---- The [[Poincare duality]] map $D$ lets us transport the [[cup product]] product from [[singular cohomology|cohomology]] to [[singular homology|homology]], and the result turns out to have geometric meaning. > [!definition] Definition. ([[intersection product]]) > Let $M$ be a $\mathbb{Z}$-[[(homological) orientation of a manifold|oriented]] [[compact]] [[manifold]] of dimension $d$; $a \in H_{k}(M)$, $b \in H_{\ell}(M)$. Define the **intersection product of $a$ and $b$** as $a \cdot b:=D_{M}\big( D_{M}^{-1}(a) \smile D_{M}^{-1}(b) \big) \in H_{k+\ell-d}(M).$The geometric meaning is that if $N,L \subset M$ are [[transverse submanifolds|transverse]] [[embedded submanifold|submanifolds]], then $[N] \cdot [L]=[N \cap L]$, where $[-]$ denotes [[The Thom Theorem for oriented manifolds|fundamental class]]. ^definition > [!NOTE] The Intersection Form. > When $k+\ell=2d$, the intersection product may be viewed as a [[bilinear map|bilinear form]] $H_{k}(M;R) \times H_{\ell}(M;R) \to R$, via the composite $H_{k}(M;R) \times H_{\ell}(M;R) \to H^{d}(M;R) \xrightarrow{D_{M}=\cdot[M]} H_{0}(M; R) \xrightarrow[\cong]{\varepsilon} R,$ > where we have noted that the [[Poincare duality|duality map]] $H^{d}(M;R) \to H_{0}(M;R)$ is just evaluation against the [[The Thom Theorem for oriented manifolds|fundamental class]] $[M]$. Specifically, in this case we may view $a \cdot b$ as an element of $R$ via $\begin{align} > a \cdot b &= \varepsilon \big(( D_{M}^{-1}(a) \smile D_{M}^{-1}(b) )[M] \big) > \end{align}$ > (or as an element of $H_{0}(M; R)$ by removing the $\varepsilon$ prefix). > > [!note] Remark. > The geometric meaning can be used to really grok how the cup product behaves on certain manifolds, by understanding intersections of submanifolds. Conversely, given the cup product structure on a manifold, this lets us say something about how two submanifolds intersect: if we find that the intersection product of transverse submanifolds nonzero, then they must intersect. The [[Lefschetz fixed point theorem]] is one application of this style of thinking. ^note > [!basicnonexample] Warning. > Note that if $N,L$ are not transverse then the assertion $[N] \cdot [L]=[N \cap L]$ becomes nonsensical. ![[Pasted image 20250523110812.png|500]] ^nonexample > [!basicexample] Example. (Another way to compute the [[cup product]] structure on $\mathbb{C}P^{n}$) > We have computed the [[cup product]] structure on $\mathbb{C}P^{n}$ in [[Gysin sequence]] and in [[the perfect Poincare pairing]]. Let us forget we have done this and compute it again... > Recalling e.g. that $\mathbb{C}P^{n}$ is [[cell complex|inductively built]] out of $\mathbb{C}P^{n-1}, \mathbb{C}P^{n-2},\dots$, [[cellular homology|we know]] that the classes $[\mathbb{C}P^{k}] \in H_{2k}(\mathbb{C}P^{n}) \cong \mathbb{Z}$ generate the [[singular homology|homology groups]]. Consider the [[dual vector space|dual classes]] $\alpha_{k} \in H^{2k}(\mathbb{C}P^{n})$, $\alpha_{k}(\mathbb{C}P^{k})=1$[^1] which generate the cohomology groups. > We would like to compute $\alpha_{k_{1}}\smile \alpha_{k_{2}}$, thus obtaining the [[cup product]] structure on $\mathbb{C}P^{n}$. To do so, we want to use the intersection product $[\mathbb{C}P^{k_{1}} \cap \mathbb{C}P^{k_{2}}]$. But care is needed to ensure that we embed $\mathbb{C}P^{k_{1}}$ and $\mathbb{C}P^{k_{2}}$ into $\mathbb{C}P^{n}$ such that their intersection is transverse: if we do the naive thing of embedding via the first $k+1$ coordinates, we'll get one contained in the other, which is definitely not transverse. Instead do this: $\begin{align} \mathbb{C}P^{k_{1}}&=\{ [z_{0}:\dots: z_{k_{1}}:0\dots:0] \in \mathbb{C}P^{n} \} \\ \mathbb{C}P^{k_{2}}&= \{ [0:\dots:0:z_{n-k_{2}}:\dots: z_{n}] \in \mathbb{C}P^{n}\}, \end{align}$ that's transverse. So $[\mathbb{C}P^{k_{1}}] \cdot [\mathbb{C}P^{k_{2}}]=[\mathbb{C}P^{k_{1}} \cap \mathbb{C}P^{k_{2}}]=[\mathbb{C}P^{k_{1}+k_{2}-n}].$Since $D^{-1}([\mathbb{C}P^{k}])=\alpha_{n-k}$, have $\begin{align} \alpha_{n-k_{1}} \smile \alpha_{n-k_{2}} &=D ^{-1}([\mathbb{C}P^{k_{1}}]) \smile D ^{-1}([\mathbb{C}P^{k_{2}}]) \\ &= D ^{-1}([\mathbb{C}P^{k_{1}}] \cdot [\mathbb{C}P^{k_{2}}]) \\ &= D ^{-1}([\mathbb{C}P^{k_{1}} \cap \mathbb{C}P^{k_{2}}]) \\ &=D^{-1}([\mathbb{C}P^{k_{1}+k_{2}-n}])\\ &=\alpha_{(n-k_{1})+ (n-k_{2})}. \end{align}$ This is the cup product structure we know: cup product of two generators is a generator in the sum of the degrees (and I guess also should check vanishing past $n+1$). ^basic-example > [!basicexample] > Take the genus-2 surface $\Sigma_{g}$ and the submanifolds depicted > > ![[Pasted image 20250523124723.png]] > > With suitable orientations: $\begin{align} > H_{0}(\Sigma_{g}) &\xrightarrow[\cong]{\varepsilon} \mathbb{Z} \\ > a_{1} \cap b_{1} & \mapsto 1 \\ > a_{2} \cap b_{2} & \mapsto 1. > \end{align}$ > We have $\begin{align} > [a_{i}] \cdot [b_{i}]=[a_{i} \cap b_{i}]&=[\text{pt}] \\ > [a_{i} ] \cdot [b_{j}] = [a_{i} \cap b_{j}] &= 0, i \neq j. > \end{align}$ > > > Pick dual classes $\alpha_{i}=D([a_{i}])$, $\beta_{i}=D([b_{i}])$, and write $\gamma=D([\text{pt}])$; $\gamma$ generates $H^{2}(\Sigma_{g})$. > > So $\alpha_{1} \smile \beta_{1}=\gamma=\alpha_{2} \smile \beta_{2}$ and $\alpha_{i} \smile \alpha_{j}=0$ for $i \neq j$. > > So the cup product *does* have geometric meaning; it is not merely an algebraic tool. That said, often our manifolds are complicated and/or high-dimensional, and working algebraically is preferable to working geometrically. ---- #### [^1]: Although in general [[singular cohomology is not dualized singular homology]], in the case of $\mathbb{C}P^{n}$ it is: since every other cellular chain group in the cellular chain complex is zero, the (co)homology groups and cellular chain groups coincide. Another way to ascertain this is via the [[universal coefficients theorem for cohomology]]. ---- #### 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 > ```