10
$\begingroup$

It´s a theorem that there exist only five platonic solids ( up to similarity). I was searching some proofs of this, but I could not. I want to see some proof of this, specially one that uses principally group theory.

Here´s the definition of Platonic solid Wikipedia Platonic solids

  • 0
    What definition of *platonic solid* do you have in mind?2012-04-06
  • 0
    (There is no proof that only uses group theory... *Any* proof depends on geometric input)2012-04-06
  • 0
    Ok I edited, you are right2012-04-06
  • 0
    @MarianoSuárez-Alvarez What is your reasoning for claiming there is no purely group-theoretic proof? It seems to me that the geometry could get in easily enough if you observe that the symmetry group of a polyhedron must be a subgroup of some symmetry group of $\mathbb R^3$, I guess $SL^\pm(3,{\mathbb R})$ or something. There might be an argument to be made that there is no purely group-theoretic proof; if so I would be interested to see it.2012-04-06
  • 1
    Platonic solids are not "group-theoretical objects", whatever that may be, so at some point or another some geometry *will* have to come in.2012-04-06
  • 0
    http://topologicalmusings.wordpress.com/2008/03/01/platonic-solids-and-eulers-formula-for-polyhedra/2012-04-06
  • 0
    The McKay correspondence gives a bijection finite subgroups of $SU(2)$ up to conjugation and affine simply laced Dynkin diagrams (which can be easily classified combinatorially). This correspondence uses representation theory, which may or may not fall under your definition of group theory. $SU(2)$ is a double cover of $SO(3)$ so this effectively classifies finite subgroups there too.2014-04-30
  • 0
    This one. http://www-history.mcs.st-and.ac.uk/~john/geometry/Lectures/L10.html2017-04-28

5 Answers 5