Cap product

(Redirected from Slant product)

In algebraic topology the cap product is a method of adjoining a chain of degree p with a cochain of degree q, such that qp, to form a composite chain of degree pq. It was introduced by Eduard Čech in 1936, and independently by Hassler Whitney in 1938.

Definition

Let X be a topological space and R a coefficient ring. The cap product is a bilinear map on singular homology and cohomology

${\displaystyle \frown \;:H_{p}(X;R)\times H^{q}(X;R)\rightarrow H_{p-q}(X;R).}$

defined by contracting a singular chain ${\displaystyle \sigma :\Delta \ ^{p}\rightarrow \ X}$ with a singular cochain ${\displaystyle \psi \in C^{q}(X;R),}$ by the formula :

${\displaystyle \sigma \frown \psi =\psi (\sigma |_{[v_{0},\ldots ,v_{q}]})\sigma |_{[v_{q},\ldots ,v_{p}]}.}$

Here, the notation ${\displaystyle \sigma |_{[v_{0},\ldots ,v_{q}]}}$ indicates the restriction of the simplicial map ${\displaystyle \sigma }$ to its face spanned by the vectors of the base, see Simplex.

Interpretation

In analogy with the interpretation of the cup product in terms of the Künneth formula, we can explain the existence of the cap product by considering the composition

${\displaystyle C_{\bullet }(X)\otimes C^{\bullet }(X){\overset {\Delta _{*}\otimes \mathrm {Id} }{\longrightarrow }}C_{\bullet }(X)\otimes C_{\bullet }(X)\otimes C^{\bullet }(X){\overset {\mathrm {Id} \otimes \varepsilon }{\longrightarrow }}C_{\bullet }(X)}$

in terms of the chain and cochain complexes of ${\displaystyle X}$, where we are taking tensor products of chain complexes, ${\displaystyle \Delta \colon X\to X\times X}$ is the diagonal map which induces the map ${\displaystyle \Delta _{*}}$ on the chain complex, and ${\displaystyle \varepsilon \colon C_{p}(X)\otimes C^{q}(X)\to \mathbb {Z} }$ is the evaluation map (always 0 except for ${\displaystyle p=q}$).

This composition then passes to the quotient to define the cap product ${\displaystyle \frown \colon H_{\bullet }(X)\times H^{\bullet }(X)\to H_{\bullet }(X)}$, and looking carefully at the above composition shows that it indeed takes the form of maps ${\displaystyle \frown \colon H_{p}(X)\times H^{q}(X)\to H_{p-q}(X)}$, which is always zero for ${\displaystyle p.

The slant product

The above discussion indicates that the same operation can be defined on cartesian products ${\displaystyle X\times Y}$ yielding a product

${\displaystyle \backslash \;:H_{p}(X;R)\otimes H^{q}(X\times Y;R)\rightarrow H^{q-p}(Y;R).}$

In case X = Y, the two products are related by the diagonal map.

Equations

The boundary of a cap product is given by :

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

Given a map f the induced maps satisfy :

${\displaystyle f_{*}(\sigma )\frown \psi =f_{*}(\sigma \frown f^{*}(\psi )).}$

The cap and cup product are related by :

${\displaystyle \psi (\sigma \frown \varphi )=(\varphi \smile \psi )(\sigma )}$

where

${\displaystyle \sigma :\Delta ^{p+q}\rightarrow X}$ , ${\displaystyle \psi \in C^{q}(X;R)}$ and ${\displaystyle \varphi \in C^{p}(X;R).}$

An interesting consequence of the last equation is that it makes ${\displaystyle H_{\ast }(X;R)}$ into a right ${\displaystyle H^{\ast }(X;R)-}$ module.