# Why are degenerate conics not projectively equivalent to nondegenerate conics?

This is what I understand about conics being projectively equivalent.

Two conics $C1=V(F)$ and $C2=V(G)$ are projectively equivalent if there is an invertible matrix $A$ such that $F(X,Y,Z)=0$ iff $G(X,Y,Z)=0$ where $[X Y Z] = [X' Y' Z']A.$

This seems as though what we are doing is using $A$ to change the coordinate system. Wouldn't this be the same as shifting/rotating/moving our coordinate axes in some way?

But then couldn't we move the coordinate axes to intersect a cone in a degenerate way? Since conics are formed from intersecting a plane with a cone and it's easy to visualize moving a plane which intersects the cone to create a (nondegenerate) hyperbola to where its intersection draws out a (degenerate) pair of lines for example.

I'm having trouble with understanding this visually.

• Remember that the matrix $A$ should be regular. A non-degenerate conic is not projectively equivalent to a degenerate conic, because there exists no such regular transformation matrix. Remember that a non-degenerate ellips, parabola and hyperbola are projectively equivalent. – Steven Van Geluwe May 6 '15 at 12:27
• A rotation is a transformation in the plane with a regular change-of-basis-matrix. – Steven Van Geluwe May 6 '15 at 12:30
• I think the problem with your nice reasoning with the cone is that you are not working in the projective plane, but in the projective space (so we need four coordinates $(X,Y,Z,T)$). – Steven Van Geluwe May 6 '15 at 12:31

One possible answer: A projective transformation preserves collinearity of points. A degenerate conic contains collinear points, a non-degenerate does not. So these two cannot be projectively equivalent.

A second answer: if you imagine the embedding in 3-space, then your projective plane can be visualized as the intersection of an affine (drawing) plane with linear spaces of suitable dimension: one-dimensional for points and two-dimensional for lines. (That visualization drops points at infinity, but that's irrelevant here.) In this setup, a non-degenerate conic is the intersection of your drawing plane with a double cone which is centered at the origin. A degenerate conic, on the other hand, is the intersection of the plane with a pair of planes through the origin. Sure, you could describe the same conic using some other cone, and then moving the coordinate system would lead to a degenerate situation. But describing the conic section in this way would depend on the representatives of the points, and therefore would not be well defined in a homogeneous world.

• You seem to have gone astray in your second answer. Where I come from, degenerate conics are by definition - on the same hand with the non-degenerate conics - the intersection of the drawing plane with a double cone. They are also well defined in homogeneous coordinates. I trust you don't mind using the standard form for degenerate cases? – Slumberland May 18 '16 at 6:04
• @Slumberland: Perhaps we can continue this discussion in chat? – MvG May 18 '16 at 15:08
• all has been cleared up for me. Sorry for my confusion and thanks again! – Slumberland May 18 '16 at 16:32

Let the drawing plane pass through the vertex of the double cone. And let us agree to call such a section "degenerate".

I say, that the resulting section is

1. a Pair of Intersecting Lines, if any part of the plane passes inside the cone.
2. a Line, if the plane touches (is tangent to) the conic surface.
3. a Point, if the only point of contact is the vertex.

Now. The algebra in your post is an abstraction. It represents a geometric transformation, and will be valid when it preserves measure in the same way. So what are the constraints, and what transformations do we agree (are we able) to carry out?

Constraint: The center of the first drawing plane is the vertex of the cone.

Transformations: Translation, rotation.
Procedure: Project the first section onto a second plane which may have a different position and orientation in space.

Begin. For simplicity, let us set up case 1: C1 is a pair of intersecting lines.
On the drawing plane, mark the vertex V. Now project. Mark V', the image of V. And let C2 be any projection where -whatever the resulting shape- the self-intersection is still visible at V': there are two, distinct (possibly curved) lines, passing through V'.

• (a) There is only one cone. Whatever else is true about C2, it is not drawn by directly intersecting the second plane with a cone.
• (b) By assumption, any arrangement where the two arms of the section do not intersect at V' is not the case. We are projecting C1, and no other figure.
• (c) There exists only one conic section in which two distinct lines -curved or straight- meet at a point: Intersecting Lines.
• (d) If the projecting lines are straight, the breadth of C1 can vanish to a single line only if the image of the whole first plane is a line... and therefore C2 is a straight line.

We now have the answer to your question. We have a case where C1, C2 must meet two distinct criteria

• The image on the second plane cannot be equivalent to a free intersection of cone and plane. The constraint C1 contains the vertex of the cone determines a property of its image (an intersection), which restricts the kind of section.

• If the image is a conic section and the intersection is visible then it can only be a pair of intersecting lines. But also, if the image is a conic section and there is only one line, it is straight.

Hence, if the first section is degenerate, it cannot project to a non-degenerate section. The two kinds cannot be equivalent under invertible projection.

Solution 2
If we already agree that straight lines project to straight lines, any algebra that allows equivalence of degenerate and non degenerate conics is against the hypothesis, and should be thrown out. This didn't seem to honestly engage your question, so I chose instead not-to-know key properties in advance.

Lastly,to the algebra. Consider a matrix solution of (the planar equation) $$Ax^2 + B x y + C y^2 + Dx + E y+F = 0$$ from 5 given points, $\;P_i = (x_i, y_i), \;\; i = 1, 2, ..., 5.\;$ We have a 5x6 matrix of known values made up of $(x_i, y_i)$, and a column vector of the coefficients: $[X]\vec{c} = 0$

It appears we can pick any value for F, push it to the RHS, leaving a 5x5 matrix $[\bar X]$ on the LHS. Then the coefficients are $$\vec{c} = [\bar X]^{-1} \left[ \begin{array}{} 1 \\ \vdots \\ 1 \end{array} \right] F$$ But this doesn't always work. Reducing the original 5x6 matrix using Gaussian Elimination will. I think you will find the degenerate conics are among those that fail this matrix inversion.

Hope that helps.