Consider a real vector space $T$ of dimension $p+q$ with a non-degenerate symmetric bilinear form, $B:T\times T\to\mathbb{R}$, with signature $(p,q)$. Define the cone $$ C=\{(x_1,\dots,x_p,x_{p+1},\dots,x_{p+q})\in T\setminus\{0\}: x_1^2+\cdots+x_p^2-x_{p+1}^2-\cdots-x_{p+q}^2=0\}, $$ which is nothing else than the set of isotropic vectors w.r.t. the bilinear form.
I want to know how many connected components $C$ has. I strongly believe that it has two, but I'd like to know how to show it.
First, note that if $\min\{p, q\} = 0$, then by definition $C$ is empty and so has zero connected components. For other signatures, the count is given by the following:
(It's perhaps hard to visualize the result in the first case, since it first occurs in dimension $4$.)
The key construction is the following: If $\min\{p, q\} \geq 1$ but $\max\{p, q\} > 1$, then $SO(p, q)$ acts transitively on $C$ with stabilizer $P_0 := SO(p - 1, q - 1) \rtimes \Bbb R^{p - 1, q - 1}$, that is, we may view $C$ as the base of a fibration $SO(p, q) \to C$ with fiber that stabilizer, in which case we have an exact sequence $$\cdots \to \pi_0(P_0) \to \pi_0(SO(p, q)) \to \pi_0(C) \to 0.$$ (This is the tail end of the long exact sequence in homotopy for this fibration, but we don't need the full power of that construction here. If you're not familiar with homotopy, the salient fact here is that $\pi_0(X)$ can be regarded as the set of path components of $X$; since $C$ is a manifold, its path components are precisely its connected components.) In particular, the map $\pi_0(SO(p, q)) \to \pi_0(C)$ is surjective; this statement is easy to see directly, too: It's just a consequence of the fact that the projection $SO(p, q) \to C$ is surjective and continuous. (In the remaining case, $p = q = 1$, the quadratic form induced by $B$ is $(x_1, x_2) \mapsto x_1^2 - x_2^2$, and so $C$ is the symmetric difference of the diagonals $\{x_1 = \pm x_2\}$. This is the union of the open rays from the origin along the diagonals, and it has $4$ components.)
Now, $\Bbb R^{p - 1, q - 1}$ is contractible, and so $P_0$ admits a deformation retract onto $SO(p - 1, q - 1) \times \{0\} \cong SO(p - 1, q - 1)$; in particular, $$\pi_0(P_0) = \pi_0(SO(p - 1, q - 1)).$$ Now, $SO(r, s)$ has one component if $\min\{r, s\} = 0$ but two components if $\min\{r, s\} > 0$: In the latter case the identity component consists of the elements that preserve both the space and time orientations of the space, and the other the elements that reverse both.
So, if $\min \{p, q\} = 1$, then $\pi_0(P_0) = \pi_0(SO(p - 1, q - 1))$ has just one element. By the exactness of the above sequence, $\pi_0(SO(p, q)) \to \pi_0(C)$ is injective and (since we showed above that it is surjective) bijective. Since $SO(p, q)$ has two components, so does $C$.
If $\min \{p, q\} > 1$, then $\pi_0(P_0) = \pi_0(SO(p - 1, q - 1))$ has two elements, and a little meditation shows that the two components of $P_0$ are subsets of different components of $SO(p, q)$. Thus, the map $\pi_0(P_0) \to \pi_0(SO(p, q))$ is surjective, and again by exactness $\pi_0(SO(p, q)) \to \pi_0(C)$ has image a single element; since this map is also surjective, $\pi_0(C)$ contains one element, that is, $C$ has a single component.
Remark In fact, it's not too hard to identify $C$ up to homeomorphism: For $p, q > 1$, $$C \cong \Bbb S^{p - 1} \times \Bbb S^{q - 1} \times \Bbb R.$$ (In the case $q = 1$, this specializes to $$C \cong (\Bbb S^{p - 1} \times \Bbb R) \amalg (\Bbb S^{p - 1} \times \Bbb R),$$ and the result for $p = 1$ is analogous. In the case $p = q = 1$, this specializes further to $$C \cong \Bbb R \amalg \Bbb R \amalg \Bbb R \amalg \Bbb R.)$$