Hatcher's Algebraic Topology - Example 0.6: Constructing $\Bbb CP^n$ from $D^{2n}$ or $D^{2n+1}$?

Question

This question is related to my previous post.

Why is the following argument wrong?

I know that $\Bbb CP^n$ can be constructed from $\Bbb S^{2n+1}\subset \Bbb C^{n+1}$ by $v\sim \lambda v$ ($\lambda\neq 0$) and I think this is equivalent to identifying the boundary of the upper hemisphere $D_+^{2n+1}\subset\Bbb S^{2n+1}$ ( which is $\Bbb S^{2n}$) by $v\sim \lambda v$ ($\lambda\neq 0$).

But Hatcher says:

It is also possible to obtain $\Bbb CP^n$ as a quotient space of the disk $D^{2n}$ under the identifications $v\sim \lambda v$ for $v\in \partial D^{2n}$ (which is $\Bbb S^{2n-1}$).

Answer 1

(Edit: in light of the discussion in the comments I included a small remark about $\mathbb{C}P^n$ below.)

If you form a quotient space of $D_+^{2n+1}$ by only identifying points on the boundary (and it results in a manifold), it will be a manifold of dimension $2n+1$ and so can't be $\mathbb{C}P^n$. When you form $\mathbb{R}P^n$ out of $D^n_+$ you're using the antipodal action on $S^n$ and restricting it to a fundamental domain, but it's more complicated with the circle action on $S^{2n+1}$. In fact, your argument implicitly assumes that the circle action on $S^{2n+1}$ preserves the subspace $S^{2n}\subset D^{2n+1}_+ \subset S^{2n+1}$, but it does not.

For all $n$ fix the isomorphism $\Phi\colon \mathbb{C}^{n+1} \cong \mathbb{R}^{2n+2}$ by defining $$\Phi (a_1 + ib_1,\dots,a_{n+1} + i b_{n+1}) = (a_1, b_1, \dots, a_{n+1}, b_{n+1}).$$

Now, $S^{2n+1}\subset \mathbb{R}^{2n+2}$ is by definition the space of all unit $(2n+2)$-tuples of real numbers, and this corresponds under $\Phi$ to the subspace of $\mathbb{C}^{n+1}$ of unit $(n+1)$-tuples of complex numbers. The upper hemisphere $D^{2n+1}_+ \subset \mathbb{R}^{2n+2}$ is the set of unit $(2n+2)$-tuples such that $x_{2n+2} \geq 0$, and $S^{2n}$ is then the subspace of unit $(2n+2)$-tuples where $x_{2n+2} = 0$.

Now let $X = \Phi^{-1}(S^{2n})\subset\mathbb{C}^{n+1}$. This is the set of unit $(n+1)$-tuples of complex numbers where $z_{n+1}$ is purely real. But if $z\in S^1$ has a non-zero imaginary component then so will $z\cdot z_{n+1}$, and so $z\cdot v \notin X$ for all $v\in X$.

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The space $\mathbb{C}P^n$ is meant to be the space of all complex lines through the origin in $\mathbb{C}^{n+1}$, which is a subspace of the form $L_v = \{ z\cdot v \mid z\in \mathbb{C} \} = \mathbb{C} \cdot \{v\}$ for some $v\neq 0\in \mathbb{C}^{n+1}$. The function $\mathbb{C}^{n+1}\setminus\{0\}\to\{\text{space of lines}\}$ given by sending $v$ to $L_v$ is surjective, but there is indeterminacy: if $v, v' \neq 0$ then $L_v = L_{v'}$ iff $z\cdot v = v'$ for some $z\neq 0\in \mathbb{C}$. Then it makes sense to call the quotient space $\mathbb{C}P^n = (\mathbb{C}^{n+1}\setminus\{0\})/(\mathbb{C}\setminus\{0\}) \cong S^{2n+1}/S^1$ "the space of all complex lines." It turns out this is a complex manifold of complex dimension $n$, and hence real dimension $2n$.

Answer 2

We know that $\mathbb C P^n$ can be represented as the quotient $S^{2n+1} / \sim$, where $(z_1,\ldots,z_{n+1}) \sim \lambda(z_1,\ldots,z_{n+1}) = (\lambda z_1,\ldots, \lambda z_{n+1})$ with $\lvert \lambda \rvert = 1$.

We know that $S = \{(z_1,\ldots,z_{n+1}) \in S^{2n+1} \mid \text{Im}(z_{n+1}) = 0, \text{Re}(z_{n+1}) \ge 0\}$ is the graph of $\phi : D^{2n} \to \mathbb C, \phi(w) = \sqrt{1-\lvert w \rvert}$. Let $\hat \phi : D^{2n} \to S, \hat \phi(w) = (w, \phi(w))$, be the induced homeomorphism.

1. Each $(z_1,\ldots,z_{n+1}) \in S^{2n+1}$ is equivalent to some $(z'_1,\ldots,z'_{n+1}) \in S$. In fact, if $z_{n+1} = 0$, then $(z_1,\ldots,z_{n+1}) \in S$, and if $z_{n+1} \ne 0$, then $\lambda = \overline{z_{n+1}}/\lvert z_{n+1} \rvert$ has absolute value $1$ and we get $\lambda z_{n+1} = \lvert z_{n+1} \rvert$, thus $\lambda (z_1,\ldots,z_{n+1}) \in S$. We conclude that $\mathbb C P^n$ can be represented as the quotient $S / \sim$.

2. Using $\hat \phi$, we see that $\mathbb C P^n$ can be represented as the quotient $D^{2n} / \equiv$, where $w \equiv w'$ iff $\hat \phi(w) \sim \hat \phi(w')$.

3. $\hat \phi(w) \sim \hat \phi(w')$ means that there exists $\lambda$ with $\lvert \lambda \rvert = 1$ such that $\hat \phi(w) = \lambda \hat \phi(w')$. This is equivalent to $w = \lambda w'$ and $\phi(w) = \lambda \phi(w')$. The latter implies $\lvert \phi(w) \rvert = \lvert \phi(w') \rvert$ and hence $\phi(w) = \phi(w')$ because both numbers are non-negative reals. If $\phi(w) > 0$, i.e. $w \notin S^{2n-1}$, we get $\lambda = 1$, thus $\hat \phi(w) = \hat \phi(w')$. That is, the equivalence class of $w \notin S^{2n-1}$ is the singleton $\{ w \}$. For $\phi(w) = 0$, i.e. $w \in S^{2n-1}$, we get $\phi(w') = 0$, i.e. $w' \in S^{2n-1}$, and $w = \lambda w'$ with any $\lambda$ such that $\lvert \lambda \rvert = 1$.

This also gives the following interesting result:

$S^{2n-1}/ \sim$ is nothing else than $\mathbb C P^{n-1}$. Thus $\mathbb C P^n$ is obtained by attaching the $2n$-cell $D^{2n}$ to $\mathbb C P^{n-1}$ via the quotient map $S^{2n-1} \to \mathbb C P^{n-1}$.

This allows to construct the complex projective spaces $\mathbb C P^n$ by successively attaching even-dimensional cells beginning with $\mathbb C P^0 = S^1/\sim$ which is a one-point space.