Re: chromatic number of a Hilbert space

David Bernier <david250@xxxxxxxxxxxx> writes:

The problem of the chromatic number of a euclidean plane asks for the
minimum number
of colours needed to colour the plane so that any two points one unit
distance apart
are coloured differently.

In other words, an admissible colouring of (the normed space) R^2 is a
mapping
f: R^2 -> N
such that if ||u-v|| =1, then f(u) != f(v) for all u, v in R^2.

#f(R^2) is then the number of colours used, and one seeks to minimize this
for admissible colourings. As a recall, at least four colours are
needed, but
seven colours are enough.

Above, one could replace 'R' by L^2([0,1]), the separable
infinite-dimensional
complex Hilbert space of square-integrable functions on [0, 1]; also, N
would
be replaced by a colour-space such as R, the set of reals.

If {e_i}_{i in N} is a Hilbert base of L^2([0,1]), then for i !=j,
< (e_i - e_j)/sqrt(2) , (e_i - e_j)/sqrt(2) > = 2/( sqrt(2) *
sqrt(2) ) = 1.
<., .> is the inner product on the Hilbert space.

So we have aleph_0 points ( the e_i/sqrt(2) ) all at unit distance
from each other.
So for this Hilbert space, at least aleph_0 colours are needed.

At the moment, I don't know whether an admissible colouring using
aleph_0 colours exists.

Yes, it does, simply because this is a separable metric space.

In any separable metric space, consider a dense sequence {x_j: j in N}.
Colour point x with colour number f(x) = min {j: d(x,x_j) < 1/2}.
Then if x and y have the same colour j,
d(x,y) <= d(x,x_j) + d(x_j,y) < 1.
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.