1
$\begingroup$

Let $B$ be a closed oriented $n$-manifold, and $\pi_N:N\to B$ be an oriented $m$-dim ball bundle, i.e. each fiber is an oriented $m$-dim ball(disk) $D^m$. We have a sphere bundle $\pi_\partial:\partial N\to B$, where $\iota: \partial N\to N$ is the boundary of $N$.

Let $\omega\in \Omega^n(N)$ be a closed form. By the pullback of $\iota$ and pushforward of $\pi_\partial$, we have a closed form $$ \omega_b=\frac{1}{\mathrm{Vol}(S^{m-1})}\pi_{\partial,*}(\iota^*\omega\wedge \mathrm{vol}_{S^{m-1}}), $$ where

  • $\mathrm{vol}_{S^{m-1}}$ denotes the standard volume form of $S^{m-1}$,
  • $\mathrm{Vol}(S^{m-1})$ denotes the standard volume of $S^{m-1}$,
  • $\pi_{\partial,*}$ denotes the pushforward map, i.e. the integral over the fiber.

Question Can we say that $\int_B\omega_b=\int_B \omega$? In particular, when $N$ is ball bundle of an oriented vector bundle over $B$, is this true?

PS: The version is modified by Daniele Tampieri, I just removed an extral word.

Improve this question
$\endgroup$
5
  • $\begingroup$ Should the second integral in the Question be over $N$, rather than $B$? $\endgroup$
    – Mark Grant
    Commented Jun 12, 2022 at 18:28
  • $\begingroup$ Sorry, since have a section $s:B\to N$, the second one means $\int_B(s^*\omega)$, and as $\omega$ is closed, we can show that the integral is indepedent of the section. @MarkGrant $\endgroup$
    – DLIN
    Commented Jun 13, 2022 at 2:24
  • $\begingroup$ Ok, thanks. But now I realise I don't understand the definition of $\omega_b$. Inside the brackets you take the wedge of a form on $\partial N$ with a form on $S^{m-1}$. How does one do this if the sphere bundle is non-trivial? $\endgroup$
    – Mark Grant
    Commented Jun 13, 2022 at 8:19
  • $\begingroup$ @MarkGrant The volume form is tangent to the fiber, we can construct it by using a fiber-wise metric. Recall tha Gysin sequence gives the isomrophism $H^{n+m-1}(\partial N)\cong H^n(B)$. $\endgroup$
    – DLIN
    Commented Jun 13, 2022 at 10:24
  • $\begingroup$ Maybe there are some additional assumptions you are implicitly making to have that map in the Gysin sequence be an isomorphism. Otherwise take for example the unit disk bundle of the tangent bundle of $S^2$; the boundary $\partial N$ is $RP^3$, and $H^2(RP^3)$ is not trivial while $H^1(S^2)$ is. Or, something that persists to real cohomology, $H^1(RP^3) = 0$ while $H^0(S^2) = Z$. Then one can cross everything with a circle to get different $H^2(\partial N)$ and $H^1(B)$. $\endgroup$
    – Aleksandar Milivojević
    Commented Jun 14, 2022 at 12:59

0

Reset to default

You must log in to answer this question.