Show that a "hairy coconut" in an odd number of dimensions cannot be combed: if n is odd, there is no...

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Problem

Show that a "hairy coconut" in an odd number of dimensions cannot be
combed: if n is odd, there is no continuous map f from the unit sphere
S in R^n to itself with f(x) perpendicular to x for every x.
[ Editor: problem statement reconstructed from the posting. ]

Show solution

Gene Miller (gene@nynexst.com) wrote:
: I think there are no combed hairy coconuts for odd dimensions,
: but I can't come up with a good reason why.

Let n be an odd integer, and let S be the unit sphere in R^n.  Suppose
there is a continuous function f: S --> S such that <x, f(x)> = 0 for
all x in S.  That is, (S,f) is a "combed hairy coconut".

                (1-t) x + t f(x)
Let F(x,t) = ---------------------- , t in [0,1].
             || (1-t) x + t f(x) ||

Then F is a homotopy from the identity to f.  (Sorry for the jargon
here.  Think of F as a family of functions indexed by t.  When t=0 we
get the identity, and when t=1 we get f.)

Similarly,
               (1-t) f(x) + t (-x)
   G(x,t) = -------------------------
            || (1-t) f(x) + t (-x) ||

is a homotopy from f to the antipodal map (x --> -x).  Therefore, the
identity and antipodal maps are homotopic.

The identity map has degree 1.  The antipodal map can be represented in
matrix form as -I, where I is the nxn identity matrix.  The determinant
of -I is (-1)^n, which is -1 if n is odd.  Consequently, the degree of
the antipodal map is -1.

However, degree is a homotopy invariant, i.e. if g and h are homotopic
then degree(g) = degree(h) [Editor: "degree(f) = degree(g)" in the
original].  This is a contradiction.

-- David Radcliffe <radcliff@csd4.csd.uwm.edu>

[ Editor: the "dimension" here is that of R^n; the sphere S itself has
dimension n-1. ]