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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.

u184
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1 Answers1

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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.

Jack Lee
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  • So if I understand correctly we always choose the action generated by the vector field and the group action to act on "opposite sides", in this case right and left respectively, because we want them to commute with each other ? – u184 Jun 21 '17 at 10:04
  • @u184: I'm not quite sure what you mean -- there's not really any choice involved. The flow of the vector field $e_1$ is completely determined by $e_1$, and $e^2$ and $e^3$ are not invariant under it. – Jack Lee Jun 21 '17 at 13:56
  • I probably mean what if I started off by right invariant vector fields instead. Wouldn't all the arguments above still hold? – u184 Jun 21 '17 at 14:23
  • @u184: See my addition above. – Jack Lee Jun 21 '17 at 23:37