Show $H_{dR}^1(S^n)=0$ for $n>1$ without de-Rham Theorem, and some similar questions.

96 Views Asked by At

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)?

1

There are 1 best solutions below

3
On

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 $\Bbb RP^n$.