In Hatcher, Poincaré duality is stated for coefficients in a ring rather than a general abelian group. I am wondering whether it also holds when taking coefficients in the circle group, which is not a ring.
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1The circle group is no ring. – Paul Frost Mar 20 '23 at 14:32
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Indeed. If it was a ring, the statement would apply. My question is: are $H_k(M,\mathbb{T})$ and $H^{n-k}(M,\mathbb{T})$ isomorphic? Or is there a counter-example? I edit my question to make it more clear. – mathieu Mar 20 '23 at 15:19
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Poincaré duality does not actually require a ring structure on the coefficients. Suppose $R$ is a ring and $M$ is an $R$-module. The scalar multiplication of $M$ then gives a cap product operation $H_i(X;R)\times H^j(X;M)\to H_{i-j}(X;M)$. When $X$ is an $R$-orientable closed $n$-manifold with fundamental class $[X]\in H_n(X;R)$, the cap product with $[X]$ is then an isomorphism of $R$-modules $H^k(X;M)\to H_{n-k}(X;M)$ for each $k$. This can be proved in essentially the same way as the special case $M=R$, which is just Poincaré duality with coefficients in a ring.
In particular, when $R=\mathbb{Z}$, this says that $H^k(X;A)\cong H_{n-k}(X;A)$ for an arbitrary coefficient abelian group $A$.
Eric Wofsey
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It is very enlightening, I was not aware of this generality. Thank you very much! – mathieu Mar 21 '23 at 08:32