Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
I think you at least want to specify that $X$ is connected, or else you could take, say, two points. –  Qiaochu Yuan Mar 30 '11 at 16:21
    
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. –  Ryan Budney Mar 30 '11 at 16:22
2  
@Ryan: But... Doesn't a homology $3$-sphere have $H_{3} \cong \mathbb{Z}$ by definition? –  t.b. Mar 30 '11 at 16:28
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? –  t.b. Mar 30 '11 at 16:36
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. –  Ryan Budney Mar 30 '11 at 17:43

3 Answers 3

up vote 4 down vote accepted

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.

share|improve this answer
    
I don't understand how the first two sentences are related. –  Qiaochu Yuan Dec 31 '11 at 9:59
    
@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. –  Greg Graviton Dec 31 '11 at 11:55

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 $ <s,t | s^{-3}(st)^2, t^{-5}(st)^2> $ that the second homology group is trivial as well.

share|improve this answer
2  
Very nice. This is the example I should have given! –  Ryan Budney Mar 30 '11 at 17:51

(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.

share|improve this answer
    
Thanks for all the clarifications :) –  t.b. Mar 30 '11 at 21:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.