Difference between the real projective plane and the complex projective plane

Question

Well the title says it all. If we consider the $P^2(\Bbb R)$ and the $P^2(\Bbb C)$, and we compare them, my guess is that it will be like a round $\Bbb R^2$ versus a sphere. I don't have very good geometric intuition, so I really can't picture this well, and I'm usually wrong about these things.

Answer 1

Although their construction are similar, these two spaces are actually quite different from the topological point of view.

Recall that $P^2(K)$ is the set of all one dimensional subspace in $K^3$, where $K = \mathbb R$ or $\mathbb C$ in this case.

For $P^2(\mathbb R)$, as all lines passes through the unit sphere $S^2 \subset \mathbb R^3$ at exactly two points $x$ and $-x$, we can also think $P^2(\mathbb R)$ as

$$P^2(\mathbb R) = \mathbb S^2 /\{\pm 1\}\ .$$

As a result, we see that $\mathbb S^2 $ is a two to one cover of $P^2(\mathbb R)$, and thus $\pi_1(P^2(\mathbb R)) = \{\pm 1\}$.

For $P^2(\mathbb C)$, one can also consider the unit sphere $\mathbb S^5$ in $\mathbb C^3 \cong \mathbb R^6$. However, now every complex one dimensional subspaces intersects the sphere in a $\mathbb S^1$. In particular, we have the following fibration

$$\mathbb S^1 \to \mathbb S^5 \to P^2(\mathbb C),$$

this shows that $P^2(\mathbb C)$ is real four dimensional (complex two dimension).

Answer 2

We think of $P^2(R)$ as a 2 dimensional affine space extended with points 'at infinity'. Instead of $S^5$ above, use a hyperplane (1,x,y) $x,y\in C$. Then, except for points at infinity, $P^2(C)$ consists of points in this 4-dimensional (real) linear manifold. The 'lines' in $P^2(C)$ are affine planes. The fact that there is only one line through 2 distinct points indicates that only certain planes deserve to be called lines. In fact these are solutions of equations of the form c = ax + by. $a,b,c\in C$ which translates into a pair of linear equations.