A question about orientation and real plane curves Let $C$ be a smooth projective curve in $\mathbb{R}P^2$ given by a polynomial equation $F(x,y,z)=0$ of even degree (so $C$ consists of closed loops). Then $C$ divides $\mathbb{R}P^2$ into 2 parts, only one of which is orientable. Given a point $P\in\mathbb{R}P^2$, is there a way (in terms of $F$, say) to determine to which part $P$ belongs?
 A: If $f$ is of degree $ d=2k$ and with real coefficient, the sign of $f$ is well defined. Indeed, if $f(x,y,z) = a$, then $f(\lambda x, \lambda y, \lambda z) = \lambda^{2k}a$ which has the same sign as $a$. So the non-orientable region will have a well-defined sign, and the orientable region will have the opposite sign. Of course you have to fix $f$ and not just the curve as $f = 0$ and $-f = 0$ define the same curve. I believe the usual convention is that the non-orientable part has negative sign. 
If you are studying the subject I would advice the notes by O. Viro which contains lot of useful informations about topology of real algebraic curves. 
Edit : let me add one remark. You said that $C$ is of even degree so consists of closed loops. If $C$ is smooth and of odd degree, it is also topologically a union of closed loops (i.e topological circles). The only difference is that the ovals of $C$ always bound a disk if the degree is even, and if the degree is odd then exactly one oval will not bound a disk. 
A: Let $\pi:S^2\to\mathbb{R}P^2$ denote the quotient, and let $U\subset\mathbb{R}P^2$ be an open subset. Then $\pi^{-1}(U)$ is the orientation double cover of $U$. Hence, as always, $U$ is orientable if and only if $\pi^{-1}(U)$ is disconnected.
