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". It is also called the fiber 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 $$\widetilde{w_i}$$ 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(i.e. $$[d,\int]=0$$), 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,


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

From this local calculation, the next formula follows easily (see Poincaré_lemma): if $$\alpha$$ is any k-form on $$M \times [0, 1],$$


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

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 (also called a chain homotopy):


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

which implies $$f_1, f_0$$ 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 $$f_t: U \to U, x \mapsto tx$$. 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 : E → B 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 $$\Omega_{vc}^*(E)$$ 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. We make $$\Omega_{vc}^*(E)$$ a right $$\Omega^*(B)$$-module by setting $$\alpha \cdot \beta = \alpha \wedge \pi^* \beta$$.

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. $$\square$$