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What is the necessary and sufficient condition for a CW-complex to have its homology groups torsion-free?

Definition from Wikipedia: An abelian group $A$ is called torsion-free if every element of $A$ except the identity is of infinite order.

Examples:

  1. $H_i(S^n)=\mathbb{Z}$ if $\in \{0, n\}$, $H_i(S^n)=\mathbb{0}$ if $i \notin \{0, n\}$, so all its homology groups all torsion-free.

  2. Projective space, $H_0(\mathbb{R}P^2)=\mathbb{Z}$, $H_1(\mathbb{R}P^2)=\mathbb{Z_2}$, $H_i(\mathbb{R}P^2)=0, i>1$. The first homology group is not torsion-free.

When reading Wikipedia, it said that the orientability of the projective spaces is obvious from its homology groups. http://en.wikipedia.org/wiki/Real_projective_space#Homology

Question: Is the orientability the necessary and sufficient condition for a CW-complex to have its homology groups torsion-free? If not which direction is implied and what other condition would be sufficient and necessary?

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    $\begingroup$ What's your definition of orientability of a cw-complex? $\endgroup$
    – user17786
    Commented Apr 22, 2013 at 19:46
  • $\begingroup$ @user17786 I thought there was a standard definition of orientability. Looking at wikipedia, they define orientability for manifolds and surfaces without the generality. So there is no general definition for orientability? $\endgroup$ Commented Apr 23, 2013 at 3:55
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    $\begingroup$ I don't know about it. Here Hatcher says that he neither doesn't know about notions of orientability for general simplicial complexes: mathoverflow.net/questions/23713/… $\endgroup$
    – user17786
    Commented Apr 24, 2013 at 19:44

2 Answers 2

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As mentioned above, there is no standard notion of "orientability" for CW-complexes. I will assume you meant manifolds instead.

  • The open Möbius band is a nonorientable manifold. It has the homotopy type of a circle, hence its homology groups are free abelian.

  • Let $M$ be a compact, non-orientable $n$-manifold. By Poincaré Duality, $\mathbb{Z}/2\mathbb{Z}$-Poincaré Duality and the Universal Coefficient Theorem, both $H_1(M,\mathbb{Z})$ and $H_{n-1}(M,\mathbb{Z})$ have nontrivial $2$-torsion.

  • $\mathbb{R} \mathbb{P}^3$ is orientable but has first homology group $\mathbb{Z}/2\mathbb{Z}$.
  • A nonorientable manifold admits a connected $2$-sheeted orientation covering. In particular its fundamental group admits an index $2$ subgroup, so for instance any simply connected manifold must be orientable. This does not mean that there must be an order $2$ element in the fundamental group of a nonorientable manifold: e.g. it follows from basic group cohomology that the fundamental group of any manifold with contractible universal cover -- so in particular, every nonorientable closed surface other than the projective plane -- has torsionfree fundamental group.
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Regarding the assertion in wikipedia, recall that a compact connected manifold (without boundary) of dimension $n$ is orientable if and only if the top homology group $H_n(X;\mathbb Z)\cong \mathbb Z$ (see in Hatcher's Algebraic topology the section entitled Orientations and homology, chapter 3.3).

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    $\begingroup$ That's true, but if it's nonorientable $H_n(X,\mathbb{Z}) = 0$. Either way it's torsionfree. $\endgroup$ Commented Apr 23, 2013 at 18:52
  • $\begingroup$ My answer was regarding the assertion of wikipedia, not the question itself, which was properly answered by yourself. I just had the feeling that the question linked that assertion with the torsion in homology, and wanted to show that they have nothing to do. $\endgroup$
    – user17786
    Commented Apr 24, 2013 at 19:39

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