I've been thinking a bit about why we define the singular homology and cohomology groups with simplices rather than, say, cubes, and it seems to me that the elementary aspects of the theory would all become more elegant if we used cubes:

Say we define $C_n(X)$ to be the free abelian group on maps $[0,1]^n \to X$, the boundary operator $\partial$ as Spivak defines it in his first differential geometry book (or probably Calculus on Manifolds), and $H_n(X)$ to be the homology groups of the resulting complex. The first thing that really needs proof is that homotopic maps $f,g: X \to Y$ induce chain homotopic maps $S(f),S(g):C_n(X) \to C_n(Y)$, and this isn't the most obvious thing in the world if we use simplices - we have to decompose $\Delta ^n \times [0,1]$ into a union of $(n+1)$-simplices. But with cubes the proof reduces to the following:

Define $P:C_n(X) \to C_{n+1}(Y)$ on singular cubes $\sigma : [0,1]^n \to X$ by $P(\sigma) = F \circ (\sigma \times id)([0,1]^{n+1})$, where $F$ is a homotopy from $f$ to $g$. Then the proof that $\partial P = S(g)-S(f)-P \partial$ follows formally from the definition of $\partial$, but is also quite clear - it says that the boundary of the singular cube $P(\sigma)$ is the top minus the bottom minus the sides (also, $P(\sigma)$ is a singular cube, and if we work with simplices, it's only a singular chain).

Next we'd want to show that homology groups can be computed using small cubes. From this we easily obtain the excision theorems and Mayer-Vietoris sequences. With simplices we have to define the barycentric subdivision, which is a beautiful geometric idea but seems to be impossible to define without some decidedly ugly notation. However, for cubes, we can use the standard subdivision of a cube into $2^n$ cubes with side lengths halved. That is, if $I_0 = [0,1/2]$ and $I_1 = [1/2,1]$ we could define for $\sigma : [0,1]^n \to X$,

$B(\sigma) = \sum_f (-1)^{\sum f(i)} \sigma | I_{f(1)} \times I_{f(2)} \times \cdots \times I_{f(n)}$,

the sum taken over functions $\{1,2,\dots , n\} \to \{0,1\}$. No inductive formula necessary for the subdivision! Showing that $B$ gives a chain map homotopic to the identity is conceptually easier than with the barycentric subdivision again since $[0,1]^n \times [0,1] = [0,1]^{n+1}$ and the subdivision of the $(n+1)$-cube is related in a relatively clear way to the subdivision of the $n$-cube.

Last note: say we want to define for a smooth manifold a map from the de Rham complex into the cochain complex Hom$(C_n(X), \mathbb{R})$. As usual, we define it as $\alpha \mapsto \big(c \mapsto \int _c \alpha\big)$. The fact that this is a chain map is exactly Stokes' theorem for cubes, which boils down to the ordinary fundamental theorem of calculus and interchanging orders of integration and could be done in an ad-hoc way in any elementary algebraic topology textbook without taking more than half a page.

To summarize, it seems like the formalism at the beginning of an algebraic topology course could be expedited and made more intuitive if we defined the homology groups using cubes. Moreover, the equivalence of the cube definition and the simplex definition is easy because we can decompose a cube into simplices. So why don't introductory algebraic topology books use cubes? Maybe people think simplices aren't that much more difficult to manipulate than cubes and there's some historical inertia, but cubes also seem to have a central role in other fields of mathematics (e.g. Whitney cubes, rectangular paths like those most books use to prove Runge's theorem, the usual proof of Stokes' theorem/Green's theorem).

Can anyone please name a place where simplices can be used but cubes can't, or give a good justification for the use of simplices instead of cubes?

  • 1
    $\begingroup$ Simplexes for simplicity ... $\endgroup$ Jul 22, 2016 at 17:34
  • 1
    $\begingroup$ Maybe it's an eye of the beholder thing - I just never found simplices to be that simple. Cubes are just so easy to work with in our usual Cartesian coordinate system, whereas simplices are best handled with barycentric coordinates, which we're not as used to. $\endgroup$ Jul 22, 2016 at 17:42
  • 1
    $\begingroup$ This was a really nice set of thoughts to read through (as well as the flagged answer). $\endgroup$ Jul 22, 2016 at 20:23

1 Answer 1


The very first thing we need is that $H_*(pt) = \Bbb Z$ in degree zero and is zero elsewhere. Your definition, as stated, does not have this. Clearly $C_*(pt) = \Bbb Z$ in all degrees, but the boundary map is always identically zero (as opposed to the simplicial case, where the boundary map alternates between being the identity and being zero). So the cohomology groups $H_n(pt) = \Bbb Z$.

The way to avoid this is to mod out by "degenerate cubes", those that only depend on $(n-1)$ of the $n$ variables in your cube. When you then define $C_n(X)/Q_n(X)$ to be the chain complex of cubes modulo degenerate cubes, you recover singular homology. If you'd like to see this theory developed, see Massey's book "Singular homology theory".

So one reason people might favor simplices is that one doesn't need to worry about modding out by degeneracy. (There's also the fact that simplices are better suited to proving that there's a natural suspension isomorphism.)

Suppose you don't mind modding out by degeneracies, but you'd really like that easy suspension isomorphism back. Why not work with some class of 'probing objects' that contains $I$, is closed under product and cone, like some set of polyhedra? You can do that (indeed, if I recall some folks already have, though I forget the reference)... or you could go even more generally and probe by maps out of smooth manifolds with corners, which not only allows an easy suspension isomorphism, an easy product map, etc, but also easy ways of doing "smooth operations" like taking inverse images etc. See Lipyanskiy, Geometric homology.

  • $\begingroup$ PS: Somehow I've never seen you around! Maybe this coming year. $\endgroup$
    – user98602
    Jul 22, 2016 at 17:51
  • $\begingroup$ Thanks, this is a nice answer - somehow I overlooked computing $H_n(pt)$. I like the idea of working with a class of polyhedra closed under products and cones, but I'm struggling to see how you could define a singular complex when you have several distinct probing objects of the same dimension. Maybe you identify $\sigma$ and $\tau$ from distinct $n$-dimensional polyhedra if there is a homeomorphism $f$ between the polyhedra with $f \circ \sigma = \tau$? $\endgroup$ Jul 22, 2016 at 18:18
  • $\begingroup$ @NoahOlander You probably should take $f$ to be a linear homeomorphism or something like that (or we lose the notion of boundary), but otherwise, that sounds like a good idea. $\endgroup$
    – user98602
    Jul 22, 2016 at 18:19
  • $\begingroup$ Right, ok. Thanks, I'll probably see you around! $\endgroup$ Jul 22, 2016 at 18:25
  • $\begingroup$ if you can get this to work, I'd love to know - I haven't ever found a good treatment of "polyhedral homology" or whatever you'd like to call it. The definition should probably mirror the above, but now a degenerate polyhedron $\sigma: P \to X$ should be a map that factors through a linear projection $P \to Q$, $Q$ some polyhedron of smaller dimension. $\endgroup$
    – user98602
    Jul 22, 2016 at 18:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .