0

Do the Betti numbers determine the topology? I.e., if we are given a list of the Betti numbers of a topological space, then are we able to find the topology? What about in the smooth manifold case? Or in the compact case?

Levi
  • 4,766
C.F.G
  • 8,523
  • 2
    If only.......... – Angina Seng Nov 10 '19 at 07:56
  • 3
    This is what Poincare discovered over 100 years ago: He first expected that for compact connected manifolds homology groups (he did not have the notion but he knew about Betti numbers and torsion numbers) determine the manifold. But then he discovered the Poincare homology sphere. This led him to state the Poincare Conjecture and geometric topology was not the same ever since. – Moishe Kohan Nov 10 '19 at 13:17
  • @MoisheKohan: Thanks for your nice historical comment. similar to Poincare argument, Homotopy groups can do this? – C.F.G Nov 11 '19 at 07:42
  • 1
    @C.F.G: If you are asking if homotopy-equivalent topological spaces are homeomorphic then the answer is obviously negative: $R^n$ is homotopy-equivalent to a point. But if you are asking this question for homotopy-equivaent closed manifolds, then it is a good question (still, negative answer), deserving a separate post. – Moishe Kohan Nov 11 '19 at 15:56
  • https://math.stackexchange.com/questions/281339/what-is-the-difference-between-homotopy-and-homeomorphism – Moishe Kohan Nov 11 '19 at 16:01
  • @MoisheKohan: I have posted a new question as you suggested. https://math.stackexchange.com/q/3437720/272127 – C.F.G Nov 16 '19 at 08:06

1 Answers1

3

In general topological spaces this does not work, consider $$X_n = \mathbb R^n \setminus \mathbb R^{n-2}.$$ Then the homotopy type of all the $X_n$ is identical, so they have the same Betti numbers (but are not homeomorphic). So a better question might be whether they determin the homotopy type. The answer is still no: consider the space $$\mathbb R \mathbb P^3 = S^3/\pm1.$$ This space falls in all of your categories, has the same Betti numbers as $S^3$, but is not homeomorphic to $S^3$ (it has nontrivial but finite fundamental group).

Levi
  • 4,766
  • Why $RP^3$ and $S^3$ have same Betti numbers? Is it true in general for $RP^n$ and $S^n$? and for $b_i^M=b_i^{M/\Gamma}$? where $\Gamma$ is a discreet subgroup of isometry group that acts properly discontinuous. – C.F.G Nov 10 '19 at 17:01
  • 2
    No, this fails for even $n$ since then $\mathbb R \mathbb P^n$ is non-orientable. If I understand it correctly, what you suggest in your second question fails for the same reason. – Levi Nov 10 '19 at 17:08