension" has a square equals to the identity, which reflects that we are not dealing with an extradimension, but with a discrete set of 2 points. The line element associated to D is thus the one associated to the extended D_E, hence the Pythagorean relation between the lines elements.

These observations are certainly very naive. However they question quite nicely (at least I hope...) the heuristic definition of the line element. It is also quite pleasant to have an example in which both the square of the Dirac operators and the square of the line elements add. However I do not know to what extend the condition square of DI = 1 could be relevant in a more general context. It looks very restrictive.