# Real & Complex Projective Space (Part 2)

We can now move on to talk about Complex Projective Space, using some of the intuition we built in the real case. Define the equivalence relation $\sim$ on $\mathbb{C}$, so that $x \sim y$ if $x = \lambda y$ with $\lambda \in \mathbb{C}^{\times}$. Then define $\mathbf{CP}^{n} = \bigg( \mathbb{C} \setminus \{ 0 \} \bigg) / \sim$. We can still think of $\mathbf{CP}^{n}$ as the space of complex lines through the origin in $\mathbb{C}^{n+1}$, but because multiplication by complex numbers “rotate” as well as “scale,” it’s not quite as easy to see what’s going on here.

In this post, we will prove a number of nice topological properties about $\mathbf{CP}^{n}$, and in the last post we will attempt a simple study of its structure, introducing important examples.

1.  $\mathbf{CP}^{n}$ is compact.

Let $\mathbf{CS}^{n} = \{ v \in \bigg( \mathbb{C}^{n+1} \setminus \{0 \} \bigg) : |v| = 1 \}$. This is the $n$-dimensional complex sphere. Next, define the projection map $\pi : \bigg( \mathbb{C}^{n+1} \setminus \{ 0 \} \bigg) \to \mathbf{CP}^{n}$. Note that for every non-zero vector $v$ in $\mathbb{C}^{n+1}$ there is (at least one) scalar $\lambda$ such that $|\lambda v| = 1$.  Thus, every equivalence class intersects the complex $n$-sphere at least once. In fact, they intersect the complex sphere in many different places, so it isn’t a transversal.

As a result, we can see that $\pi(\mathbf{CS}^{n}) = \mathbf{CP}^{n}$. Since $\mathbf{CS}^{n}$ is compact, and $\pi$ is continuous (by the construction of the quotient space), so is $\mathbf{CP}^{n}$. $\square$

2. $\pi$ is open.

It will suffice to show that $[U]$ (the equivalence class of $U$) is open for any open $U \subseteq \bigg( \mathbb{C}^{n+1} \setminus \{ 0 \} \bigg)$. This is because $\pi(U)$ is open iff $\pi^{-1}(\pi(U))$ is open (by the construction of the quotient topology), and $\pi^{-1}(\pi(U)) = [U]$. Write $[U] = \displaystyle\bigcup_{\lambda \in \mathbb{C}^{n+1} \setminus \{ 0 \} } \lambda U$. $\lambda U = \{ \lambda u : u \in U \}$ is open, so $[U]$ is the union of open sets, and hence is open itself. $\square$

Before we move on, we need to prove two simple theorems about quotient spaces.

Theorem 1: Let $X$ be a topological space and let $\sim$ be an equivalence relation on $X$. Let $\pi : X \to X/ \sim$ be the projection map. A function $f$ from $X/ \sim$ to a topological space $Y$ is continuous if and only if $f \circ \pi$ is continuous.

Proof: Since the composition of continuous functions is continuous, the continuity of $f$ implies the continuity of $\pi$. Conversely, if $f \circ \pi$ is continuous, let $V$ be an open subset of $Y$. Then $(f \circ \pi)^{-1}(V) = \pi^{-1}(f^{-1}(V))$ is an open subset of $X$. By the definition of the quotient topology, $f^{-1}(V)$ is an open subset of $X/\sim$. Thus $f$ is continuous. $\square$

Theorem 2: Let $f$ be a continuous function form a topological space $X$ to a topological space $Y$. Let $\sim$ be an equivalence relation on $X$ such that $f$ is constant on each equivalence class. Then there exists a continuous function $g: X/\sim \to Y$ such that $f = g \circ \pi$.

Proof: Consider the following diagram. Define the value of $g$ on an equivalence class to be the value of $f$ at any element of the equivalence class. Then $g$ is well defined and $g \circ \pi = f$. The continuity of $g$ follows from Theorem 1. $\square$

3.The equivalence of $\mathbf{CP}^{n}$ with $\mathbf{CS}^{n}/\sim_{1}$.

Define $\sim_{1}$ on $\mathbf{CS}^{n}$ as follows: $x \sim_{1} y$ if $\exists \lambda \in \mathbb{C}, |\lambda| = 1$ and $x = \lambda y$. Let $\pi_{1} : \mathbf{CS}^{n} \to \mathbf{CS}^{n} / \sim_{1}$ be the natural projection map. We want to show that $\mathbf{CS}^{n}/\sim_{1} \cong \mathbf{CP}^{n}$. By Theorem 2, we need to construct a map from $\mathbf{CS}^{n} / \sim_{1}$ to $\mathbf{CP}^{n}$ that respects $\sim_{1}$.

Observe that $\varphi$ is a bijection, as $x \sim_{1} y$ iff $\pi(x) = \pi(y)$. By construction, $\varphi$ is continuous. We want to show it is open, and hence a homeomorphism. We know that $U \subseteq \mathbf{CS}^{n}/\sim_{1}$ is open iff $\pi^{-1}(U)$ is open in $\mathbf{CS}^{n}$. To see that $\varphi$ is open, note that $\varphi(U) = \pi(\pi_{1}^{-1}(U))$ is open iff $\pi^{-1}(\pi(\pi_{1}^{-1}(U)))$ is open in $\bigg( \mathbb{C}^{n+1} \setminus \{ 0 \} \bigg)$, which is true iff $[\pi_{1}^{-1}(U)]$ is open.

Hence, it will suffice to show that if $V \subseteq \mathbf{CS}^{n}$ is relatively open, so is $[V]$. To show this, construct a map $f : \big( \mathbb{C}^{n+1} \setminus \{ 0 \} \big) \to \mathbf{CS}^{n}$ by $x \to \frac{x}{|x|}$. This map normalizes a vector, sending it to sphere. Note that $f$ is continuous, so that if $V \subseteq \mathbf{CS}^{n}$ is open, $f^{-1}(V)$ is open. But $[V] = f^{-1}(V)$, so we are done. $\square$

4. $\mathbf{CP}^{n}$ is Hausdorff.

We will work in $\mathbf{CS}^{n} /\sim_{1} \cong \mathbf{CP}^{n}$, identifying $\pi_{1}$ with $\pi$. Let $x,y \in \mathbf{CS}^{n}$ such that $\pi(x) \neq \pi(y)$. Let $E = \{ \lambda x : |\lambda| = 1\}$ and $F = \{ \lambda y : |y| = 1 \}$. Then $E \cap F = \emptyset$. Moreover, $E$ and $F$ are closed, hence compact. Let $d = d(E,F) > 0$ be the distance between $E$ and $F$, and define $U = [B(x,d/2)]$ and $V = [B(y,d/2)]$. These are the equivalence classes of open balls around $x$ and $y$ in $\mathbf{CS}^{n}$, and are open themselves.

We claim that $U \cap V = \emptyset$. Suppose, to the contrary, that $w \in U \cap V$. Then $w = \lambda_{1} x^{'}$ with $x^{'} \in B(x,d/2) \cap \mathbf{CS}^{n}$ and $|\lambda_{1}| = 1$. Similarly, $w = \lambda_{2} y^{'}$ with $y^{'} \in B(y,d/2) \cap \mathbf{CS}^{n}$ and $|\lambda_{2}| = 1$.

$|\lambda_{1} x - \lambda_{2}y| \leq |\lambda_{1} x - \lambda_{1}x^{'}| + |\lambda_{1}x^{'} - \lambda_{2}y| = |\lambda_{1}| |x - x^{'}| + |\lambda_{2}||y^{'} - y| < d/2 + d/2 = d$

This contradicts the fact that $d$ is the infimum of the distances between points in $E$ and $F$.

By construction, $\pi(U) \cap \pi(V) = \emptyset$, both are open, and $\pi(x) \in \pi(U)$ and $\pi(y) \in \pi(V)$, so $\mathbf{CP}^{n}$ is Hausdorff. $\square$.

In summary, we have shown that complex projective space is compact and hausdorff, and the projection map is a homeomorphism. Moreover, we can study complex projective space by working only with a quotient space of the complex sphere, which is simpler to work with.