During one of my meetings with my advisor, I overheard her give a problem to another grad student, which I decided to do as a warmup to my research day. It ended up having a cute solution, so I wanted to jot it down.
Question: If \(M\) is an \(\mathbb{Z}_2\)-homology \(S^3\), what could we say about its integral homology?
We will make direct use of the universal coefficient theorem in homology:
\[ 0 \to H_n(M; \mathbb{Z}) \otimes \mathbb{Z}_2 \to H_n(M; \mathbb{Z}_2)\to \text{Tor}(H_{n-1}(M;\mathbb{Z}); \mathbb{Z}_2)\to 0 \]
and a little bit of Poincare duality. To start, let’s show that \(H_1(M)\) must be a direct sum of cyclic groups of odd order. Start by assuming that
\[H_1(M) \cong \mathbb{Z}^{f} \oplus \bigoplus_j\mathbb{Z}_{1_j}\]
Applying the UCT to when \(n=2\), as well as the fact that \(M\) is a \(\mathbb{Z}_2\)-homology sphere, implies that \( \text{Tor}(H_{1}(M;\mathbb{Z});\mathbb{Z}_2) =0\). But now we can use our presentation of \(H_1(M)\) and properties of the Tor functor to say that
\[0= \text{Tor}(H_{1}(M;\mathbb{Z});\mathbb{Z}_2) \cong \bigoplus_j \ker (\times 1_j:\mathbb{Z}_2 \to \mathbb{Z}_2)\]
So if we want the entire term to vanish, we better have the kernels all vanish, which happens precisely when the \(1_i\)’s are odd numbers.
Next, we can show that \(f=0\) by applying the UCT to \(n=1\). Since \(H_1(M; \mathbb{Z}_2) = 0\), we have that \(H_1(M; \mathbb{Z})\otimes \mathbb{Z}_2 = 0\), which implies that the free part of \(H_1(M)\) vanishes.
Finally, we can use Poincare duality to show that in fact \(H_2(M; \mathbb{Z})\) must be trivial.
Poincare duality implies that \(H_2(M; \mathbb{Z})\cong H^1(M; \mathbb{Z})\), and the latter is isomorphic to \(\text{Hom}(H_1(M;\mathbb{Z}),\mathbb{Z} )\) via the UCT in cohomology. But since we have already deduced that \(H_1\) is only torsion, that hom set must totally vanish as well.
To conclude, we might note that these obstructions are sharp, in the sense that these are the best conditions we could put on \(H_1\). For example, \(L(p,1)\) for odd \(p\) satisfies \(H_1(L(p,1))\cong \mathbb{Z}_p\).
Answer: \(M\) must be a homology \(L(p,1)\) for odd \(p\).
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