Homotopy type of $\Bbb CP^2 - \Bbb RP^2$

Question

Consider the complex projective plane $\Bbb CP^2$. We can embed $\Bbb RP^2$ in $\Bbb CP^2$ in a natural way, namely, $[x_0:x_1:x_2]\in \Bbb RP^2 \mapsto [x_0:x_1:x_2] \in \Bbb CP^2$. We can thus consider $\Bbb RP^2$ as a subspace of $\Bbb CP^2$. On the other hand, let $Q=\{[z_0:z_1:z_2]\in \Bbb CP^2 : z_0^2+z_1^2+z_2^2 =0\}$. Clearly $Q$ does not intersect $\Bbb RP^2$, so we have $Q \subset \Bbb CP^2-\Bbb RP^2$. Does $\Bbb CP^2 - \Bbb RP^2$ deformation retracts onto $Q$?

Actually I want to show that $\Bbb CP^2-\Bbb RP^2$ is homotopy equivalent to $S^2$ in two steps, by first showing that $\Bbb CP^2-\Bbb RP^2$ is homotopy equivalent to $Q$ and then showing $Q$ is homotopy equivalent to $S^2$. Though I have no idea with the latter one, I want to manage the former one first.

What I know about these is, that if we let $K=\{(z_0,z_1,z_2)\in \Bbb C^3-0:z_0^2+z_1^2+z_2^2=1\}$, then the canonical projection $K\to \Bbb CP^2-Q$ is a surjective, two-to-one, covering map, and that $K$ is homeomorphic to $TS^2$ and hence deformation retracts onto $S^2$. Thus $\Bbb CP^2-Q$ is homotopy equivalent to $\Bbb RP^2$. (But I'm not sure that these informations is helpful for the above questions)

Any hints or ideas please? (I hope some elementary approaches, if possible.)

Answer 1

It is easier to retract the corresponding sets in $\mathbb{C}^3$. There, $\tilde{Q}$ given by $z_0^2+z_1^2+z_2^2=0$ when written in coordinates $z_j=x_j+i y_j$ becomes $|\vec{x}|^2-|\vec{y}|^2=0, \vec{x}\cdot \vec{y}=0$, that is the set of orthogonal vectors of equal length. The lift of $\mathbb{R}P^2$ is the set $P$ of parallel vectors. The complement of this $P$ retracts to $\tilde{Q}$ by, say, taking any pair of non-parallel vectors $(\vec{x},\vec{y})$ and rotating $\vec{y}$ in the plane spanned by $(\vec{x},\vec{y})$ until it is orthogonal to $\vec{x}$ (i.e. retracting the set of directions in this plane not equal to $\pm \vec{x}$ to the two directions orthogonal to $\vec{x}$, aka retracting $S^1\setminus\text{{two points}}$ to two points ) and then rescaling resulting $\vec{y}$ to have the same norm as $\vec{x}$. To do this in a way that descends to the projective space is a bit harder, so the following is proposed as an alternative.

Consider vectors $\vec{z}=\vec{x}+i\vec{y}$ in $\mathbb{C}^3=\mathbb{R}^6$ and the functional $|\vec{x}\times \vec{y}|^2$. It has gradient $2 ( |\vec{y}|^2 \vec{x}-(\vec{x}\cdot \vec{y})\vec{y}, |\vec{x}|^2 \vec{y}-(\vec{x}\cdot \vec{y})\vec{x})$. We restrict to the sphere $S$ given by $|\vec{z}|^2=|\vec{x}|^2+|\vec{y}|^2=1$ and find that the critical points of the restriction are given by Lagrange condition $$( |\vec{y}|^2 \vec{x}-(\vec{x}\cdot \vec{y})\vec{y}, |\vec{x}|^2 \vec{y}-(\vec{x}\cdot \vec{y})\vec{x}) \sim (\vec{x}, \vec{y}).$$

This holds either when $\vec{x}\cdot \vec{y}=0$ and $|\vec{x}|^2=|\vec{y}|^2$ (maximum, area is 1) OR when $x$ is parallel to $y$ (minimum, area is zero).

Thus the gradient flow of this functional (with respect to the restriction of the standard metric) retracts $S\setminus \{ x \text{ is parallel to } y\}$ to the set $S\cap \{\vec{x}\cdot \vec{y}=0\text{ and } |\vec{x}|^2=|\vec{y}|^2\}$.

The functional and the standard metric are both invariant under the equivalence relation $\vec{z}\sim e^{i\theta}\vec{z}$ (for the metric observe that the map $\vec{z} \to e^{i\theta}\vec{z}$ is block diagonal with 2 by 2 rotation matrix in each block; for the functional $([\cos \theta] \vec{x}-[\sin\theta] \vec{y} ) \times ([\sin \theta] \vec{x}+[\cos\theta] \vec{y} )=(\cos^2\theta+\sin^2 \theta)(\vec{x}\times \vec{y})$).

Hence the flow descends to the flow on equivalence classes. But $ \left(S\setminus \{ x \text{ is parallel to } y\}\right)/\sim$ is $\mathbb{C}P^2\setminus \mathbb{R}P^2$ and $\left(S\cap \{\vec{x}\cdot \vec{y}=0\text{ and } |\vec{x}|^2=|\vec{y}|^2\}\right)/\sim$ is $Q$.

This can probably rewritten in terms of momentum maps etc., which, if possible, would make it more illuminating, but less "elementary".