## A Proof

#### John Baez

#### November 26, 2005

A lot of people have asked me about the proof of this statement:
Since K is a division algebra, we can find a [linear operator]
mapping any point on the unit sphere to any other point. The only way the
unit sphere in K can have this much symmetry is if the norm on K comes
from an inner product.

So, here's one way to prove it.
Suppose |·| is a norm on R^{n}, let G be the
group of linear transformations of R^{n} that preserves
this norm, let S be the unit sphere for this norm, and suppose
elements of G can be found that map any given point on S to any
other given point. We wish to show that |·| comes from
an inner product on R^{n}.

First, note that G is compact subgroup of GL(n). It's a standard
fact that any compact subgroup G of GL(n) preserves some inner product
on R^{n} - to see this, just average an arbitrary inner
product with respect to Haar measure on G. So, G preserves some
inner product on R^{n}. Call the
norm associated to this inner product ||·||.
Take a point with |x| = 1 and rescale the norm ||·||
so that also ||x|| = 1.
Let S' be the unit sphere for ||·||.

By assumption G acts to map x to all other points in S, but we have just
seen that G preserves the norm ||·||, so G can only map x to points
within S'. So, S is a subset of S'. But, both S and S' are homeomorphic to
the (n-1)-sphere, so S must equal S'. So, |·| must equal ||·||.
So, |·| comes from an inner product.

There might be an easier way.

© 2005 John Baez

baez@math.removethis.ucr.andthis.edu