In differential geometry, the integration along fibers of a k-form yields a -form where m is the dimension of the fiber, via "integration".
Definition
Let be a fiber bundle over a manifold with compact oriented fibers. If is a k-form on E, then for tangent vectors wi's at b, let
where is the induced top-form on the fiber ; i.e., an -form given by: with the lifts of to E,
(To see is smooth, work it out in coordinates; cf. an example below.)
Then is a linear map . By Stokes' formula, if the fibers have no boundaries, the map descends to de Rham cohomology:
This is also called the fiber integration.
Now, suppose is a sphere bundle; i.e., the typical fiber is a sphere. Then there is an exact sequence , K the kernel,
which leads to a long exact sequence, dropping the coefficient and using :
- ,
called the Gysin sequence.
Example
Let be an obvious projection. First assume with coordinates and consider a k-form:
Then, at each point in M,
- [1]
From this local calculation, the next formula follows easily: if is any k-form on
where is the restriction of to .
As an application of this formula, let be a smooth map (thought of as a homotopy). Then the composition is a homotopy operator:
which implies induces the same map on cohomology, the fact known as the homotopy invariance of de Rham cohomology. As a corollary, for example, let U be an open ball in Rn with center at the origin and let . Then , the fact known as the Poincaré lemma.
Given a vector bundle π : E → B over a manifold, we say a differential form α on E has vertical-compact support if the restriction has compact support for each b in B. We written for the vector space of differential forms on E with vertical-compact support.
If E is oriented as a vector bundle, exactly as before, we can define the integration along the fiber:
The following is known as the projection formula.[2] We make a right -module by setting .
Proof: 1. Since the assertion is local, we can assume π is trivial: i.e., is a projection. Let be the coordinates on the fiber. If , then, since is a ring homomorphism,
Similarly, both sides are zero if α does not contain dt. The proof of 2. is similar.
See also
Notes
- ^ If , then, at a point b of M, identifying 's with their lifts, we have:
and so
Hence,
By the same computation, if dt does not appear in α.
- ^ Bott−Tu 1982, Proposition 6.15. harvnb error: no target: CITEREFBott−Tu1982 (help); note they use a different definition than the one here, resulting in change in sign.
References