$\mathbb{CP}^n-\mathbb{RP}^n$ deformation retracts onto a quadric

Question

Consider the natural embedding of $\mathbb{RP}^n$ into $\mathbb{CP}^n$, and let $Q$ be the quadric $z_0^2+\cdots +z_n^2=0$ in $\mathbb{CP}^n$. How can we show that $\mathbb{CP}^n-\mathbb{RP}^n$ deformation retracts onto $Q$? The answer to this question handles the case $n=2$, and I'm trying to generalize it.

Let $p:\mathbb{C}^{n+1}-0\to \mathbb{CP}^n$ be the natural projection. One way would be to find a deformation retraction of $p^{-1}(\mathbb{CP}^n-\mathbb{RP}^n)$ onto $p^{-1}(Q)$, that descends to a deformation retraction of $\mathbb{CP}^n-\mathbb{RP}^n$ onto $Q$. If we identify $\mathbb{C}^{n+1}$ with $\mathbb{R}^{n+1}\times \mathbb{R}^{n+1}$ by $(x_0+iy_0,\dots,x_n+iy_n)\to (x_0,\dots,x_n,y_0,\dots,y_n)$, then $p^{-1}(Q)$ is exactly $\{(x,y):|x|^2=|y|^2, \langle x,y\rangle=0\}$, and $p^{-1}(\mathbb{RP}^n)$ is exactly $\{(x,y): \{x,y\} \text{ is linearly dependent over $\mathbb{R}$}\}$. So, following the linked answer, if we take $(x,y)$ in the complement of $p^{-1}(\mathbb{RP}^n)$, then $x$ and $y$ are not parallel, so we may rotate $y$ in the plane generated by $x$ and $y$ until it becomes orthogonal to $x$, and then scale so that $x$ and $y$ have the same norm. But I have no idea to write this in formula (we have to choose the direction of rotation, and is this well defined?), so I cannot check whether this retraction descends to $\mathbb{CP}^n$. (Or maybe I should try this on $S^{2n+1}$ instead of $\mathbb{C}^{n+1}-0$, so that we don't need to care about norms.)

Other way would be defining a suitable function and show that its gradient flow gives a deformation retraction, but still I can't see what function to use. (Maybe the function $S^{2n+1}\to \mathbb{R}$ given by $(x,y)\mapsto \langle x,y\rangle^2$?)

Edit: I'm looking for references, and I'm trying to check whether Remark 2.8 of this paper, the first two paragraphs of Section 2 of this paper, and 3.1.2 of this paper are relevant.

Edit. Here is an argument showing that the complement $\mathbb{CP}^n-Q$ is diffeomorphic to the tangent bundle $T\mathbb{RP}^n$ such that $\mathbb{RP}^n \subset \mathbb{CP}^n-Q$ corresponds to the zero section. Note that the affine quadric $\tilde{Q}:=\{z_0^2+\cdots+z_n^2=1\}\subset \mathbb{C}^{n+1}$ is a double cover of $\mathbb{CP}^n-Q$. On the other hand, consider the tangent bundle $TS^n=\{(x,y)\in \mathbb{R}^{n+1}\times \mathbb{R}^{n+1}: |x|=1, \langle x,y\rangle=0\}$ and view $T\mathbb{RP}^n$ as the quotient $TS^n/(x,y)\sim (-x,-y)$. Identifying $\mathbb{C}^{n+1}$ with $\mathbb{R}^{2n+2}$ as above, we see that $\tilde{Q}$ is diffeomorphic to $TS^n$: the map $\tilde{Q}\to TS^n$, $(x,y)\to (x/|x|,y)$ has inverse $TS^n\to \tilde{Q}$, $(x,y)\to (\sqrt{1+|y|^2} x,y)$. So we have a series of diffeomorphisms $\mathbb{CP}^n-Q \approx \tilde{Q}/\mathbb{Z}_2 \approx TS^n/ \mathbb{Z}_2 \approx T\mathbb{RP}^n$ and clearly the zero section of $T\mathbb{RP}^n$ corresponds to the standardly embedded $\mathbb{RP}^n$ in $\mathbb{CP}^n$. So, we have $\mathbb{CP}^n= Q\cup (\mathbb{CP}^n-Q)=Q\cup T\mathbb{RP}^n$ and $\mathbb{CP}^n-\mathbb{RP}^n=Q\cup (T\mathbb{RP}^n-\mathbb{RP}^n)$. But this does not directly gives a deformation retraction of $\mathbb{CP}^n-\mathbb{RP}^n$ to $Q$. But, if we can show that the complement $\mathbb{CP}^n -D(T\mathbb{RP}^n)$ of the unit disk bundle is a tubular neighborhood of $Q$, then we can conclude that $\mathbb{CP}^n-\mathbb{RP}^n$ retracts to $Q$. I got stuck here.

Answer 1

You can probably show that $\mathbb{CP}^n\backslash D(T\mathbb{RP}^n)$ is a tubular neighbourhood of $Q$ but it is also possible to build a deform retract by using flows. Let the real analytic function $$ f : [z_0:\cdots:z_n] \longmapsto \frac{\displaystyle \left|\sum_{k = 0}^n z_k^2\right|^2}{\displaystyle \left(\sum_{k = 0}^n |z_k|^2\right)^2}. $$ Notice that $0 \leqslant f \leqslant 1$, $f^{-1}\{0\} = Q$ and $f^{-1}\{1\} = \mathbb{RP}^n$. In particular, $Q$ and $\mathbb{RP}^n$ and critical points of $f$ since they are extrema.

Claim : $f$ has no other critical points.

By symmetry, we only need to verify it in the chart $\{z_0 \neq 0\} \cong \mathbb{C}^n$ and in this chart, we have $$ f(z) = \frac{\displaystyle \left|1 + \sum_{k = 1}^n z_k^2\right|^2}{\displaystyle \left(1 + \sum_{k = 1}^n |z_k|^2\right)^2} = \frac{\displaystyle \left(1 + \sum_{i = 1}^n z_i^2\right)\left(1 + \sum_{j = 1}^n \overline{z_j}^2\right)}{\displaystyle \left(1 + \sum_{k = 1}^n z_k\overline{z_k}\right)^2} $$ Thus, its derivative with respect to $z_m$ is given by \begin{align*} \left(1 + \sum_{k = 1}^n z_k\overline{z_k}\right)^4\frac{\partial f}{\partial z_m}(z) & = 2z_m\left(1 + \sum_{j = 1}^n \overline{z_j}^2\right)\left(1 + \sum_{k = 1}^n z_k\overline{z_k}\right)^2\\ & - \left(1 + \sum_{i = 1}^n z_i^2\right)\left(1 + \sum_{j = 1}^n \overline{z_j}^2\right)2\overline{z_m}\left(1 + \sum_{k = 1}^n z_k\overline{z_k}\right)\\ & = 2\left(1 + \sum_{j = 1}^n \overline{z_j}^2\right)\left(1 + \sum_{k = 1}^n z_k\overline{z_k}\right)\\ & \times \left(z_m\left(1 + \sum_{k = 1}^n z_k\overline{z_k}\right) - \overline{z_m}\left(1 + \sum_{i = 1}^n z_i^2\right)\right) \end{align*} We deduce that if $\frac{\partial f}{\partial z_m}(z) = 0$, then either $1 + \sum_{j = 1}^n \overline{z_j}^2 = 0$ (and in this case, $z \in Q$) or, $$ z_m\left(1 + \sum_{k = 1}^n |z_k|^2\right) = \overline{z_m}\left(1 + \sum_{i = 1}^n z_i^2\right) $$ If all the $z_m$ are zero, then $z \in \mathbb{RP}^n$. Else, we can find an $m$ such that $z_m \neq 0$. If $\frac{\partial f}{\partial z_m}(z) = 0$, then, by taking the module of the above equality, $$ 1 + \sum_{k = 1}^n |z_k|^2 = \left|1 + \sum_{i = 1}^n z_i^2\right|, $$ which can only be true if all the $z_i^2$ are real non-negative i.e. the $z_i$ are real. It proves the claim.

---

Now, let us consider the gradient flow of $f$, $$ \left\{ \begin{array}{rcl} \phi(x,0) & = & x,\\ \frac{\partial\phi}{\partial t}(x,t) & = & \nabla f(\phi(x,t)). \end{array} \right. $$ The gradient can be computed thanks to the Fubini-Study metric on $\mathbb{CP}^n$ for example. The fixed points of $\phi(\cdot,t)$ for each $t$ are the critical points of $f$ i.e. $Q$ and $\mathbb{RP}^n$. And we can show the following.

Claim : For all fixed $x$, $t \mapsto \phi(x,t)$ converges at $t \rightarrow +\infty$ and $t \rightarrow -\infty$. Moreover, the function $$ \phi : \mathbb{CP}^n \times [-\infty,+\infty] \rightarrow \mathbb{CP}^n $$ is continuous at each $(x,t)$ if $-\infty < t < +\infty$, each $(x,+\infty)$ if $x \notin Q$ and each $(x,-\infty)$ if $x \notin \mathbb{RP}^n$. Moreover, for all $x \notin Q$, $\phi(x,+\infty) \in \mathbb{RP}^n$ and for all $x \notin \mathbb{RP}^n$, $\phi(x,-\infty) \in Q$.

In particular, $$ \phi : \mathbb{CP}^n\backslash Q \times [0,+\infty] \longrightarrow \mathbb{CP}^n\backslash Q $$ is a deformation retract on $\mathbb{RP}^n$ and $$ \phi : \mathbb{CP}^n\backslash\mathbb{RP}^n \times [-\infty,0] \longrightarrow \mathbb{CP}^n\backslash\mathbb{RP}^n $$ is a deformation retract on $Q$.

---

I won't detail all the proof because it would be too long but I will give the main arguments. The proof is the same at $+\infty$ and at $-\infty$ so I will only show it at $-\infty$ (since it is the case that interess you anyway). First of all, since $f$ is analytic and $\mathbb{RP}^n$ and $Q$ are compact, it satisfies a Łojasiewicz gradient inequality.

There are $C > 0$, $\delta > 0$ and $\frac{1}{2} \leqslant \theta < 1$ such that for all $x \in \mathbb{CP}^n$, $$ f(x) \leqslant \delta \quad \Longrightarrow \quad \|\nabla f(x)\| \geqslant Cf(x)^{1 - \theta}. $$ Once more, the norm is computed thanks to the Fubini-Study metric.

Let $g : t \mapsto f(\phi(x,t))$ so $$ g'(t) = \|\nabla f(\phi(x,t))\|^2. $$ $g$ is a non-decreasing function of $[0,1]$ so it converges at $-\infty$. In particular, its derivative converges toward $0$. By a properness argument, we deduce that $\phi(x,t)$ approaches at $-\infty$ a critical point of $f$. It implies that $g$ converges at $-\infty$ toward a critical value of $f$.

If $x \notin Q$, $g(0) = f(x) < 1$ so $\lim_{-\infty} g = 0$. Therefore, there is a $t_0 \leqslant 0$ depending on $x$ such that $f(\phi(x,t_0)) \leqslant \delta$. For all $t \geqslant t_0$, $f(\phi(x,t)) \leqslant \delta$ so we may apply the Łojasiewicz gradient inequality. \begin{align*} \frac{\partial}{\partial t}(f(\phi(x,t))^{1 - \theta}) & = (1 - \theta)f(\phi(x,t))^{-\theta}\left<\nabla f(\phi(x,t)),\frac{\partial\phi}{\partial t}(x,t)\right>\\ & = (1 - \theta)f(\phi(x,t))^{-\theta}\|\nabla f(\phi(x,t))\|^2\\ & \geqslant C(1 - \theta)\|\nabla f(\phi(x,t)\|\\ & = C(1 - \theta)\left\|\frac{\partial\phi}{\partial t}(x,t)\right\|. \end{align*} Therefore, if $d$ is the Riemannian distance on $(\mathbb{CP}^n,\omega_{FS})$ and $t_1 \leqslant t_2 \leqslant t_0$, \begin{align*} d(\phi(x,t_1),\phi(x,t_2)) & \leqslant \int_{t_1}^{t_2} \left\|\frac{\partial\phi}{\partial s}(x,s)\right\| \, ds\\ & \leqslant \frac{1}{C(1 - \theta)}\int_{t_1}^{t_2} \frac{\partial}{\partial s}(f(\phi(x,s))^{1 - \theta}) \, ds\\ & = \frac{1}{C(1 - \theta)}(f(\phi(x,t_1)^{1 - \theta} - f(\phi(x,t_2)^{1 - \theta})\\ & = \frac{1}{C(1 - \theta)}(g(t_1)^{1 - \theta} - g(t_2)^{1 - \theta})\\ & \underset{t_1,t_2 \rightarrow -\infty}{\longrightarrow} 0. \end{align*} By completeness of $\mathbb{CP}^n$, $\phi(x,t)$ converges when $t \rightarrow -\infty$.

To show the continuity, we use similarly the Łojasiewicz gradient inequality. Here is an article of Lerman doing it : https://arxiv.org/abs/math/0410568. He shows it with the gradient of the norm square of a moment map but all what is required is the Łojasiewicz gradient inequality. The arguments are the same.