Number of connected components of $\{x \in \mathbb{R}^n : x_1^2 + ... + x_p^2 - x_{p+1}^2 - ... - x_n^2 \neq 0\}$ Let $n \geq 3$, $p \in [1,n-1]$ and $S_p = \{x \in \mathbb{R}^n : x_1^2 + ... + x_p^2 - x_{p+1}^2 - ... - x_n^2 \neq 0\}$.  How many connected components does $S_p$ have?
For $p = 1$ or $n-1$, it is clear the answer is $3$.  I am unsure for other values of $p$.
 A: By scaling we can deformation retract this space onto the disjoint union of the "unit spheres"
$$S_{+} = \{ x_1^2 + \dots + x_p^2 - x_{p+1}^2 - \dots - x_n^2 = 1 \}$$
$$S_{-} = \{ x_1^2 + \dots + x_p^2 - x_{p+1}^2 - \dots - x_n^2 = -1 \}$$
which I personally find easier to think about. Write $v = (x_1, \dots x_p), w = (x_{p+1}, \dots x_n)$ so we can write $S_{+} = \{ \| v \|^2 = \| w \|^2 + 1 \}$. For any fixed value of $w$ the set of possible values of $v$ is a sphere of radius $\sqrt{\| w \|^2 + 1}$, so the projection $(v, w) \to w$ has connected fibers for $p \ge 2$, and $w$ can take any value in $\mathbb{R}^{n-p}$ so the base is connected too. So $S_{+}$ is connected if $p \ge 2$.
Similarly $S_{-} = \{ \| w \|^2 = \| v \|^2 + 1 \}$ so by the same argument with $v$ and $w$ switched, $S_{-}$ is connected if $p \le n-2$.
So for $p \in [2, n-2]$ there are two connected components, $S_{+}$ and $S_{-}$. When $p = 1$, $S_{+}$ disconnects into two pieces and when $p = n-1$, $S_{-}$ disconnects into two pieces. Looking at picture of the hyperboloids of one and two sheets might be helpful here.
With a bit more effort we should in fact get a deformation retract of $S_{+}$ onto the $p-1$-sphere $\{ \| v \| = 1 \}$ and of $S_{-}$ onto the $(n-p-1)$-sphere $\{ \| w \| = 1 \}$.
