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Principal Vectors -- Worawut Wisutmethangoon
sigmaCompact non Loc.Compact Topol. Group -- bergmann@marte.mat.uniroma1.it (Bergmann Wolfgang)
Re: Odering Combinations -- Frank Ruskey
Re: sigmaCompact non Loc.Compact Topol. Group -- vladimir.pestov@mcs.vuw.ac.nz (V Pestov)
Fourcoloring triangular graphs -- gnalle
Re: program for finitely generated abelian groups -- "Michael C. Sullivan"
Re: local divisibility by p -- Antoine.Chambert-Loir@ens.fr (Antoine Chambert-Loir)
Re: Principal Vectors -- rumynin@math.umass.edu (Dmitriy Rumynin)
Complexity of Subgroup Generation -- Luis Goddyn
Re: topology question -- rld@math.ohio-state.edu (Randall Dougherty)
Q about division algebras -- Paul Gunnells
Re: topology question -- dhartley2@aol.com (DHartley2)
This Week's Finds in Mathematical Physics (Week 104) -- baez@math.ucr.edu (John Baez)

Articles

Principal Vectors
Worawut Wisutmethangoon
Mon, 09 Jun 1997 05:57:51 -0500
Hello,
	I have a question hopefully any of you can help.
As you all know :
If we have a square matrix A, we can always find another square
matrix X such that
	   X(-1) * A * X = J
where J is the matrix with Jordan canonical form.  Column vectors
of X are called principal vectors of A.
	(If J is a diagonal matrix, then the diagonal members are
the eigenvalues and column vectors of X are eigenvectors.)
	It is also known that if A is real and symmetric matrix, 
then we can find X such that X is "orthogonal" and J is diagonal.
The question :
	Are there any less strict conditions of A so that we can 
guarantee X orthogonal, with J not necessarily a diagonal ?
	I would appreciate any answers and/or pointers to any
references.
Thanks,
Worawut W.
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sigmaCompact non Loc.Compact Topol. Group
bergmann@marte.mat.uniroma1.it (Bergmann Wolfgang)
9 Jun 1997 14:38:40 GMT
Question : 
Does anybody have an example of topological group G which is Hausdorff,
sigma-compact, complete (that is, every Cauchy net in its left (or right)
uniformity converges) and NOT locally compact?
For example, G=(X,+) where X is some real or complex infinite-dimensional
Banach space should not work, because here compact sets are of 1st category.
On the other hand, G=(X',+) with the weak topology sigma(X',X) is
sigma-compact (Banach-Alaoglu thm.), but it is not complete
G = Q = (rational numbers) is also sigma-compact, not locally compact, but
again not complete.
Thank you for any suggestion. 
		Wolfgang R. Bergmann 
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Re: Odering Combinations
Frank Ruskey
Mon, 09 Jun 1997 10:35:56 -0700
Truman Prevatt wrote:
> 
> Let N > M and consider the set S of all combinations of M ojects taken N
> at a time.  Represent an element of S as follows.  Assume M bins numbered
> 1 through M and N balls.  An element of S is an arrangement where N bits
> have balls and M-N are empty.   What I am looking for as an fast algorithm
> to uniquely encode the set of all such elements in S to the integers 1 to
> binomial coefficient of M choose N as a function of which bins contain
> balls.  Since M choose N in my case is quite large, a table lookup is
> prohibitive.
> 
> Does anyone know of such an algorithm.
> 
> Truman
> 
> 
Let your N elements be a[1],a[2],...,a[N], where a[1] < a[2] < ... <
a[N].
Compute 1+sum{j=1,2,...,N}C(a[j]-1,j), where C(n,k) is "n choose k".
The necessary binomial coefficients can be computed on the fly so that
the whole process takes O(n) arithmetic operations.
PS: The underlying ordering of subsets is known as "colex order".
-- 
Frank Ruskey                     e-mail: fruskey@csr.uvic.ca
Dept. of Computer Science        fax:    250-721-7292
University of Victoria           office: 250-721-7232
Victoria, B.C. V8W 3P6 CANADA    WWW: http://www.csc.uvic.ca/~fruskey
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Re: sigmaCompact non Loc.Compact Topol. Group
vladimir.pestov@mcs.vuw.ac.nz (V Pestov)
Tue, 10 Jun 1997 09:09:32 +1200
In article <5nh4hg$mtg$2@srv.caspur.it>, bergmann@marte.mat.uniroma1.it
(Bergmann Wolfgang) wrote:
> Question : 
> 
> Does anybody have an example of topological group G which is Hausdorff,
> sigma-compact, complete (that is, every Cauchy net in its left (or right)
> uniformity converges) and NOT locally compact?
Such topological groups exist in abundance. 
The easiest way to generate them is to use Markov's construction of free
topological groups. Recall that the free topological group, $F(X)$, on 
a compact topological space $X$ contains $X$ as a closed topological 
subspace in a universal way, that is, every continuous map, $f$, from 
$X$ to an arbitrary topological group $G$
extends in a unique fashion to a continuous homomorphism 
$\bar f\colon F(X)\to G$. The group $F(X)$ is algebraically free over $X$.
It is known to be Weil-complete (= complete in its left --- or right ---
uniformity) if $X$ is compact (Graev, 1948). 
At the same time, it is easy to prove that $F(X)$ is never locally 
compact -- unless $X$ (and $F(X)$) is discrete. Thus, the free 
topological group on any infinite compact space provides an example of 
the kind you wish.
There also exists an abelian version of this construction and, more
generally, the free topological group in any given variety of groups.
For references on the subject, you may wish to consult my recent survey
article: 
V. Pestov, Universal arrows to forgetful functors from categories
of topological algebra, Bull. Austral. Math. Soc. 48 (1993), 209--249.
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Fourcoloring triangular graphs
gnalle
Tue, 10 Jun 1997 01:07:34 -0700
Here goes a short proof that any triangular graph can be fourcolored.
I think it is easier to understand, than the proof presented in the
appendix. I have made this proof myself so I hope it is correct 
In the following I will try to prove that fourcoloring of triangular
graphs consequence of the theory of decomposible graphs 
(used in statics) For a good book read steffen Lauritzen: graphical
models.
In this book it is proven that any triangular graph G(V,E) is
decomposible. That is 
1	The graph is complete		or
2	There exists a decomposition (A,B,C), where
	G(A \cup C, E \cap (A \cup C)^2) and
        G(B \cup C, E \cap (B \cup C)^2) are decomposible
Given a graph G=(V,E) a decomposition is a tripple of disjoint sets
(A,B,C), so that
1	A \cup B \cup C = V
2	C is perfect (E=V\times V)
3	C separates A from B
In a planar triangular graph there will exist a decomposition (A,B,C)
as the graph is planar C can have only 4 vertices. We can now prove
the fourcoloring theorem for triangular graphs by induktion.
The two graphs
(A \cup C, E \cap (A \cup C)^2) and
(B \cup C, E \cap (B \cup C)^2) can be fourcolored and the colors can 
be chosen so that the vertices in C are colored in the same way. As
C separates A from B this can be used to fourcolor G(V,E)
Finally I will schetch the proof that any triangular graph is
decomposible:
If the graph is not perfect there will exist two vertices a, b
that are not adacent. Given two noadjacent vertices there will 
exist a minimal separating set, C. That is C separates a from b
but if c \in C then C\{c} does not separate a from b
Now let A be the set of vertices that C does not separate from a
and let B=V\{C\cup A}. To prove that (A,B,C) is a decomposition we only
have to prove that C\times C is a subset of E. As C is minimal we have
that for each C\in C We can find a path (a,..,c,..,b) containing only
one element in c. If we chose two vertices c1,c2 \in C this give us
a path a..c1..b..c2..a. Trwowing away redundant vertives we get a cycle
c1....c2....c1. As the graph is triangular this cycle has a corde, that
we can use to make the cycle shorter, and as C separates A from B this
gives us a new cycle c1...c2...c1. We end up with a cycle c1, a',c2,
b',c1. As the graph is triangular there must be a corde (c1,c2).
Repeating this argument we find C\times C in E, so (A,B,C) is a
decomposition.
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Re: program for finitely generated abelian groups
"Michael C. Sullivan"
Mon, 09 Jun 1997 11:25:38 -0500
Michael C. Sullivan wrote:
> 
> I am looking for a simple free program for
> calculating canonical forms of matrix presentations
> of finitely generated abelian groups. I.e. I have
> some nxn integer matrix A. The group Z^n / AZ^n
> has a standard from Z^m (+) Z_p1 (+) Z_p2 (+) ... Z_pk.
> 
> Please respond with e-mail. I know of something called
> Magma, but do have access to it.
> 
> Mike Sullivan
It turns out this is given by the Smith Normal Form 
which can be done with the Maple ismith command in the
linalg package.
Thanks much to all those who replied. 
Mike Sullivan
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Re: local divisibility by p
Antoine.Chambert-Loir@ens.fr (Antoine Chambert-Loir)
9 Jun 1997 16:43:28 GMT
In article <5na30i$85m@agate.berkeley.edu>,
Kevin M. Buzzard  wrote:
>Let X be a scheme. We have the structure sheaf, a sheaf of rings O_X on X.
>Let p be a prime. Let I denote the *presheaf* of ideals defined on open
>sets U by J(U)=pO_X(U). One might ask whether J is a sheaf.
I have the impression that the subscheme of A^2 over Z
defined by the equation
    p (x^2 - y^2) = 0
gives an example of a scheme where J is not a sheaf.
Indeed, they are two open subsets U and V defined respectively
by x and y invertible.
The functions y/x in U and x/y in V glue in U inter V once
multiplied by p, since
   p (y/x - x/y)=p(y^2-x^2)/xy = 0.
But there doesn't exist a polynomial f(x,y)
such that
   p y/x = f(x,y) mod p(x^2-y^2)
since it would give   p y - x f(x,y)    divisible by p(x^2-y^2)
which is forbidden by the linear term  p y.
--
Antoine Chambert-Loir
ENS-DMI, 45 rue d'Ulm, 75230 Paris Cedex 05
Tel : (33) 01 44 32 20 56   Fax : (33) 01 44 32 20 80
http://www.dmi.ens.fr/~chambert
-- 
Antoine
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Re: Principal Vectors
rumynin@math.umass.edu (Dmitriy Rumynin)
9 Jun 97 16:57:27 GMT
Worawut Wisutmethangoon (wisutmet@cae.wisc.edu) wrote:
: If we have a square matrix A, we can always find another square
: matrix X such that
: 	   X(-1) * A * X = J
: 	
: where J is the matrix with Jordan canonical form.  Column vectors
: of X are called principal vectors of A.
: 	(If J is a diagonal matrix, then the diagonal members are
: the eigenvalues and column vectors of X are eigenvectors.)
: 	It is also known that if A is real and symmetric matrix, 
: then we can find X such that X is "orthogonal" and J is diagonal.
: The question :
: 	Are there any less strict conditions of A so that we can 
: guarantee X orthogonal, with J not necessarily a diagonal ?
 It does not seem that there is an easy generalization.
You consider the adjoint action of the orthogonal group
O(n) on the space of matrices M_n. The orbits of this action
may be parameterized using continious and discrete data.
So, you have to figure out what data it is and which ones
correspond to the Jordan form. The problem looks quite classical
and it may have been done in Invariant theory. Ask 
an expert in invariant theory.
 If you look only at the orbits of O(n) in O(n) then the story
is more familiar to me. Conjugacy classes are known in the classical
groups and you may find them in Humphreys, "Conjugacy classes
in algebraic groups" or in the older book by Steinberg on
the same subject.
Dmitriy
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Complexity of Subgroup Generation
Luis Goddyn
Mon, 09 Jun 1997 10:47:09 -0700
Does anyone know anything about the computational complexity of the
following:
Instance:  The Caley table (multiplication table) of a group G
           of order n, and a subset S of G.
Ouput:     A listing of the subgroup  generated by S.
Is there a polytime algorithm? 
Any references would be appreciated.
Thank you in advance
---                                                      _____._____
Luis Goddyn                 email: goddyn@math.sfu.ca   /|   /|\   |\
Dept. of Math. and Stats.    http://www.sfu.ca/~goddyn /_|__/ | \__|_\
Simon Fraser University       tel: (604) 291-4699      | |__|_|_|__| | 
Burnaby BC  V5A 1S6           msg:          -3331      | /  | ^ |  \ | 
CANADA                        fax:          -4947      |/___|/ \|___\|
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Re: topology question
rld@math.ohio-state.edu (Randall Dougherty)
9 Jun 1997 02:33:56 GMT
In article <339830D8.26C6@cs.keele.ac.uk>,
John G. Stell  wrote:
>Let X be a topological space, and A an open set in X.
>An open set, B, is said to be "well within A" if the closure of
>B is a subset of A.
>Now use W(A) for the set of all open sets well within A.
>
>The space X is said to be regular, if for every open set A,
>A is the union of all the elements of W(A).
>This is one of the common usages of regular, some books 
>use T_3 for this, but there is no agreement on how these terms are used.
>
>In a regular space, it is true that for any opens A and B we have
>
>   If W(A) \subseteq W(B) then A \subseteq B        ------(*)
>
>My first question then is 
>
>     are there spaces which are not regular but which
>     do satisfy condition (*)?
Yes.  Consider the unit square [0,1]x[0,1] with its usual topology,
and then enlarge the topology so that the set
    X = {(x,y) : x > 0 or x = y = 0}
is open.  (In other words, a set is open in the new topology
iff it is of the form (X intersect A) union B, where A and B are
open in the old topology.)  The new topology is not regular, because no
set in W(X) contains the point (0,0).  But (*) holds: if we have
open sets A and B such that A is not a subset of B, then either
A contains a point other than (0,0) which is not in B, in which case
a small neighborhood of that point is in W(A) but not in W(B), or
A contains (0,0) and B doesn't, in which case, for small positive d,
the interior of the triangle with vertices (0,0), (d,d), and (d,2d)
is in W(A) but not in W(B).
>The second question involves a modification involving considering only
>the regular opens well within an open set. I'll go into more details
>if anyone thinks they may be able to help.
The above-mentioned triangular set is regular open; does that take
care of this unstated question?
Randall Dougherty                       rld@math.ohio-state.edu
Department of Mathematics,  Ohio State University,  Columbus, OH 43210  USA
"I have yet to see any problem, however complicated, that when looked at in the
right way didn't become still more complicated."  Poul Anderson, "Call Me Joe"
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Q about division algebras
Paul Gunnells
Sun, 08 Jun 1997 18:44:40 -0400
Hello everyone:
My question is about division algebras, and the reduced norm.  I don't
know if this is a silly question or not, as I have very little
expertise with these objects.  I wasn't able to find an answer in
standard places like Pierce and Weil, so I have a feeling the answer
is no.  But I thought I should at least ask.
Let D be a division algebra a number field K, and let L be a splitting
field for D.  That is, D \otimes L is isomorphic to a matrix algebra
M_n(L).  Then the reduced norm of an element d of D is defined to be
the determinant of the matrix corresponding to d.  This is an element
of K, and the reduced norm takes the role of the usual norm map for
finite field extensions.
Now, there is another description of the norm map in field theory
using conjugates.  Specifically, the norm of an algebraic number is
the product of its conjugates.  This is also true for the reduced norm
in the most famous noncommutative division algebra, the classical
quaternions.  My question is, is there a general technique to compute
the reduced norm using "conjugates"?
Better still, the norm map for K over Q can be computed using a
universal polynomial.  For K \otimes R is isomorphic to R^r \times C^s,
and there is a polynomial function f: R^r \times C^s -> R such that on
k \otimes 1, we get the absolute value of the norm of k.  Is there
such a set up for D \otimes R?
-- 
Paul Gunnells
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Re: topology question
dhartley2@aol.com (DHartley2)
Sun, 8 Jun 1997 14:02:52 -0400
In article <339830D8.26C6@cs.keele.ac.uk>, "John G. Stell"
 writes:
>Can anyone help with the following query (which arises in connection
>with some work on the theory of geographical information systems)?
>Let X be a topological space, and A an open set in X.
>An open set, B, is said to be "well within A" if the closure of
>B is a subset of A.
>Now use W(A) for the set of all open sets well within A.
>The space X is said to be regular, if for every open set A,
>A is the union of all the elements of W(A).
>This is one of the common usages of regular, some books 
>use T_3 for this, but there is no agreement on how these terms are used.
>In a regular space, it is true that for any opens A and B we have
>   If W(A) \subseteq W(B) then A \subseteq B        ------(*)
>My first question then is 
>     are there spaces which are not regular but which
>     do satisfy condition (*)?
>The second question involves a modification involving considering only
>the regular opens well within an open set. I'll go into more details
>if anyone thinks they may be able to help.
>------------------------------------------------------------------------
>Dr. John Stell
>Department of Computer Science         email     john@cs.keele.ac.uk
>Keele University
>Keele, Staffordshire,                  telephone +44 1782 584083
>ST5 5BG                                fax       +44 1782 713082
>U. K.
Here is a weaker result which may be of use.
Define UW(A) to be the union of W(A), and introduce the stronger condition
(**):
    For any open A, B in X
          If UW(A) \subseteq UW(B) then A \subseteq B.
Clearly regularity => (**) => (*), and the question is equivalent to does
(*) => regularity?  In a Hausdorff space, (**) does imply regularity
Proof.  Consider  any open B in X and a point x in B. Let A = B - {x}. If
y
is in UW(B), y!=x, then there's an open set such that y in V, cl V within
B.
By the Hausdorff condition, we can get V such that x is not in cl V. Hence
cl
V within A, and so y in UW(A). I.e. UW(B) - {x} \subseteq UW(A). Since B
is
not contained within A, by (**) we know UW(B) not \subseteq UW(A). Hence x
is
in UW(B). Thus B = UW(B) and X is regular.
Hope this helps,
David Hartley
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This Week's Finds in Mathematical Physics (Week 104)
baez@math.ucr.edu (John Baez)
8 Jun 1997 18:10:57 -0700
This Week's Finds in Mathematical Physics - Week 104
John Baez
A couple of weekends ago I flew up to Corvallis, Oregon to an AMS
meeting.  The AMS, in case you're unfamiliar with it, is the American
Mathematical Society.  They have lots of regional meetings with special
sessions on various topics.  One reason I went to this one is that there
was a special session on octonions, organized by Tevian Dray and Corinne
Manogue.  
After the real numbers come the complex numbers, and after the complex
numbers come the quaternions, and after the quaternions come the
octonions, the most mysterious of all.  The real numbers, complex
numbers, and quaternions have lots of applications to physics.  What
about the octonions?  Aren't they good for something too?  This
question has been bugging me for a while now.  
In fact, it bugs me so much that I decided to go to Corvallis to look
for clues.  After all, in addition to Tevian Dray and Corinne Manogue -
the former a mathematician, the latter a physicist, both deeply
interested in octonions - a bunch of other octonion experts were going
to be there.  One was my friend Geoffrey Dixon.  I told you about him in
"week53".  He wrote a book on the complex numbers, quaternions and
octonions and their role in physics.  He has a theory of physics in
which these are related to electromagnetism, the weak force, and the
strong force, respectively.  It's a bit far out, but far from crazy!
In fact, it's fascinating.
After writing about his book I got in touch with him in Cambridge,
Massachusetts.  I found out that his other main interest in life,
besides the octonions, is the game Myst.  This is probably not a
coincidence.  In both the main question is "What the heck is really
going on here?"  He has Myst all figured out, but he loves watching
people play it, so he got me to play it for a while.  Someday I will buy
a CD-ROM drive and waste a few weeks on that game.  Anyway, I got to
know him back in the summer of 1995, so it was nice to see him again in
Corvallis.
Another octonion expert is Tony Smith.  He too has a far-out but
fascinating theory of physics involving octonions!  I wrote about his
stuff in "week91".  I had never met him before the Corvallis conference,
but I instantly recognized him when I met him, because there's a picture
of him wearing a cowboy hat on his homepage.  It turns out he always
wears that hat.  He is a wonderful repository of information concerning
octonions and other interesting things.  He is also a very friendly and
laid-back sort of guy.  He lives in Atlanta, Georgia.  
I also met another octonion expert I hadn't known about, Tony Sudbery,
from York.  (The original York, not the "new" one.)  He gave a talk on
"The Exceptions that Prove the Rule".  The octonions are related to a
host of other mathematical structures in a very spooky way.  In all
sorts of contexts, you can classify algebraic structures and get a nice
systematic infinite list together with a finite number of exceptions.
What's spooky is how the exceptions in one context turn out to be
related to the exceptions in some other context.  These relationships
are complicated and mysterious in themselves.  It's as if there were a
hand underneath the water and all we see is the fingers poking out here
and there.  There seems to be some "unified theory of exceptions"
waiting to be discovered, and the octonions must have something to do
with it.  I figure that to really understand what the octonions are good
for, we need to understand this "unified theory of exceptions".
Let's start by recalling what the octonions are!
I presume you know the real numbers.  The complex numbers are things
like
a + bi
where a and b are real.  We can multiply them using the rule 
i^2 = -1
They may seem mysterious when you first meet them, but they lose their
mystery when you see they are just a nice way of keeping track of
rotations in the plane.
Similarly, the quaternions are guys like 
a + bi + cj + dk 
which we can multiply using the rules
i^2 = j^2 = k^2 = -1
and 
ij =  k,  jk =  i,  ki =  j
ji = -k,  kj = -i,  ik = -j
They aren't commutative, but they are still associative.  Again they may
seem mysterious at first, but they lose their mystery when you see that
they are just a nice way of keeping track of rotations in 3 and 4
dimensions.  Rotations in more than 2 dimensions don't commute in
general, so the quaternions had *better* not commute.  In fact, Hamilton
didn't invent the quaternions to study rotations --- his goal was merely
to cook up a "division algebra", where you could divide by any nonzero
element (see "week82").  However, after he discovered the quaternions,
he used them to study rotations and angular momentum.  Nowadays people
tend instead to use the vector cross product, which was invented later
by Gibbs.  The reason is that in the late 1800s there was a big battle
between the fans of quaternions and the fans of vectors, and the
quaternion crowd lost.  For more on the history of this stuff, see:
1) Michael J. Crowe, A History of Vector Analysis, University of Notre
Dame, Notre Dame, 1967.
Octonions were invented by Cayley later on in the 1800s.  For these,
we start with *seven* square roots of -1, say e1 up to e7.  To learn 
how multiply these, draw the following diagram:
                      e5
                  e2      e3
                      e4 
           e7         e1         e6
Draw a triangle with e5, e6, and e7 as vertices, draw a line from each
vertex to the midpoint of the opposite edge, and draw a circle
containing e1, e2, and e3.  Draw arrows on the edges of the triangle
going around clockwise, draw arrows on the circle also going around
clockwise, and draw arrows on the three lines pointing from each vertex
of the triangle to the midpoint of the opposite edge.  Come on, DO it!
I'm doing all this work for you... you should do some, too.
Okay.  Now you have your very own octonion multiplication table.  Notice
that there are six lines and a circle in your picture.  Each one of
these gives us a copy of the quaternions inside the octonions.   For
example, say you want to multiply e1 and e5.  You notice that the
the vertical line says "e5, e4, e1" on it as we follow the arrow down.
Thus, just as for i, j, and k in the quaternions, we have
e4 e1 =  e5,   e5 e4 =  e1,   e1 e5 =  e4
e1 e4 = -e5,   e4 e5 = -e1,   e5 e1 = -e4
So in particular we have e1 e5 = e4.  
In case you lose your octonion table, don't worry: you don't really need
to remember the *names* of those 7 square roots of -1 and their
positions on the chart.  You just need to remember the geometry of the
chart itself.  Names are arbitrary and don't really matter, unless
you're talking to someone else, in which case you have to agree on them.
If you want to see spiffy high-tech octonion multiplication tables,
check out the following websites:
2) Tony Smith, http://galaxy.cau.edu/tsmith/TShome.html 
3) Geoffrey Dixon, http://www.7stones.com (Warning: to really get into
this you need to have at least Netscape 3.0 with Java and Shockwave
stuff.)  
What's so great about the octonions?  They are not commutative, and
worse, they are not even *associative*.  What's great about them is that
they form a division algebra, meaning you can divide by any nonzero
octonion.  Better still, they form a "normed" division algebra.  Just as
with the reals, complexes, and quaternions, we can define the norm of
the octonion
x = a0 + a1 e1 + a2 e2 + a3 e3 + a4 e4 + a5 e5 + a6 e6 + a7 e7 
to be
|x| = sqrt(a0^2 + a1^2 + a2^2 + a3^2 + a4^2 + a5^2 + a6^2 + a7^2).
What makes them a "normed division algebra" is that
|xy| = |x||y|.
It's a wonderful fact about the world that the reals, complexes,
quaternions and octonions are the *only* normed division algebras.
That's it!
However, the octonions remain mysterious, at least to me.  They are
related to rotations in 7 and 8 dimensions, but not as simply as one
might hope.  After all, rotations in *any* number of dimensions are
still associative.  Where is this nonassociative business coming from?
I don't really know.  This question really bugs me.
A while ago, in "week93", I summarized a paper by John Schwarz on
supersymmetric Yang-Mills theory and why it works best in dimensions 3,
4, 6, and 10.  Basically, only in these dimensions can you cook up
spin-1/2 particles that have as many physical degrees of freedom as
massless spin-1 particles.  I sort of explained why.  This in turn
allows a symmetry between fermions and gauge bosons.  I didn't explain
how *this* works... it seems pretty tricky to me... but anyway, it works.
So far, so good.  But Schwarz wondered: is it a coincidence that the
numbers 3, 4, 6, and 10 are just two more than the numbers 1, 2, 4, and
8 - the dimensions of the reals, complexes, quaternions, and octonions?
Apparently not!  The following papers explain what's going on:
4) Corinne A. Manogue and Joerg Schray, Finite Lorentz transformations
automorphisms, and division algebras, Jour. Math. Phys. 34 (1993),
3746-3767. 
Corinne A. Manogue and Joerg Schray, Octonionic representations of
Clifford algebras and triality, preprint available as hep-th/9407179.
5) Anthony Sudbery, Division algebras, (pseudo)orthogonal groups and
spinors, Jour. Phys. A 17 (1984), 939-955.  
Anthony Sudbery, Seven types of incongruity, handwritten notes.
Here's the basic idea.  Let 
R = real numbers
C = complex numbers
H = quaternions                         
O = octonions
Let SO(n,1) denote the Lorentz group in n+1 dimensions.  Roughly
speaking, this is the symmetry group of (n+1)-dimensional Minkowski
spacetime.  Let so(n,1) be the corresponding Lie algebra (see "week63"
for a lightning introduction to Lie algebras).  Then it turns out that:
sl(2,R) = so(2,1)
sl(2,C) = so(3,1)
sl(2,H) = so(5,1)
sl(2,O) = so(9,1)
This relates reals, complexes, quaternions and octonions to the Lorentz
group in dimensions 3, 4, 6, and 10, and explains the "coincidence"
noted by Schwarz!  But it requires some explanation.  Roughly speaking,
if SL(2,K) is the group of 2x2 matrices with determinant 1 whose entries
lie in the division algebra K = R, C, H, O, then sl(2,K) is defined to
be the Lie algebra of this group.  This is simple enough for R or C.
However, one needs to be careful when defining the determinant of a 2x2
quaternionic matrix, since quaternions don't commute.  One needs to be
even more careful in the octonionic case.  Since octonions aren't even
associative, it's far from obvious what the group SL(2,O) would be, so
defining the Lie algebra "sl(2,O)" requires a certain amount of finesse.
For the details, read the papers.
As Corinne Manogue explained to me, this relation between the octonions
and Lorentz transformations in 10 dimensions suggests some interesting
ways to use octonions in 10-dimensional physics.  As we all know, the
10th dimension is where string theorists live.  There is also a nice
relation to Geoffrey Dixon's theory.  This theory relates the
electromagnetic force to the complex numbers, the weak force to the
quaternions, and the strong force to octonions.  How?  Well, the gauge
group of electromagnetism is U(1), the unit complex numbers.  The gauge
group of the weak force is SU(2), the unit quaternions.  The gauge group
of the strong force is SU(3)....
Alas, the group SU(3) is *not* the unit octonions.  The unit octonions
do not form a group since they aren't associative.  SU(3) is related to
the octonions more indirectly.  The group of symmetries (or technically,
"automorphisms") of the octonions is the exceptional group G2, which
contains SU(3).  To get SU(3), we can take the subgroup of G2 that
preserves a given unit imaginary octonion... say e1.   This is how Dixon
relates SU(3) to the octonions.
However, why should one unit imaginary octonion be different from the
rest?  Some sort of "symmetry breaking", presumably?  It seems a bit ad
hoc.  However, as Manogue explained, there is a nice way to kill two
birds with one stone.  If we pick a particular unit imaginary octonion,
we get a copy of the complex numbers sitting inside the octonions, so we
get a copy of sl(2,C) sitting inside sl(2,O), so we get a copy of
so(3,1) sitting inside so(9,1)!  In other words, we get a particular
copy of the good old 4-dimensional Lorentz group sitting inside the
10-dimensional Lorentz group.  So fixing a unit imaginary octonion not
only breaks the octonion symmetry group G2 down to the strong force
symmetry group SU(3), it might also get us from 10-dimensional physics
down to 4-dimensional physics.
Cool, no?  There are obviously a lot of major issues involved in turning
this into a full-fledged theory, and they might not work out.  The whole
idea could be completely misguided!  But it takes guts to do physics, so
it's good that Tevian Dray and Corinne Manogue are bravely pursuing this
idea.
Upon learning that there is a deep relation between R, C, H, O and the
Lorentz group in dimensions 3, 4, 6, 10, one is naturally emboldened to
take seriously a few more "coincidences".  For example, in "week82" I
described the Clifford algebras C_n --- i.e., the algebras generated by
n anticommuting square roots of -1.  These Clifford algebras are
relevant to n-dimensional *Euclidean* geometry, as opposed to the
Clifford algebras relevant to n-dimensional *Lorentzian* geometry, which
appeared in "week94".  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(8)
C_7  R(8) + R(8)
C_8  R(16)
where K(n) stands for n x n matrices with entries taken from K = R, C,
or H, and "+" stands for "direct sum".  Note that C_8 is the same as 16
x 16 matrices with entries taken from C_0.  That's part of a general
pattern called "Bott periodicity": in general, C_{n+8} is the same as 16
x 16 matrices with entries taken from C_n.
Now consider the dimension of the smallest real representation of C_n.
It's easy to work this out if you keep in mind that the smallest
representation of K(n) or K(n) + K(n) is on K^n - the vector space
consisting of n-tuples of elements of K.   We get
Dimension of smallest real representation
C_0  1
C_1  2
C_2  4
C_3  4
C_4  8
C_5  8
C_6  8
C_7  8
C_8  16
Note that it increases at n = 1, 2, 4, and 8.  These are the dimensions
of R, C, H, and O.  Coincidence?  
No!  Indeed, C_n has a representation on a k-dimensional real
vector space if and only if the unit sphere in that vector space,
S^{k-1}, admits n linearly independent smooth vector fields.  So the
above table implies that:
The sphere S^0 admits 0 linearly independent vector fields.  
The sphere S^1 admits 1 linearly independent vector fields.  
The sphere S^3 admits 3 linearly independent vector fields.  
The sphere S^7 admits 7 linearly independent vector fields.  
These spheres are the unit real numbers, the unit complex numbers, the
unit quaternions, and the unit octonions, respectively!  If you know
about normed division algebras, it's obvious that these sphere admit the
maximum possible number of linear independent vector fields: you can
just take a basis of vectors at one point and "left translate" it to get
a bunch of linearly independent vector fields.  
Now --- Bott periodicity has period 8, and the octonions have dimension
8.  And as we've seen, both have a lot to do with Clifford algebras.  So
maybe there is a deep relation between the octonions and Bott
periodicity.  Could this be true?  If so, it would be good news, because
while octonions are often seen as weird exceptional creatures, Bott
periodicity is bigtime, mainstream stuff!
And in fact it *is* true.  More on Bott periodicity and the octonions
coming up next Week.
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