14
$\begingroup$

Is it true that a compact polyhedron X with trivial homology groups (except $H_{0}(X)$ of course) is necessarily contractible? If yes, what is the approach in proving it? If not, do you see a counter-example?

  • 0
    I think you at least want to specify that $X$ is connected, or else you could take, say, two points.2011-03-30
  • 0
    The Poincare homology sphere is a counter-example which is a manifold. It's also called the Poincare Dodecahedral Space. There's a Wikipedia page for it.2011-03-30
  • 0
    @Qiaochu: or use reduced homology groups in the formulation of the question instead of homology.2011-03-30
  • 2
    @Ryan: But... Doesn't a homology $3$-sphere have $H_{3} \cong \mathbb{Z}$ by definition?2011-03-30
  • 1
    @Ryan: In other words: what was wrong with Mariano's now deleted answer along the lines: If $X$ is connected and *simply connected* then it follows from Hurewicz and Whitehead that the inclusion of any point is a homotopy equivalence. What am I missing?2011-03-30
  • 0
    @Theo: I don't understand the relationship between your two comments. The Poincare dodecahedral space is not simply connected (but its fundamental group does have trivial abelianization).2011-03-30
  • 0
    @Qiaochu: I'm sorry, I didn't make myself clear. I agree with what you say, but I don't understand how Ryan's example should be a counterexample to the question because it has $H_3 = \mathbb{Z}$. Mariano's argument missed the assumption "simply connected" but otherwise it seemed fine to me (I was typing up the exact same thing and got immensely confused). I made the bold assumption that Mariano's deletion was a reaction to Ryans comment and I guess that's what "in other words" was referring to.2011-03-30
  • 0
    Well, you could take the mapping cone of a homology isomorphism $f:PHS\rightarrow S^3$. Maybe that doesn't fit what's meant by "polyhedron" here, though.2011-03-30
  • 1
    @Theo: ah, right. I forgot about $H_3$. I'll give a proper example as an answer, if someone else doesn't beat me to it.2011-03-30
  • 0
    @Ryan: Thanks! Now I'm happy again :)2011-03-30
  • 0
    @Liudas: please do not use answers to make comments.2011-03-30
  • 0
    Thank you all. You have been of much help. Now I have a counter-example and a proof (for spaces with trivial $\pi_{1}$)! And indeed, I forgot to mention $X$ was at least path-connected.2011-03-30
  • 0
    @Theo, I deleted the answer because I was sure (hah!) that I did not need the simply connectedness but could not think of a reason... Of course, I did need it.2011-03-30

3 Answers 3

3

(This is the deleted answer the comments refer to; it was missing the hypothesis about simple-connectedness)

Using the Hurewicz theorem, you deduce at once that such a polyhedron [Edit: if it is simply connected] has trivial homotopy groups, so that it is weakly homotopy equivalent to a point. Since it is a CW-complex, then Whitehead's theorem tells you that the polyhedron is in fact contractible.

15

The 2-skeleton of the Poincare homology sphere, also describable as the presentation complex of the binary icosahedral group, provides a counterexample to your original question. The fundamental group is of order 120 and is perfect, which implies that that $H_1$ is trivial. You can check from the group presentation $ $ that the second homology group is trivial as well.

  • 2
    Very nice. This is the example I should have given!2011-03-30
11

To sum up the comments: when Poincaré worked on the beginnings of algebraic topology, he originally thought that a space with trivial homology groups must be contractible. (More precisely, he thought that having the homology group of a 3-sphere implies being a 3-sphere.) However, he soon found a counterexample, the Poincaré homology sphere, which led him to the construction of the fundamental group.

When taking the fundamental group into account, the statement is indeed true: if a space has trivial fundamental group and trivial higher homology groups, then it must be contractible. This is a consequence of Whitehead's theorem and the Hurewicz map.

  • 1
    I don't understand how the first two sentences are related.2011-12-31
  • 0
    @QiaochuYuan: If I were to conjecture that homology is a complete invariant, then this would imply both sentences. As to Poincaré's thinking, I am only aware of evidence that backs up the sentence in parentheses, however.2011-12-31