Are conic sections equivalent in the REAL projective plane?

Question

To transform between the unit ellipse (circle) $x^2 + y^2 =1$ and the unit hyperbola $x^2 - y^2=1$, we can use the simple change of coordinates $y_1 = iy_2$ in the complex projective plane. However, this change of coordinates is obviously not available in the real projective plane, so it is no longer obvious to me whether hyperbolas and ellipses are equivalent in the real projective plane.

Questions:

1. What is the definition of a projective change of coordinates in $\mathbb{RP}^2$? Is it the same as the definition for $\mathbb{CP}^2$ but with all real coefficients?

2. Are all non-degenerate conic sections (i.e. ellipses, hyperbolas, and parabolas), still equivalent in $\mathbb{RP}^2$, the same way they are equivalent in $\mathbb{CP}^2$?

My Research So Far:
This video on YouTube certainly seems to suggest that ellipses and hyperbolas are equivalent in the real projective plane. https://www.youtube.com/watch?v=lDqmaPEjJpk

Likewise, this webpage seems to say that all non-degenerate conics are projectively equivalent in the real projective plane. However, they say that, given an old conic section $Q$, a new conic section $Q'$, and an invertible linear transformation, that $Q'=Q\circ M$ or $Q'= -Q \circ M$. Why can't we just say that $-Q \circ M = Q \circ (-M)$ and note that $-M$ is also an invertible linear transformation? Do we have to use a different definition besides "invertible linear transformation of the homogeneous coordinates" for projective changes of coordinates in the real projective plane? The fact that sign issues are relevant seems encouraging at least since it seems connected to the problem of transforming hyperbolas into and from ellipses using real coefficients mentioned at the beginning. http://www.math.poly.edu/courses/projective_geometry/chapter_five/node4.html

The answer to this question What shape do we get when we shear an ellipse? And more generally, do affine transformations always map conic sections to conic sections? seems related to my confusion, because it states that "Since the sign of the discriminant $B^2-4AC = -4\det M$ determines the type of conic section, and the transformation $\det M \to (\det T)^2\det M$ preserves the sign, all linear and affine transformations of the plane map conics to conics of the same type (ellipses to ellipses, parabolas to parabolas, and hyperbolas to hyperbolas)" where $T$ is the transformation matrix, and $M$ is the matrix of coefficients, which is admittedly confusing given that above $M$ was the transformation matrix (not the coefficient matrix) and $Q$ was the matrix of coefficients of the quadratic form corresponding to the conic section.

So it seems like whether or not all conic sections are equivalent in the real projective plane comes down to what definition of "projective change of coordinates we use", because using the direct analog of the complex projective definition seems to make the statement fail to be true.

Answer 1

What is the definition of a projective change of coordinates in $ℝℙ^2$? Is it the same as the definition for $ℂℙ^2$ but with all real coefficients?

Well, what is your definition of that “change of coordinates” in $ℂℙ^2$? Usually the term I'd prefer would be “change of basis”. A projective basis in the projective plane can be given by four points, and with respect to that basis all points in the plane can be expressed using homogeneous coordinate vectors. The convention is to associate the four points with the homogeneous coordinates $[1,0,0]$, $[0,1,0]$, $[0,0,1]$ and $[1,1,1]$.

If this is what you have in mind, then a change of basis would be the application of a projective transformation, i.e. the multiplication of point vectors with an invertible $3\times3$ matrix over the base field, be it $\mathbb C$, $\mathbb R$ or something else.

While you just multiply such a matrix with the vector of a point, a conic as a quadratic form is usually described by a symmetric matrix, and you have to apply the transformation matrix from both sides. To be more specific, assume that $p$ is on the original conic $C$. That means $p^T\cdot C\cdot p=0$. The transformed point $p'=M\cdot p$ lies on the transformed conic $C'$ if $p'^T\cdot C'\cdot p'=0$. Which you can achieve by taking $C'=(M^{-1})^T\cdot C\cdot(M^{-1})$ since that way the matrices $M$ and $M^{-1}$ will cancel out.

So you can say that two conics $A$ and $B$ are projectively equivalent if there exists a matrix $M$ such that $M^T\cdot A\cdot M=B$.

Now consider the eigendecomposition of such a conic matrix $A$. A real symmetric matrix is always decomposable ^[1], so you always have $A=Q\cdot\Lambda\cdot Q^T$ with $\Lambda$ diagonal. If you interpret $Q^T$ as your projective transformation, that means every real conic is projectively equivalent to one which can be expressed by a diagonal matrix.

In the next step, you can apply a scaling transformation to simplify this further. For example, assume that $\lambda_1$ is the first eigenvalue, the first diagonal element of $\Lambda$. Then you can multiply with a diagonal matrix which has $1/\sqrt{\lvert\lambda_1\rvert}$ as its first entry. That way you can bring that first entry to $\pm1$. Except if it is equal to zero, in which case scaling it to magnitude one won't work. Do this scaling for all three elements and you end up with a matrix with has diagonal elements taken from $\{-1,0,1\}$.

Now you can do some permutations of coordinates, to rearrange elements. And you can also replace $A$ by $-A$ and still describe the same conic. So at the end of the day, you can say that every real conic is up to projective transformations equivalent to one where the diagonal elements are one of the following:

• $(1,1,-1)$, e.g. $x^2+y^2-1=0$. That's your regular real non-degenerate conic.
• $(1,1,1)$, e.g. $x^2+y^2+1=0$. That's a non-degenerate conic which contains no real points. Purely complex despite the real coefficients.
• $(1,-1,0)$, e.g. $(x+y)(x-y)=0$. A pair of real lines.
• $(1,1,0)$, e.g. $x^2+y^2=0$. A pair of conjugate complex lines, with their real point of intersection as the only real point in the conic.
• $(1,0,0)$, e.g. $x^2=0$. A single real line with multiplicity 2.

Are all non-degenerate conic sections (i.e. ellipses, hyperbolas, and parabolas), still equivalent in $ℝℙ^2$, the same way they are equivalent in $ℂℙ^2$?

All non-degenerate conics which do contain real points are projectively equivalent. Ellipses, hyperbolas and parabolas do fall into this category. However there is also a class of quadratic forms which has no real solutions, and if you consider these as conics as well, then no, they are not equivalent to the former.

Over the complex numbers, the above classification would become even simpler, because multiplication by $i$ from both sides will convert between $-1$ and $+1$. So you'd only distinguish by how many zeros there are. On the other hand, the statement that every symmetric matrix is diagonalizable holds only over the reals, so that part of the argument may need more work.

To gain some intuition as to why ellipses, parabolas and hyperbolas are the same, here is something in the spirit of this answer of mine. Imagine a conic, e.g. an ellipse. You can view a projective transformation as one of two things: either a transformation which takes your ellipse somewhere else, or as the inverse transformation applied not to the ellipse but to the reference frame, and to the line at infinity in particular. If you move the line at infinity so that in its new position it intersects the ellipse, then you've essentially turned the ellipse into a hyperbola. A hyperbola is just a non-degenerate conic which intersects the line at infinity in two distinct points. Between ellipse and hyperbola there is the parabola, where the line at infinity merely touches the conic in a single point. It's also easy to imagine a projective transformation taking the line at infinity to such a tangential position.

If you are restricted to affine transformations, then the line at infinity stays where it is, and so does the kind of conic you have. So the key point in the question What shape… is that it only considers affine transformations. The matrix $Q$ used in the diagonalization above will in general not be affine.