Register to reply

The Relation between Cup Product and Wedge Product

by lavinia
Tags: product, relation, wedge
Share this thread:
lavinia
#1
Jan22-13, 05:09 PM
Sci Advisor
P: 1,716
On a smooth triangulation of a manifold differential forms can be viewed as real cochains by integration. The wedge product of two forms gives another real cochain. So does their cup product.

- are they cohomologous?

- Is there a limiting process that relates them?
Phys.Org News Partner Science news on Phys.org
Hoverbike drone project for air transport takes off
Earlier Stone Age artifacts found in Northern Cape of South Africa
Study reveals new characteristics of complex oxide surfaces
mathwonk
#2
Jan24-13, 04:35 PM
Sci Advisor
HW Helper
mathwonk's Avatar
P: 9,453
isn't this the content of theorem 14.28 of Bott-Tu, and the assertions on page 192?

I.e. they say that the deRham cohomology with wedge products, is isomorphic as an algebra to the Cech cohomology with cup products, and they seem to say that this also holds for singular and Cech cohomologies with cup products.

Isn't it more or less Fubini's theorem? i.e. to compare the integration action of a wedge product on a cell, with the cup product value?

Have you tried these "obvious" approaches and run into difficulties checking them out?
lavinia
#3
Jan25-13, 04:43 PM
Sci Advisor
P: 1,716
Quote Quote by mathwonk View Post
isn't this the content of theorem 14.28 of Bott-Tu, and the assertions on page 192?

I.e. they say that the deRham cohomology with wedge products, is isomorphic as an algebra to the Cech cohomology with cup products, and they seem to say that this also holds for singular and Cech cohomologies with cup products.

Isn't it more or less Fubini's theorem? i.e. to compare the integration action of a wedge product on a cell, with the cup product value?

Have you tried these "obvious" approaches and run into difficulties checking them out?
My thought was that one approximates the integral by the product of the integrals on the faces of the cube. Then subdivision should improve the approximation. The integral of the wedge product would be the limit.

Before taking any limits, are the wedge and cup products - viewed as smooth real valued cochains- cohomologous?

I don't know anything about Czech cohomology. Thanks for the info.

mathwonk
#4
Jan26-13, 10:45 PM
Sci Advisor
HW Helper
mathwonk's Avatar
P: 9,453
The Relation between Cup Product and Wedge Product

well i am embarrassed to realize i know diddle about cech cohomology myself after all these years, but i want to recommend it very much to you as a beautiful and natural theory.

start from an open cover, in particular look at a polyhedron like an icosahedron. to each vertex associate the open faces (and edges) adjacent to that vertex (the "star" of the vertex). this is the canonical open cover. each open set is thought of as a vertex.

then intersect two such open sets and consider that as a 1 simplex, i.e. an edge.
notice that in our example the intersection of two open sets gives the open faces adjacent to the edge joining the two vertices defining the original open stars.

intersecting three such open stars associated to the vertices of a face, gives that open face. thus abstractly, a k simplex is the intersection of any k+1 open sets of the original open cover.


this is a beautiful construction since it allows one to consider functions defined on the open sets associated to the k simplices. this allows a natural way to use say smooth functions as coefficients for ones cohomology rather than just integers.

since one can consider constant functions as well as smooth functions as coefficients, this allows cech cohomology to be used as a bridge between singular and derham cohomology. this is the approach in bott tu.


Register to reply

Related Discussions
Wedge product and cross product Atomic, Solid State, Comp. Physics 0
Wedge Product Calculus 5
Wedge product Differential Geometry 8
Wedge Product Linear & Abstract Algebra 3
Wedge Product/Cross Product? General Math 2