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Let $V_f$ be the zero set of a quadratic $z_1^2+\dots +z_{n}^2$ in $\mathbb CP^{n}$. I would like to show that

$P^{n}(\mathbb C) \setminus V_f$ is diffeomorphic to the total space of the tangent bundle $T (\mathbb RP^{n})$.


Some of my observations:

Let $\tilde{V_f}$ be the preimage of $V_f$ under the projection $S^{2n+1} \to \mathbb CP^n$. Rewriting the condition that

letting $\mathbf z \in \mathbb C^{n+1}$ be a vector $z_n^2+\dots +z_0^2=0$ in real terms, we get that $(|x|^2-|y|^2)+2i \langle x, y \rangle=0$, so $|x|^2=|y|^2$ and the inner product is zero.

However, since $|x|^2+|y|^2=1$ if it is to be on $S^{2n+1}$, we get that $|x|$ and $|y|$ are both $\frac{1}{\sqrt{2}}$.

Hence, a diffeomorphism $\tilde{V_f} \to V_2(\mathbb R^{n+1})$ (where the codomain is the set of orthonormal $2$-frames in $\mathbb R^{n+1}$) is given by $(x,y) \mapsto (\sqrt{2}x,\sqrt{2}y)$.

From this, we can conclude that $V_f \cong G_2(\mathbb R^{n+1})$.

Is there a way to conclude from here?

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  • $\begingroup$ does the quadratic involves n+1 complex coordinates or n? $\endgroup$ – Elad Jul 30 '18 at 7:15
  • $\begingroup$ I think I have a nice hint to build an explicit homeomorphism. The condition that the inner product is $0$ can be equivalent to a statement about the normal bundle of $\mathbb{R}P^n$ $\endgroup$ – Elad Jul 30 '18 at 8:25
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Let $\pi:S^{2n+1}\rightarrow \mathbb{C}P^n$ be the canonical projection and let $\tilde V_f=\pi^{-1}(V_f)\subseteq S^{2n+1}$ be the inverse image of $V_f\subseteq \mathbb{C}P^n$. As you have shown, $\tilde V_f\cong V_2(\mathbb{R}^{n+1})$, and $\pi$ restricted to this space is an $S^1$-principal fibration over $V_f\cong Gr_2(\mathbb{R}^{n+1})$. Let

$$F=\{(z_0,\dots,z_n)\in S^{2n+1}\setminus\tilde V_f\mid z_0^2+\dots+z_n^2=1\}$$

be the Milnor fibre of the polynomial $f$. Notice that $\pi|_F:F\rightarrow \mathbb{C}P^n\setminus V_f$ is surjective.

Writing $\underline z=(z_0,\dots,z_n)\in S^{2n+1}\subseteq \mathbb{C}^{n+1}$ as its real and imaginary parts $\underline z=\underline x+i\underline y$, we define a map

$$\tilde \Psi:F\rightarrow TS^n,\qquad \underline x+i\underline y\mapsto\left(\frac{\underline x}{|\underline x|},\underline y\right),$$

where we have identified the tangent bundle $TS^n=\{(\underline x,\underline y)\in S^{n}\times \mathbb{R}^{n+1}\mid \underline x\cdot \underline y=0\}$. We see that this map is well-defined using the calculation you have already supplied. In fact, writing

$$F=\{(\underline x,\underline y)\in\mathbb{R}^{n+1}\times\mathbb{R}^{n+1}\mid \underline x^2=\underline y^2+1,\;\underline x\cdot \underline y=0\}$$

we see quite easily that $\tilde \Psi$ is one-to-one and onto. It is clearly smooth and its smooth inverse is easy to write down, so it is seen to be a diffeomorphism.

We observed before that $\pi|_F:F\rightarrow \mathbb{C}P^n\setminus V_f$ is surjective, although it is not difficult to see that it fails to be $S^1$-principal. What is true, however, is that if we consider the $S^1$ orbit in $S^{2n+1}\setminus\tilde V_f$ of a point $\underline z\in F$, then we find that

$$(S^1\cdot \underline z)\cap F=\{\pm\underline z\}$$

consists of exactly two points. In fact $\pi^{-1}(\pi(\underline z))\cap F=\{\pm\underline z\}$.

The point is that $F$ is closed under the restriction of the $S^1$-action to its $\mathbb{Z}_2$-subgroup and we have a principal fibration

$$\mathbb{Z}_2\hookrightarrow F\xrightarrow{\pi|} \mathbb{C}P^n\setminus V_f.$$

Now recall that map $\tilde \psi:F\rightarrow TS^n$. it is clear from its definition that this map is $\mathbb{Z}_2$-equivariant with respect to the natural $\mathbb{Z}_2$-action on $TS^n$ induced by the tangent map of the antipodal map on $S^n$. Thus there is an induced map of $\mathbb{Z}_2$-orbit spaces

$$\Psi:\mathbb{C}P^n\setminus V_f\rightarrow T\mathbb{R}P^n\cong (TS^n)/\mathbb{Z}_2.$$

Since the fibre-preserving map $\tilde\Psi$ is a diffeomorphism, so too is its quotient $\Psi$, which is therefore the map you have been looking for.

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  • $\begingroup$ Very nice answer! $\endgroup$ – Jason DeVito Jul 30 '18 at 14:39
  • $\begingroup$ Typo after “as you have shown”? $\endgroup$ – Andres Mejia Jul 30 '18 at 15:58
  • $\begingroup$ Beautiful answer. Super well written and extremely lucid. $\endgroup$ – Andres Mejia Jul 30 '18 at 16:03
  • $\begingroup$ @tyrone this is a separate question, but do you have a quick way to see that $\mathbb CP^n \setminus V_f$ has the homotopy type of $\mathbb RP^n$ without explicitly seeing this diffeomorphism? $\endgroup$ – Andres Mejia Jul 31 '18 at 18:46
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    $\begingroup$ @Tyrone I am not sure if this is still of any interest to you at all, but there is a satisfactory answer given here[(sciencedirect.com/science/article/pii/0001870880900511) it turns out to generalize to the fact that the complement of a degree $d$-polynomial retracts onto the $n+1$-skeleton of the classifying space for $\mathbf Z/d$ (lens space.) $\endgroup$ – Andres Mejia Feb 17 '19 at 22:20

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