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Does there exist any no-where vanishing smooth $1$-form on $S^2$. I , think there is such one. For example, consider the smooth $1$-form $\omega=dx+dy+xdz$ on $\Bbb R^3$ consider the pull-back of $\omega$ w.r.t. the inclusion map $i:S^2\rightarrow \Bbb R^3$. Then I claim that $i^*\omega$ is a no-where vanishing smooth $1$-form on $S^2$. My argument is the following:

Obviously $i^*\omega$ is a smooth $1$-form on $S^2$. I have to only show that it is no-where vanishing. To show this it is enough to show that , for fix a point $(a,b,c)\in S^2$ , we have to find a $(x,y,z)\in \Bbb R^3$ with $ax+by+cz=0$ and $x+y+az\not =0$.

  1. First of all consider the point $(a,b,c)=(0,1,0)$ of $S^2$, then $(x,y,z)=(1,0,0)$ will do the job. Similarly for $(0,-1,0)\in S^2$.

  2. Now assume one of $a$ or $c$ or both $a$ and $c$ is not zero. First assume $a=0$ and $c\not =0$, then choose $(x,y,z)\in \Bbb R^3$ such that $by+cz =0$ and $x+y\not =0$. So we are done.

  3. Next assume $a\not=0$ and $c=0$, then we can find $(x,y,z)\in \Bbb R^3$ with $ax+by=0$ and $x+y+az\not= 0$.

  4. Finally assume $ac\not =0$.

4.1.) Then consider the subcase $a+b=0$ , $ac\not=0$, choosing $(x,y,z)=(1,1,0)$ we are done.

4.2.) Next consider the subcase $a+b\not =0$ and $ac\not =0$, choose $(x,y,z)\in \Bbb R^3$ such that $x=y$ and $(a+b)x+cz=0$ and $2x+az\not =0$, this is possible as last two distinct line may intersect at most one point.

Is my argument correct ? Thanks in advance.

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    $\begingroup$ There is no such $1$-form. Since $\omega_p \neq 0$ for all $p \in S^2$, $\ker \omega_p \subset T_p M$ is a $1$-dimensional subspace. Then $\ker \omega \subset TS^2$ is a $1$-dimensional subbundle. Since $S^2$ is simply connected, any line bundle over $S^2$ is trivial. Take a smooth section of that trivial line bundle; that gives a nowhere vanishing tangent vector field of $S^2$. This contradicts the hairy ball theorem. $\endgroup$ – Balarka Sen Nov 26 '18 at 12:24
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The lines $(a+b)x + cz=0$ and $2x+az = 0$ will not be distinct if $(a+b, c)$ is a constant multiple of $(2, a)$, so there is a problem in case 4.2 of the proof. (In fact, the result is false -- there is no such one-form -- as you can find elsewhere on this site.)

Conceretly, solutions to the system $a^2 + ba = 2c$ and $a^2 + b^2 + c^2 = 1$ provide a counterexample. This system is a little messy to solve out algebraically, but a CAS gives a formula for all the complex solutions and there is a real solution e.g. near $(-0.881, 0.428, .2)$.

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