## Thursday, February 22, 2007

### the Rosetta Stone of noncommutative geometry

Inspired by Alain's post, I would like to draw attention to a kind of dictionary or rather analogy between commutative concepts and their noncommutative counterparts. (for more on the role of analogies in mathematics and the intricacies of this idea see this 1940 letter of Andre Weil to his sister; my title is inspired by this letter too). I say analogy because there can be no exact correspondence between commutative and noncommutative worlds as the latter is much richer and more complicated. In fact people are usually confused about this. The perfect correspondence between commutative algebra and classical geometry is not what is being discussed here. We are now seeing only bits and parcels of this noncommutative landscape, but it is important to communicate the visible parts that we can see so far. I hope, together with contributions by others, we can eventually discuss at least the important elements of this `dictionary' in this blog.

In any case, Alain mentioned that the spectrum of ab and ba are the same except possibly for 0 and this played a role in Dirac's paper. Let me add this, and this will be my first entry into
this dictionary. The Spectrum can be regarded as the noncommutative analogue of the Range of a function. So let us make a small table:

Commutative .........................................................................Noncommutative
functions f: X \to C .................operators on Hilbert space; elements of an algebra
pointwise multiplication fg.....................................ab (composition)
range of a function................................spectrum of an operator

Let me be more specific and recall some basic facts. Let A be an algebra with a unit (for example A can be: the algebra of, say continuous, functions on a space; the algebra of all n by n matrices; or the algebra of operators on Hilbert space etc.) The spectrum of an element a \in A, denoted Spect (a) is the set of all numbers (scalars) \lambda such that a-\lambda 1 is not invertible.

Example 1. Let A= C(X) be the algebra of continuous functions on a compact space X, and let f \in A. Obviously Spect (f) ={ f(x); x \in X }= Range (f).
This is one reason we should think of spectrum as a noncommutative analogue of the range of a function i.e. the set of values attained by a function (a classical observable).

Example 2. For A=M_n (C) the algebra of n by n matrices the spectrum of a matrix is of course nothing but the set of its good old eigenvalues.

Classically, we have Range (fg) = Range (gf)
(because in fact fg=gf!). Now there is an almost perfect analogue of this fact in the noncommutative world. I said almost because there is a nuance: Spect (ab) and Spect (ba) are the same except for 0!

Here is a nice exercise in noncommutative algebra: show that in any unital algebra A, (ab-1) is invertible iff (ba-1) is invertible. From this of course follows that Spect(ab)\0 = Spect (ba)\0.
The fact that the two spectra are not exactly the same is in fact a good thing and the discrepancy is responsible for the index of Fredholm operators. Indeed assuming there is a b such that both (1-ba) and (1-ab) are trace class operators, it follows that a is Fredholm, and its index is given by index (a)= Tr (1-ba)- Tr (1-ab).
There are several other similar formulas for Fredholm index, e.g. the McKean-Singer formula, derived from the same principle.

CarlBrannen said...

Hmmm. If a or b is has an inverse then it's easy. For example,
if s(ab-1) = 1, and a has an inverse, then
(a^{-1}sa)(ba-1) = 1.

Funny thing is that I work in a non commutative geometry, Schwinger's measurement algebra. In that algebra, nothing much of interest is invertible.

Elements of the algebra correspond to Stern-Gerlach filters. Multiplication means putting the output of one filter into the input of another. Addition means combining two outputs. 0 is a beam stop, and 1 is a free (unimpeded) beam with all polarities present.

The quantum states (and the models of Stern-Gerlach filters) are the primitive idempotents, which of course are not invertible.

CarlBrannen said...

I wandered back, and this time read through the Weil letter. It reminds me that I should comment on the relationship between ideals / primitive idempotents and quanta, and what this means physically.

Quantum mechanics is usually written in a linear form where one looks for eigenstates |v), of operators O, with eigenvalues v:

O |v) = v |v).

The quanta show up in that the spectrum of the operator is quantized.

In the measurement algebra, the states are also operators. This can be accomplished from spinors by making density matrices,

\rho_v = |v)(v|,

but one does not need spinors (or matrix representations), one can work entirely in the operators.

In the measurement algebra, the states are then the primitive idempotents. The quanta show up because of the structure of the ideals.

The algebra one prefers to use for this are Clifford algebras associated with the manifold of spacetime, as used by Hestenes or the Cambridge geometry group.

A lot of QFT is about writing propagators. Physically, we define a particle as something that propagates without changing. This allows us to break propagators in two:

G(x,x") = \int G(x,x') G(x',x") dx'

The above has spacetime integration going on, but if we ignore all that (and the continuous degrees of freedom such as energy and momentum that are associated), we can focus just on the multiplication (and look at the discrete degrees of freedom).

Then the equation becomes
G = G^2
which is just the equation that defines G as an idempotent. That G is primitive follows from our desire to split things up as much as possible, and to avoid things like 0 or 1 as solutions.

The usual linear way of writing QM gives a natural tool for perturbation theory, but it is weak in non perturbational areas. The advantage of primitive idempotents is that they are inherently defined as nonlinear objects, but they still can be easily analyzed.

For example, suppose we have a bound state with three quantum objects, and the three quantum objects are represented (in free form) by primitive idempotents. We want to model the bound state by making a single primitive idempotent out of the individual primitive idempotents.

The natural way of doing this turns out to be placing the propagators in a 3x3 array M and requiring M^2 = M. For example, the (12) entry means a particle initially of type 2 has an interaction and finally becomes a particle of type 1. For this we ignore the gauge boson. It doesn't go far and gets absorbed.

The equation M^2 = M then becomes a requirement that the conglomeration retain its character as it propagates, that is, as it moves forward in time.

If we can represent the various propagators as complex numbers, the problem becomes one of solving for complete sets of primitive idempotents among the 3x3 complex matrices. There will be exactly three primitive idempotent solutions to this, and these will correspond to the three particle generations.

William said...

Invertibility of the operators isn't important - what is important is invertibility of (a - lambda) for some complex number lambda. Many operators of interest aren't invertible, but for a bounded operator there will always be some lambda for which (a - lambda) is invertible.

If you need a hint on the exercise, it's just an algebraic trick relating (1 - ab)^-1 to (1 - ba)^-1. Try writing out (1 - ab)^-1 as a power series.

Doug said...

Has anyone researched the type of geometry [NCG or CG] used by and / or during the evolution of chordate semicircular canals?

These bilateral 3-tori structures are nearly orthogonal, but can vary by 10-15 degrees.

They seem to function as angular accelerometers that might be more consistent with NGC than CG.
http://hyperphysics.phy-astr.gsu.edu/hbase/sound/eari.html#c3

Masoud said...

I think William's hint is very relevant here and is a good way of guessing the formula for the inverse of (1-ba) starting with the inverse of (1-ab). The formula for the index has a mild extension to
index (a)= Tr (1-ba)^k- Tr (1-ab)^k
where this time we only assume (1-ba)^k and (1-ab)^k are trace class
operators.....

Carl Brannen said...

The primary problem in finding an consistent preon model of the standard model is defining a bound state that one can calculate.

Infinite series are great for pertubation theory, but I'm not so sure that they are a good thing to study when what you really want is the non perturbational solutions (all of them) to a problem.

Dong Ta said...

This is the second time I look at this post and I find it particularly entertaining. I was discouraged at the first time very much due to the presentation, which uses latex code to replace actual mathematical symbols.

So why don't you have a look at this instruction and embed Latex to this blog?

Dong Ta said...

I forgot to put the link to the instruction page. Here it is: http://wolverinex02.googlepages.com/emoticonsforblogger2

masoud said...

Dear Dong Ta,
Thanks for your comment and the link! I will try to use it in my next post.