I managed to screw up this statement: ----- SO(3) / I = S^3 / (2I). Strikingly, 2I is the *only* subgroup of S^3 for which SO(3) is a homology 3-sphere. ----- which should read instead: ----- SO(3) / I = S^3 / (2I). Strikingly, 2I is the *only* subgroup of S^3 for which S^3 / (2I) is a homology 3-sphere. ----- (And by S^3, I mean the double cover of P^3 = SO(3) both as a manifold and as a group, making S^3 into a Lie group (also known as Spin(3)). —Dan -----
I don't know much about this, but it's so interesting that I thought I'd share it with math-fun.
The rotation group of 3-space R^3, called SO(3), is homeomorphic to P^3, (real) projective 3-space. Its unique invariant metric, up to scaling, is that of P^3 induced from
P^3 = S^3 / (x ~ -x)
which just means this metric on P^3 is inherited from the 3-sphere S^3, and in fact is locally isometric to it.
Now suppose H is a finite subgroup of SO(3). Then H is known to be isomorphic to one of the following:
* a cyclic group C_n of order n,
* a dihedral group of order 2n,
* the tetrahedral group T of order 12 (also known as A_4),
* the octahedral group O of order 24,
* the icosahedral group I of order 60 (also known as A_5).
Since SO(3) is a Lie group*, any finite subgroup G of SO(3) gives rise to the left coset space
SO(3) / H = {g H | g in SO(3)}
whose elements by this definition are indeed copies of the subgroup K, translated on the left by some element g of SO(3). Unless H is a normal subgroup (the only nontrivial one is {-I, I}), this is not a group.
But suppose H is a discrete subgroup of SO(3) with the property that its commutator subgroup is the whole group:
[H, H] = H.
The word for this is: H is a "perfect" group.
Then interesting things happen! The coset space SO(3) / H is what's called a homology sphere: A 3-dimensional manifold with the same homology groups as the 3-sphere S^3 (H_0(S^3) = H_3(S^3) = Z and H_1(S^3) = H_2(S^3) = 0).
And if you take H = I, promoting it by a mere one place in the alphabet, then SO(3) / H = SO(3) / I is a homology sphere whose fundamental group is the binary icosahedral group 2I, whose order is 120.
Not surprisingly we can get the same coset space by factoring the double cover of SO(3) = P^3 out by the double cover of I, namely 2I:
SO(3) / I = S^3 / (2I).
Strikingly, 2I is the *only* subgroup of S^3 for which SO(3) is a homology 3-sphere.
And there is not enough space in this forum to go into detail, but the so-called "Poincaré dodecahedral space" S^3 / (2I), seems to be uniquely responsible for the disruption of low-dimensional topology in 4 dimensions. Where as you may know there are uncountably many smooth structures on R^4, while every other R^n has just one.