Given a quadratic form $C(x)=x_1 ^2+x_2 ^2 + x_3^2$ in $\mathbb{C}[x_1,x_2,x_3],$ we have a conic $$C=\{C(x)=0\} = \{[x_1:x_2:x_3]: x_1 ^2+x_2 ^2 + x_3^2 = 0\}$$ in $\mathbb{P(C^3)}$, given in homogeneous co-ordinates, as usual. I want to understand why is $C \cong $ $\mathbb{P(C^2)}$. What is a right way to approach this? BTW, I know that $\mathbb{P(C^2)} \cong S^2$, if that helps.
Obviously, in $C$, one has only two free variables. In particular, $$x_3=\pm i\sqrt{x_1 ^2+x_2 ^2}.$$ So, I can define a map $$\psi: C \to \mathbb{P(C^2)}, \ [x_1:x_2:x_3]\mapsto [x_1:x_2].$$ This is well-defined I think. But this is not injective, clearly.
PS: I also want to have intuition for this.
In order to avoid $i$'s (and use standard notations) switch to $$C: x_0^2 - x_1^2 - x_2^2 = 0$$ When $x_0=1$ ( in the affine plane) this is the circle $x_1^2 + x_2^2=1$ and the trig substitution $t= \tan \frac{\theta}{2}$ for $x_1= \cos \theta$, $x_2=\sin \theta$ (or, if you want, the projection of the circle from the point $(-1,0)$ to the tangent line at $(1,0)$) gives \begin{eqnarray} x_1 = \frac{1-t^2}{1+t^2} \\ x_2 = \frac{2t}{1+t^2} \end{eqnarray} Let now $t = a \colon b\,$; get the map $$\mathbb{P}^1 \ni [a\colon b] \mapsto [x_0\colon x_1\colon x_2] = [a^2 + b^2:a^2 - b^2: 2 a b]\in C$$
The inverse map $C \to \mathbb{P}^1$ is achieved using two rational maps $\mathbb{P}^2 \to \mathbb{P}^1$ \begin{eqnarray} [x_0\colon x_1 \colon x_2] = [a^2 + b^2:a^2 - b^2: 2 a b] \mapsto [2 ab \colon 2b^2] = [x_2 \colon (x_0 - x_1)] \\ [x_0\colon x_1 \colon x_2] = [a^2 + b^2:a^2 - b^2: 2 a b] \mapsto [2 a^2 \colon 2 ab] = [(x_0 + x_1) \colon x_2] \end{eqnarray}
Added: For the quadric in $\mathbb{P}^n$ $$C: x_0^2 - x_1^2 - \cdots - x_n^2=0$$
the map $$\mathbb{P}^{n-1} \ni [a_0 \colon \ldots a_{n-1}] \mapsto [ (a_0^2 + \cdots + a_{n-1}^2)\colon (a_0^2 - \cdots - a_{n-1}^2)\colon 2 a_0 a_1\colon \ldots \colon 2 a_0 a_{n-1}] \in C$$ gives an isomorphism ( from the stereographic projection).