Conic in the projective plane

Question

An imaginary conic belongs to the projective plane? Because the projective plane has only elements of $\mathbb{R^3}$ and then I can not understand that classification of conics. There are imaginary elipes as parallel imaginary lines (degenerate conic) how do you explain this? Or is that imaginary conic means that it does not satisfy a given equation in real numbers?.

Now any conic in $\mathbb{P^2}$ as long as they are not imaginary can be associated to conics in $\mathbb{R^2}$ by means of a function that already is known. But how can I represent an imaginary conic? Or is it simply abstraction?

Answer 1

To me, a conic in the projective plane is a symmetric matrix $M\in\mathbb R^{3\times 3}$ which encodes the set of points

$$\{x\in\mathbb P\,\mid\,x^TMx=0\}$$

Just as with homogeneous coordinates for points and lines, multiples of that matrix $M$ describe the same conic, and the null matrix does not represent any conic.

Now take the identity matrix for example.

$$(X,Y,Z)\cdot\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix} \cdot\begin{pmatrix}X\\Y\\Z\end{pmatrix}=X^2+Y^2+Z^2=0$$

This equation has no solutions in the reals, since the null vector does not represent any point. So this is a purely imaginary conic. It's also non-degenerate, as the determinant of the matrix is nonzero. For a pair of imaginary lines, you could use

$$(X,Y,Z)\cdot\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix} \cdot\begin{pmatrix}X\\Y\\Z\end{pmatrix}=X^2+Y^2=0$$

This has a single real solution, namely $[0:0:1]$. The determinant is zero, so this conic factors into a pair of lines, $[1:i:0]$ and $[1:-i:0]$ in this case.

$$(1X+iY+0Z)(1X-iY+0Z)=X^2+Y^2$$

If you only consider the real projective plane, you might not see these as lines of your plane.

Actually, the above is only half the story. Projective geometry has a nice duality where one can exchange the roles of points and lines and end up with the same structure. In this sense, it makes sense to speak about a conic not only in terms of its incident points, but also in terms of its incident lines, which are the tangents in the non-degenerate case. For non-degenerate conics one can obtain the matrix of the dual conic from that of the primal by computing the inverse. For degenerate matrices that computation would entail a division by zero, but if the primal matrix has rank two (i.e. it factors into two distinct lines), then one can still obtain the matrix of the dual as the adjunct of the primal. If the primal has rank one (a double line), it does not contain sufficient information to unambiguously reconstruct the dual from that.

So more correctly I'd say that to me a conic is a pair of symmetric and non-null matrices such that their product is any multiple of the unit matrix, possibly null. Then one of the matrices describes the set of incident points and the other the set of incident lines. In this view I follow Perspectives on Projective Geometry by Richter-Gebert.