From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Numbers's space with three dimensions
Date: 5 Aug 1996 18:29:31 GMT

In article <32030749.4A3E@mail.mcnet.ch>,
Sylvestre Blanc  <marblanc@mail.mcnet.ch> wrote:

>What I would like to know is precisely a demonstration of the 
>impossibility of making a field in R^3.

Since there are fields  k  which allow a field structure on  k^3, you
need to use some special features of  R. One possibility is the
topological nature of the real numbers. There are a number of issues
one must check carefully but it's easy to see why this proof works:

Assuming the usual distributive laws, a product structure on  E=R^3
would be a bilinear map  E x E --> E; in particular, it's continuous in
both variables. Use this product to define the continuous function
	F(v,w) = (v*w)/|v*w|
This is defined wherever  v*w  is not zero. If you expect  E  to be
a _field_ (or more generally a division ring), then this product should
never be zero unless one of the factors is, so in particular if we
take  v  to be a fixed nonzero element of  E  and restrict  w  to
lying in the unit sphere  S  in  E, then   f(w) = F(v,w)  is a
continuous function from  S  to  S.

Next consider the effect of varying  v: if we replace  v  by a nearby
vector  v'  we'd get a different map  f'  from  S  to  S. Indeed, if
you gradually change  v  to  v'  then for each  w  in  S  we would see
the images  f(w)  gradually change to  f'(w)  along a little curve.
Use these curves to draw small direction vectors at each point of  S
showing the direction in which these curves point.

(Caution: this only gives a vector if there really is a _curve_ drawn;
this requires that  f(w)  be a different point from  f'(w), i.e., that
we not have
	(v*w)/|v*w| = (v'*w)/|v'*w|
i.e.,   (|v'*w|v - |v*w|v') * w = 0.
Since  w  isn't zero, the invertibility assumption shows we would have
	|v'*w| v = |v*w| v'
so that in particular,  v  and  v'  point in the same direction. To
prevent this from happening, we might for example insist for example that 
v  and  v'  also be on the sphere  S  (and not be antipodal).)

Now here's the "gotcha": you have just described a vector field on the
sphere  S  which is nonzero at each point. That's forbidden by the
"hairy ball theorem": you can't comb the hair on a hairy billiard ball.
This contradiction prohibits the existence of the field structure on  E. 


Comments: 
	1. The same proof works to show there are no field structures on
any odd-dimensional Euclidean space except  R  itself.
	2. The proof does not use commutativity of  '*', nor even 
associativity or the existence of an identity! In fact, it doesn't
even really need bilinearity, just the differentiability of the product
structure.
	3. If you drop continuity and only ask if there is _some_ field
structure on  R^3  which includes  R  as a subfield, you need only put
the elements of  R^3 - R  in one-to-one correspondence with the points
of  C - R  and thus make  R^3  into a field (isomorphic to the complex
numbers). But clearly that takes the fun out of the problem.
	4. More sophisticated topological arguments are needed to
show that the dimension of the space  E  must actually be a _power_ of 2,
and then further, more subtle, arguments limit it to  1, 2, 4, or 8.
Stronger results are possible if you insist on a notion of 'field' closer
to the commutative axioms.
	5. As I posted earlier, you can get some more information from
index/products.html
[That's an updated URL -- djr 1999/01]

dave