Cup product

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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.

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[edit] 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 cp is a p-cochain and dq 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 cp and dq 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) \otimes H^q(X) \to H^{p+q}(X).

[edit] 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.

[edit] 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).

[edit] Other definitions

[edit] 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.

[edit] Cup product and geometric intersections

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.

[edit] See also

[edit] References

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