# How can we rigorously cellulate spaces we can't visualise? Difficulty finding nontrivial cellulations of the $n$-cube, $n$-torus

In the interests of understanding cellular homology better I decided to try to verify for myself the identity: $$\large H_k((S^1)^n)\cong\Bbb Z^{\binom{n}{k}}$$

Using cellular homology. To do that, I must first cellulate the $$n$$-torus. Note: the only thing I know about cellular homology is its raw definition in terms of computing degrees of maps related to the CW complex - I do not yet know, and am not currently interested in, more efficient abstract methods of deriving this identity.

I asked around, and I was told that one can cellulate the $$n$$-torus by first cellulating the $$n$$-cube (which is 'easy') and using the quotient: $$[0,1]^n/_\sim\twoheadrightarrow(S^1)^n$$Which glues opposite faces, to see 'what happens' to the CW-complex for the cube and obtain a CW-complex for the $$n$$-torus. More formally we could take the map: $$(x_j)_{j=1}^n\mapsto(\exp(2\pi i\cdot x_j))_{j=1}^n$$

However, cellulating the $$n$$-torus has not proven easy for little old me, probably because I haven't yet committed to a full study of algebraic topology. I've just been browsing the surface, as my recent question history attests to. There are trivial cellulations, e.g. use $$[0,1]^n\cong D^n$$ and the standard cellulation of $$D^n$$ that uses one $$0$$-cell, one $$(n-1)$$-cell and one $$n$$-cell. That is not useful for my purposes, since there are insufficiently many distinct cells to track what happens to the cube when gluing opposite faces - since one cell represents all faces simultaneously. No, I expect I need to cellulate the $$n$$-cube on every edge, every face, every cubical face, every hypercubical face,...

I am reasonably confident that there are $$2^{n-k}\binom{n}{k}$$ $$k$$-cells of an $$n$$-cube when one uses one cell per face, where a '$$k$$-face' is some subspace obtained by taking $$k$$ products of $$[0,1]$$ with $$n-k$$ products with elements of $$\{0,1\}$$. E.g. a $$2$$-face of a $$5$$-cube could be $$\{0\}\times[0,1]\times\{1\}\times[0,1]\times\{0\}$$. However, I need to systematically describe the attaching maps in such a way that I can be absolutely sure I get the $$n$$-cube as a result.

For example, even if we know the $$3$$-cube has eight $$0$$-cells, twelve $$1$$-cells we can't just link the eight vertices willy-nilly. I will draw a correct diagram of the $$1$$-skeleton on the left, and an incorrect one on the right:

The purpose of these diagrams is to show that, without prior visual knowledge, it's quite easy to write down an incorrect cellulation. I could easily write down similar diagrams for purported $$1$$-skeleta of a $$4$$-cube, but I'd have no way of knowing if they were right. I could probably compute the homology or something similar to show incorrect diagrams are incorrect, but this is circular w.r.t my purposes and also doesn't tell me how to do it correctly - by 'no way', I mean that I personally do not know a way. There must be some way, and this is the essence of my question!

To impose some sanity, I've ensured every vertex has three edges coming out of it, since I "know" this should be a rule. How can I possibly know this for sure? It is 'obvious' that the $$n$$-cube cellulated in this way must have $$n$$ $$1$$-cells attached to every $$0$$-cell, but I wouldn't know how to prove that. Moreover, I need more than this: I need systematic rules for how many $$2$$-cells should be associated to every $$1$$-cell, I need rules for attaching $$1$$-cells such that attaching the correct $$2$$-cells is even possible... and I need rules telling me that this cellulation will, in the $$n$$th stage, produce (a homeomorph of) the $$n$$-cube. I haven't tried to visualise my "incorrect" diagram, but I'm almost certain it is not identifiable with the $$1$$-skeleton of the $$3$$-cube. At least, I'm certain I could attach $$2$$-cells to it in a 'random' way so as to make something that is not equal to the $$3$$-cube.

Here's the rub: I can visualise $$0,1,2$$ and $$3$$-cubes very easily, so identifying which attaching maps are correct is trivial. However, I can't (and I expect very few can) visualise a $$4$$-cube in a coherent enough way to tell at a glance whether or not a given $$1$$ or $$2$$-skeleton is the $$1,2$$-skeleton of a $$4$$-cube. I defy anyone to do this for, say, a $$10$$-cube, on a purely visual basis. But! I know this should be somehow 'easy' - I was told it was! My question is:

• How can we specify what the correct attaching maps should be when it is impossible to visualise the space (the arbitrary $$n$$-cube) we desire? How can we guarantee the end result is indeed an $$n$$-cube?

I'd like to recall that this was all with the purpose of having a concrete cellulation, that distinguishes every face, so that I could diagrammatically take quotients to find a cellulation of the $$n$$-torus. I would be very interested to hear about any techniques whatsoever that help us find (high-resolution) cellulations of a given space, but that's probably very hard in general: just understanding how to do this for a cube, a torus, or any similarly 'fundamental' shape would be nice. I've tagged the question with combinatorics because I suspect there might be combinatorical ways to systematically see how the $$(k+1)$$-cells should fit to the $$k$$-cells, how to systematically index everything so that it is admissible in a formal proof.

When dealing with CW complexes, it is useful to know that the product of a $$k$$-cell and an $$n$$-cell is a $$(k+n)$$-cell. (See Cartesian product of two CW-complexes, for example.) So if you want to find a CW decomposition of $$(S^1)^n$$, one option is to use the two-cell description of a circle (one $$0$$-cell and one $$1$$-cell) and take its $$n$$-fold product with itself. If we describe the cells of the circle as $$e^0$$ and $$e^1$$, and if we further label the cells of the $$i$$th circle in the product as $$e^0_i$$ and $$e^1_i$$, then the typical cell in $$(S^1)^n$$ will look like $$e_0^{i_0} \times \dots \times e_n^{i_n}$$ where each $$i_j$$ is either 0 or 1. The dimension of this cell is $$\sum i_j$$, so you will get $$\binom{n}{k}$$ $$k$$-cells, one for each choice of $$k$$ indices $$i_j$$ which will value $$1$$.
The same sort of argument will work for an $$n$$-cube, using its description as an $$n$$-fold product of the unit interval with itself, but the boundary maps will be nonzero and the computations will be harder. Fortunately an $$n$$-cube is contractible, so you can get the homology groups from that.
• +1 Thanks for pointing me to this. I'll wait a little longer before accepting just to let more of the community give their ideas about any other methods of figuring out cellulations or cellulation 'rules' e.g. determining correct $n$-cube diagrams. Commented Nov 13, 2022 at 23:28