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.

—Dan

_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun