Subject: Re: two postings
From: Tom Goodwillie <tomg@math.brown.edu>
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 j<n let F(j,n) consist of the point at infinity and all points 
(v_1,...,v_n) in nV such that the dimension of the linear span of 
the vectors {v_{j+1},...,v_n} is no more than 1.

F(j,n) is preserved by the G-action. F(n-1,n) is all of S^{nV}. F(1,n) 
contains all of the non-free part of the G-action.

Identify S^{kV} with a subspace of S^{nV} by 
(v_1,...,v_k)=(v_1,...,v_k,0,...,0). 
We will also use the resulting filtration of F(j,n) by subspaces F(j,k).

I claim:


(1) F(1,n)/G is contractible.

(2) For 1<j<n, (F(j,n)/F(j-1,n))/G is a certain suspension of a certain 
truncated 
real projective space:
Sigma^{j+1}(RP^{n+j-2}/RP^{2j-2}). In particular its homology away from 
the 
prime 2 looks like that of a point if n+j is even or an (n+2j-1)-sphere if 
n+j is odd.

(3) The homology of (F(n-1,n)/F(1,n))/G is the direct sum of the homology 
of (F(j,n)/F(j-1,n))/G for 1<j<n.


To see (1), one can check that F(1,1)/G is a closed 1-disk and that for 
each k>1 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 i<j.

The group H acts trivially on the coordinates z_i and t
and acts like -1 on s_i, x_i, y_i, and m.

Thus this H-space is the product of

- an (n-j-1)-sphere with antipodal action (unit sphere in the space with 
coordinates (s_1,...,s_{n-j}))
- a (2j-1)-dimensional vector space with -1 action (coordinates 
(x_1,y_1,...x_{j-1},y_{j-1},m))
- an open (j+1)-disk with trivial action (coordinates (z_1,...,z_{j-1},t, 
and length of the vector (s_1,...,s_{n-j})).

Now add the point at infinity and make the H-orbit space, which is also 
the 
G-orbit space of the induced G-space F(j,n)/F(j-1,n). It comes out to be 
the 
(j+1)-fold suspension of the Thom space of a vector bundle over the 
projective 
space RP^{n-j-1}, namely the vector bundle which is the direct sum of 2j-1 
copies 
of the tautological line bundle. This can also be written as

Sigma^{j+1}(RP^{n+j-2}/RP^{2j-2}).


For (3), observe that there is a cell structure here: The G-space 
F(n-1,n)-F(1,n) 
can be decomposed into free G-cells F(j,k)-(F(j-1,k)UF(j,k-1)), one for 
each j and k 
such that1<j<k<=n. In the orbit space (F(n-1,n)/F(1,n))/G, then, there is 
a (non-basepoint) 
cell for each such j and k. Call it E(j,k). Its dimension is k+2j-1. In 
the resulting 
cellular chain complex for (F(n-1,n)/F(1,n))/G there is not much 
happening: The cell E(j,k) 
is in the subcomplex F(j,k)/G, so its boundary cannot involve cells 
E(j',k') with j'>j or k'>k 
(they are not in F(j,k)/G); but it cannot involve those with j'<j and k<=k 
either, because these 
belong to F(j-1,k)/G and so have dimension <  dimE(j,k)-1.

    Tom Goodwillie


Now look at the points of F(j,k) that are not in F(j-1,k)UF(j,k-1). These 
exist if j<k. 
I claim that they form a single free G-cell, so that they contribute a 
single "top cell" 
of F(j,k)/G, of dimension 2j+k-1. To see this, note that these are the 
points (v_1,...,v_k) such that

v_k is not 0
v_{j+1} through v_{k-1} are multiples of v_k
v_j is independent of v_k.

This G-space is induced from the trivial subgroup: it is GxZ where Z is 
the open cell 
consisting of the points (v_1,...,v_k) such that

v_k has the form (0,0,t>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<k<=n

In fact, in the cellular chain complex the only possible boundary terms 
are those 
relating the top cell of F(j,k)/G to that of F(j-1,k)/G. The reason is the 
following:

Any given cell is the top cell of some F(j,k). In computing the boundary 
of that cell 
we need only look at cells of one dimension less that happen to be in that 
subcomplex 
F(j,k), and there is only one of these, none if j=k-1.