Take the 2-minute tour ×

I've heard it said that the reason why the homology groups of a space are a computable invariant is because they are a stable invariant in the sense that they are stable under suspension.

I'm familiar with the standard computation which shows that for reduced homology ${\tilde H_n}(S X)\simeq {\tilde H_{n-1}}(X)$ for all $n$, where $S X$ denotes the suspension of a topological space $X$; just let $A$ be $S X$ with the top vertex removed and $B$ be $S X$ with the bottom vertex removed, and apply Mayer-Vietoris, noting that $A$ and $B$ are contractible, and that $A \cap B$ is homotopy equivalent to $X$.

Is this computation the sense in which my first sentence should be understood? That it enables us to compute the homology of certain spaces (such as spheres) which are built out of each other by suspension, by some kind of inductive argument? But other than the obvious example of spheres, are there many other examples of spaces whose homology can be computed by taking suspensions in this way? Are there other reasons why this "stability" of homology groups under suspension makes it much more computable than other invariants?

share|improve this question
Saying homology is "computable" is at least in part a deception. Homology is quite computable for a simplicial complex. But given a garden-variety space with no additional structure, even determining $H_0$ can be difficult. And stability under suspension is not a major prize -- stable homotopy theory is stable with respect to suspension but it's not all that easy to compute, for example the stable homotopy groups of spheres. –  Ryan Budney Oct 7 '10 at 5:33
As a rather simplistic example, $H_0$ of the topological group of diffeomorphisms of $D^4$ -- computing this would be a major feat. –  Ryan Budney Oct 7 '10 at 5:57
...or the $H_0$ of the set of zeroes of $\zeta$ not on the critical line! –  Mariano Suárez-Alvarez Oct 7 '10 at 6:16
(and one can even cook examples of non-computable $H_0$'s, with some help from the halting problem and friends, of course) –  Mariano Suárez-Alvarez Oct 7 '10 at 6:17
I'm not really a topologist, but my impression was that ordinary homology is "more computable" than homotopy because it satisfies excision (allowing more combinatorial attacks), and "more computable" than K-theory because it satisfies the dimension axiom (meaning relatively small spaces have homology with low complexity). –  S. Carnahan Oct 7 '10 at 8:36

1 Answer 1

up vote 8 down vote accepted

Stability alone surely does not give you much. It gives you just the values on the spheres if you have the value on $S^0$. But many interesting spaces (eg all smooth manifolds) are CW-complexes and can be built out of spheres. And if you have a way to handle cofiber sequence $X\to Y \to CX\cup_X Y$ with your theory, you have some hope to compute something. In essence, you have then (togehter with homotopy invariance) exactly the notion of a reduced homology theory. And homology theories are comparatively computable.

But as said already in the comments above, that might be deceptive. Because

1) you need to know the value on $S^0$. This is, of course, the case in singular homology, but in a general theory stable under suspension it is far from true. Eg the stable homotopy groups of spheres are unknown in general (but far better known than the unstable ones).

2) not every space has a CW-structure and even if it has one, you need not know it. For example, you can write down smooth manifolds by giving it as a zero-set of suitable polynomial equations. It might be quite hard to compute its homology then.

3) even if you have a CW-structure and can compute singular homology of the space, you may not be able to compute other homology theories of it (even if you know their coefficients), although these are stable under suspension, of course, as well; you need to go to the Atiyah-Hirzebruch spectral sequence, which need not collapse.

share|improve this answer
Re (2): such CW-structures are computable. Of course this is not an easy computation. But "computable" and "tractable" are very different animals. –  Thierry Zell Oct 7 '10 at 12:33
I think the OP might even mean tractable in a computational way. Glad people mentioned $\pi^S_0$. –  Sean Tilson Oct 8 '10 at 11:15
@Thierry: in which way are they computable? Is there an algorithm (in some sense) which produces out of a given set of equations a CW-structure? –  Lennart Meier Oct 8 '10 at 12:14

Your Answer


