Integration along fibers explained

In differential geometry, the integration along fibers of a k-form yields a

(k-m)

-form where m is the dimension of the fiber, via "integration". It is also called the fiber integration.

Definition

Let

\pi:E\toB

be a fiber bundle over a manifold with compact oriented fibers. If

\alpha

is a k-form on E, then for tangent vectors wi's at b, let

(\pi*\alpha)b(w1,...,wk-m)=

\int
\pi-1(b)

\beta

where

\beta

is the induced top-form on the fiber

\pi-1(b)

; i.e., an

m

-form given by: with

\widetilde{wi}

lifts of

wi

to

E

,

\beta(v1,...,vm)=\alpha(v1,...,vm,\widetilde{w1},...,\widetilde{wk-m

}).

(To see

b\mapsto(\pi*\alpha)b

is smooth, work it out in coordinates; cf. an example below.)

Then

\pi*

is a linear map

\Omegak(E)\to\Omegak-m(B)

. By Stokes' formula, if the fibers have no boundaries(i.e.

[d,\int]=0

), the map descends to de Rham cohomology:

\pi*:\operatorname{H}k(E;R)\to\operatorname{H}k-m(B;R).

This is also called the fiber integration.

Now, suppose

\pi

is a sphere bundle; i.e., the typical fiber is a sphere. Then there is an exact sequence

0\toK\to\Omega*(E)\overset{\pi*}\to\Omega*(B)\to0

, K the kernel,which leads to a long exact sequence, dropping the coefficient

R

and using

\operatorname{H}k(B)\simeq\operatorname{H}k+m(K)

:

\operatorname{H}k(B)\overset{\delta}\to\operatorname{H}k+m+1(B)\overset{\pi*}\operatorname{H}k+m+1(E)\overset{\pi*}\operatorname{H}k+1(B)

,called the Gysin sequence.

Example

Let

\pi:M x [0,1]\toM

be an obvious projection. First assume

M=Rn

with coordinates

xj

and consider a k-form:

\alpha=f

dx
i1

\wedge...\wedge

dx
ik

+gdt\wedge

dx
j1

\wedge...\wedge

dx
jk-1

.

Then, at each point in M,

\pi*(\alpha)=\pi*(gdt\wedge

dx
j1

\wedge...\wedge

dx
jk-1

)=\left(

1
\int
0

g(,t)dt\right)

{dx
j1

\wedge...\wedge

dx
jk-1
}.[1]

From this local calculation, the next formula follows easily (see Poincaré_lemma#Direct_proof): if

\alpha

is any k-form on

M x [0,1],

\pi*(d\alpha)=\alpha1-\alpha0-d\pi*(\alpha)

where

\alphai

is the restriction of

\alpha

to

M x \{i\}

.

As an application of this formula, let

f:M x [0,1]\toN

be a smooth map (thought of as a homotopy). Then the composition

h=\pi*\circf*

is a homotopy operator (also called a chain homotopy):

d\circh+h\circd=

*
f
1

-

*:
f
0

\Omegak(N)\to\Omegak(M),

which implies

f1,f0

induce 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

ft:U\toU,x\mapstotx

. Then

\operatorname{H}k(U;R)=\operatorname{H}k(pt;R)

, the fact known as the Poincaré lemma.

Projection formula

Given a vector bundle π : EB over a manifold, we say a differential form α on E has vertical-compact support if the restriction

\alpha|
\pi-1(b)
has compact support for each b in B. We write
*(E)
\Omega
vc
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:

\pi*:

*(E)
\Omega
vc

\to\Omega*(B).

The following is known as the projection formula.[2] We make

*(E)
\Omega
vc
a right

\Omega*(B)

-module by setting

\alpha\beta=\alpha\wedge\pi*\beta

.

Proof: 1. Since the assertion is local, we can assume π is trivial: i.e.,

\pi:E=B x Rn\toB

is a projection. Let

tj

be the coordinates on the fiber. If

\alpha=gdt1\wedge\wedgedtn\wedge\pi*η

, then, since

\pi*

is a ring homomorphism,

\pi*(\alpha\wedge\pi*\beta)=\left(

\int
Rn

g(,t1,...,tn)dt1...dtn\right)η\wedge\beta=\pi*(\alpha)\wedge\beta.

Similarly, both sides are zero if α does not contain dt. The proof of 2. is similar.

\square

See also

References

Notes and References

  1. If

    \alpha=gdt\wedged

    x
    j1

    \wedge\wedged

    x
    jk-1
    , then, at a point b of M, identifying
    \partial
    xj
    's with their lifts, we have:

    \beta(\partialt)=\alpha(\partialt,

    \partial
    x
    j1

    ,...,

    \partial
    x
    jk-1

    )=g(b,t)

    and so

    \pi*(\alpha)b(\partial

    x
    j1

    ,...,

    \partial
    x
    jk-1

    )=\int[0,\beta=

    1
    \int
    0

    g(b,t)dt.

    Hence,

    \pi*(\alpha)b=\left(

    1
    \int
    0

    g(b,t)dt\right)d

    x
    j1

    \wedge\wedged

    x
    jk-1

    .

    By the same computation,

    \pi*(\alpha)=0

    if dt does not appear in α.
  2. note they use a different definition than the one here, resulting in change in sign.