1-Forms on $SO(3)$ and $S^2$ The is a mistake somewhere in the following reasoning and I can't seem to detect which argument is wrong. 
Consider the lie group $SO(3)$ with $e_{1},e_{2},e_{3}$ as left invariant vector fields. Each of which generates an $S^1$ action. Quotienting by any one of these $S^1$ actions (say $e_1$), I should get a 2-sphere, $S^2$. Now the dual 1-forms, $e^2$ and $e^3$ are left invariant 1-forms and horizontal with respect to this quotient. So they should pass to the quotient space i.e. $S^2$. Now my problem is that these 1-forms are globally well-defined and nowhere vanishing on $S^2$ which contradicts the Hairy ball theorem.
 A: Left-invariant vector fields are invariant under left multiplication. However, the flow of a left-invariant vector field acts by right multiplication: For example, if $\theta^i_t$ represents the flow of $e_i$, then
$$
\theta^i_t(x) = R_{\exp t e_i}(x) = x \cdot \exp t e_i.
$$
(See my Introduction to Smooth Manifolds (2nd ed.), Proposition 20.8(h).) The $1$-forms $e^2$ and $e^3$ are not invariant under right multiplication, so they do not descend to the quotient.
Added in response to comment: If you start with a right-invariant vector field, say $v$, then it is true that the left-invariant $1$-forms $e^2$ and $e^3$ are invariant under the flow of $v$. But that's not enough for them to descend to the quotient. Here's a useful lemma -- see if you can prove it:
Lemma. If $M$ is a smooth manifold and $v$ is a vector field on $M$ that generates a free $\mathbb S^1$ action, then a differential form $\eta$ on $M$ is a lift of a form on $M/\mathbb S^1$ if and only if $v\mathbin{\lrcorner} \eta\equiv 0$ and $\mathscr L_v\eta \equiv 0$ (where $\lrcorner$ denotes interior multiplication and $\mathscr L$ denotes the Lie derivative).
In the case at hand, we do have $\mathscr L_v e^2=\mathscr L_v e^3=0$. But $e^2(v)$ and $e^3(v)$ are not identically zero. 
