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I want to prove the following:

Show that $H^2(\mathbb R^3 \setminus \{0\})$ (the second de Rham cohomology space) does not vanish. Hint: Use the form $\omega = i_V \lambda_3$ with the standard volume form $\lambda_3 = dx_1 \wedge dx_2 \wedge dx_3$ and the vector field $V: \mathbb R^3 \setminus \{0\} \to \mathbb R^3$, $V(x) := x/ \lVert x \rVert_2^3$.

I already proved that $d\omega = 0$ using $i_V\lambda_3 = x_1/ \lVert x \rVert_2^3 dx_2 \wedge dx_3 - x_2/ \lVert x \rVert_2^3 dx_1 \wedge dx_3 + x_3/ \lVert x \rVert_2^3 dx_1 \wedge dx_2$
and then calculating the exterior derivative (is there a faster way?).

Now I want to calculate $$\int_{S^2}\omega$$ but I don't know how to do that. The interior product $i_V$ really confuses me here. Any help appreciated.

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1 Answer 1

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The faster way to see $\omega$ is closed is to pull back by a spherical coordinate parametrization of $\Bbb R^3-\{0\}$. With $g(\rho,\phi,\theta) = \rho(\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi)$, you can easily check that $g^*\omega = \sin\phi\,d\phi\wedge d\theta$. Of course, $g^*(d\omega) = dg^*\omega = 0$.

Indeed, this then gives you an easy way to compute $\displaystyle\int_{S^2}\omega$, since the unit sphere is given precisely by $\rho=1$. Just parametrize (almost all) the sphere by $(0,\pi)\times (0,2\pi)$ in $\phi\theta$-space.

Alternatively, note from your formula that on the unit sphere we have $$\omega = x_1\,dx_2\wedge dx_3 + x_2\,dx_3\wedge dx_1+x_3\,dx_1\wedge dx_2,$$ and this is precisely the (standard) area $2$-form on the sphere (because — again on the unit sphere — $V$ is the outward-pointing unit normal vector to the sphere).

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  • $\begingroup$ Thanks for the quick answer! The problem I have is: Why is $V$ the outward pointing unit normal vector if we raise it to the third power? If it was just $x / \lVert x \rVert $ I would see it. $\endgroup$
    – Staki42
    Feb 6, 2018 at 22:37
  • $\begingroup$ I specifically said on the unit sphere :) So $\|x\|=1$. Otherwise, of course, you're completely correct! $\endgroup$ Feb 6, 2018 at 22:37
  • $\begingroup$ Oh, I see. Very trivial. But why do we actually raise the norm to the third power? Is there reason? I think this whole argument should work by only taking the first power, wouldn't it? $\endgroup$
    – Staki42
    Feb 6, 2018 at 22:38
  • $\begingroup$ You only get a closed $2$-form when you use the inverse-square radial vector field for $V$. If you know some physics, this is the usual gravitational or electrostatic force due to a point mass/charge. You can see this easily by following up on my spherical coordinates formula for $\omega$. If there's a nontrivial power of $\rho$ in there, you'll in fact get a $d\rho$ contribution when you take the exterior derivative. $\endgroup$ Feb 6, 2018 at 22:40
  • $\begingroup$ I understand it now I think. Thanks for explaining :) $\endgroup$
    – Staki42
    Feb 6, 2018 at 23:05

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