Each particle's position gets a coordinate, so the ambient space is @@1@@, below.
We can factor off an @@0@@, which we compute below.
Let's draw @@0@@ and the unit circle in @@1@@. The @@2@@ plane minus the origin is homotopy equivalent to the circle. (You can homotope the radial direction away.)
That circle is also the generator of @@3@@!
Now let's see what that generator looks like as a motion of particles. I've offset it a bit from the obvious @@0@@ interval, so that the particles start distinct.
Note that this is the distinguishable points version. The indistinguishable points version should have more generators, corresponding to the elements of the symmetric group. One of them is special though (shown below), since three times it is the image of the one above.
Now we have @@1@@.
Again, though, there is an "all equal" vector @@2@@ which is parallel to all of the planes determined by the above equations. Let's set up some coordinates so that we can easily project the whole thing out.
The four matrices, can then be rewritten in terms of the 3-space spanned by @@0@@.
The kernels of the new matrices are 1-dimensional, so we find a basis for them.
We know that when we remove these lines and quotient by the radial distance, we will get a 8 punctured sphere. Draw it for kicks.
I'm having a difficult time seeing an obvious way to draw the 8 generating loops based at a common point. Maybe if I stereographically project?
A set of loops going through the base point, representing generators.
Bring those loops back up to the sphere!
Now bring them back up to 4-space and draw them as components!
note that there are pairs of intersections which do not give rise to a triple point.
Approximately, anyway, since we don't connect to the base point.
For kicks, let's look at an animation of the curve simplifying.
Now we have @@1@@.
Again, though, there is an "all equal" vector @@2@@ which is parallel to all of the 3-planes determined by the above equations. Let's set up some coordinates so that we can easily project the whole thing out.
Looking at the above graphs, we see that in the excluded locus, @@0@@, there are @@1@@ corresponding to the locus of 4-particle interactions.
This gives us a cell complex for @@2@@ with 40 1-chains and 10 0-chains. The chain complex for @@3@@ is then @@4@@ Since we know that @@5@@.
Now, by Alexander duality, @@6@@.
See the following computation of the homology groups for a higher number of particles: The Homology of "k-Equal" Manifolds and Related Partition Lattices Bjorner A., Welker V.
In particular, when they get to 6 particles, you pick up 2nd homology. When you get 9 particles, you get 3rd homology. Next 12 and 4th homology. This seems to be related to commutation relations from 3-interactions which involve disjoint sets of particles. i.e. 6 particles split as (012)(345), 9 particles as (012)(345)(678), etc.
This indicates that these higher homology classes should be representable as tori of the appropriate dimension.