Cup product: Definition from Answers.com

In mathematics, specifically in algebraic topology, the cup product is a method of adjoining two cocycles of degree p and q to form a composite cocycle of degree p + q. This defines an associative (and distributive) graded commutative product operation in cohomology, turning the cohomology of a space X into a graded ring, H^∗(X), called the cohomology ring. The cup product was introduced in work of J. W. Alexander, Eduard Čech and Hassler Whitney from 1935–1938, and, in full generality, by Samuel Eilenberg in 1944.

1 Definition
2 Properties
3 Interpretation
4 Examples
5 Other definitions
- 5.1 Cup product and differential forms
- 5.2 Cup product and geometric intersections
6 Massey products
7 See also
8 References

Definition

In singular cohomology, the cup product is a construction giving a product on the graded cohomology ring H^∗(X) of a topological space X.

The construction starts with a product of cochains: if c^p is a p-cochain and d^q is a q-cochain, then

$(c^p \smile d^q)(\sigma) = c^p(\sigma \circ \iota_{0,1, ... p}) \cdot d^q(\sigma \circ \iota_{p, p+1 ,..., p + q})$

where σ is a (p + q) -singular simplex and $\iota_S , S \subset \{0,1,...,p+q \}$ is the canonical embedding of the simplex spanned by S into the $(p + q)$ -standard simplex.

Informally, $\sigma \circ \iota_{0,1, ..., p}$ is the p-th front face and $\sigma \circ \iota_{p, p+1, ..., p + q}$ is the q-th back face of σ, respectively.

The coboundary of the cup product of cocycles c^p and d^q is given by

$\delta(c^p \smile d^q) = \delta{c^p} \smile d^q + (-1)^p(c^p \smile \delta{d^q}).$

The cup product of two cocycles is again a cocycle, and the product of a coboundary with a cocycle (in either order) is a coboundary. Thus, the cup product operation passes to cohomology, defining a bilinear operation

$H^p(X) \times H^q(X) \to H^{p+q}(X).$

Properties

The cup product operation in cohomology satisfies the identity

$\alpha^p \smile \beta^q = (-1)^{pq}(\beta^q \smile \alpha^p)$

so that the corresponding multiplication is graded-commutative.

The cup product is functorial, in the following sense: if

$f\colon X\to Y$

is a continuous function, and

$f^*\colon H^*(Y)\to H^*(X)$

is the induced homomorphism in cohomology, then

$f^*(\alpha \smile \beta) =f^*(\alpha) \smile f^*(\beta),$

for all classes α, β in H ^*(Y). In other words, f ^* is a (graded) ring homomorphism.

Interpretation

It is possible to view the cup product $\smile \colon H^p(X) \times H^q(X) \to H^{p+q}(X)$ as induced from the following composition:

$\displaystyle C^\bullet(X) \times C^\bullet(X) \to C^\bullet(X \times X) \overset{\Delta^*}{\to} C^\bullet(X)$

in terms of the chain complexes of $X$ and $X \times X$ , where the first map is the Künneth map and the second is the map induced by the diagonal $\Delta \colon X \to X \times X$ .

This composition passes to the quotient to give a well-defined map in terms of cohomology, this is the cup product. This approach explains the existence of a cup product for cohomology but not for homology: $\Delta \colon X \to X \times X$ induces a map $\Delta^* \colon H^\bullet(X \times X) \to H^\bullet(X)$ but would also induce a map $\Delta_* \colon H_\bullet(X) \to H_\bullet(X \times X)$ , which goes the wrong way round to allow us to define a product. This is however of use in defining the cap product.

Bilinearity follows from this presentation of cup product, i.e. $(u_1 + u_2) \smile v = u_1 \smile v + u_2 \smile v$ and $u \smile (v_1 + v_2) = u \smile v_1 + u \smile v_2$ .

Examples

As singular spaces, the 2-sphere S² with two disjoint 1-dimensional loops attached by their endpoints to the surface and the torus T have identical cohomology groups in all dimensions, but the multiplication of the cup product distinguishes the associated cohomology rings. In the former case the multiplication of the cochains associated to the loops is degenerate, whereas in the latter case multiplication in the first cohomology group can be used to decompose the torus as a 2-cell diagram, thus having product equal to Z (more generally M where this is the base module).

Other definitions

Cup product and differential forms

In de Rham cohomology, the cup product of differential forms is also known as the wedge product, and in this sense is a special case of Grassmann's exterior product.

Cup product and geometric intersections

The linking number can be defined in terms of a non-vanishing cup product on the complement of a link. The complement of these two linked circles deformation retracts to a torus, which has a non-vanishing cup product.

When two submanifolds of a smooth manifold intersect transversely, their intersection is again a submanifold. By taking the fundamental homology class of these manifolds, this yields a bilinear product on homology. This product is dual to the cup product, i.e. the homology class of the intersection of two submanifolds is the Poincaré dual of the cup product of their Poincaré duals.

Similarly, the linking number can be defined in terms of intersections, shifting dimensions by 1, or alternatively in terms of a non-vanishing cup product on the complement of a link.

Massey products

Massey products generalize cup product, allowing one to define "higher order linking numbers", the Milnor invariants.

Main article: Massey product

The cup product is a binary (2-ary) operation; one can define a ternary (3-ary) and higher order operation called the Massey product, which generalizes the cup product. This is a higher order cohomology operation, which is only partly defined (only defined for some triples).

References

James R. Munkres, "Elements of Algebraic Topology", Perseus Publishing, Cambridge Massachusetts (1984) ISBN 0-201-04586-9 (hardcover) ISBN 0-201-62778-0 (paperback)

Glen E. Bredon, "Topology and Geometry", Springer-Verlag, New York (1993) ISBN 0-387-97926-3

Allen Hatcher, "Algebraic Topology", Cambridge Publishing Company (2002) ISBN 0-521-79540-0

This entry is from Wikipedia, the leading user-contributed encyclopedia. It may not have been reviewed by professional editors (see full disclaimer)

Donate to Wikimedia

	How many cups are there in 1 cup? Read answer...
	Which cup can hold heat better a plastic cup a coffee cup or a glass cup? Read answer...
	What are the kinds of cups? Read answer...

	13 of a cup is how many cups?
	Shourtning cups to onces to cups?
	How much is 35g of flour in cups?

		Sci-Tech Dictionary. McGraw-Hill Dictionary of Scientific and Technical Terms. Copyright © 2003, 1994, 1989, 1984, 1978, 1976, 1974 by McGraw-Hill Companies, Inc. All rights reserved. Read more
		Wikipedia. This article is licensed under the Creative Commons Attribution/Share-Alike License. It uses material from the Wikipedia article "Cup product". Read more

Cup product

Contents