the combinatorial object "the n-cube" is just length n bitstrings, neighbor means one bit flip away. It can be thought of as being embedded in the geometric
n cube.
the recursive matrices as they are defined are, if you ignore the signs, the incidence matrices of the n cubes. This is only mentioned in the final line of the paper (which is not totally unreasonable). The rest of the result is just some pretty basic facts in linear algebra, but you still have to be comfortable with linear algebra in order to see why they might be natural constructions.
How is the bitstring 'embedded' in the n-cube? It must be a linearisation from a 1-dimensional string to vertices of the n-dimensional object but...? What does 'embedded' even mean, mathematically.
Edit: neighbouring vertices can't be collapsed to a 1-dim string such that each bit x in the string abuts bit y if x is a neighbouring vertex of y (neighbouring = 1 edge away). Works for 1 and 2 dim cubes (line and square) fails for 3. Maybe am missing something.
'incidence matrices' something is incident on something else? It occurs to me that in some sense the recursive matrices might represent the n-cube, each sub-matrix being an n-1 cube. If this is so it is starting to make a little sense.
But that's not the point, which is, how are you seeing this so clearly? It's like a pea-souper over at my end.
> Edit: neighbouring vertices can't be collapsed to a 1-dim string such that each bit x in the string abuts bit y if x is a neighbouring vertex of y (neighbouring = 1 edge away). Works for 1 and 2 dim cubes (line and square) fails for 3. Maybe am missing something.
It sounds like you're missing something. A 3-dimensional cube has 2^3 = 8 vertices:
x y z
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
Since there are three dimensions, each vertex has three neighbors. If you imagine the vertex at (0, 1, 1), you can travel across the x-dimension to the neighbor at (1, 1, 1), or across the y-dimension to the neighbor at (0, 0, 1), or across the z-dimension to the neighbor at (0, 1, 0). Obviously, you represent this by flipping the x-, y-, or z-bit in the corresponding bitstring representation. The bitstring representation itself is blindingly straightforward: the vertex at (0, 1, 1) has the representation 011, etc.
It's not really clear to me where you're becoming confused?
> What does 'embedded' even mean, mathematically.
In this case, it means that a complex structure (the cube) includes a simpler structure (the collection of discrete bitstrings). You can analyze the bitstrings by themselves if you want to, or you can imagine them as being the vertices of a cube that may or may not have properties other than its vertices, but however you describe the cube, it will include the bitstrings.
> It's not really clear to me where you're becoming confused?
Got it. I was interpreting the bitstring as 1 bit mapped to each vertex of the cube, I so badly, utterly misinterpreted it (my string would be 2^n length, not n-length). A simple example and the intent here is totally clear. I guess the problem is at least partly I tend to read what I expect, not what's there. Sigh.
> In this case, it means that a complex structure (the cube) includes....
Righto, a model of interest, a *-morphism or bijection between the string-set and the cube (where * = iso/homo/auto/whatever, it's been a long time)
the recursive matrices as they are defined are, if you ignore the signs, the incidence matrices of the n cubes. This is only mentioned in the final line of the paper (which is not totally unreasonable). The rest of the result is just some pretty basic facts in linear algebra, but you still have to be comfortable with linear algebra in order to see why they might be natural constructions.