Quotiented manifold homeomorphic to a complex projective space?

Question

I define an action on $\mathbb{C}-0 × \mathbb{C^2}-(0,0)$ by $(x,y,z) \mapsto ((1/a)x,ay,az)$ when $a$ is a non zero complex number, I get a manifold by quotienting. Taking element from this quotiented manifold and quotienting back in the projective,I thought it was homeomorphic to a part of the projective space. But now , I'm not sure. I'd like to know if the quotiented manifold is homeomorphic to a part of the projective space $\mathbb{C}P^2$?

Answer 1

$\newcommand{\Cpx}{\mathbf{C}}$Throughout, define an action of $\Cpx^{\times}$ on $\Cpx^{3}$ (or various invariant open subsets) by $$ \phi_{a}(x, y, z) = (x/a, ay, az). $$ (Terms such as "axis" or "plane" refer to complex subspaces.)

Consider the complement of the $x$-axis and the $(y, z)$-plane, $$ U = (\Cpx\setminus\{0\}) \times (\Cpx^{2}\setminus\{(0,0)\}) = \{(x, y, z) : x \neq 0, (y, z) \neq (0, 0)\}; $$ the complement of the $(y, z)$-plane, $$ U' = (\Cpx\setminus\{0\}) \times \Cpx^{2} = \{(x, y, z) : x \neq 0\}; $$ and $V = \Cpx^{3}\setminus\{(0,0,0)\}$.

The orbits of $\phi$ are the origin; the $x$-axis (minus the origin) $\{(x, 0, 0) : x \neq 0\}$; lines (minus the origin) in the $(y, z)$-plane; and "hyperbolas" $\{(x, y, z) \in U : xy = x_{0}y_{0}, xz = x_{0}z_{0}\}$.

The orbit of $\phi$ through an arbitrary point $(x_{0}, y_{0}, z_{0})$ in $U'$ hits the plane $\{x = 1\}$ precisely when $a = x_{0}$, i.e., at $(1, x_{0}y_{0}, x_{0}z_{0})$. Since this plane is clearly a slice of the action, $$ U'/\phi \simeq \Cpx^{2}. $$ (Restricting, $U/\phi \simeq \Cpx^{2}\setminus\{(0,0)\}$.)

So, what if we remove only the origin?

As a set, the quotient $V/\phi$ is (in bijective correspondence with) the complex projective plane. As a topological space, by contrast, $V/\phi$ is not Hausdorff: If $x_{\infty}$ denotes the image of the $x$-axis, then every neighborhood of $x_{\infty}$ meets every neighborhood of each point in the image of the $(y, z)$-plane.

The "right" example to consider for intuition is perhaps to let the multiplicative group of non-zero reals act on the real plane minus the origin by $$ \phi_{a}(x, y) = (x/a, ay). $$ The orbits are the two coordinate axes, and the hyperbolas $xy = \text{const}$. The quotient (as a set) is the vertical line $\{x = 1\}$ with one additional point $x_{\infty}$ corresponding to the $y$-axis, but as a topological space the quotient is a line with two origins: Every neighborhood of $x_{\infty}$ meets every neighborhood of the origin, a.k.a., the image of the $x$-axis.