This post is based upon a question that I have encountered in Arnold’s ODE a while ago:

How many connected components does X^2 - Y^2 - Z^2 = \text{c}W^2 have in \mathbb{RP}^3?

Today, we will address the specific case of when c = 0, i.e.:

X^2 - Y^2 - Z^2 - W^2 = 0 (*).

Let’s first consider this in \mathbb{R}^4. After learning a bit about Lorentzian geometry in my General Relativity course, it is now easy for me to see the geometric object defined by (*). This is the zero set (or the null cone) of the Minkowski inner product on \mathbb{R}^4. This surface in \mathbb{R}^4 is a cone (more specifically, 2 cones) with its vertex at the origin. I will attach a picture of the null-cone of the 3-dimensional Minkowski space:

Screenshot taken from Desmos 3D Graphing Calculator

\mathbb{RP}^3 is the space of lines passing through the origin of \mathbb{R}^4. Geometrically, the curve for this equation in \mathbb{RP}^3 is given by the set of lines passing through the origin intersecting with the null-cone (including the points at infinity on the cone). So it seems intuitive (at least in \mathbb{R}^3) that this curve will have only one connected component.

Let’s make this more rigorous.

Claim: the curve defined by (*) in \mathbb{RP}^3 is diffeomorphic to S^2.

Let v = [X, Y, Z, W] be a solution to (*). Equivalently, we can consider v to be a vector (up to scaling) of the null-cone of the 4-dim. Minkowski space. We take for granted the following fact:

v can be decomposed as v = v_{+} + v_{-} where v_{+} is a space-like vector and v_{-} is a time-like vector and both are non-zero.

This decomposition is geometrically intuitive as the time-like vectors lie in the time-cones, which are inside the null-cones, and the space-like vectors like outside the null-cones. But it seems there are infinitely many ways to decompose, how do we define the most natural decomposition?

Since v is defined up to scaling, v_{+}, v_{-} are both defined up to scaling. Thus, we can normalize the 2 vectors; we denote the two new vectors to be u, w, respectively. The dimension of the time-like subspace is 1 and the dimension of the space-like subspace is 3 because the Minkowski product has signature (1, 3), and so:

u \in S^0 and w \in S^2

It follows that (u, w) \in S^0 \times S^2.

S^0 \times S^2 is 2 disjoint copies of S^2.

Notice that we identify (u, w) and (-u, -w) as the same point. Hence, we are really dealing with the space:

(S^0 \times S^2)/ \{ \text{(u, v) ~ (-u, -w)} \} \cong S^2, which is what we want.

It is not difficult to see that the map that arises from this process is a diffeomorphism.

Since the curve is diffeomorphic to S^2, which is connected, the curve is connected.

Leave a Reply

Is this your new site? Log in to activate admin features and dismiss this message
Log In