2

I was wondering about which groups can be the simplicial homology groups of a topological space? For example, is it possible to construct a topological space $X$ such that $H_1(X;\mathbb{Z}) = \mathbb{Z}/3\mathbb{Z}$ and $H_2(X;\mathbb{Z}) = \mathbb{Z}/5\mathbb{Z}$?

What about replacing $3$ and $5$ by arbitrary positive integers $m$ and $n$?

Thanks in advance.

mtsecco
  • 113
  • 2
    Google "Moore space". – Pedro Nov 11 '16 at 01:52
  • Since it's homology, Moore spaces are also a good thing to look up. – J126 Nov 11 '16 at 01:55
  • @PedroTamaroff Arnt they for homotopy ? – Rene Schipperus Nov 11 '16 at 01:55
  • Yes, sure. Moore space it is. – Pedro Nov 11 '16 at 01:56
  • Thanks for the answer. I googled it, but the Moore Space gives $H_n(X) = G$ for any abelian group $G$, but $H_{n-1}(X) = 0$, for example. Is possible to make for instance $H_n(X) = G_1$ and $H_{n-1}(X) = G_2$, where $G_1$ and $G_2$ are abelian groups? – mtsecco Nov 11 '16 at 02:02
  • 1
    I was thinking here. Please, correct me if I'm wrong. If I take the disjoint union of two Moore Spaces, for example $X_1$ with $H_n(X_1) = G_1$ and $X_2$ with $H_{n-1}(X_2) = G_2$, do I get the desired example? – mtsecco Nov 11 '16 at 02:04
  • It's better to take the wedge of the Moore spaces than the disjoint union, but if you don't care what happens to $H_0$ then disjoint union will work. – Matt Samuel Nov 11 '16 at 02:31
  • It's indeed better to take the wedge sum, because it maintains the space connected. I have a final question: is the Moore space simply connected? – mtsecco Nov 11 '16 at 02:38
  • Not if $H_1$ is nontrivial. But for higher Moore spaces you should be able to choose them simply connected. – Matt Samuel Nov 11 '16 at 02:49
  • Thank you very much! You helped me a lot. – mtsecco Nov 11 '16 at 02:50
  • You're welcome. – Matt Samuel Nov 11 '16 at 02:51
  • Just a very last point. There are spaces which have $H_1=0$ but $\pi_1$ is non-trivial - see perfect groups. – Dan Rust Nov 11 '16 at 13:06
  • @Dan That's why I only said you should be able to choose them simply connected. I meant to say probably. Off the top of my head I can't rule out acyclic Moore spaces with nontrivial fundamental group. – Matt Samuel Nov 14 '16 at 01:42
  • @MattSamuel I'm guessing that taking the universal cover of a Moore space isn't necessarily a Moore space? – Dan Rust Nov 14 '16 at 02:04
  • @Dan I don't think it is necessarily. In any case Moore spaces in Hatcher are constructed as CW complexes with cells attached to destroy the extra homology classes, probably along the way you can attach 2-cells to kill the loops. (I'm by no means an expert by the way. Took two courses in algebraic topology 10 years ago with Hatcher, but also read the book outside of class often). – Matt Samuel Nov 14 '16 at 02:11

0 Answers0