Smooth no-where vanishing form

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.

• 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. – Balarka Sen Nov 26 '18 at 12:24

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 $$(-0.881, 0.428, .2)$$.