A vector field on the projective plane with one critical point, a sink I'm studying (for my own, this is not a homework question) vector fields on the real 2d projective plane.

Is there a continuous vector field on $\mathbb R P^2$ (the 2d-sphere in $\mathbb R^3$ with antipodal points identified) with a single critical point that is a sink, such that all integral curves end at the critical point? (The critical point might be called a global attractor.)

By the Poincare-Hopf index theorem, the sum of the indexes of any continuous vector field with finitely many critical points on the projective plane must add to one, so the vector field I want would not contradict Poincare-Hopf (since a sink has index 1, and 1 is the Euler Characteristic of $\mathbb RP^2$.) However, I suspect there is no vector field like the one I want. Is there a vector field on $\mathbb R P^2$ like the one I want, or can it be proved there is no such vector field?
 A: There is no such vector field.
Suppose there were such a field, say, $X$: Then, there is a unique vector field $\tilde X$ on $S^2$ that pushes forward to $X$ via the canonical quotient $\pi : S^2 \to \Bbb R P^2$. The preimage of the sink $\ast$ of $X$ is a pair of points, $\pm p$, by construction both sinks for $\tilde X$.
On the other hand, for any sink $s$ of a continuous vector field $\tilde X$, the set $U_s$ of points whose $\tilde X$-integral curves converge to $s$ is open. Since every $X$-integral curve converges to $\ast$, every $\tilde X$-integral curve converges to one of $\pm p$. Thus, $U_p, U_{-p}$ are disjoint, open, and nonempty, and their union is $S^2$, i.e., they comprise a separation of $S^2$. But $S^2$ is connected. $\,\square$
One the other hand, if we relax the condition that all integral curves converge to the sink, we can exhibit an example for which there is a single maximal integral curve that doesn't converge to the sink.
Example A vector field $\tilde X$ on $S^2 \subset \Bbb R^3$ descends to a vector field on $\Bbb R P^2$ iff it is odd, that is, iff $\tilde X_{-p} = -\tilde X_p$ for all $p$ in $S^2$. (Here, as usual, we canonically identify $T_{\pm p} \Bbb R^3 \supset T_{\pm p} S^2$ with $\Bbb R^3$ itself.)
Computing directly gives that the continuous (in fact, smooth) vector field $\tilde X$ on $S^2$ defined by $$\tilde X_{(x, y, z)} := (-y (1 - z^2) - x z^2, x (1 - z^2) - y z^2, (1 - z^2) z)$$ is odd and vanishes only at $(0, 0, \pm 1)$, so it descends to a vector field $X$ on $\Bbb R P^2$ that vanishes only at $[0, 0, 1]$.

The integral curve of $\tilde X$ through a point on the equator $E := \{z = 0\}$ traces the equator and so the integral curve of $X$ in the projective line $[E]$ traces that line. On the other hand, integrating the $z$-component of $\tilde X$ gives that the integral curve of $\tilde X$ through any point not on $E$ has $z$-component $z(t) = \pm (1 + C e^{-2t})^{-1 / 2}$ for some constant $C$ and sign $\pm$. In particular, the curve approaches $(0, 0, \pm 1)$, and so the integral curve of $X$ through any point not on $[E]$ approaches $[0, 0, 1]$.
