Using Poincaré duality to show a closed manifold is a homology sphere

Question

Suppose that $M$ is an orientable, compact, $(n-2)$-connected, $(2n-3)$-dimensional smooth manifold, where $n$ is a natural number.

I want to show that $M$ is a homology sphere if and only if the reduced homology group $\tilde{H}_{n-1} = 0$.

Supposedly this can be done using Poincaré duality. My first thought is to use the Hurewicz isomorphisms to show that $\tilde{H}_{k} = 0$ for $k=1, \ldots, n-2$.

How can we use duality to show that higher homology groups vanish as well?

Answer 1

As is, the statement is false as stated. For example, take $M = [0, 1]$, then $H^1(M; \mathbb{Z}) = 0 \neq \mathbb{Z} = H^1(S^1; \mathbb{Z})$. However if we also assume that $M$ has no boundary (i.e. $M$ is closed), then the statement is true.

As $M$ is connected, $H_0(M; \mathbb{Z}) \cong \mathbb{Z}$. By the Hurewicz Theorem, $H_k(X; \mathbb{Z}) = 0$ for $k = 1, \dots, n-2$.

As $M$ is a closed, orientable manifold, Poincaré duality tells us that $H^i(M; \mathbb{Z}) = H_{2n-3-i}(M; \mathbb{Z})$ for every $i$, in particular $H^{2n - 3}(M; \mathbb{Z}) \cong \mathbb{Z}$ and $H^i(M; \mathbb{Z}) = 0$ for $i = n - 1, \dots, 2n - 4$.

By the Universal Coefficient Theorem, $H^i(M; \mathbb{Z}) \cong \operatorname{Hom}(H_i(M; \mathbb{Z}), \mathbb{Z})\oplus\operatorname{Ext}(H_{i-1}(M; \mathbb{Z}), \mathbb{Z})$. Recall the following facts:

• $\operatorname{Hom}(\mathbb{Z}_n, \mathbb{Z}) = 0$ and $\operatorname{Hom}(\mathbb{Z}, \mathbb{Z}) = \mathbb{Z}$,
• $\operatorname{Ext}(\mathbb{Z}_n, \mathbb{Z}) = \mathbb{Z}_n$ and $\operatorname{Ext}(\mathbb{Z}, \mathbb{Z}) = 0$.

So for $i = n - 1, \dots, 2n - 4$, we see that $H_i(M; \mathbb{Z})$ has rank zero and $H_{i-1}(M; \mathbb{Z})$ is torsion-free. Therefore $H_{n-1}(M; \mathbb{Z}) = \dots = H_{2n-4}(M; \mathbb{Z}) = 0$ and $H_{n - 2}(M; \mathbb{Z})$ is torsion free (in fact, we already know it is zero from the above).

As the homology groups of a manifold vanish for degrees greater than the dimension, we see that

$$H_i(M; \mathbb{Z}) = \begin{cases} \mathbb{Z} & i = 0, 2n-3\\ 0 & \text{otherwise}. \end{cases}$$

Therefore $M$ is a $(2n - 3)$-dimensional integral homology sphere.

By shifting dimensions, we obtain a slightly more natural statement:

An $n$-connected, $(2n + 1)$-dimensional closed manifold is an integral homology sphere.

Note, I dropped the orientability assumption because for $n > 0$ it follows from the fact that the manifold is simply connected, while for $n = 0$ it follows from the classification of one-dimensional manifolds (in fact, it follows from the classification that the manifold in question is $S^1$).