# Are complex projective spaces orientable?

I know that $\mathbb{RP}^1$ is oriented (since it is essentially the projectively extended real line), but $\mathbb{RP}^2$ is not (as the non-orientable surface of genus 1 it is arguably the simplest non-orientable surface). According to Wikipedia, for real projective spaces, this pattern continues -- orientable for $n$ odd and non-orientable for $n$ even.

However, in contrast, Wikipedia says nothing about the orientability of complex projective spaces. I know that $\mathbb{CP}^1$ is homeomorphic to the sphere (provided $\mathbb{R}^2 \sim \mathbb{C}$) which is orientable, and that the real projective space of dimension $1$ is orientable. On the other hand, complex dimension $1$ is in some sense more like real dimension $2$, and the real projective surface $\mathbb{RP}^2$ is non-orientable.

My questions:

1. Is orientability invariant under homeomorphisms? Or just diffeomorphisms? In the former case we could use the orientability of $S^2$ and the homeomorphism with $\mathbb{CP}^1$; in the latter case we would still have more work to do if we were to claim that $\mathbb{CP}^1$ is orientable.

2. Are all complex projective spaces orientable? (E.g. not just the complex projective line $\mathbb{CP}^1$, but also the complex projective plane $\mathbb{CP}^2$.) Wikipedia says that all complex manifolds are oriented (not just orientable), which would imply that they are. Is this claim about complex manifolds true? (I don't know complex manifold theory, thus why I am asking.)

• To point (2), look at the complex structure -- multiplication by $i$ should give a way of picking a consistent orientation. (I don't remember all the details offhand) – Neal Aug 25 '16 at 2:28
• 1) Yes. Though it is quite likely that whatever homeomorphism you have in mind is actually a diffeomorphism. 2) Yes. – user98602 Aug 25 '16 at 2:40

Every complex manifold is orientable, as every complex vector space as a canonical orientation as a real space. Namely if $V$, is a complex vector space, and $B= (u_1,...,u_n)$ is a base (over C), then $B^*=(u_1,...,u_n, iu_1,...iu_n)$ is a base over $\bf R$. Note that if $B'$ is another base over $C$ and $l$ the unique linear map $C$ map such that $lB=B'$, $lB^*=B'^*$, and the determinant of $l$, viwed as a $R$ linear map is the squre of the modulus of the determinant of $l$, hence positive. Thus the orientation given by $B$ is the same than that given by $B'$, and complex linear maps preserves this orientation.
1) An orientation on a topological manifold $X^n$ is a choice of generator $X_p$ of each local homology group $H_n(X, X\setminus, p, \mathbb{Z}) = \mathbb{Z}$ compatible with the inclusions $(X, X\setminus U) \to (X, X\setminus p)$ for neighborhoods $U$ of $p$. For a smooth, rather than merely topological, manifold, unraveling the definition above gives the usual definition of orientability in terms of the triviality of $\det \bigwedge^n T^*X$.
In this particular case, an easy way of proving that $\mathbb{CP}^n$ is orientable is noting that $\pi_1 \mathbb{CP}^n = 0$. That follows immediately from the cellular approximation theorem, for example, or the fibration $S^1 \to S^{2n+1} \to \mathbb{CP}^n$, where $S^1\subset \mathbb{C}$ acts on $S^{2n+1}\subset \mathbb{C}^{n+1}$ by $z.(w_0, \dots, w_n) = (zw_0, \dots, zw_n)$.
2) More generally, the result holds for any almost-complex manifold; that is, any real manifold $X$ with a map $J:TX \to TX$ such that $J_p:T_p X \to T_p X$ with $J^2 = -1$. (Any complex manifold is also almost-complex; take $J$ to be multiplication by $i$ on each fiber.) To orient $X$, take a complex basis $v_1, \dots, v_n$ of $T_p X$, and let $v_1, \dots, v_n, Jv_1, \dots, Jv_n$ be its corresponding orientation as a real vector space. Alternatively, for a smooth complex manifold, consider the top form $\omega = dz_1 \wedge d\overline{z}_1 \wedge \dots dz_n \wedge d\overline{z}_n$ for a local coordinate chart $(z_1, \dots, z_n)$.