From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: hairy ball (boule chevelue)
Date: 19 Jan 1999 18:18:57 GMT
Newsgroups: sci.math
Keywords: Number, construction of independent vector fields on an n-sphere
In article <36A4446B.B6C2494F@pop.hit.fi>,
Pertti Lounesto wrote:
>
>
>Ed Hook wrote:
>> I'm reasonably certain that one can find linearly independent
>> vector fields on far more spheres than just S^1, S^3 and S^7.
>
>Sure. What is the largest number of linearly independent
>vector fields on S^5? What about S^9? Is there a patterns?
>Adams answered that problem, too, if I remember.
J. F. Adams, "Vector Fields on Spheres", Annals of Math 75 (1962) 603-632.
The maximum number p of continuous, linearly-independent vector fields on
the (n-1)-sphere in R^n (with the usual topology) is
p = 2^c - 1 + 8 d
where n = (2a+1) 2^( c+4d ) and c= 0,1,2,3. I guess it's fair to say that
this many vector fields were known to exist by computations with "linear"
vector fields over R, C, H, and O as far back as Hurwitz; what Adams
did was rule out the possibility of having any more.
So on S^5 and S^9 (n=6 or 10, c=1, d=0) we have p=1.
This information from Atiyah's reviews of Adams's papers and other
related reviews in Steenrod's compilation of MR reviews in topology, 1967.
dave
==============================================================================
From: hook@nas.nasa.gov (Ed Hook)
Subject: Re: hairy ball (boule chevelue)
Date: 23 Jan 1999 00:42:36 GMT
Newsgroups: sci.math
In article <36A779F5.7EC1@hit.fi>,
Pertti Lounesto writes:
|> Edward C. Hook wrote:
|> > the existence of 2^c + 8d - 1 fields is a result about
|> > modules over Clifford algebras ...
|>
|> What result?
One way to get a bunch of linearly independent orthonormal vector
fields on S^{n-1} is to start with an "orthogonal multiplication"
m: R^k x R^n --> R^n, which is defined to be a bilinear map with
the property that |m(x,y)| = |x| |y| for all x \in R^k, y \in R^n
(where |.| denotes the appropriate Euclidean norm). If m is such
a map, then m(x,.): R^n --> R^n is an orthogonal map if x \in S^{k-1}
and m:(.,y): R^k --> R^n is an isometry if y \in S^{n-1}. It's always
possible to assume that m(e_k,x) = x where e_i denotes the i'th
standard basis vector -- these are called "normalized" orthogonal
multiplications. Then
Theorem: If there is an orthogonal multiplication m: R^k x R^n --> R^n,
then there are k-1 linearly independent orthonormal vector fields on
S^{n-1}.
To prove it, assume (WLOG) that m is normalized. Then u_i: S^{n-1} --> R^n
defined by u_i(x) = m(e_i,x) can be viewed as an orthonormal field of
vectors tangent to the sphere. And, for each x \in S^{n-1}, the vectors
u_1(x), u_2(x), ... , u_{k-1}(x) are linearly independent (because
m(.,x): R^k --> R^n is an isometry for x \in S^{n-1}).
It turns out that the set of vector fields that the above construction
produces from a given orthogonal multiplication necessarily have some
further properties. Namely, each u_i: R^n --> R^n is, in fact, an
element of the orthogonal group O(n), u_i^2 = -1 and u_i u_j + u_j u_i = 0
for i != j. Moreover, given such a set of u_i, one gets a corresponding
normalized orthogonal multiplication m: R^k x R^n --> R^n such that
m(e_i,x) = u_i(x) ( i < k), m(e_k,x) = x.
(Basically, a bilinear m as above is an orthogonal multiplication iff
m(x,.) is in O(n) whenever x \in S^{k-1}. If x = Sum x^i e_i, then
m(x,.) = Sum x^i u_i in Gl(n,R) lies in O(n) iff m(x,.)m(x,.)* = 1,
'*' denoting transpose. Computing,
m(x,.)m(x,.)* = (Sum x^i u_i)(Sum x^j u_j)* = (Sum x^i u_i)(Sum x^j u_j*)
= Sum (x^i)^2 u_i u_i* + Sum x^i x^j (u_i u_j* + u_j u_i*)
i i R^n
can be viewed as describing R^n as a module over the Clifford algebra C_{k-1}
of the quadratic form f(x,y) = -(x,y) on R^{k-1}. So, if there exist k-1
linearly independent orthonormal vector fields on S^{n-1} "due to Clifford
algebra considerations", then R^n is a C_{k-1}-module. In particular, if
you can manufacture k-1 vector fields on S^{n-1} in this way but _not_ k,
then R^n is a C_{k-1}-module but _not_ a C_k-module. At this point, my
memory is getting a bit shaky, but I believe that you then reason that
_odd_ factors in n can be ignored (since R^n will be a direct sum of
irreducible C_-modules, all of which are even-dimensional over
R ==> if n = 2^r*odd, R^n is a C_k-module iff R^{2^r} is). Letting b_k
be the least n such that R^n is a C_{k-1}-module, it turns out that the
sequence (b_k) can be explicitly calculated (mostly because it turns out
to be periodic with period 8, which is some sort of algebraic analogue
of Bott periodicity) -- in particular, each b_k is a power of 2 and the
sequence increases monotonically. So, if r(n) is defined to be that integer
such that the above Clifford module approach gives r(n)-1 vector fields
on S^{n-1} and no more, then it turns out that b_r(2^m) = 2^m. This allows
you to calculate the function r and the result is:
Theorem. If n = 2^c(n) 16^d(n) (odd) [0 <= c(n) <= 3], then
r(n) = 2^c(n) + 8d(n) .
--
Ed Hook | Copula eam, se non posit
MRJ Technology Solutions, Inc. | acceptera jocularum.
NAS, NASA Ames Research Center | I can barely speak for myself, much
Internet: hook@nas.nasa.gov | less for my employer