Why Complex Projective Space Is Hausdorff
A quotient of a Hausdorff space need not be Hausdorff, so explicit projective charts do not settle the separation question by themselves. For complex projective space, normalize each nonzero vector to unit length. The remaining ambiguity is multiplication by a unit complex number, giving an action of a compact group on a sphere.
We use compactness to separate distinct orbits by neighbourhoods in the quotient. The argument has two parts: prove the separation result for compact group actions, then compare the sphere quotient with the usual definition of projective space.
1. Proof that \(\mathbb CP^n\) is Hausdorff
We prove the separation property directly from the quotient construction. We first collect the compactness and separation facts used in the quotient argument.
Lemma 1.
Let \(X\) be a Hausdorff space, and \(K\) be a compact subset of \(X\). Then, \(K\) is closed.
Let \(X\) be a compact space and \(Y\) be a Hausdorff space. Suppose \(f: X \to Y\) is a bijective continuous function. Then, \(f\) is a homeomorphism.
Any compact Hausdorff space is normal, i.e. for any given disjoint closed subsets \(E\) and \(F\), there exist open neighbourhoods \(U\) of \(E\) and \(V\) of \(F\) such that \(U \cap V = \emptyset\).
Let \(X\) and \(Y\) be compact spaces. Then, the product space \(X \times Y\) is also compact.
Proof of Lemma 1. For part 1, fix \(x\notin K\). For every \(y\in K\), choose disjoint open neighbourhoods \(U_y\ni x\) and \(V_y\ni y\). A finite collection \(V_{y_1},\ldots,V_{y_N}\) covers \(K\). Then \(U=\bigcap_iU_{y_i}\) is an open neighbourhood of \(x\) disjoint from \(K\). Thus the complement of \(K\) is open. The intersection must use the neighbourhoods of \(x\), not the neighbourhoods covering \(K\).
For part 2, a closed subset \(A\) of compact \(X\) is compact: append \(X\setminus A\) to any relative open cover and use a finite subcover of \(X\). Its continuous image \(f(A)\) is compact, and hence closed in Hausdorff \(Y\) by part 1. Thus the continuous bijection \(f\) is closed, which means its inverse is continuous.
For part 3, let \(E,F\) be disjoint closed subsets of compact Hausdorff \(X\). Both are compact. For each \(e\in E\), the preceding finite-neighbourhood argument gives disjoint open sets \(U_e\ni e\) and \(V_e\supset F\). Choose \(U_{e_1},\ldots,U_{e_N}\) covering \(E\), and put
Both are open, contain their respective closed sets, and are disjoint: a point in \(U_{e_i}\) cannot be in \(V\subseteq V_{e_i}\).
For part 4, let an open cover of \(X\times Y\) be given. For each fixed \(x\), choose product neighbourhoods \(U_{x,y}\times V_{x,y}\) lying in cover elements and covering \(\{x\}\times Y\). Compactness of \(Y\) gives finitely many \(V_{x,y_i}\) covering \(Y\). Their corresponding \(U_{x,y_i}\) have an intersection \(U_x\) containing \(x\), and finitely many original cover elements cover the strip \(U_x\times Y\). Finally choose finitely many \(U_x\) covering compact \(X\). The associated finite collections cover \(X\times Y\). ◻
Lemma 2. Let \(X\) be a compact, Hausdorff space and \(G\) be a compact topological group. Suppose \(G\) acts on \(X\) continuously, i.e. the function \(m : G \times X \to X : (g,x) \mapsto gx\) is continuous. Let \(\pi: X \to X/G\) be the quotient map. Then,
\(\pi\) is an open mapping;
\(\pi\) is a closed mapping;
the orbit \(X/G\) is Hausdorff.
Proof of Lemma 2. We first show that \(\pi\) is an open mapping: to do so, we need to show that \(\pi(U)\) is open in \(X/G\) for any open set \(U \subset X\): equivalently, we need to show that \(\pi^{-1} (\pi(U))\) is open in \(X\). We claim that
First, for any \(g \in G\), \(g:X\to X\) is a homeomorphism, therefore \(g(U)\) is open in \(X\) for any open subset \(U \subset X\). Second, to see that one inclusion holds: pick any \(a \in \pi^{-1} (\pi(U))\); therefore, \(\pi(a) \in \pi(U)\), i.e. there exists some \(b \in U\) such that \(\pi(a) = \pi(b)\). Therefore, \(a = gb \in g(U)\) for some \(g \in G\). As such, it must be that \(a \in \bigcup_{g \in G} g(U)\). Third, for the reverse inclusion: pick any \(c \in \bigcup_{g \in G} g(U)\). Then, it must be that \(c \in g(U)\) for some \(g \in G\), implying \(c = gd\) for some \(d \in U\). Since \(\pi(gd) = \pi(d)\), it follows that \(\pi(c) = \pi(d) \in \pi(U)\), from which the desired inclusion follows. The claim is therefore proven, and from here so is (1).
Next we show \(\pi\) is a closed mapping: note that because \(X\) is compact, any closed subset of \(X\) is also compact. So, let \(A\) be a closed subset of \(X\), implying \(A\) is compact. Since any product of compact spaces is compact, \(G \times A\) is compact. Since \(G(A) = m(G, A)\) and \(m\) is continuous, \(G(A)=\bigcup_{g \in G} g(A)\) is compact. From here, note that \(\pi^{-1}(\pi(A)) = G(A)\) is a compact subset of a Hausdorff space \(X\), it is therefore closed. This proves (2).
To prove part 3, distinct orbits \(Gx,Gy\) are disjoint compact sets: each is the continuous image of compact \(G\). By normality choose disjoint open \(U,V\subset X\) containing them. Since \(\pi\) is closed, the sets
are open. The first consists of orbits entirely contained in \(U\), and the second of orbits entirely contained in \(V\). Thus \([x]\in\widetilde U\), \([y]\in\widetilde V\), and the two sets are disjoint. This uses the closed complements of \(U,V\); an open neighbourhood itself need not be closed. Therefore \(X/G\) is Hausdorff. ◻
Remark 3. Let \(\mathbb S^1\) act on \(\mathbb S^{2n+1}\) via \(\mathbb S^1 \times \mathbb S^{2n+1} : (\lambda,x) \mapsto \lambda x\). This action is continuous: the function \(\eta : \mathbb C \times \mathbb C^{n+1} \to \mathbb C^{n+1} : (\lambda, x) \mapsto \lambda x\) is continuous. The action of \(\mathbb S^1\) on \(\mathbb S^{2n+1}\) is the restriction of \(\eta\) to \(\mathbb S^1 \times \mathbb S^{2n+1}\), which is therefore continuous. Therefore, identifying two points \(x,y \in \mathbb S^{2n+1}\) via \(\lambda \in \mathbb S^1\) such that \(x = \lambda y\) defines equivalence classes. The quotient space of \(\mathbb S^{2n+1}\) modulo this relation gives meaning to the notation \(\mathbb S^{2n+1} / \mathbb S^1\). We denote the equivalence class of \(x \in \mathbb S^{2n+1}\) (modulo this relation as described) by \([x]_{\mathbb S^1}\), and \(q : \mathbb S^{2n+1} \to \mathbb S^{2n+1} / \mathbb S^1\) is the quotient map.
Lemma 4. The complex projective space \(\mathbb CP^n\) is homeomorphic to \(\mathbb S^{2n+1}/\mathbb S^1\).
Proof of Lemma 4. Let \(\pi:\mathbb C^{n+1}\setminus\{0\}\to\mathbb CP^n\) be the quotient by nonzero complex scalar multiplication, and \(q:S^{2n+1}\to S^{2n+1}/S^1\) the circle quotient. Normalization \(r(z)=z/\|z\|\) is continuous. If \(z=\mu z'\) for \(\mu\ne0\), then
so \(f([z])=[r(z)]_{S^1}\) is well-defined. Conversely, if \(f([z])=f([z'])\), then \(r(z)=\lambda r(z')\) for some \(|\lambda|=1\), and
This is a nonzero complex multiple, proving injectivity. Every unit vector is fixed by \(r\), proving surjectivity. The diagram records \(f\circ\pi=q\circ r\).
For any open \(W\subset S^{2n+1}/S^1\),
is open. By the definition of the quotient topology for \(\pi\), this says \(f^{-1}(W)\) is open, so \(f\) is continuous. No inverse function \(\pi^{-1}\) exists or is needed.
To check the inverse continuously, the restriction \(\pi|_{S^{2n+1}}\) is constant on circle orbits, so it descends to \(g:S^{2n+1}/S^1\to\mathbb CP^n\), \(g([u]_{S^1})=[u]\). The identity \(g\circ q=\pi|_{S^{2n+1}}\) and the quotient topology for \(q\) prove that \(g\) is continuous. Directly \(g\circ f=\mathrm{id}\) and \(f\circ g=\mathrm{id}\). Thus \(f\) is a homeomorphism. As a further check, \(\mathbb CP^n\) is compact because \(\pi|_{S^{2n+1}}\) is a continuous surjection from the compact sphere; this compactness argument does not presuppose Hausdorffness. ◻
Lemma 5. Let \(X,Y\) be topological spaces, and \(f: X \to Y\) is a homeomorphism. Then, \(X\) is Hausdorff if and only if \(Y\) is Hausdorff.
Proof. For the “only if” direction, assume \(X\) is Hausdorff; then, by the surjectivity of \(f\), for any \(y_1,y_2 \in Y\) there must exist \(x_1,x_2 \in X\) such that \(y_1=f(x_1)\) and \(y_2=f(x_2)\). By injectivity, \(x_1 \neq x_2\). Since \(X\) is Hausdorff, there exists open neighbourhoods \(U_1\) of \(x_1\) and \(U_2\) of \(x_2\) such that \(U_1 \cap U_2 = \emptyset\). Now, let \(V_1 = f(U_1)\) and \(V_2 = f(U_2)\). Because \(f\) is a homeomorphism, it is therefore an open mapping, and \(V_i\) are open neighourhoods of \(y_i\) for \(i = 1,2\). We claim that \(V_1 \cap V_2 = \emptyset\). If not, then there exists some \(w \in V_1 \cap V_2\); by the surjectivity of \(f\), there exists some \(x_w \in X\) such that \(w = f(x_w)\). But this means \(x_w \in f^{-1}(V_1 \cap V_2) \subset U_1 \cap U_2 = \emptyset\), which is absurd. This concludes the proof. (Note that the “if” argument is identical, because \(f\) is a homeomorphism.) ◻
Proof that \(\mathbb CP^n\) is Hausdorff. \(\mathbb CP^n\) is homeomorphic to \(\mathbb S^{2n+1}/\mathbb S^1\) by Lemma 4, and \(\mathbb S^{2n+1}/\mathbb S^1\) is Hausdorff by Lemma 2, part 3. Therefore, an application of Lemma 5 implies \(\mathbb CP^n\) is Hausdorff. ◻
Reading: Carmo (1992), Riemannian geometry.