Diffeomorphism between $\mathbf S^2 \times \mathbf S^2$ and complex projective hypersurface I have a problem while studying differential geometry and I have a question.
I want to show that $\{[{z_0:z_1:z_2:z_3]:z_0^2+z_1^2+z_2^2+z_3^2=0}\}$ and $\mathbf S^2 \times \mathbf S^2$ are diffeomorphic. How can I construct a diffeomorphism? I tried to solve this problem by using the fact that $\mathbb {CP}^1$ and $\mathbf S^2$ are diffeomorphic, but I failed.
($\mathbb {CP}^1$ is complex projective line.)
Because homogeneous coordinates are used in hypersurface or complex projective space, it is too difficult for me to construct diffeomorphism and I am not yet familiar with it.
 A: Here's an elementary argument. The quadric surface in $\Bbb CP^3$ is biholomorphically (in fact, linearly) the same as the quadric surface
$$S=\{[z_0,z_1,z_2,z_3]: z_0z_3-z_1z_2=0\}.$$
(Make the linear change of variables $T(z_0,z_1,z_2,z_3) = (z_0+iz_3,z_1+iz_2,-(z_1-iz_2),z_0-iz_3)$.)
Now you can see that $S$ is biholomorphically $\Bbb CP^1\times\Bbb CP^1$ in several ways. For example, take
$$\phi([s_0,s_1],[t_0,t_1]) = [s_0t_0,s_0t_1,s_1t_0,s_1t_1],$$
and note that the image of $\phi$ lies in $S$.  Now just argue explicitly that it is all of $S$.
A: Here's a geometric explanation I like.
Writing $z=x+iy$, the equation $z\cdot z=0$ is equivalent to $x\cdot x=y\cdot y$ and $x\cdot y=0$.  WLOG we can assume $x,y$ are unit vectors, so $(x,y)$ can be interpreted as a unit tangent vector of $S^3$, which in turn corresponds to an oriented 2D subspace $\mathrm{span}\{x,y\}$ of $\mathbb{R}^4$, or in other words a point on the oriented Grassmanian $\widetilde{\mathrm{Gr}}_2\mathbb{R}^4$. Observe that applying a phasor $e^{i\theta}$ to $z=x+iy$ corresponds to rotating the orthonormal frame $(x,y)$ within $\mathrm{span}\{x,y\}$ along the orientation. Thus, this complex projective variety can be reinterpreted as the oriented Grassmanian (as a real manifold).
Any oriented 2D subspace $\mathrm{span}\{x,y\}$ (with $\{x,y\}$ orthonormal) corresponds to a pure blade $x\wedge y\in\Lambda^2\mathbb{R}^4$. These pure blades can be interpreted as the solutions to $\eta\wedge\eta=0$ and $\|\eta\|^2=1$. The exterior square decomposes as $\Lambda^2\mathbb{R}^4=\Lambda^2_L\oplus\Lambda_R^2$ into two 3D eigenspaces (with eigenvalues $\pm1$) with respect to the Hodge star operator $\star$ (interpreting $2$-vectors as elements of $\mathfrak{so}(4)$, these are exactly the left-isoclinic vs right-isoclinic infinitessimal rotations), and the aforementioned  unit $2$-vectors $\eta$ correspond to $S^2\times S^2$ within this decomposition.
