Subject: Re: two postings
From: Tom Goodwillie
Date: Fri, 3 Mar 2006 01:23:46 -0500
I figured out the homology of the orbit space for G=SU(2) acting
on the product G^n by conjugation.
First, it's enough to solve the problem for G acting on the n-fold
smash product of G by conjugation. The reason is that the orbit
space of G acting on an n-fold product of based G-spaces stably
splits as a wedge of orbit spaces for G acting on k-fold smash
products, one for each set of k elements in {1,..,n}. This does
not use anything hard:
(i) If a based space A has B as a retract then A is stably equivalent
to the wedge of B and A/B (if the inclusion is nice).
(ii) If a based G-space A contains a based G-space B as a retract,
then the orbit space A/G has the orbit space B/G as a retract, so by
(i) A/G is stably the wedge of B/G and (A/G)/(B/G)=(A/B)/G.
(iii) Apply (ii) repeatedly to get the desired statement.
So now look at that smash product.
SU(2) is a 3-sphere, and SU(2) with the given action is S^V, the one-point
compactification of the vector space V=R3 with SU(2) acting linearly.
Of course, the action factors through SO(3).
So now I am thinking of G=SO(3) acting on the smash product of n copies
of S^{V}, in other words on S^{nV} where nV is the direct sum of n copies
of the standard 3-dimensional representation V of SO(3).
I will use a filtration of S^{nV} by subspaces F(j,n).
For j1 the quotient
(F(1,k)/G)/(F(1,k-1)/G)=(F(1,k)/F(1,k-1))/G is a closed (k+1)-disk.
For (2), consider the difference set F(j,n)-F(j-1,n). This consists of the
n-tuples (v_1,...,v_n) such that
v_j is not 0
{v_{j+1}, ...., v_n} spans a line in R3 different from the line spanned by
v_j.
This G-space is induced from the subgroup T=SO(2), in the sense that it is
G\times_T Y
for some T-space Y, namely the set of all (v_1,...,v_n) such that
v_j has the form (0,0,t>0)
{v_{j+1}, ..., v_n} spans a line in R3 different from 0x0xR.
This T-space in turn is induced from H, the subgroup of order 2 in T. The
H-space in
question is the set of all (v_1,...,v_n) such that
v_j has the form (0,0,t>0)
{v_{j+1}, ..., v_n} spans a line in 0xRxR different from 0x0xR.
Such a line is given by (0,s,ms) for some m.
Write v_{j+i}=(0,s_i,ms_i) for i<=n-j.
Write v_i=(x_i,y_i,z_i) for ij or k'>k
(they are not in F(j,k)/G); but it cannot involve those with j'0)
v_{j+1} through v_{k-1} are multiples of v_k
v_j has the form (0,x,y>0).
So we have a cell structure for the orbit space (R(n)/G)/R(1)/G), which we
can use to
compute the homology of the equivalent space R(n)/G=S^{nV}/G. There is one
cell, of
dimension 2j+k-1, for every j and k with 2<=j