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I am not very good at working with $2$-forms, though I'm hoping to learn. I found this problem and have trouble starting it. I also do not know what a restriction would look like.

Is the $2$-form $\omega = z\, dx \wedge dy$ an exact form in $\mathbb{R}^3$? Is the restriction of $\omega$ to the surface $N = \{(x,y,z): z -x^2 - y^2 = 1\}$ exact?

Any suggestions are appreciated!

phenomenalwoman4
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    $d\omega = dx\wedge dy\wedge dz \neq 0$, so it's not closed, hence not exact in $\mathbb{R}^3$. – Daniel Fischer Jul 08 '13 at 21:39
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    For the 2nd question, note that $x$, $y$ give global coordinates on $N$. In these coordinates, the restriction of $\omega$ is just $(x^2+y^2+1)dx\wedge dy=d((x^3/3+x+y^2x)dy)$. – Start wearing purple Jul 08 '13 at 22:27

1 Answers1

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Another, computation-less way to answer the second question:

  1. A $k$-form on $k$-dimensional manifold is necessarily closed, because there are no nontrivial $(k+1)$-forms there.
  2. Hence, the restriction of $\omega$ to $N$ is closed.
  3. Since $N$ is diffeomorphic to $\mathbb R^2$ (being the graph of a smooth function), its de Rham cohomology is trivial: in other words, all closed forms are exact.

But the computation $(x^2+y^2+1)dx\wedge dy=d((x^3/3+x+y^2x)dy)$ by O.L. may get you there faster.

And there is nothing to add to the answer to 1st question "$d\omega = dx\wedge dy\wedge dz \neq 0$" given by Daniel Fischer.

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