At 09:21 03/06/02 +0200, you wrote:
Dear Alejandro, unfortunately I cvould not find the discussion on Baez's
server. I scanned the last two or three month. May be, it occurred under a
diferent label. Could you please give me a more specific hint?
Manfred,
Actually there was problems with the thread, a message was completely lost. So
better I put here the relevant messages, then including also the email exchange.
=========================================================
Hello everyone. Given two manifolds X and Y, the latter being equipped with a metric g, and an embedding f: X > Y, one can pull the metric g back to X in the standard way. This in turn induces a measure on X. My questions are then: 1) How does this work within the framework of noncommutative geometry? 2) Does the construction generalise to discrete and noncommutative spaces? The embedding creates a homomorphism from the algebra of functions on Y to those on X, so that this would seem to be the basic datum in the noncommutative case, but what about the Hilbert spaces H, derivative operators D, and other pieces of the noncommutative geometry picture? In particular, I am interested in the case in which X is discrete and Y is a manifold with metric, and even more particularly in the measure on X induced by the pulled back metric/operator D, if it exists. Thanks very much. Ian Jermyn. Ian.Jermyn@sophia.inria.fr
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Hello Ian. You are in France, so you have a lot of experts around to ask, happy you! Exactly as you describe, the embedding X>Y creates the dual homomorphism F(Y)>F(X) in the function algebra. Which of course is the only datum we have in the noncommutative setting. The Dirac operator and the differentials [D,A] are defined on F(Y) so they will have a well defined action in F(X) through the homomorphism, will them? At first glance, my only doubt resides in the differential algebra; remember you have quotiented through an ideal to get the good representation of this algebra; I do not know if you need to recalculate again. About the metric g, it is not between us anymore. In some algebras one can use a set of coordinate functions X_i and then imagine the operator via g_ij [F,X_i][F,X_j], F = sign[D]. This is not very useful because the direct analysis of D is a powerful tool, even to get Riemaniann curvature, etc.
> The embedding creates a homomorphism from the algebra of functions on Y to
> those on X, so that this would seem to be the basic datum in the
> noncommutative case, but what about the Hilbert spaces H, derivative
> operators D, and other pieces of the noncommutative geometry picture? In
> particular, I am interested in the case in which X is discrete and Y is a
> manifold with metric, and even more particularly in the measure on X induced
> by the pulled back metric/operator D, if it exists.
Hmm, I am *very* interested on this. Can you detail, here or privately? Alejandro
==============================================================
Hi Alejandro. Thanks for your reply. I have inserted my comments below.
"alejandro.rivero" wrote in message
news:...
> "Ian Jermyn" wrote in message
news:...
>
> The Dirac
> operator and the differentials [D,A] are defined on F(Y) so they
> will have a well defined action in F(X) through the homomorphism,
> will them?
Well, the Dirac operator, F(Y) and [D, a], a in F(Y) are represented as operators on a
Hilbert space H(Y) that is part of the data for Y being a manifold with metric. What I do
not see how to do is this: in the general case the data one has are the algebra
homomorphism f*: F(Y) > F(X), the Hilbert space for Y, H(Y), a representation of F(Y) in
H(Y), p(Y), and the operator D(Y) in H(Y) (maybe some other operators to, like a grading
and a real structure, and all satisfying certain conditions). As far as I know, at this
stage one does not have a separate Hilbert space for X, and certainly not the other
operators. So one must construct from the data: a Hilbert space H(X), a representation
p(X) of F(X) in H(X), an operator D(X) and so on. While it is easy to write down the type
of relations that these quantities must satisfy (basically a bunch of commutative
squares), I do not see how to construct them in general.
In the commutative case, one has that H(Y) is the spinor bundle, and this can be pulled
back from Y to X using f, and I think then that one can construct the relevant
representation (although I am not sure). For the Dirac operator it is even less clear:
intuitively speaking the Dirac operator, being a derivative, takes into account two
"neighbouring" points of Y, and so it is hard to see that selecting (in the discrete case)
a finite subset of points can lead to a well defined operator. In any case, this
construction uses the underlying space to create the pullback bundle.
In the general case, one can imagine completing the following square with arrows from
F(X) > A and from L(H(Y)) to A, where L indicates the space of linear operators
f*
F(X) < F(Y)

 p(Y)

V
A L(H(Y))
using a pushout in the category of C* algebras (or something), if it exists, although
whether the bottom left hand corner can then be used to generate a Hilbert space
representation of A without further input I am not sure.
> At first glance, my only doubt resides in the differential
> algebra; remember you have quotiented through an ideal to get the good
> representation of this algebra; I do not know if you need to
> recalculate again.
I do not know this either, but until I have the basic data straight, I am not going to
think about it too hard!
> > The embedding creates a homomorphism from the algebra of functions on Y to
> > those on X, so that this would seem to be the basic datum in the
> > noncommutative case, but what about the Hilbert spaces H, derivative
> > operators D, and other pieces of the noncommutative geometry picture? In
> > particular, I am interested in the case in which X is discrete and Y is a
> > manifold with metric, and even more particularly in the measure on X induced
> > by the pulled back metric/operator D, if it exists.
>
> Hmm, I am *very* interested on this. Can you detail, here or privately?
I posted again to the group in reply to my first post :), but it has not yet appeared: I
include it below. It is about another puzzle I have in the discrete case. By the way, what
do you do, and what is your interest in these things?
 Begin post 
Excuse me for replying to my own post, but having thought about the issue a
little more, there is something else that puzzles me in the case of discrete
spaces, at least when they are finite.
Given a metric on a mdimensional manifold, one can define an mform/measure
by starting from any function and applying the Hodge star. The dual of the
unit function is distinguished however because it is covariantly constant in
the metric connection: "parallel transport preserves distances and volumes"
or something like that. It is this measure that arises in the
noncommutative definition of integration using the Dixmier trace.
In the case of a discrete space, unless I have misunderstood something, the
derivative operator D does not enter into the formula for integration
because the geometry is 0dimensional, and the Dixmier trace is replaced by
the ordinary trace in the representation of the function algebra. Since, at
least in the standard constructions I have seen, the representation for
finite discrete spaces is in a finitedimensional Hilbert space that is the
direct sum of subspaces (of perhaps differing dimensions) corresponding to
each point of the discrete space, this seems to imply that the measure of
each point in the discrete space must be a positive integer.
This puzzles me. I could understand if the only measure generated was the
uniform measure, since I could believe (although I have not tried to show)
that this might be "covariantly constant" in an appropriate sense, and
therefore be the natural correspondent to the measure in the manifold case.
The idea that any integral measure is allowed but not others is however
mysterious to me.
Thanks for listening. I would appreciate any comments, even if just to point
out errors.
Ian.
 End post 
Cheers Alejandro.
I look forward to hearing what you think.
Yours,
Ian.
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Hi again,
You are right, in the literature there is not actual examples of embeddings
and induced metrics; the closest thing is the proof of the index theorem,
as they need to cast Bott periodicity and that is done by mapping an
space into other and then getting an induced map in the Ktheory. Very
far from your requeriments. I keep thinking; I *should* to know how to
do it. Till tomorrow,
Alejandro
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Hi Alejandro. Thanks for your email. I work at INRIA in the field of image
processing/computer vision, although I used to be in physics  QFT/TQFT and so on. You are
right I have some good groups nearby  just no contact points.
Your ideas sound interesting, although they are far from my current research as you can
imagine. Regarding your comments about discrete spaces: I am always talking about Connes'
approach to discrete/noncommutative spaces, and my comments are directly based on the
example given on page 15 of the paper that you cite. There, the representation space is a
direct sum of a finite number of finite dimensional Hilbert spaces, with the algebra
represented diagonally. Whatever choice you make, the trace of the operator corresponding
to a function $a \in C^{N}$ will be of the form $\sum_{j} n_{j}a_{j}$ where the $n_{j}$
are integers. This corresponds to an integral measure as I said in my post (still yet to
appear!).
Looking forward to your post.
Yours,
Ian
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From: alejandro.rivero (rivero@sol.unizar.es)
Subject: Re: Noncommutative geometry and metric pullback
Newsgroups: sci.physics.research
View this article only
Date: 20020523 21:37:26 PST
"Ian Jermyn" wrote in message news:...
Well, really I am aswering the second post from Ian, which has not appeared
yet, waiting in moderation queue, I guess.
>In the case of a discrete space, unless I have misunderstood something, the
>derivative operator D does not enter into the formula for integration
>because the geometry is 0dimensional, and the Dixmier trace is replaced by
>the ordinary trace in the representation of the function algebra. Since, at
Just a point here; it is not that it is replaced... it is directly got
because a way to calculate the Dixmier trace is as limit of the Trace
of some heat kernel (of D^2); in the 0dim case this limit is still
the usual Trace.
>least in the standard constructions I have seen, the representation for
>finite discrete spaces is in a finitedimensional Hilbert space that is the
>direct sum of subspaces (of perhaps differing dimensions) corresponding to
>each point of the discrete space, this seems to imply that the measure of
>each point in the discrete space must be a positive integer.
>This puzzles me. I could understand if the only measure generated was the
>uniform measure, since I could believe (although I have not tried to show)
And it is; it it were not the uniform, it could not be a trace, as one
still needs Tr(B A B^1)= Tr(A).
The formalim with a discrete sum of subspaces is not the one from
measure theory, but the full theory of manifolds; there integration
should proced, I believe, by joining Cycles and Cocycles. On the
contrary, in the measuretheoretic setting there are not a "dirac
operator", nor dimensions, etc.
Said that, I have been thinking in the generic problem of discrete integrals
and yes, it seems a lot of work is still to be published.
Our main problem seems to be that the algebra A describe the value of
functions at some points, and the operator D describes the distances
between those points. So they are in principle unrelated; one can measure
A by just summing the points, or one can try to relate the discrete
points to the distances; this is the theory of numerical integration and
somehow Connes' theory should be fitted to it.
Some adhoc procedures have been tryed. Some authors use the one
dimensional formula for cero dimensional triples, and in this way the
Dirac operator contributes to the integral. I also have made calculations
with other workaround: Find a X such that [D,X]=1, then use it as a
Robinson infinitesimal and define A'=[D,A]/[D,X]; Now one can put other
measures by using this A', ie to calculate Tr(B A'); sometimes it
works.
At the end of the day we need to understand what is a change of
variables (change of chart) for a cerodimensional manifold and what
is a (local?) diffeomorfism for a cerodimensional manifold. The
classification of diffeomorphism should be too the classification
of integration methods, that is Brouder' clue IMO.
To say in in other way: while Tr(A Id) is naive "rectangle" integration,
Tr (A F) with F=diag( 14/45, 64/45, 24/45, 64/45, ...) will be
Bode integration, and so on... As far as I know, nothing has been
published on this.
Yours,
Alejandro
============================================================
To: "Ian Jermyn"
Subject: RE: Noncommutative geometry and metric pullback
Ian,
I have sent you another message to the newsgroup, but it is frozen
somewhere. Untill it appears, let me to point the main remarks:
1)In measure theory it seems that the only measure is the uniform,
as you take Tr(A), no Tr(\pi(A)). In manifold theory, you take a
representation \pi(A), but then one must try to follow the rules of
manifold integration (which are unclear in the discrete case).
2) Tr(A) is forced to us from the definition of Dixmier Trace:
a) because of the theorem giving as the Tr_w(D) as a limit of
heat kernel calculations of D^2.
b) because it is still a trace, thus Tr(B A B^{1}) = Tr(A) implies you
can only use the standard Trace, then the uniform measure.
3) The puzzle is that A give values at points, while D gives distances
between points. Giving a D, we should have a numerical integration method
over A, I think, but nobody teach us how to do it. Surely some approximation
to non commutative diffeomorfisms will do, as it is the way to get
Runge Kutta methods (Brouder teach us this!). But also unorthodox methods
can be essayed for concrete problems. For instance, find X such that
[D,X]=1 and then define A' as the Robinsonlike cocient [D,A]/[D,X]
Yours truly,
Alejandro
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================================================================
=================================================================
From then, I have not have more messages from Ian or the newsgroup. I
supposse now that a number analyst expects to find, given a lattice of points, a
set of integration weights defined via Vandermonde matrix. This does not
happen in NCG measure. On other hand, the set of weights could be
derived also from RungeKutta order conditions for a trivial integration, but
again the relevant theorems of NCG are not well defined for dimension zero.
Feel free to forward this thread to anyone who could be interested.
Yours
Alejandro