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Jun 14, 2022 at 12:59 comment added Aleksandar Milivojević 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 $H^1(B)$.
Jun 13, 2022 at 10:24 comment added DLIN @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 $H^{n+m-1}(\partial N)\cong H^n(B)$.
Jun 13, 2022 at 8:19 comment added Mark Grant 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?
Jun 13, 2022 at 7:46 history edited DLIN CC BY-SA 4.0
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Jun 13, 2022 at 2:24 comment added DLIN 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
Jun 12, 2022 at 18:28 comment added Mark Grant Should the second integral in the Question be over $N$, rather than $B$?
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Jun 12, 2022 at 5:58 history edited Daniele Tampieri CC BY-SA 4.0
Minor formatting and Math Jaxing (use of `\mathrm`)
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Jun 12, 2022 at 2:39 history asked DLIN CC BY-SA 4.0