From: Robin Chapman <rjc@maths.ex.ac.uk>
Newsgroups: sci.math
Subject: Re: Non-zero vector fields on surface of spheres
Date: Mon, 09 Nov 1998 14:05:43 GMT

In article <724qqt$408$1@vixen.cso.uiuc.edu>,
  c-blair@staff.uiuc.edu (Charles Blair) wrote:
>    There is a theorem that any vector field tangent to the surface
> of an odd-dimensional sphere must be zero someplace.  This is given
> in Hirsch's book on differential topology, among other places.
>
>     I'm looking for a simple proof or even plausibility argument
> that is willing to assume without proof the Brouwer fixed-point
> theorem.  (i.e, the algebraic topology concepts get buried inside
> the Brouwer result)

I like Milnor's proof (from the Monthly c. 20 years ago).
It doesn't use Brouwer at all (in fact Milnor deduces Brouwer from it).

By some standard Weierstrass approximation handwaving we can assume
such a vector fields is smooth and has magnitude 1 everywhere. Consider
S = S^n as the unit sphere in R^{n+1}. The vector field is a map
f: S -> S with v orthogonal to f(v) everywhere. Extend f to all of
R^{n+1} by setting f(kv) - kf(v) for k >= 0. For each t > 0 define a map
G_t on R^{n+1} by G_t(v) = v + t f(v). Note that |G_t(v)| = sqrt(1+t) |v|.
For small enough t, G_t is one-to-one (look at its effect on S) and is an open
map away from the origin so is onto. On an annulus A = {v : a < |v| < b}
it muliplies the volume by (1+t)^{(n+1)/2}. But by computing the Jacobian
one sees that the volume of G_t(A) is a polynomial in t.  This is
only possible if n+1 is even.

Robin Chapman                           + "They did not have proper
SCHOOL OF MATHEMATICal Sciences         -  palms at home in Exeter."
University of Exeter, EX4 4QE, UK       +
rjc@maths.exeter.ac.uk                  -  Peter Carey,
http://www.maths.ex.ac.uk/~rjc/rjc.html +    Oscar and Lucinda, chapter 20

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From: hook@nas.nasa.gov (Edward C. Hook)
Newsgroups: sci.math
Subject: Re: Non-zero vector fields on surface of spheres
Date: 18 Nov 1998 23:17:22 GMT

In article <724qqt$408$1@vixen.cso.uiuc.edu>,
             c-blair@staff.uiuc.edu (Charles Blair) writes:
|>    There is a theorem that any vector field tangent to the surface
|> of an odd-dimensional sphere must be zero someplace.  This is given
|> in Hirsch's book on differential topology, among other places.
|> 

  Actually, the result is slightly different.

  Theorem. S^n admits a nowhere-zero tangent vector field iff n is odd.

  Using the usual picture, a field of tangent vectors to S^n is equivalent
  to a map f: S^n --> R^{n+1} such that <x,f(x)> = 0 for all x \in S^n
  (where <.,.> is the Euclidean inner product). If the tangent vectors
  never vanish, then you can assume that f: S^n --> S^n.

  The proof of sufficiency is easy. Basically, you think of S^{2k-1} as
  the unit sphere in C^k -- then multiplication by the scalar i gives
  a map of the desired sort. Or you can just write f down explicitly:

      f(x1, x2, ... , x{n-1}, xn) = (-x2, x1, ... , -xn, x{n-1}) .

  To see the necessity, suppose that you have f: S^n --> S^n of the
  appropriate sort. Then, for each x \in S^n, x and f(x) are linearly
  independent, so F(x,t) = (1-t)x + tf(x) (with t \in [0,1], x \in S^n)
  never vanishes. Normalizing, you obtain a homotopy from the identity
  map on S^n to the map f. Changing the definition just slightly, you
  can also produce a homotopy from f to the antipodal map, A: S^n --> S^n,
  with A(x) = -x. So, the existence of f implies that the identity map
  on S^n is homotopic to the antipodal map. At this point, the easiest 
  way to proceed is to note that homotopic maps induce the same
  homomorphism on (say) integral homology groups, the identity map
  induces the identity homomorphism and the antipodal map induces the
  homorphism that multiplies by (-1)^{n+1}. This will only work out right
  if (-1)^{n+1} = 1, i.e., if n is odd.

|>     I'm looking for a simple proof or even plausibility argument
|> that is willing to assume without proof the Brouwer fixed-point
|> theorem.  (i.e, the algebraic topology concepts get buried inside
|> the Brouwer result)

  I've spent some time thinking about this and have come to believe
  that what you're asking can't be done. For starters, the Brouwer 
  Fixed Point Theorem holds for _every_ disk -- there's no difference
  between what it says in odd dimensions and what it says in even 
  dimensions, so it doesn't look hopeful as a vehicle for proving a
  result that holds in the one case and _not_ in the other. Another way
  of looking at this is to note that the Brouwer result is a corollary
  of the fact that you can't retract the disk onto its boundary, which
  follows since id: S^n --> S^n is not null-homotopic. The only bit of
  algebraic topology that you need for that result is that (say)
  H_n(S^n;Z) = Z. But the result about vector fields really needs something
  stronger -- at the very least, you need to be able to compute the degree
  of a nice map f: S^n --> S^n so that you can tell various ones of them
  apart.

  In fact, you might want to redirect your efforts toward finding a
  plausible discussion of the concept of degree and what you can do
  with it ...

-- 
 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