# topology of $\mathbb{P}_2 (\mathbb{C}) \setminus \mathbb{P}_2 (\mathbb{R})$?

What is the topology of $\mathbb{P}_2 (\mathbb{C}) \setminus \mathbb{P}_2 (\mathbb{R})$? For example what is the homology of this manifold with coefficients in $\mathbb{Z}$. I know that this is known but I cant find a good reference for it. Can anyone give me a reference? Thanks

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Have you tried the long exact sequence of the pair $(\mathbb{P}^2(\mathbb{C}), \mathbb{P}^2(\mathbb{C})\setminus \mathbb{P}^2(\mathbb{R})$? – Liviu Nicolaescu Apr 12 '12 at 19:40
Why are you interested in that space? – Fernando Muro Apr 12 '12 at 19:52

$H_i (CP^2 \setminus RP^2)\cong H^{4-i}(CP^2, RP^2)$ by Poincare-Alexander-Lefschetz duality (Bredon, Topology and Geometry, Theorem 8.3 on p. 351). The latter can be computed using the long exact sequence. $H^4 = Z$, $H^0=H^1=0$ is immediate. The piece

$$0 \to H^2 (CP^2,RP^2) \to H^2 (CP^2) \to H^2 (RP^2) \to H^3 (CP^2 , RP^2) \to 0$$

needs an extra argument. The map $Z=H^2 (CP^2 ) \to H^2 (RP^2)=Z/2$ is onto because the tautological complex line bundle restricts to the complexification of the real tautological line bundle, whose first Chern class generates $H^2 (RP^2)$. Thus $H_2 (CP^2 \setminus RP^2)=Z$ and $H_1 (CP^2 \setminus RP^2)=0$.

Because the first map in the above sequence is multiplication by $2$ (after identification with $Z$), it follows that the inclusion $H_2 (CP^2 \setminus RP^2)\to H_2 (CP^2)$ takes a generator to twice a generator. A generator of $H_2 (CP^2-RP^2)$ can be represented by an embedded sphere as follows. Take a quadric $Q \subset CP^2$ without real point, for example the one defined by the homogeneous equation $z_{0}^{2}+z_{1}^{2}+z_{2}^{2}=0$. By the degree genus formula, $Q$ has genus $0$, hence is a sphere. It lies in $CP^2 - RP^2$, and because its fundamental class is twice a generator of $H_2 (CP^2)$, it must represent a generator of $H_2 (CP^2 - RP^2)$.

In fact, $CP^2-RP^2$ is diffeomorphic to the normal bundle of $Q$, see Tom's comment below.

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Write a non-zero complex vector as $x+iy$ where $x$ and $y$ are real vectors not both zero. The corresponding point in $\mathbb CP^n$ is in $\mathbb RP^n$ if and only if $x$ and $y$ are linearly dependent. It is in the quadric hypersurface $Q$ given by $z_0^2+\dots +z_n^2=0$ if and only if $x$ and $y$ are orthogonal and of equal length. From this you can work out that $\mathbb CP^n-\mathbb RP^n$ is a vector bundle neighborhood of $Q$ and that $\mathbb CP^n-Q$ is a vector bundle neighborhood of $\mathbb RP^n$. – Tom Goodwillie Apr 13 '12 at 2:43
Alternatively, use that $SO(n+1)$ acts on $\mathbb{C}P^n$ with cohomogeneity 1. The special orbits are $\mathbb{R}P^n$ and $Q$. By the theory of cohomogeneity 1 manifolds, removing one special orbit leaves a vector bundles over the other. – Johannes Nordström Apr 13 '12 at 7:26

There is a beautiful (and elementary) paper by V. I. Arnold, which discusses this and generalizations.

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Thanks Igor, Can you please send the name of the paper since I cant open your link. – user13559 Apr 14 '12 at 16:49
A little surprising that you can't open it, but anyhow: Relatives of the quotient of the complex projective plane by complex conjugation (google will dredge it up, I don't know where, and even whether it appeared). – Igor Rivin Apr 14 '12 at 17:17