# Finitely vanishing vector field on the real projective plane

I am trying to find a vector field on \begin{align} \mathbb{P}_\mathbb{R}^2 := \mathbb{R}^3 \setminus \{(0,0,0)\} \big/ \sim, \end{align} where $$\sim$$ is the equivalence relation for $$z, z'\in \mathbb{R}^3$$ defined by \begin{align} z \sim z' \iff z = \lambda z', \qquad \lambda \in \mathbb{R}\setminus {0}. \end{align}

Specifically, I am looking for a vector field that vanishes only at finitely many points.

I'm not sure how to start - I can't find any charts that make the answer apparent to me. I'd appreciate if anyone could explain how to go about this problem.

• Hint: If you think of $\mathbb{RP}^2$ as $\mathbb S^2/\sim$ with $x\sim -x$, then if you can find a vector fields $X$ on $\mathbb S^2$ with $X(-x) = A_*X(x)$, where $A (x) =-x$, then this vector fields $X$ descend to one on $\mathbb{RP}^2$. Commented Dec 2, 2021 at 17:00
• Thank you for the hint. To clarify the notation: is $A_*X(x)$ the pushforward of the vector field X, given the map $A: \mathbb{S}^2/\sim \rightarrow \mathbb{S}^2/\sim: x \mapsto -x$? Also, can this be done with a single chart, or will I need 2? Commented Dec 2, 2021 at 17:26
• No, $A :\mathbb S^2 \to \mathbb S^2$. In my argument I did not refer to any chart at all.. Commented Dec 2, 2021 at 17:28
• You will need to construct such a vector field on $S^2$ first, but it is quite easy (consider some rotations) Commented Dec 2, 2021 at 17:48
• I think I see what you mean now - I think I will give it a try and return if I get stuck Thank you! Commented Dec 2, 2021 at 17:51