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Consider a conic $X^2 + Y^2 - Z^2 = 0$ in $\mathbb{C}P^2$. In an affine chart $Z \neq 0$ it is supposed to look like a circle (however it looks in $\mathbb{C}^2$), but the deceptiveness of imagining it as something analogous to a real circle broke down when I realized that in this chart it has two points at infinity, namely $A = [1, i, 0]$ and $B = [1, -i, 0]$.

Moreover, it turns out that these points at least lie in the same analytic connected component as $C = [1, 0, 1]$, as can be seen by considering another affine chart $X \neq 0$, where by letting $y = Y/X$ and $z = Z/X$ we readily get $A = (i : 0)$, $B = (-i : 0)$, and $C = (0 : 1)$, so there is an obvious analytically smooth path $y = ki$, $z = \sqrt{1 - k^2}$ ($-1 \leq k \leq 1$) connecting the three points.

At this point cannot imagine what the complex flat conic looks like and how to visualize it. In the affine chart $Z \neq 0$ when $X/Z$ and $Y/Z$ are real it forms a familiar circle, but at the same time when, say, $Y/Z$ is purely imaginary, it now looks like a hyperbola, and when both are purely imaginary, no points satisfy the equation - and all this is just one chart! I assume that in other charts the picture is kind of similar: we get either real circles or real hyperbolas along different axes (and I assume that at some clever sections we get real parabolas), but visualizing it is hard.

UPD: the added complication is that my naive attempts at deforming a circle to a hyperbola fail: assuming $x = \hat x$, $y = e^{i\pi t/2} \hat y$, where $\hat x, \hat y \in \mathbb{R}$ the equation $x^2 + y^2 = 1$ becomes $\hat x^2 + e^{i \pi t}\hat y^2 = 1$, and this equation only has solutions when $t$ is half-integer.

My question is;

Is there a trick to understanding the conic better? Maybe something involving 3-spheres?

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Why should 3-spheres come into the picture? The "curve" you describe is a 2-real-dimensional surface embedded in a 4-real-dimensional space. Every non-singular projective conic, such as the one you describe, is in fact isomorphic to the projective line (via stereographic projection), so this is topologically a sphere. –  Zhen Lin Jan 8 '12 at 8:24
    
Try to parametrize it with a map $\mathbb CP^1\to\mathbb CP^2$. –  Mariano Suárez-Alvarez Jan 8 '12 at 8:25
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1 Answer

up vote 4 down vote accepted

Your problem
Strangely but very satisfactorily all non degenerate quadrics in $\mathbb P^2(\mathbb C)=\mathbb CP^2$ are isomorphic as algebraic varieties.
"Non degenerate" means that if your conic has equation $\Sigma a_{ij}X^iX^j=0\quad (a_{ij}=a_{ji}) \;$, then the matrix $(a_{ij})$ has maximal rank i.e. rank $3$ .
As a consequence all these conics are also holomorphically, real-analytically , $C^\infty$ and topologically isomorphic.
And the common model is $\mathbb P^1(\mathbb C)$, as pointed out by Zhen and Mariano.
This is a spectacular example of the superiority of algebraic geometry over an algebraically closed field and in projective space, a point of view that slowly developed over the centuries (think Desargues, Poncelet, von Staudt, ...)
Of course you must not be deceived by the real affine picture: all these conics are irreducible and have only one connected component in the Zariski or analytic topology ( I emphasize this aspect because you write a little ambiguously : "Moreover, it turns out that these points at least lie in the same analytic connected component ..." )

A generalization
Elementary quadratic form theory at the undergraduate level teaches us that over an algebraically closed field of characteristic $\neq 2$ all non degenerate quadratic forms $\Sigma a_{ij}X^iX^j \quad (a_{ij}=a_{ji}) \;$ in $n+1$ variables $X_0,...,X_n$ are congruent i.e. become identical after a change of variables .
(Non degenerate again means that the matrix $(a_{ij})$ has maximal rank $n+1$).
Thus all the quadrics $\Sigma a_{ij}X^iX^j =0$ are isomorphic to the one given by $\Sigma _{i=0}^n (X^i)^2=0$.
So that closes the subject, right?
Wrong! There is a very interesting geometry of subvarieties of these quadrics, starting with the linear spaces they contain. The last chapter of Griffiths-Harris's Principles of algebraic Geometry is devoted to these questions.
To give just a simple example of what I mean, a non degenerate quadric in $\mathbb P^3(\mathbb C)$ is the simplest smooth surface on which there are smooth curves of all genera.

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