A.Connes's proof of Morley's theorem
- Dear Hyacinthists,
may be you will be interested by the following - I try to translate a
cotation from a french maths forum -:
" In the special book for the 40 th birthday of IHES, we can find a nice
proof of Morley's theorem due to Alain Connes.
Let k be a commutative field with characteristic 0 ,
G the group of the affine mappings from k to k , ( x -> a x + b with a<>0),
H the morphism f -> a from G to k*,
T = ker H (translations)
and for f in G - T, fix(f) = b/(1-a) the fixed point of f.
Theorem :Let f , g, h in G such as f.g, g.h, h.f and f.g.h are not in T
and j = H(f.g.h) .
i) and ii) are equivalent where :
i) (f^3).(g^3).(h^3) = 1
ii) (j^3) = 1 and fix(f.g) + j fix(g.h) + (j^2) fix(h.f) = 0
(It can be proved with an easy computation)
Appication : Morley's theorem :
k is the complex field :
angle A = 3a , angle B= 2b ,angle C=2c
f = rotation with center A, angle 2a
g = rotation with center B, angle 2b
h = rotation with center C, angle 2c
p = fix (fg) , q = fix (gh) , r = fix (hf) are the common points of the
As f^3.g^3.h^3 =1 ( because f^3, g^3, r^3 are products of reflexions w.r.t.
two sides), we get
p+jq+j2^r = 0 which means that p,q,r are the vertices of an equilateral
Very nice; don't you think so.
Friendly from France. Jean-Pierre