Good points, Tom and James! Let X (like the S^3 in question) be the abstract union of unit n-dimensional cubes Q = [0, 1]^n, where any two intersect only along entire common k-faces, for some k. Let Vert(X) denote its set of vertices (i.e., coming from the n-cubes Q). By definition, a "subcube" of X is any k-dimensional cubical face of one of its component n-cubes Q, for any k. I've been trying to think of what additional constraint(s) will fix this problem of too many examples. A natural condition is to require that: (*) Any subcube is uniquely determined by its vertices. (See below for another related condition.) Assuming this condition: ------------------------ This rules out the examples of Tom and James, and leads to the 6 square faces of a cube as the unique smallest possible cubulated S^2. (Sorry for having overlooked the need for this, in my headlong, willy-nilly rush to post my half-baked idea. I hope this time it's at least 3/4-baked.) ----- Question: --------- Under the additional constraint of I) above, what is a (the?) cubulated 3-sphere S^3 with the fewest cubes Q = [0, 1]^3 ??? ----- —Dan ————— *PS Also consider this condition: (**) X embeds in some Euclidean space R^n with Vert(X) = X ∩ Z^n. I.e., there is a mapping f : X —> R^n that is a topological equivalence with its image f(X), such that the vertices of X are precisely X ∩ Z^n. It seems clear to me that (**) implies (*). --------- QUESTION: Does (*) imply (**) ??? --------- —D. --------------------------------------------------------------------------- James Buddenhagen wrote: ----- More generally for 2D squares, let n be any positive integer greater than 2. Form a (say regular) n-gon as an equator of S^2 and join all vertices to both the north and south pole. Now erase the equatorial polygon and you are left with a topological S^2 composed of n topological squares. On Tue, Apr 30, 2019 at 10:38 PM Tom Karzes <karzes@sonic.net> wrote:
For the case with 2D squares, can't you join three squares at a common vertex, as in the corner of a cube, and then do the same thing with the opposite corners to obtain a sphere from 3 squares?
Also, can't you join two squares at an edge to obtain a rectangle, then join two rectangles (as in the two square case) to obtain a topological sphere from 4 squares? Or is there a constraint that I missed that these examples don't satisfy?
Tom
Dan Asimov writes:
Suppose the sphere S^2 is made of (2D) square pieces of rubber sewn together abstractly along common edges. (I.e., the result is topologically equivalent to a sphere.)
According to this rule, there is a trivial case where only 2 squares are sewn together, by identifying corresponding edges — this is indeed a sphere topologically. The fact that [in 3-space any two planar squares with the same boundary are the same] is not our concern.
It's seems clear that — except for this trivial example — the smallest number of squares in such a thing is 6.
Now suppose the 3-sphere S^3 = {x in R^4 | ||x|| = 1} is, similarly, built abstractly out of (solid) cubes, copies of Q = [0,1]^3, that are allowed to intersect only along entire common square faces, edges, or vertices.
Puzzle: ------- What is the smallest number of cubes with which it's possible to build something topologically equivalent to S^3 this way? ------
I think I know, but I haven't tried to prove it yet.