Is Klein bottle an algebraic variety?

Question

Is Klein bottle an algebraic variety? I guess no, but how to prove. How about other unorientable mainfolds?

If we change to Zariski topology, which mainfold can be an algebraic variety?

Answer 1

Any complex manifold is not merely an orientable manifold but an oriented manifold. Hence the Klein bottle cannot be a complex manifold (and so not complex algebraic).

Indeed, consider the holomorphic tangent bundle $T M$ of a complex manifold $M$. We define an orientation as follows: take a complex basis $e_1, \ldots, e_n$, and declare the real basis $e_1, \ldots, e_n, i e_1, \ldots, i e_n$ to be positively oriented. One can check that this is independent of the choice of complex basis, so this defines a global orientation of $M$.

Answer 2

In the introduction (second page) of this paper of Biswas and Huisman, it is explained that any non-orientable compact topological surface $X$ is real algebraic (i.e. there exists a real smooth algebraic surface $S$ whose real points $S(\mathbb R)$ endowed with the natural differential structure is diffeomorphic to $X$).

In the case of Klein bottle, the corresponding algebraic surface is simply the blowup of $\mathbb P^2(\mathbb R)$ along a real point (I don't know topology enough to see why this is true). They also prove this algebraic surface is unique (in some sense, because blowing-up further a non-real point doesn't change the real points but change the algebraic surface).