Integration along fibers

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".

Definition

Let \pi: E \to B 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(w_1, \dots, w_{k-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{w_i} the lifts of w_i to E,

\beta(v_1, \dots, v_m) = \alpha(v_1, \dots, v_m, \widetilde{w_1}, \dots, \widetilde{w_{k-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 \Omega^k(E) \to \Omega^{k-m}(B). By Stokes' formula, if the fibers have no boundaries, the map descends to de Rham cohomology:

\pi_*: \operatorname{H}^k(E; \mathbb{R}) \to \operatorname{H}^{k-m}(B; \mathbb{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 \to K \to \Omega^*(E) \overset{\pi_*}\to \Omega^*(B) \to 0, K the kernel, which leads to a long exact sequence, dropping the coefficient \mathbb{R} and using \operatorname{H}^k(B) \simeq \operatorname{H}^{k+m}(K):

\cdots \rightarrow \operatorname{H}^k(B) \overset{\delta}\to \operatorname{H}^{k+m+1}(B) \overset{\pi^*} \rightarrow \operatorname{H}^{k+m+1}(E) \overset{\pi_*} \rightarrow \operatorname{H}^{k+1}(B) \rightarrow \cdots,

called the Gysin sequence.

Example

Let \pi: M \times [0, 1] \to M be an obvious projection. First assume M = \mathbb{R}^n with coordinates x_j and consider a k-form:

\alpha = f \, dx_{i_1} \wedge \dots \wedge dx_{i_k} + g \, dt \wedge dx_{j_1} \wedge \dots \wedge dx_{j_{k-1}}.

Then, at each point in M,

\pi_*(\alpha) = \pi_*(g \, dt \wedge dx_{j_1} \wedge \dots \wedge dx_{j_{k-1}}) = \left( \int_0^1 g(\cdot, t) \, dt \right) \, {dx_{j_1} \wedge \dots \wedge dx_{j_{k-1}}}.[1]

From this local calculation, the next formula follows easily: if \alpha is any k-form on M \times I,

\pi_*(d \alpha) = \alpha_1 - \alpha_0 - d \pi_*(\alpha)

where \alpha_i is the restriction of \alpha to M \times \{i\}.

As an application of this formula, let f: M \times [0, 1] \to N be a smooth map (thought of as a homotopy). Then the composition h = \pi_* \circ f^* is a homotopy operator:

d \circ h + h \circ d = f_1^* - f_0^*: \Omega^k(N) \to \Omega^k(M),

which implies f_1, f_0 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 f_t: U \to U, x \mapsto tx. Then \operatorname{H}^k(U; \mathbb{R}) = \operatorname{H}^k(pt; \mathbb{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 written \Omega_{vc}^*(E) 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_*: \Omega_{vc}^*(E) \to \Omega^*(B).

The following is known as the projection formula.[2] We make \Omega_{vc}^*(E) a right \Omega^*(B)-module by setting \alpha \cdot \beta =  \alpha \wedge \pi^* \beta.

Proposition  Let \pi: E \to B be an oriented vector bundle over a manifold and \pi_* the integration along the fiber. Then

  1. \pi_* is \Omega^*(B)-linear; i.e., for any form β on B and any form α on E with vertical-compact support,
    \pi_*(\alpha \wedge \pi^* \beta) = \pi_* \alpha \wedge \beta.
  2. If B is oriented as a manifold, then for any form α on E with vertical compact support and any form β on B with compact support,
    \int_E \alpha \wedge \pi^* \beta = \int_B \pi_* \alpha \wedge \beta.

Proof: 1. Since the assertion is local, we can assume π is trivial: i.e., \pi: E = B \times \mathbb{R}^n \to B is a projection. Let t_j be the coordinates on the fiber. If \alpha = g \, dt_1 \wedge \cdots \wedge dt_n \wedge \pi^* \eta, then, since \pi^* is a ring homomorphism,

\pi_*(\alpha \wedge \pi^* \beta) = \left( \int_{\mathbb{R}^n} g(\cdot, t_1, \dots, t_n) dt_1 \dots dt_n \right) \eta \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

Notes

  1. If \alpha = g \, dt \wedge d x_{j_1} \wedge \cdots \wedge d x_{j_{k-1}}, then, at a point b of M, identifying \partial_{x_j}'s with their lifts, we have:
    \beta(\partial_t) = \alpha(\partial_t, \partial_{x_{j_1}}, \dots, \partial_{x_{j_{k-1}}}) = g(b, t)
    and so
    \pi_*(\alpha)_b(\partial_{x_{j_1}}, \dots, \partial_{x_{j_{k-1}}}) = \int_{[0, 1]} \beta = \int_0^1 g(b, t) \, dt.
    Hence, \pi_*(\alpha)_b = \left( \int_0^1 g(b, t) \, dt \right) d x_{j_1} \wedge \cdots \wedge d x_{j_{k-1}}. By the same computation, \pi_*(\alpha) = 0 if dt does not appear in α.
  2. Bott−Tu 1982, Proposition 6.15.; note they use a different definition than the one here, resulting in change in sign.

References

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