This site is being phased out.

Cap product

From Mathematics Is A Science
Jump to navigationJump to search

By the Kunneth theorem, $$H_{k}(M\times U)=\sum_{i+j=k}H_{i}(M)\otimes H_{j}(U),k=0,1,2,...$$ The (homology) cap product is the homomorphism $$\frown:(H^{\ast}(M)\otimes H_{\ast}(M))_{k}\rightarrow H_{k}(M)$$ given on the chain level by $$f\frown c=(1\times f)\Delta c,$$ where $\Delta$ is a diagonal approximation.

Then $a\in H_{k}(M)$ and $f\in H^{k}(M)$ are called Poincare dual if $f\frown a=1.$ In particular, the homology fundamental class $O_{M}$ and the cohomology fundamental class $\overline{O}_{M}$ of manifold $M$ are dual.

For an arbitrary finitely generated chain complex $C$ we use $\Delta : C → C⊗C$ a chain-diagonal: $$(\epsilon⊗1)\Delta = (1⊗\epsilon)\Delta,$$ where the map $\epsilon : C_0 → R$ is the augmentation map defined by $$\epsilon \big( n_1 \sigma _1 + ... + n_k\sigma _k \big)= n_1 + ... + n_k.$$ Then we use the same formula $$f\frown c=(1\times f)\Delta c,$$ to define the cap product as a chain map: $$\frown:(C^{\ast} \otimes C)_{k}\rightarrow C_{k}.$$

Alternatively we have the composition $$\frown: C \otimes C^* \overset{\Delta \otimes \mathrm{Id}}{\longrightarrow} C \otimes C \otimes C^* \overset{\mathrm{Id} \otimes \varepsilon}{\longrightarrow} C,$$ where $$\varepsilon \colon C_p \otimes C^q \to R$$ is the evaluation map (which is always $0$ except for $p=q$).

Suppose $$\sigma \in C_{p+q},\psi \in C^q,\varphi \in C^p.$$

$$\partial(\sigma \frown \psi) = (-1)^q(\partial \sigma \frown \psi - \sigma \frown \delta \psi). $$

$$f( \sigma ) \frown \psi = f(\sigma \frown f^* (\psi)). $$

$$\psi(\sigma \frown \varphi) = (\varphi \smile \psi)(\sigma).$$


Compare to cup product.