How many sheets can a hyperboloid have in $n$-dimensions? For the equation describing a hyperboloid in an indefinite number of dimensions centered at $v$:
$$ (x-v)^TA(x-v)=1$$
I read that the two sheeted hyperboloid $A$ has one positive eigenvalue and two negative, vice versa for the one sheeted. Since an $n\times n$ matrix can have up to $n$ eigenvalues, how do you determine how many sheets a hyperboloid has in $n$ dimensions?
 A: Similarly.
Say that $A$ has $k$ positive and $n-k$ negative eigenvalues. After a change of coordinates, you can rewrite your equation as
$$
x_1^2+x_2^2+\dots+x_k^2-x_{k+1}^2-\dots-x_n^2=1
$$
or equivalently
$$
x_1^2+x_2^2+\dots+x_k^2=1+x_{k+1}^2+\dots+x_n^2
$$
Notice that we can choose $x_{k+1},\dots,x_n$ arbitrarily; having done so, $x_1,\dots,x_k$ are constrained to lie on a $(k-1)$-sphere of radius $\sqrt{1+x_{k+1}^2+\dots+x_n^2}$.
This gives a diffeomorphism of this hyperboloid to $\Bbb{R}^{n-k} \times S^{k-1}$. If $k>1$, both factors are connected spaces, and so the hyperboloid is also connected. If $k=1$, $S^0=\{1,-1\}$ has two connected components, so the hyperboloid has two sheets.
That is, even in higher dimensions, a hyperboloid will have either two sheets (if it only has one positive eigenvalue) or one sheet (if it has multiple positive eigenvalues). However, the number of positive eigenvalues will determine its topology; a hyperboloid with $3$ positive and $1$ negative eigenvalues is certainly a different shape from one with $2$ positive and $2$ negative eigenvalues, even though both of them are connected!
