Is a set of cubes a topological space? Suppose you have a set of identical cubes; if you arrange eight of these in a stack there is only one arrangement which is the shape of a cube and an infinite number of arrangements which aren't a cubic shape. However, the cubic stack of eight cubes is equivalent to a ball which is sliced into eight equal parts, and this is also equivalent to an octahedron sliced through its edges into the same number of parts. The slices are plane in each case, and in each case there are three orthogonal cutting planes. You get, under rotations along slices, a permutation group which is \( S_8 \). Well, you do if you label the eight parts with numbers or colors. The cubic stack has elements that can have three labels each, preserving this means labelling the vertices of the sliced octahedron. With these labels you have a way to track the orientation of each element in the stack, you get something called the wreath product of \( S_8 \) with \( C_3 \). The topology depends on the slices, and that they pass through the centre of each geometrical object ( a deep slice) at right angles to each other. In fact the "stack" can have any shape at all as long as its sliceable in the prescribed way.
An observation: with a sliced 3-ball there are three ways to form a set of four quadrants of a sphere by removing one slice, or by gluing the ball back together along one slice. If you have a real Rubik's sphere you can use tape to stick pairs of octants back together. 'Permutations' are now (mathematically and physically) restricted to rotations of 180°. With four quadrants you have a 2-dimensional projection (onto an abstract plane surface) along with transpositions of pairs of quadrants of a disc. Is this a representation of the symmetric group on 4 letters?
According to Louis Kauffman, it's quite common to start with a geometric object and end up using it as the basis for a more general kind of object. But, staying with the sphere and its symmetries for the moment: we can consider a plane that passes through the centre of a sphere is akin to a line that passes through the centre of a circle; we can just alter the number of dimensions and give everything one more dimension (or one less). However, there is only one other line in the plane which is perpendicular to a first line, and which intersects the first line at the centre of a circle. In three spatial dimensions, there are two other planes that pass through the centre of a sphere and are all perpendicular. A first plane through the sphere which divides it into two disjoint regions is any one of an infinite number of planes which pass through the centre. Choosing a first cutting plane fixes something: the orientation of all other perpendicular planes to this first plane, which also pass through the centre. Choosing a first line in the case of the circle leaves one and only one other possible line such that both lines intersect at the centre of the circle and are mutually perpendicular. Hence we can project at most two planes cutting through the sphere onto an external plane, if we associate lines to planes and circles to spheres. Lastly, let's note that any labeling or coloring scheme is really a way to keep track of all the rotations, so it's a kind of coordinate system or a set of abstract indices; with a 'colorless' sliced cube or octahedron, or the sphere in between, nothing changes under rotations. For example suppose the sphere is sliced in two and one side is colored black, the other white. This is equivalent to not coloring it as far as rotations of one half against the other, so what's needed is a way to distinguish this, and that amounts to a system of coordinates: markings perhaps, along the boundaries of each hemisphere. It seems it can be quite, well, arbitrary, as long as it can measure change in the relative angle between halves of the sphere. One boundary could have a sequence of fixed marks, the other a fixed 'pointer', that is to say, circular or polar systems of coordinates, aka Euclidean vector spaces. The three different ways to fix "motion" along one of three perpendicular planes in 3-space (by gluing octants of the sliced sphere together along common boundaries) correspond to the three projections each having only two perpendicular lines through their circles; the projections are subspaces.
What are the open sets? A topological space is a set with a privileged collection of subsets known as open sets, satisfying the usual requirements that the empty set and the entire set are open, and the open sets are stable under arbitrary unions and finite intersections. If you tell me what your open sets are, I'll tell you if you have a topological space. http://en.wikipedia.org/wiki/Topological_space
A permutation space is open because there is no limit on the number of compositions that give a particular permutation of n distinguished objects. The open sets are the space itself: \( S_8 \wr (C_3 = Z/3Z) \), and any subspace, like \( S_4 \), that being the subspace you get when you suppress motion over one slice. But the question "is a set of cubes a topological space" does have an answer which depends on certain "packing constraints" say, and on how the cubes are labeled, so the answer (if there is one) depends on geometrical and algebraic notions.
This seems to be about geometry as much as about some algebra that can be defined on that geometry. Looking again at plane slices, through a sphere (a geometrically symmetric object) there are an infinite number of planes passing through the centre; but then there are also an infinite number of planes through any point in the interior. If you slice a sphere with three orthogonal planes intersecting a point which is off-centre, do you still get rotations? Absolutely, it doesn't matter where the three planes intersect, as long as each section of the sphere is an object with three internal faces; the shape and size of the external faces isn't important. So the shape of the disc in the plane and where the two cutting lines intersect is also not relevant. The relevant feature is the set of internal 'surfaces' for the sphere or the disc.
If we want to look at sets, and in particular open sets, then an empty set is needed as well. In a permutation space, "the" empty set would be one which contains no permutations; the identity is a permutation so can't be in this (proposed) empty set. To "have" an identity, parts of the geometric space being considered, a sliced ball, have to be distinguished, which means labeling them with different numbers or other symbols, using colors, etc. and there is a certain amount of freedom in how the sets of symbols and/or colors are mapped to the different parts. Nonetheless, not identifying any parts will mean no identity permutations can exist over all the parts, despite their being related by sharing common boundaries. But wait, there is an "external" identity which is an element of the group of rotations of a whole cube in 3-space. Clearly, an unmarked cube will look the same in any orientation, and if this "cube" is actually a cubic stack of eight smaller cubes none of which has any external marking, then rotating one half against the other such that the shape of a cube is preserved will also look the same, so all the rotations map to the empty set of permutations. What happens if you mark one of the eight parts is that now you can distinguish this part in one of eight different positions, under whole or partial rotations. Any finite length sequence of rotations over whole or part of the stack of cubes leaves the marked element in one of eight positions. So all that must mean that the cube-shaped stack of cubes (or other appropriately sliced object) has two kinds of identity: a geometric identity, or a choice of orientation of the whole object which is left unchanged by any operation, and an algebraic identity, which is the same object but with different parts distinguished from each other. The difference between them being that, you can rotate one half of a 2x2x2 stack of cubes which is colored such that any subsequent rotation of the whole cube leaves the stack rotated (internally). The internal rotation is "quotiented by" the external group of rotations.
Ok, I remember once asking a lecturer about how you fix some object (the example was an equilateral triangle) so you get its symmetry group. Well, it's not hard; if the triangle is not fixed in place but allowed to rotate and translate freely, it can be in a position which isn't a permutation (of any of the objects "features") relative to any other position/orientation. You have to fix the triangle, any choice is valid but the orientation can't change, or rather, it can change only in steps--a rotation under symmetry takes vertices to vertices and edges to edges "in the same place". For a pair of equilateral triangles you have two choices of 'global' orientation, both objects are fixed in the plane if you can draw a pair of Cartesian axes anywhere and define fixed distances for each vertex of the pair of triangles from an origin. You then have the same group of symmetries over both figures. This applies to any number of triangles in general position in the plane (not intersecting), indeed to any regular polygon. All of that suggests the map from a sliced ball/cube/octahedron has to fix a few things. Fixing a slice could mean suppressing rotations on that slice, or just using a subset of slices so you have a projection along their geometric intersection onto the centre of a disc, correspondingly quartered; you have the same symmetry group over a 2-dimensional and a 3-dimensional figure, the 2-dimensional representations are just those surfaces of the sliced 3-ball which are fixed along the line of projection.
Another look at braid groups. A braid group on n strands has "generators" \( \sigma_i \) which cross the ith strand over the i+ 1th strand, so there are n-1 generators over n strands. A generator corresponds to a single transposition, and when the braid group is quotiented by an equivalence relation the quotient space is a permutation group. The required equivalence is \( \sigma^2_i = 1 \). The strands in a braid group are disposed linearly, the nth strand isn't "next to" the 1st strand, so transposing (crossing) these two strands is equivalent to a larger composition of generators. Permutation spaces are linear; any symmetric group has a matrix representation and matrices are linear elements. So although you can permute sets such as quadrants of a disc (or 3-ball), there can only be n-1 fundamental transpositions over the set; in other words, it looks like there are four of these transpositions (generators with the crossover "modded out") but the fourth one is a composition and not fundamental. Ergo, for a string like abcd with transpositions, there is no mathematical sense in the notion that a can be transposed with b unless this via a composition of other transpositions; this relation is fixed when you label the quadrants with characters like abcd, choosing which is the first and which the last element (or strand).
. . ."that a can be transposed with b" should read "that a can be transposed with d", above. Sorry about that. Anyhoo, because of the equivalence \( \sigma^2_i = 1 \), multiplying both sides by \( \sigma^{-1}_i \) on the left (or on the right): \( \sigma^2_i\sigma^{-1}_i = (1)\sigma^{-1}_i \) \(=> \sigma_i = \sigma^{-1}_i \) So the quotient space has a left over right crossing of two strands equivalent to a right over left crossing. So that makes \( \sigma^2_i \) look like a type II Reidemeister move, you can pull the strands straight and they uncross: Please Register or Log in to view the hidden image!. So with three strands you expect there are equivalent type III Reidemeister moves, and that's what happens. Because of the equivalence relation, "crossed strands" can be pulled apart as if they can pass through each other; but of course this is also because a permutation group doesn't have crossings.
More musings: the stack of cubes or a (larger) sliced cube are the same object, so is a sliced octahedron and both are the same object as a sliced ball; there are continuous functions that take this object to the different shapes. In fact it doesn't matter what shape is sliced up, it just has to be homeomorphic to a 3-ball. The surface is sliced as well (since it "comes along for the ride"), so you have a sliced 2-sphere. Some questions about symmetry: what happens to the symmetry of a sphere when you slice through it? Topologically you separate the sphere or ball into two distinct parts. What if the slice removes a single point from the sphere? A hemisphere or its topological equivalent is an inflated disc. Slice the sphere again at right angles to the first slice, now what?
