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Question Without using de Rham's theorem, prove:

(1) Show $H_{dR}^1(S^n)=0$ for $n>1$.

(2) Use (1) to show $H_{dR}^1(RP^n)=0$

(3) a n-form $\Omega$ is exact on $S^n$ if and only if $$ \int_{S^n}\,\Omega =0$$

(4) Use (3) to show that every smooth 2-form $\omega$ on $RP^2$ has the form $d \alpha $ for some smooth 1-form $\alpha$

I know I can prove $H_{dR}^1(S^n)=0$ by de-Rham theorem and Universal coefficient theorem. But perhaps we are expected to use some simpler tool to prove it.

Moreover, What I am really wondering is how to use (1) to prove (2) and use (3) to prove (4)?

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  • $\begingroup$ (1) is likely asking you to prove directly that every closed $1$-form is also exact. As for (2), note that thanks to the canonical projection $\pi \colon S^n \to \Bbb{RP}^n$ you can lift every function defined on $\Bbb{RP}^n$ to a function defined on $S^n$. In particular, this holds for $1$-forms, so you're done if you can show that $\omega \circ \pi$ is closed when $\omega$ is closed. $\endgroup$
    – A.P.
    May 24, 2015 at 16:21
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    $\begingroup$ Of course, a quick option for (1) is to use the Mayer-Vietoris sequence for the de Rham cohomology, if you have that tool at your disposal. $\endgroup$
    – A.P.
    May 24, 2015 at 16:26

1 Answer 1

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When $n>1$, $S^n$ is simply connected, and $H^1_{dR}(M) = 0$ for any simply connected manifold $M$. (Hint: For any closed $1$-form $\omega$ and any closed curve $\gamma$, we have $\displaystyle\int_\gamma\omega = 0$.)

Now, in order to deal with $\Bbb RP^n$, consider the $\Bbb Z/2\Bbb Z$ action given by the deck transformations. Show that any exact invariant $1$-form comes from an invariant function and hence descends to an exact $1$-form on $\Bbb RP^n$.

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  • $\begingroup$ Thank you! But could you please give more detail about how to show any exact invariant 1-form comes from an invariant function? $\endgroup$
    – Hang
    May 25, 2015 at 8:53
  • $\begingroup$ Sure. Let $\alpha$ be the antipodal map of the sphere. If $\omega=df$ is an invariant exact $1$-form, we have $\alpha^*\omega = \omega$, so $d(f\circ\alpha) = df = \omega$. Thus, if we set $g = \frac12(f + f\circ\alpha)$, we see that $g$ is $\Bbb Z/2\Bbb Z$-invariant and $dg = \omega$. $\endgroup$
    – Ted Shifrin
    May 25, 2015 at 11:49
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    $\begingroup$ This is a great answer. Thanks for the answer - I learned a lot. $\endgroup$
    – Bombyx mori
    May 23, 2016 at 5:06

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