----- $R$ is a [[ring]] (say, a [[PID]]). > [!proposition] Proposition. ([[the perfect Poincare pairing]]) > Let $M$ be a $d$-dimensional [[compact]] $R$-[[(homological) orientation of a manifold|oriented]] [[manifold]]. Consider the [[bilinear map|pairing]] $\begin{align} \langle -,- \rangle : H^{k}(M; R) \otimes H^{d-k} (M,R) & \to R \\ [\varphi] \otimes [\psi] & \mapsto (\varphi \smile \psi)[M]. \end{align}$ If $H_{*}(M;R)$ is [[free module|free]] (so that [[universal coefficients theorem for homology|UCT]] applies), then $\langle -,- \rangle$ is [[nondegenerate bilinear form|nondegenerate]] (in fact, [[perfect pairing|perfect]].) ^proposition > [!note] Remark. > This is a corollary of [[Poincare duality]]. [[Poincare duality]] concerns the [[cap product]], not the [[cup product]], but since the two determine each other we can use [[Poincare duality]] to draw conclusions about the [[cup product]]. Here, the punchline is that while we have seen before examples where we can figure out if a [[cup product]] vanishes, this result tells us that certain cup products are *not* zero. That's new! ^note > [!intuition] > Informally, the idea is that if $a \in H^{k}(M)$ is nonzero, then there exists some $b \in H^{d-k}(M)$ with $a \smile b \neq 0$. This in turn puts a lot of constraint on what the [[cup product]] can be. ^intuition > [!basicexample] Example. (Computing (again) the [[singular cohomology|cohomology ring]] for $\mathbb{C}P^{n}$) > Let's forget that [[Gysin sequence|we have already computed]] the [[singular cohomology|cohomology ring]] structure on [[complex projective space]] $\mathbb{C}P^{n}$, and do it again. > > We know $H_{*}(\mathbb{C}P^{n}, \mathbb{Z}) = \begin{cases} > \mathbb{Z} & *=2i, 0 \leq i \leq n \\ > 0 & \text{else.} > \end{cases}$ > Also, we know $\mathbb{C}P^{n}$ is $\mathbb{Z}$-[[(homological) orientation of a manifold|oriented]] (as is any complex [[manifold]]). Suppose for induction that we know $H^{*}(\mathbb{C}P^{n-1})\cong\mathbb{Z}[x] / x^{n}, x \text{ has degree } 2.$ > As $\mathbb{C}P^{n}$ is obtained from $\mathbb{C}P^{n-1}$ by [[cell complex|attaching a]] $2$-cell, the [[inclusion map|inclusion]] $\mathbb{C}P^{n-1} \hookrightarrow \mathbb{C}P^{n}$ [[homomorphism on cohomology induced by a cochain map|induces]] an [[isomorphism]] on [[singular cohomology|cohomology]] in degrees lt;2n$ (e.g. per [[how singular homology interacts with cell complexes]]). Since $\langle -,- \rangle :H^{2}(\mathbb{C}P^{n}) \otimes H^{2n-2}(\mathbb{C}P^{n}) \xrightarrow{- \smile -}H^{2n}(\mathbb{C}P^{n}) \xrightarrow{-([\mathbb{C}P^{n}])}\mathbb{Z}$ > is [[nondegenerate bilinear form|nondegenerate]], $\langle x, - \rangle : \mathbb{Z} \to \mathbb{Z}$ is an [[isomorphism]], so it must map generators to generators; in particular, $\langle x,x^{n-1} \rangle = \pm 1$. Now, by definition $\langle x, x^{n-1} \rangle = (\underbrace{ x \smile x^{n-1} }_{ x^{n} })[\mathbb{C}P^{n}]$ > which implies $x^{n}$ must [[submodule generated by a subset|generate]] $H^{2n}(\mathbb{C}P^{n})$: after all, something is a generator of $H^{2n}(\mathbb{C}P^{n})$ (viewed as $\mathbb{Z}$) iff it evaluates against the [[The Thom Theorem for oriented manifolds|fundamental class]] to $\pm 1$.[^1] > > [!proof]- Proof. ([[the perfect Poincare pairing]]) > ~ > > Consider > $\begin{align} > H^{k}(M; R) & \xrightarrow[\cong, \text{UCT}]{h} \text{Hom}_{R}\big( H_{k} (M;R), R\big) \\ > & \xrightarrow[\cong]{D_{M}^{\vee}} \text{Hom}_{R}\big( H^{d-k}(M; R), R \big) \\ > &= \big(H^{d-k}(M; R)\big)^{\vee} > \end{align}$ > where $\text{UCT}$ gives the stated isomorphism because $H_{*}$ are all free. This map is $[\varphi] \mapsto \big([\psi] \mapsto [\varphi](D_{M}([\psi]))\big) = [\varphi]([M] \frown [\psi])=([\psi] \smile [\varphi])[M]$ > which is exactly $\langle [\psi], [\varphi] \rangle$. > > The other [[adjoint]] is similar. ----- #### [^1]: Well, in general for $M$ [[connected]] in addition to the other assumptions within this note, [[Poincare duality]] gives an [[isomorphism]] ($\text{dim }M=d$) $\begin{align} H^{d}(M;R) &\to H_{0}(M;R) \cong R \\ \alpha & \mapsto [M] \frown \alpha = \alpha([M]). \end{align}$\mathbb{C}P^{n}$ is connected; here, $R=\mathbb{Z}$. In other words, evaluating a top-degree cohomology class against the fundamental class is an isomorphism in situations such as this. ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```