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.)
Best Answer
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".