Consider the projective plane $\mathbb{R}P^2$ and a symmetric matrix $B \neq 0$ of a bilinear form that defines a quadric $Q := \{ [v] \in \mathbb{R}P^2 : v^tBv = 0\}$.

Is the following ok? And for the affine part I would need help/tips. I am very new to this stuff and do not really know how to handle these "equivalences" and keep projective, affine, euclidian geometry/maps apart.

First classify equivalent quadrics under projective transformations:

Sylvester's law of inertia allows to diagonalize $B$ using $S^{t}BS$ where $S \in Gl(\mathbb{R},3)$ so that $B$ has only diagonal elements in $\{0,1,-1\}$. Since $S \in Gl(\mathbb{R},3)$ it is a projective map, and hence the resulting quadrics are equivalent under projective transformations. We have the following 5 equivalence classes of quadrics represented by there diagonal elements: $(1,1,1),\ (1,1,-1),\ (1,1,0),\ (1,-1,0),\ (1,0,0)$.

Second assume now the projective plane $\mathbb{C}P^2$. Then in analogy we represent the 3 equivalence classes by there occuring diagonal elements $(1,1,1),\ (1,1,0),\ (1,0,0)$ after transformation. So here Sylvester provides us only with diagonal elements $\{0,1\}$ left and hence the number of classes of quadrics reduces.

Third assume again the projective plane $\mathbb{R}P^2$ but now consider only equivalence using affine transformations.

Here there should now be more than 5 cases, since less matrices for diagonalzation of $B$ are allowed, namely only those of affine transformations of the form $\begin{pmatrix}A & a \\ 0\ \ 0 & 1\end{pmatrix}$ where $A \in Gl(\mathbb{R},2), a \in \mathbb{R^2}$.

But I do not know how to work this out now..help? For me this looks weird..


An affine transformation is a projective transformation which fixes the line at infinity. So start with the five classes from $\mathbb R\mathrm P^2$, and examine each further.

  1. $(1,0,0)$ is a double line. This splits into two cases: either the line in question is the line at infinity, or it is a finite line.
  2. $(1,-1,0)$ consists of two different lines. Here you have three cases to consider: either one of the lines is the line at infinity, or neither is but they still intersect at infinity (i.e. are parallel), or they intersect in a finite point.
  3. $(1,1,0)$ are two complex conjugate lines intersecting in a real point. Neither complex line can be the line at infinity, since that is a real line. So you have two cases: the point of intersection is either finite or infinite.
  4. $(1,1,-1)$ is the classical non-degenerate real conic. It intersects the ine at infinity in either zero or one or two real points, which corresponds to the classification in ellipse, parabola and hyperbola.
  5. $(1,1,1)$ has no real points. So in particular, it has no real points in common with the line at infinity.

For the real cases (1, 2, 4) I'm fairly convinced of my result: given two quadrics from the same equivalence class, I could always find an affine transformation to transform one into the other. For the complex cases (3, 5) I was not as sure, since intuition fails me. So in those cases you might want a more reliable proof.

  • $(1,1,1)$: Complex non-degenerate quadric.

    Any real line intersects the quadric in two complex conjugate points. So intersect your quadric with $z=0$ (the line at infinity) and with $y=0$ to obtain two pairs of points. Doing so for two quadrics of the same kind, you obtain two pairs of points, and there exists a unique projective transformation between these two which maps corresponding pairs to one another. This projective transformation will be affine, since the line at infinity will be fixed due to the intersection points with $z=0$. The matrix will also be real, since the conjugate of the matrix can be determined using the conjugates of the defining pairs, but each pair is its own conjugate, since it consists of conjugate points.

  • $(1,1,0)$: Two complex lines with real finite point of intersection.

    The same argument as above still holds, except you should not use the line $y=0$ but instead a line which avoids the real point of intersection so you get two distinct complex conjugate intersections.

  • $(0,1,1)$: Two complex lines with real infinite point of intersection.

    In this case you only intersect with a single finite line, one which avoids the real point of intersection and yields a single pair of complex conjugate points. The third defining point for the transformation will be the point of intersection at infinity, and the fourth point can be an arbitrary real point at infinity. The matrix mapping corresponding points will again fix the line at infinity, since two points at infinity are mapped to two points at infinity. It will also be real, since the one pair is its own complex conjugate, and the other points are real already, so each of them is its own conjugate. Furthermore, the quadric is really mapped to its image since each quadric is already defined by its class, the point of intersection at infinity and the two complex points of intersection.

So now that the more doubtful cases have been verified, I'm convinced of my classification. You have $2+3+2+3+1=11$ classes to consider in total.


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