Why the choice of the simplex for defining homology?

Question

This question arose out of my attempts to understand another question. The most popular construction for the chain complex for defining singular homology uses the $n$-simplex.

But it is also possible to use other spaces. For example, one can use the $n$-cubes instead, as done in certain books. Also it occurs to me that one can use the discs $D^n$, with orientation specified at the boundary. I haven't checked all the details; but I am hopeful that it can be made to work and made to prove that this homology is the same as the homology constructed with cubes or with simplexes.

So, why is the simplex the most used choice? Granted, it has a certain symmetry in all directions and so it is aesthetically somehow more satisfying. But are there other reasons?

Re to Greg Kuperberg: The details go roughly like the following. An $n$-disc is the unit ball in $\mathbb R^n$. The boundary map is $D^n$ going to $S^{n-1}$ on the boundary, but so that this boundary is a union of two discs $D^{n-1}$ but equipped with opposite orientation, ie the parts above and below the equator. Orientation for an $n$-disc is a choice of direction into the center, or going away from the center. Orientation for a $1$-disc is just a choice of direction. A bit more checking is needed to fix the orientation, but I am hopeful that it can be done. Assuming this, and after extending to chains, clearly the images are contained in kernels. Thus homology can be defined. It will be more difficult to prove that this is (after sorting out minor discrepancies) essentially the same as the homology given by simplexes. There is no nice barycentric subdivision, which was a problem. But, since the $n$-disc is homeomorphic to the $n$-simplex, I was hoping that at least an ugly proof of equivalence could be constructed.

Answer 1

There is a specific reason that singular homology with simplices is simpler than singular homology with cubes. You would like the homology of a point to be "trivial" according to the Eilenberg-Steenrod axioms. (That is, $H^0 = \mathbb{Z}$ and the others are trivial.) However, if you look carefully at the chain complex of maps from cubes to a point, it is not true. The homology groups have to be corrected by hand by annihilating certain degenerate chains. If instead you carefully use the standard ordered simplex --- not just an abstract simplex floating in space but one with numbered vertices --- then it works automatically. On the other hand, products of chains look a bit simpler at first with cubes. They are only barely simpler though, because of the magic of simplicial sets. So on balance simplices are nicer than cubes.

Likewise, at the entirely rigorous level, defining singular homology with "disks" is not really a complete proposal, and not necessarily a simple proposal if you were to flesh it out. Do you mean maps from "all" balls, or some specific collection of balls, or just one standard ball? What is the boundary operator?

You can think of singular homology as a conversion from topology to combinatorics, attained by building a simplicial set (which is a generalization of a simplicial complex) from a topological space X, and then taking the simplicial homology of the combinatorial object. There is a specific way to do that with simplices, and an analogue with cubes that gives you a cubical complex. For instance, in the case of simplices, even if you start with a point, you get an infinite-dimensional complex with one simplex in each dimension. Presumably you have in mind bulding some CW complex from a topological space X, and then taking its CW homology. But what do you really want to connect to what?

It is also true that once you have a definition of homology, you can use a mapped-in disk to define a cycle or a relative cycle, which is then unique up to homology. But that is not the same as defining homology.

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The question was extended to include a more precise description of the boundary operator in the proposed disk-based definition of homology. Namely, the proposal is that the boundary a mapped-in disk $D^n$ is a formal sum of two mapped-in disks $D^{n-1}$, whose restriction to the equator is the same mapped-in sphere $S^{n-2}$. I don't know what it means to equip these maps with an orientation; one thing that it could mean is to take the formal difference rather than the formal sum of the two hemispherical maps.

Either way, I don't think that the homology groups that result are correct. In this theory, a mapped-in line segment is only homologous to another mapped-in line segment with the same endpoints; there is no way to split the line segment into two line segments. I think that if the space is $\mathbb{R}^n$, then the resulting CW complex is weakly homotopy equivalent to a complete graph whose vertex set is $\mathbb{R}^n$ with the discrete topology. This isn't what you want.

Answer 2

One general answer to these kinds of questions is:

in principle one can base homotopy theory and related topics (such as homology) on shapes other than simplices. But the theory for other shapes is much less developed and may be less elegant, for reasons such as Greg mentions in his comments. It may still be equally valid in the end.

Among the few texts that seriously venture into homotopy theory based on cubical shapes is

Jardine, Cubical homotopy theory: a beginning http://hopf.math.purdue.edu/Jardine/cubical2.pdf