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How real are the "Virtual" partticles?
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Dave L. Renfro
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Joined: 29 Apr 2005
Posts: 570

PostPosted: Thu Mar 24, 2005 9:51 pm    Post subject: Re: fractional iteration of functions Reply with quote

qmagick@yahoo.com
[sci.math.research: January 28, 2005 14:00:05 +0000 (UTC)]
http://mathforum.org/epigone/sci.math.research/dralyoysli

wrote (in part):

Quote:
Wow, first, thanks for all the responses. I now have more
then enough references to investigate. I have gotten a good
response on this question. Second, I would like to respond
to Mr. Geisler's last comment about axiomatic basis for
function iteration. I think that will be the goal of the
paper I write. Well, at least an axiomatic basis for well
behaved functions over the complex plane.

Here's another reference that you might want to look at.
(I didn't see it among those suggested in this thread.)

Daniel S. Alexander, "A History of Complex Dynamics from
Schr÷der to Fatou and Julia", Aspects of Mathematics E 24,
Friedr. Vieweg & Sohn, Braunschweig, 1994.
[MR 95d:01014; Zbl 788.30001]
http://www.emis.de/cgi-bin/MATH-item?0788.30001

The Zbl review is especially long (the URL above takes you
to a publically available webpage). Two additional reviews
that I know of are:

Theodore W. Gamelin, Historia Mathematica 23 (1996), 74-84

Robert B. Burckel, SIAM Review 36(4) (Dec. 1994), 663-664.

Alexander's book is useful for its survey of early work
on what you're interested in. For example, Section 2.2
"Analytic Iteration", is preceded by this paragraph:

"Before reviewing the responses of Korkine and Farkas to
Schr÷der's study of functional equations it will be useful
to first say a few words about analytic iteration, and
then to briefly outline the respective approaches of
Schr÷der, Korkine and Farkas to this problem." (p. 24)


Dave L. Renfro
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korifey
science forum beginner


Joined: 24 Mar 2005
Posts: 1

PostPosted: Thu Mar 24, 2005 9:51 pm    Post subject: Re: Set Family on a Graph Reply with quote

Aldar Chan đ╔█┼ď:
Given a graph G=(V, E) where V and E are the sets of vertices
and edges respectively. Also given a set P which could be a set
of secret keys. I want to assign a subset of P to each vertex of
G in such a way that given any two vertices v1 and v2 (both in
V), the assigned subsets for them, denoted by p1 and p2
respectively, would have a non-empty intersection which is not
covered by the union of the assigned subsets of a bounded
number of any other vertices. What is the minimum size of P
needed for that and what is the maximum size of the subset
assigned to a node? Is there an efficient algorithm which can
do that? Any related literature appeared before?


Ϥ╠
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mdshafri
science forum beginner


Joined: 24 Mar 2005
Posts: 1

PostPosted: Thu Mar 24, 2005 9:51 pm    Post subject: Re: Hellinger Distance Reply with quote

How to set the range of d value to represent the measurement.
We could be set 0 for no changes between two discrete distributions C
and H. What happen if the value of C far much bigger than H.
Please anybody advice me !!

On 19 Apr 1997 13:46:35 -0500, Herman Rubin wrote:
Quote:
In article <335237AD.59E2@dcs.rhbnc.ac.uk>,
Peter Burge <peteb@dcs.rhbnc.ac.uk> wrote:
Please could someone give me a reference for a measure
known as the Hellinger Distance between two
discrete distributions C and H. In Latex,

d=\sum_{i=0}^{K} (\sqrt{C_{i}} - \sqrt{H_{i}})^{2}

This expression may be lacking additional terms.

It is not lacking any terms. For general measures, although
it is not likely to be of much use unless they are finite,
it is

d = \int (sqrt(dF) - sqrt(dG))^2.

That this is well-defined can be seen by using as a base
measure H = F + G. It is the supremum of the discrete
versions obtained by using finite partitions.

The real introduction of this into mathematics was by
Kakutani, who gives credit to Hellinger in a footnote.
--
This address is for information only. I do not claim that these
views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette
IN47907-1399
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX:
(765)494-0558
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Annales de Toulouse
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Joined: 13 Jun 2005
Posts: 3

PostPosted: Thu Mar 24, 2005 9:51 pm    Post subject: Annales de Toulouse, 1/2005 Reply with quote

The issue 1/2005 of the Annales de la faculte des sciences de Toulouse
has appeared.

Contents:

Michel Hickel
Sur quelques aspects de la geometrie de l'espace des arcs traces sur un
espace analytique

Laurent Bernis
Solutions stationnaires des equations de Vlasov-Poisson a symetries
cylindriques

Duc Tai Trinh
Coefficients de Stokes du modele cubique : point de vue de la
resurgence quantique

Mohammad Daher
Translations mesurables et ensembles de Rosenthal

Andrzej J. Maciejewski, Maria Przybylska
Differential Galois approach to the non-integrability of the heavy top
problem

The abstracts and some full texts can be downloaded at:
http://picard.ups-tlse.fr/~annales (french version)
http://picard.ups-tlse.fr/~annales/index_en.html (english version)
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Urs Schreiber
science forum Guru Wannabe


Joined: 04 May 2005
Posts: 127

PostPosted: Thu Mar 24, 2005 9:51 pm    Post subject: fibered categories vs. 2-sheaves Reply with quote

Ettore Aldrovandi <ealdrov@zeno.math.fsu.edu> wrote in message news:<20050202181810.GC24685@math.fsu.edu>...

Quote:
In article <ctbmgs$b8s$1@news.ks.uiuc.edu>,
Urs Schreiber <Urs.Schreiber@uni-essen.de> wrote:
"John Baez" <baez@math.removethis.ucr.andthis.edu> schrieb im Newsbeitrag
news:ctas57$1p2$1@news.ks.uiuc.edu...


2-bundles are meant to be an alternative to gerbes: although
I've done my best to hide it above, a gerbe is really more
like a categorified *sheaf* than a bundle. And, just as a
bundle has a sheaf of sections, we're hoping that a 2-bundle
has a stack of sections, which in certain cases will be a
gerbe. That's one of the things we need to figure out,
though.

First, by definition, a gerbe is a stack is a fibered category. The concept
"fibered category" is a categorification of the concept "presheaf". But not
its full categorification. I am wondering if the full categorification of
the concept "presheaf" has been studied before, and under which name.

More precisely, what I am talking about is this:

A presheaf over a topological space X is a morphism in Cat,
namely a (contravariant) functor from the category O(X) of open
subsets of X to Set.

Hi, you are maybe rigidifying the starting situation a bit too
much? A gerbe is a locally non-empty and locally connected stack
in groupoids. In particular, it is a fibered category. Now, this
does not make it a contravariant functor in Cat,


Yes. Please note that I did not claim that it is a functor in Cat!

I said that a *presheaf* is a contravariant functor and hence a
morphism in Cat. Then I said that a gerbe is almost but not quite the
*categorification* of this.

There is a principle "categorification by internalization" that
suggests that the categorification of a mathematical concept which is
a morphism in Cat should be a morphism in 2Cat, the 3-category of
2-categories (some flavor of it, at least).

Therefore I said that the categorification of "presheaf" should be
something like a 2-functor from the 2-category O(S) of open
sub-2-spaces to Cat, the 2-category of categories. This is what I
wanted to call a "2-presheaf".

There is an issue here with how to precisely define O(S), but in
general this definition does reproduce the freedom of having natural
transformations of the kind that you are referring to:


Quote:
if p: G --> X is a fibered category over X, say G is a gerbe, but
id doesn't matter here, if a, b, c are objects of X, then you
have the corresponding fiber categories G(a), G(b), G(c).If

i:b-->a

is a morphism in X, then there is a corresponding "restriction"
functor

i^*: G(a) --> G(b).

and if now j: c --> b is another morphism with the corresponding
functor

j^*: G(b) --> G(c)

then there is only a natural transformation

(ij)^* ==> j^*i^*

between the two resulting functors from G(a) to G(c).


Yes, the same holds true for the 2-sheaves that I tentatively talked
about. Here there are 2-morphisms in O(S), namely natural
transformations between "inclusion"-functors and these are taken to
2-morphisms in Cat, namely natural transformations between functors
between the G(a), G(b), G(c).

My point was that a fibered category is much like a 2-presheaf but
with the 2-morphisms in the source ignored. But thanks for your remark
about "lax functors". Maybe my question is clarified by noting how a
fibered category is to be thought of not as the categorification of a
presheaf, but as some sort of "laxification".

If you think I don't make sense please let me know!
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Herman Rubin
science forum Guru


Joined: 25 Mar 2005
Posts: 730

PostPosted: Thu Mar 24, 2005 9:51 pm    Post subject: Re: de Finetti's theorem Reply with quote

In article <4xk44kwf6yd0@legacy>, Ruitao Zhang <rzhang@jhsph.edu> wrote:
Quote:
de Finetti theorem said that if an infinite sequence is exchangeable
then there exist a unique probability measure such that the de finetti
hold. Is this probability measure the prior measure. In Bayesian we
choose start from any prior measure. Here it seems that the prior is
unique. I don't understand it.

The de Finetti theorem is a probability theorem, not a
statistical method. In probability, the measure is
unique. The mixing measure is the tail measure.

For example, if one has a Polya urn scheme with b black
balls and w white balls, the distribution of draws will
be the same as if the balls drawn were independent
Bernoulli trials with the probability of black drawn
from a Beta(b,w) distribution.

Quote:
For infinite population with exchangeability, we can use relative
frequency of an event to estimate the the probability of this event.
But, For the finite population, why we need exchangeability?

For finite populations, exchangeability does not even
give as much. The trials given the whole finite
population are not independent, as the "draws" are
without replacement.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
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W. Dale Hall
science forum Guru


Joined: 29 Apr 2005
Posts: 350

PostPosted: Thu Mar 24, 2005 9:51 pm    Post subject: Re: orientability of the universal bundle Reply with quote

philippe wrote:
Quote:
Does anyone know for which $n$ the manifold {(x,D) in Rn * G1(Rn)
such that x in D } is orientable AS A MANIFOLD (and not as a vector
bundle) (here, G1(Rn) is the set of all 1-subspaces of Rn, i.e. the
projective space of Rn) ???

P.S. For n=2, it is not orientable, because it is the moebius strip.


I'll use the notation RP^(n-1) for G1(Rn), and refer to it as "real
projective (n-1) space".

Your space, which I'll denote X, is orientable iff n is odd.

To prove this, note that the sphere bundle of the canonical line
bundle over RP^(n-1) has total space equal to the (n-1) dimensional
sphere, S^(n-1), and its projection is the canonical double cover
f: S^(n-1) --> RP^(n-1). Further, if this projection is used to
attach an n-cell D^n to RP^(n-1), yielding the complex:

D^n \cup_f RP^(n-1)

then the result is RP^n, alias real projective n-space. The normal
bundle of RP^(n-1) in RP^n is the canonical line bundle over RP^(n-1),
so the space X is homeomorphic to a tubular neighborhood of RP^(n-1)
in RP^n. I'll identify X with that tubular neighborhood.

If RP^n is orientable, then X (being an open submanifold) must also be,
and if X is orientable, then RP^n must be orientable, since RP^n \ X is
contractible (it's a disc), .

Finally, RP^n is easily shown to be orientable iff n is odd: RP^n is
the quotient of S^n by the antipodal map, and that map is orientation-
preserving (allowing one to produce an orientation on the quotient) iff
the map x |--> -x is orientation-preserving on R^(n+1).

Dale
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Ruitao Zhang
science forum beginner


Joined: 24 Mar 2005
Posts: 1

PostPosted: Thu Mar 24, 2005 9:51 pm    Post subject: Re: de Finetti's theorem Reply with quote

de Finetti theorem said that if an infinite sequence is exchangeable
then there exist a unique probability measure such that the de finetti
hold. Is this probability measure the prior measure. In Bayesian we
choose start from any prior measure. Here it seems that the prior is
unique. I don't understand it.
For infinite population with exchangeability, we can use relative
frequency of an event to estimate the the probability of this event.
But, For the finite population, why we need exchangeability?
Thanks very much for your help

Ruitao
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Ettore Aldrovandi
science forum beginner


Joined: 24 Mar 2005
Posts: 1

PostPosted: Thu Mar 24, 2005 9:51 pm    Post subject: a post that didn't appear Reply with quote

In article <ctbmgs$b8s$1@news.ks.uiuc.edu>,
Urs Schreiber <Urs.Schreiber@uni-essen.de> wrote:
Quote:
"John Baez" <baez@math.removethis.ucr.andthis.edu> schrieb im Newsbeitrag
news:ctas57$1p2$1@news.ks.uiuc.edu...


2-bundles are meant to be an alternative to gerbes: although
I've done my best to hide it above, a gerbe is really more
like a categorified *sheaf* than a bundle. And, just as a
bundle has a sheaf of sections, we're hoping that a 2-bundle
has a stack of sections, which in certain cases will be a
gerbe. That's one of the things we need to figure out,
though.

First, by definition, a gerbe is a stack is a fibered category. The concept
"fibered category" is a categorification of the concept "presheaf". But not
its full categorification. I am wondering if the full categorification of
the concept "presheaf" has been studied before, and under which name.

More precisely, what I am talking about is this:

A presheaf over a topological space X is a morphism in Cat,
namely a (contravariant) functor from the category O(X) of open
subsets of X to Set.

Hi, you are maybe rigidifying the starting situation a bit too
much? A gerbe is a locally non-empty and locally connected stack
in groupoids. In particular, it is a fibered category. Now, this
does not make it a contravariant functor in Cat, because the
restriction functors commute only up to natural
transformation. If memory serves, this is called a
"pseudo-functor" in SGA1. "Lax-functor" is also used, I think.

Note also that the "base" can be any site, I believe. Of course
so is your O(X) when you consider it as a category. At any rate,
what I'm saying is this:

if p: G --> X is a fibered category over X, say G is a gerbe, but
id doesn't matter here, if a, b, c are objects of X, then you
have the corresponding fiber categories G(a), G(b), G(c).If

i:b-->a

is a morphism in X, then there is a corresponding "restriction"
functor

i^*: G(a) --> G(b).

and if now j: c --> b is another morphism with the corresponding
functor

j^*: G(b) --> G(c)

then there is only a natural transformation

(ij)^* ==> j^*i^*

between the two resulting functors from G(a) to G(c).

--
Ettore Aldrovandi
Department of Mathematics http://www.math.fsu.edu/~ealdrov
Florida State University aldrovandi at math.fsu.edu
Tallahassee, FL 32306-4510, USA +1 (850) 644-9717 (FAX: 4053)
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Peter Spellucci
science forum Guru


Joined: 29 Apr 2005
Posts: 702

PostPosted: Thu Mar 24, 2005 9:51 pm    Post subject: Re: Defining an Infinitely Differentiable Function Reply with quote

In article <1107281374.556304@athnrd02>,
Ioannis <jgal@ath.forthnet.gr> writes:
Quote:
I am in possession of a sequence of functions {S(m,z):NxD->C}_{m \in N},
and their series valid in some domain D < C:

S(1,z)=sum(a_{1,i}*z^i,i=0..+inf),
S(2,z)=sum(a_{2,i}*z^i,i=0..+inf),
...
S(m,z)=sum(a_{m,i}*z^i,i=0..+inf),
...

I am interested in defining (in D), S(r,z), r \in R (resp S(w,z) w \in
C by analytic continuation), so that S(r,z) (resp S(w,z)) is C^{oo} with
respect to r (resp w).

The a_{m,n} obey a recursion:

a_{m,n}={1, if n=0,
1/n!, if m=1,
{sum(j*a_{m,n-j}*a_{m-1,j-1},j=1..n)}/n, otherwise}

This recursion doesn't seem to have a closed form. If it had, I could
use the analytic continuation (for fixed n_0) of its closed form, to
define S(w,z).

I tried using various functions (including polynomial and linear
interpolation) to interpolate between the coefficients {a_{m,n_0}_{m \in
N} for fixed n_0, but the resultant function's derivative is always
discontinuous at m \in N.

So that I don't waste any more time with this, my question is, is there
any guarantee that S(w,z) can be defined to be C^{oo} with respect to r
(or w), by interpolating between the coefficients (vertically) or should
I be looking at other methods?

Thanks much in advance,
--
I. N. G. --- http://users.forthnet.gr/ath/jgal/


only a vague idea: consider the mapping [1,infty]->[0,1]
z=1/(x+1)
for the variable "r" (i.e. the "m" in a_{m,n})
apply this to the "function" a_{m,n} with respect to the variable m, n
fixed.
define a sequence of finite fourierseries interpolating the first 1,...M
a{i,n}
for every n separately and then analyze the convergence of the function
sequence defined by this. then consider the convergence of the corresponding
S(f(m,n),z) series defined by this, f(r,n) being the fourierseries with
the variable "m", n fixed, (after backtransformation to [1,infty[)
and f(r,n) replacing the coefficient a_{m,n}
your recursion looks as if the a_{m,n} decay fast enough in order to be able to
show that the final series obtained this way is absolutely convergent.
hth
peter
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Robin Chapman
science forum Guru Wannabe


Joined: 25 Mar 2005
Posts: 254

PostPosted: Thu Mar 24, 2005 9:51 pm    Post subject: Re: Orthogonal latin squares of even side Reply with quote

Simone Severini wrote:

Quote:
Would you explain to me a more-or-less general (possibly simple!)
method to construct orthogonal latin squares of even side?

At

http://www-math.cudenver.edu/~wcherowi/courses/m6406/cslne.html

there is a short account by Bill Cherowitzo of a construction by Zhu Lie.

--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"Elegance is an algorithm"
Iain M. Banks, _The Algebraist_
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Axel Vogt
science forum addict


Joined: 03 May 2005
Posts: 93

PostPosted: Thu Mar 24, 2005 9:51 pm    Post subject: Re: The curse of dimensionality for integration Reply with quote

On Thu, 19 Feb 2004 06:38:52 +0000 (UTC), Greg Kuperberg wrote:
Quote:
In article <4033DA60.7060503@univie.ac.at>,
Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:
Numerically, integration is simpler than differentiation in one
dimension, but in higher dimension, integration suffers from the
curse
of dimensionality while differntiation doesn't. In particular, it is
very hard to get accurate integrals in dimensions >100, say.

Readers may be interested in my new paper in which I fight the
curse of dimensionality for numerical integration in high dimensions:

a
href="http://front.math.ucdavis.edu/math.NA/0402047">http://front.math.ucdavis.edu/math.NA/0402047</a

Granted, it's impossible to completely defeat the curse. I derive
good methods for functions that are well-approximated by low-degree
polynomials.

Is there an example available to look at for computing a cumulative
multivariate normal distribtuion?

--

use mail ńt axelvogt dot de
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John Ryskamp
science forum beginner


Joined: 24 Mar 2005
Posts: 1

PostPosted: Thu Mar 24, 2005 9:51 pm    Post subject: Re: Brouwer's characterization of Cantor set Reply with quote

The most important response of Brouwer to Cantor was Brouwer's
formulation of an infinite ordinal number. He described this in his
1912 lecture. However, it is based on that idea that Cantor proved
the well-ordering of the ordinal numbers. As Garciadiego has shown,
not only did Cantor not do so, but also, he never claimed to have done
so, and never used the term infinite ordinal number. The term
infinite ordinal number has no meaning and you should examine your
interests in light of what Garciadiego has to say about the math
history of Brouwer's era.

On Thu, 09 Dec 2004 18:09:40 +0000,
=?ISO-8859-1?Q?Jos=E9_Carlos_Santos?= wrote:
Quote:
On 08-12-2004 12:46, Jorge Buescu wrote:

Brouwer has given an equivalent characterization of the Cantor set
as
a perfect, totally disconnected, compact Hausdorff space with a
countable
base of clopen subsets.

However, I can't seem to find a suitable reference for this fact
(MathSciNet does not extend that far). But this is probably on
some
appropriate General Topology books. Can anyone point one out?

There's a proof in Willi Rinow's Lehrbuch der Topologie. It's the
last
theorem of section 24.

Best regards,

Jose Carlos Santos
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G A Edgar
science forum beginner


Joined: 24 Mar 2005
Posts: 1

PostPosted: Thu Mar 24, 2005 9:51 pm    Post subject: Re: Brouwer's characterization of Cantor set Reply with quote

In article <cvkodf$vq5$1@news.ks.uiuc.edu>, John Ryskamp
<philneo2001@yahoo.com> wrote:

Quote:
The most important response of Brouwer to Cantor was Brouwer's
formulation of an infinite ordinal number. He described this in his
1912 lecture. However, it is based on that idea that Cantor proved
the well-ordering of the ordinal numbers. As Garciadiego has shown,
not only did Cantor not do so, but also, he never claimed to have
done
so, and never used the term infinite ordinal number.

???
"Ueber unendliche, lineare Punktmannigfaltigkeiten, 5"
Math. Annalen 21 (1883) 545-586

available on-line:

http://www-gdz.sub.uni-goettingen.de/cgi-bin/digbib.cgi?PPN235181684_0021


I guess you could say Cantor did not use the term "infinite ordinal
number"
because, writing in German, he used "unendlichen Zahlen" (see page
547)...
But somehow I do not think that is what Garciadiego had in mind.
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John Baez
science forum Guru Wannabe


Joined: 01 May 2005
Posts: 220

PostPosted: Thu Mar 24, 2005 9:51 pm    Post subject: Re: This Week's Finds in Mathematical Physics (Week 211) Reply with quote

Also available at http://math.ucr.edu/home/baez/week211.html

March 6, 2005
This Week's Finds in Mathematical Physics - Week 211
John Baez

The last time I wrote an issue of this column, the Huyghens probe was
bringing back cool photos of Titan. Now the European "Mars Express"
probe is bringing back cool photos of Mars!

1) Mars Express website, http://www.esa.int/SPECIALS/Mars_Express/index.html

There are some tantalizing pictures of what might be a "frozen sea" -
water ice covered with dust - near the equator in the Elysium Planitia region:

2) Mars Express sees signs of a "frozen sea",
http://www.esa.int/SPECIALS/Mars_Express/SEMCHPYEM4E_0.html

Ice has already been found at the Martian poles - it's easily visible there,
and Mars Express is getting some amazing closeups of it now - here's a
here's a view of some ice on sand at the north pole:

3) Glacial, volcanic and fluvial activity on Mars: latest images,
http://www.esa.int/SPECIALS/Mars_Express/SEMLF6D3M5E_1.html

What's new is the possibility of large amounts of water in warmer parts of
the planet.

Now for some math. It's always great when two subjects you're interested in
turn out to be bits of the same big picture. That's why I've been really
excited lately about Bott periodicity and the "super-Brauer group".

I wrote about Bott periodicity in "week105", and about the Brauer group
in "week209", but I should remind you about them before putting them together.

Bott periodicity is all about how math and physics in n+8-dimensional space
resemble math and physics in n-dimensional space. It's a weird and wonderful
pattern that you'd never guess without doing some calculations. It shows up
in many guises, which turn out to all be related. The simplest one to verify
is the pattern of Clifford algebras.

You're probably used to the complex numbers, where you throw in just *one*
square root of -1, called i. And maybe you've heard of the quaternions, where
you throw in *two* square roots of -1, called i and j, and demand that they
anticommute:

ij = -ji

This implies that k = ij is another square root of -1. Try it and see!

In the late 1800s, Clifford realized there's no need to stop here. He
invented what we now call the "Clifford algebras" by starting with the
real numbers and throwing in n square roots of -1, all of which anticommute
with each other. The result is closely related to rotations in n+1
dimensions, as I explained in "week82".

I'm not sure who first worked out all the Clifford algebras - perhaps it was
Cartan - but the interesting fact is that they follow a periodic pattern.
If we use C_n to stand for the Clifford algebra generated by n anticommuting
square roots of -1, they go like this:

C_0 R
C_1 C
C_2 H
C_3 H + H
C_4 H(2)
C_5 C(4)
C_6 R(Cool
C_7 R(Cool + R(8)

where

R(n) means n x n real matrices,
C(n) means n x n complex matrices, and
H(n) means n x n quaternionic matrices.

All these become algebras with the usual addition and multiplication of
matrices. Finally, if A is an algebra, A + A consists of pairs of guys
in A, with pairwise addition and multiplication.

What happens next? Well, from then on things sort of "repeat" with period 8:
C_{n+8} consists of 16 x 16 matrices whose entries lie in C_n!

So, you can remember all the Clifford algebras with the help of this
eight-hour clock:


0

R

7 1

R+R C





6 R H 2





C H+H

5 3

H

4


To use this clock, you have to remember to use matrices of the right size to
get C_n to have dimension 2^n. So, when I write "R + R" next to the "7" on
the clock, I don't mean C_7 is really R + R. To get C_7, you have to take
R + R and beef it up until it becomes an algebra of dimension 2^7 = 128. You
do this by taking R(Cool + R(Cool, since this has dimension 8 x 8 + 8 x 8 = 128.

Similarly, to get C_{10}, you note that 10 is 2 modulo 8, so you look at
"2" on the clock and see "H" next to it, meaning the quaternions. But to get
C_{10}, you have to take H and beef it up until it becomes an algebra of
dimension 2^{10} = 1024. You do this by taking H(16), since this has
dimension 4 x 16 x 16 = 1024.

This "beefing up" process is actually quite interesting. For any associative
algebra A, the algebra A(n) consisting of n x n matrices with entries in A
is a lot like A itself. The reason is that they have equivalent categories
of representations!

To see what I mean by this, remember that a "representation" of an algebra
is a way for its elements to act as linear transformations of some vector
space. For example, R(n) acts as linear transformations of R^n by matrix
multiplication, so we say R(n) has a representation on R^n. More generally,
for any algebra A, the algebra A(n) has a representation on A^n.

More generally still, if we have any representation of A on a vector space V,
we get a representation of A(n) on V^n. It's less obvious, but true, that
*every* representation of A(n) comes from a representation of A this way.

In short, just as n x n matrices with entries form an algebra A(n) that's a
beefed-up version of A itself, every representation of A(n) is a beefed-up
version of some representation of A.

Even better, the same sort of thing is true for maps between representations
of A(n). This is what we mean by saying that A(n) and A have equivalent
categories of representations. If you just look at the categories of
representations of these two algebras as abstract categories, there's no
way to tell them apart! We say two algebras are "Morita equivalent" when
this happens.

It's fun to study Morita equivalence classes of algebras - say algebras over
the real numbers, for example. The tensor product of algebras gives us a way
to multiply these classes. If we just consider the invertible classes, we get
a *group*. This is called the "Brauer group" of the real numbers.

The Brauer group of the real numbers is just Z/2, consisting of the classes
[R] and [H]. These correspond to the top and bottom of the Clifford clock!
Part of the reason is that

H tensor H = R(4)

so when we take Morita equivalence classes we get

[H] x [H] = [R]

But, you may wonder where the complex numbers went! Alas, the Morita
equivalence class [C] isn't invertible, so it doesn't live in the Brauer
group. In fact, we have this little multiplication table for tensor prod
algebras:


tensor R C H
----------------------
R | R C H
|
C | C C+C C(2)
|
H | H C(2) R(4)


Anyone with an algebraic bone in their body should spend an afternoon
figuring out how this works! But I won't explain it now.

Instead, I'll just note that the complex numbers are very aggressive and
infectious - tensor anything with a C in it and you get more C's. That's
because they're a field in their own right - and that's why they don't
live in the Brauer group of the real numbers.

They do, however, live in the *super-Brauer* group of the real numbers,
which is Z/8 - the Clifford clock itself!

But before I explain that, I want to show you what the categories of
representations of the Clifford algebras look like:

0

real vector spaces

7 1
split real vector spaces complex vector spaces




6 real vector spaces quaternionic vector spaces 2




complex vector spaces split quaternionic vector spaces
5 3


quaternionic vector spaces

4


You can read this information off the 8-hour Clifford clock I showed you
before, at least if you know some stuff:

A real vector space is just something like R^n
A complex vector space is just something like C^n
A quaternionic vector space is just something like H^n

and a "split" vector space is a vector space that's been written as the direct
sum of two subspaces.

Take C_4, for example - the Clifford algebra generated by 4 anticommuting
square roots of -1. The Clifford clock tells us this is H + H. And if you
think about it, a representation of this is just a pair of representations of
H. So, it's two quaternionic vector spaces - or if you prefer, a "split"
quaternionic vector space.

Or take C_7. The Clifford clock says this is R + R... or at least Morita
equivalent to R + R: it's actually R(Cool + R(Cool, but that's just a beefed-up
version of R + R, with an equivalent category of representations. So, the
category of representations of C_7 is *equivalent* to the category of split
real vector spaces.

And so on. Note that when we loop all the way around the clock, our
Clifford algebra becomes 16 x 16 matrices of what it was before, but this
is Morita equivalent to what it was. So, we have a truly period-8 clock
of categories!

But here's the really cool part: there are also arrows going clockwise and
counterclockwise around this clock! Arrows between categories are called
"functors".

Each Clifford algebra is contained in the next one, since they're built
by throwing in more and more square roots of -1. So, if we have a
representation of C_n, it gives us a representation of C_{n-1}. Ditto
for maps between representations. So, we get a functor from the category
of representations of C_n to the category of representations of C_{n-1}.
This is called a "forgetful functor", since it "forgets" that we have
representations of C_n and just thinks of them as representations of C_{n-1}.

So, we have forgetful functors cycling around counterclockwise!

Even better, all these forgetful functors have "left adjoints" going
back the other way. I talked about left adjoints in "week77",
so I won't say much about them now. I'll just give an example.

Here's a forgetful functor:

forget complex structure
complex vector spaces ---------------------------> real vector spaces

which is one of the counterclockwise arrows on the Clifford clock.
This functor takes a complex vector space and forgets your ability to multiply
vectors by i, thus getting a real vector space. When you do this to C^n,
you get R^{2n}.

This functor has a left adjoint:

complexify
complex vector spaces <-------------------------- real vector spaces

where you take a real vector space and "complexify" it by tensoring it with
the complex numbers. When you do this to R^n, you get C^n.

So, we get a beautiful version of the Clifford clock with forgetful functors
cycling around counterclockwise and their left adjoints cycling around
clockwise! When I realized this, I drew a big picture of it in my math
notebook - I always carry around a notebook for precisely this sort of thing.
Unfortunately, it's a bit hard to draw this chart in ASCII, so I won't
include it here.

Instead, I'll draw something easier. For this, note the following mystical
fact. The Clifford clock is symmetrical under reflection around the
3-o'clock/7-o'clock axis:


0

real vector spaces

7 1

split real vector spaces complex vector spaces
\
\
\
\
6 real vector spaces \ quaternionic vector spaces 2
\
\
\
\
complex vector spaces split quaternionic vector spaces
5 3


quaternionic vector spaces

4

It seems bizarre at first that it's symmetrical along *this* axis instead
of the more obvious 0-o'clock/4-o'clock axis. But there's a good reason,
which I already mentioned: the Clifford algebra C_n is related to rotations in
n+1 dimensions.

I would be very happy if you had enough patience to listen to a full
explanation of this fact, along with everything else I want to say. But
I bet you don't... so I'll hasten on to the really cool stuff.

First of all, using this symmetry we can fold the Clifford clock in half...
and the forgetful functors on one side perfectly match their left adjoints
on the other side!

So, we can save space by drawing this "folded" Clifford clock:


split real vector spaces

| ^
forget splitting | | double
V |

real vector spaces

| ^
complexify | | forget complex structure
v |

complex vector spaces

| ^
quaternionify | | forget quaternionic structure
v |

quaternionic vector spaces

| ^
double | | forget splitting
v |

split quaternionic vector spaces


The forgetful functors march downwards on the right, and their
left adjoints march back up on the left!

The arrows going between 7 o'clock and 0 o'clock look a bit weird:


split real vector spaces

| ^
forget splitting | | double
V |

real vector spaces


Why is "forget splitting" on the left, where the left adjoints belong, when
it's obviously an example of a forgetful functor?

One answer is that this is just how it works. Another answer is that it
happens when we wrap all the way around the clock - it's like how going from
midnight to 1 am counts as going forwards in time even though the number is
getting smaller. A third answer is that the whole situation is so symmetrical
that the functors I've been calling "left adjoints" are also "right adjoints"
of their partners! So, we can change our mind about which one is
"forgetful", without getting in trouble.

But enough of that: I really want to explain how this stuff is related
to the super-Brauer group, and then tie it all in to the *topology* of Bott
periodicity. We'll see how far I get before giving up in exhaustion....

What's a super-Brauer group? It's just like a Brauer group, but where we
use superalgebras instead of algebras! A "superalgebra" is just physics
jargon for a Z/2-graded algebra - that is, an algebra A that's a direct
sum of an "even" or "bosonic" part A_0 and an "odd" or "fermionic" part A_1:

A = A_0 + A_1

such that multiplying a guy in A_i and a guy in A_j gives a guy in A_{i+j},
where we add the subscripts mod 2.

The tensor product of superalgebras is defined differently than for algebras.
If A and B are ordinary algebras, when we form their tensor product, we
decree that everybody in A commutes with everyone in B. For superalgebras
we decree that everybody in A "supercommutes" with everyone in B - meaning
that

ab = ba

if either a or b are even (bosonic) while

ab = -ba

if a and b are both odd (fermionic).

Apart from these modifications, the super-Brauer group works almost like the
Brauer group. We start with superalgebras over our favorite field - here
let's use the real numbers. We say two superalgebras are "Morita equivalent"
if they have equivalent categories of representations. We can multiply
these Morita equivalence classes by taking tensor products, and if we just
keep the invertible classes we get a group: the super-Brauer group.

As I've hinted already, the super-Brauer group of the real numbers is Z/8 -
just the Clifford algebra clock in disguise!

Here's why:

The Clifford algebras all become superalgebras if we decree that all the
square roots of -1 that we throw in are "odd" elements. And if we do this,
we get something great:

C_n tensor C_m = C_{n + m}

The point is that all the square roots of -1 we threw in to get C_n
*anticommute* with those we threw in to get C_m.

Taking Morita equivalence classes, this mean

[C_n] [C_m] = [C_{n+m}]

but we already know that

[C_{n+8}] = [C_n]

so we get the group Z/8. It's not obvious that this is *all* the super-Brauer
group, but it actually is - that's the hard part.

Now let's think about what we've got. We've got the super-Brauer group,
Z/8, which looks like an 8-hour clock. But before that, we had the categories
of representations of Clifford algebras, which formed an 8-hour clock with
functors cycling around in both directions.

In fact these are two sides of the same coin - or clock, actually. The
super-Brauer group consists of Morita equivalence classes of Clifford
algebras, where Morita equivalence means "having equivalent categories
of representations". But, our previous clock just shows their categories
of representations!

This suggests that the functors cycling around in both directions are secretly
an aspect of the super-Brauer group. And indeed they are! The functors going
clockwise are just "tensoring with C_1", since you can tensor a representation
of C_n with C_1 and get a representation of C_{n+1}. And the functors going
counterclockwise are "tensoring with C_{-1}"... or C_7 if you insist, since
C_{-1} doesn't strictly make sense, but 7 equals -1 mod 8, so it does the
same job.

Hmm, I think I'm tired out. I didn't even get to the topology yet! Maybe
that'll be good as a separate little story someday. If you can't wait,
just read this:

4) John Milnor, Morse Theory, Princeton U. Press, Princeton, New Jersey, 1963.

You'll see here that a representation of C_n is just the same as a vector
space with n different anticommuting ways to "rotate vector by 90 degrees",
and that this is the same as a real inner product space equipped with a map
from the n-sphere into its rotation group, with the property that the north
pole of the n-sphere gets mapped to the identity, and each great circle
through the north pole gives some action of the circle as rotations. Using
this, and stuff about Clifford algebras, and some Morse theory, Milnor gives a
beautiful proof that

Omega^8(SO(infinity)) ~ SO(infinity)

or in English: the 8-fold loop space of the infinite-dimensional rotation
group is homotopy equivalent to the infinite-dimensional rotation group!

The thing I really like, though, is that Milnor relates the forgetful functors
I was talking about to the process of "looping" the rotation group. That's
what these maps from spheres into the rotation group are all about... but I
want to really explain it all someday!

I learned about the super-Brauer group here:

5) V. S. Varadarajan, Supersymmetry for Mathematicians: An Introduction,
American Mathematical Society, Providence, Rhode Island, 2004.

though the material here on this topic is actually a summary of some
lectures by Deligne in another book I own:

6) P. Deligne, P. Etingof, D.S. Freed, L. Jeffrey, D. Kazhdan, J. Morgan,
D.R. Morrison and E. Witten, Quantum Fields and Strings: A Course For
Mathematicians 2 vols., American Mathematical Society, Providence, 1999.
Notes also available at http://www.math.ias.edu/QFT/

Varadarajan's book doesn't go as far, but it's much easier to read, so I
recommend it as a way to get started on "super" stuff.

-----------------------------------------------------------------------
Previous issues of "This Week's Finds" and other expository articles on
mathematics and physics, as well as some of my research papers, can be
obtained at

http://math.ucr.edu/home/baez/

For a table of contents of all the issues of This Week's Finds, try

http://math.ucr.edu/home/baez/twf.html

A simple jumping-off point to the old issues is available at

http://math.ucr.edu/home/baez/twfshort.html

If you just want the latest issue, go to

http://math.ucr.edu/home/baez/this.week.html
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