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In this question, Jrrow assumes $\int_N\omega_0\neq 0$ Why is $\deg(f)$ well-defined?

What if $\int_N\omega_0\neq 0$?

My book does not seem to address this explicitly. If $\int_N\omega_0 = 0$, then $\omega_0$ is an exact $n$-form because $H^n(N) \cong \mathbb R$, but then what? If $\int_N\omega_0\neq 0$ is somehow deduced from the assumptions, then please explain how.

My book is From Calculus to Cohomology by Ib Madsen and Jørgen Tornehave. I have a feeling that the diagram here should omit the zero element for each of the 4 vector spaces.

Update: I think I figured it out.

  • For nonzero integrals only, we have a unique $\deg_{\text{nonzero}}(f)$.

  • For zero only integrals, we have that any real number could be $\deg{\text{zero}}(f)$ including $\deg_{\text{nonzero}}(f)$.

  • Therefore, for both nonzero and zero integrals, define $\deg(f) := \deg_{\text{nonzero}}(f)$.

1 Answers1

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If $M$ is orientable without a boundary, there exists a volune form $\omega_0$ such that $\int_M\omega_0\neq 0$, the map defined on the space of $n$-forms$\Omega^n(M)$ by $I(\omega)=\int_M\omega$ is linear and surjective. The Stokes theorem implies that its kernel contains $d\Omega^{n-1}(M)$ the space of exact $n$-forms, you deduce a surjective linear map $[I]:H^n(M)=\Omega^n(M)/d\Omega^{n-1}(M)\rightarrow \mathbb{R}$ which is an isomorphism since $H^n(M)=\mathbb{R}$. This implies that if $I(\omega)=0$, $[I]([\omega])=0$ where $[\omega]$ is the cohomology class of $\Omega$, since $[I]$ is an isomorphism, $[\Omega]=0$ which is equivalent to saying that $\omega$ is exact.

Tsemo Aristide
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  • I already know $I(\omega)=0$ implies $\omega$ is exact, but then what's the degree of $f$? It looks like every real number will satisfy $\int_N(f^{*}0)=\deg(f)\int_M 0$ –  Apr 13 '19 at 13:56
  • Also, what's the point of "volume form" in your answer? Why not just "orientation form" ? –  Apr 13 '19 at 14:01
  • The volume form enables to show that $I$ is surjective. – Tsemo Aristide Apr 13 '19 at 14:01
  • Why is the degree of $f$ well-defined even if we can have $I(\omega)= 0$ please? This is my main question. For $I(\omega) \ne 0$, we must have $\deg(f)$ to exist, be unique and hold for every $\omega$ provided $I(\omega) \ne 0$. However, for $I(\omega) = 0$, I think we can have every real number be the degree of $f$ for that particular $\omega$ and therefore degree of $f$, while it exists even if $I(\omega) = 0$, is not, I believe, unique, if we indeed allow $I(\omega) = 0$. –  Apr 13 '19 at 14:38
  • To be clear, I'm asking how $\deg(f)$ is well-defined if we do not exclude $I(\omega)=0$. In the other question, $I(\omega)=0$ is excluded. In the textbook, $I(\omega)=0$ is not explicitly excluded. I conclude either 1. the book implicitly excludes $I(\omega)=0$ or 2. the book doesn't exclude $I(\omega)=0$, but then $\deg(f)$ is still well-defined. If (1), then how do we deduce that we must have $I(\omega) \ne 0$ ? If (2), then how is $\deg(f)$ still well-defined? I think $\deg(f)$ is allowed to be a different number if $I(\omega)=0$ from the unique number one gets when $I(\omega) \ne 0$ –  Apr 13 '19 at 14:45
  • How can an orientation form that is not a volume form not enable showing that I is surjective? Namely that volume forms are never exact? –  Apr 13 '19 at 14:48
  • Ok. I think I got it now and edited question. Am I correct please? –  Apr 30 '19 at 06:21