Definitions and questions related to projective space $\mathbb{R}P^3$

Question

I have the following questions regarding the definition of a quadric in a real projective space.

1. What is the precise definition on a quadric of signature (1,1) in the projective space $\mathbb{R}P^3$?

2. Why is any quadric in $\mathbb{R}P^3$ foliated by twofamilies of lines? What are these lines precisely? I need a thorough explanation or an online reference.

3. a) What is meant by: two subsets of the real projective space $\mathbb{R}P^3$ areprojectively equivalent?

   b) Why are any two such quadrics of signature (1,1) projectively equivalent?

A detailed answer will be highly appreciated. Thank you!

Answer 1

I have an idea what this signature might be: A quadric in $\mathbb R\mathrm P^3$ can be modeled as a symmetric $4\times4$ matrix $A$. More verbosely, it is the set of all points $p$ which satisfy the equation $p^T\cdot A\cdot p=0$. Using projective transformations you can make $A$ diagonal, and the sign pattern on that diagonal can be called its signature. Since your signature counts for $(1,1)$ only add up to two, this would imply two zeros on that diagonal, which indicates a highly degenerate quadric. But I'm just guessing here.

If my above assumptions are correct, you'd have the conic $x^2-y^2=0$ or any projectively transformed version thereof. This equation can equally be written as $(x+y)(x-y)=0$, so it factors into two planes. Each plane is a family of lines. I guess that could be the foliation you ask about.

I'd call two subsets of a projective space projectively equivalent if there is a projective transformation which maps one onto the other. For any matrix $A$ there is a projective matrix which makes it diagonal, you should be able to find a suitable transformation using the eigenvectors of the matrix. If you take the signature to be the sign pattern of the diagonal form, then you automatically get this euqivalence from the definition. Otherwise, if you e.g. define it using eigenvalues, then the work done to show that you can always reach diagonal form should help in proving the existence of a transformation between any two quadrics of the same signature.

If someone reads this and knows that this is mostly wrong, please let me know. I'm guessing a lot, since I know most of these concepts pretty well, but not using these names (mainly due to having encountered them in a different language).

Answer 2

There are several fishy things here. First of all, since there are four homogeneous coordinates when we work in $\mathbb RP^3$, the signature of the quadric (i.e., the signature of a nondegenerate (i.e., maximal rank) quadratic form whose zero-set is the quadric) must be $(4,0)$, $(3,1)$, $(2,2)$, $(1,3)$, or $(0,4)$. The last two are redundant (because we can multiply through by $-1$ without changing the quadric). Now $x_0^2+x_1^2+x_2^2+x_3^2=0$ represents the empty set in $\mathbb RP^3$. So this leaves us with two (up to linear change of coordinates in $\mathbb R^4$):

$-x_0^2+x_1^2+x_2^2+x_3^2=0$ — this is a sphere, and is not a ruled surface.

$-x_0^2-x_1^2+x_2^2+x_3^2=0$ — this looks like a hyperboloid of one sheet in $\mathbb R^3\subset\mathbb RP^3$, but the closed surface is a torus in $\mathbb RP^3$; this surface is doubly-ruled.

It is a matter of linear algebra (either giving an orthonormal basis for a bilinear form or applying the spectral theorem) to see why any two quadrics with the same signature are projectively isomorphic.

Now, if the quadratic form has rank $<4$, but still has mixed type, then the quadric surface will be singular. For example, $-x_1^2+x_2^2+x_3^2=0$ is a cone over a nonsingular conic in $\mathbb RP^2$ (and all such are projectively isomorphic). $-x_2^2+x_3^2=0$ consists of two planes in $\mathbb RP^3$, etc.