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

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?

• What's your definition of orientability of a cw-complex? Commented Apr 22, 2013 at 19:46
• @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? Commented Apr 23, 2013 at 3:55
• 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/… Commented Apr 24, 2013 at 19:44

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.

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).

• That's true, but if it's nonorientable $H_n(X,\mathbb{Z}) = 0$. Either way it's torsionfree. Commented Apr 23, 2013 at 18:52
• 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. Commented Apr 24, 2013 at 19:39