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Ordinary Minkowski space is $\mathbb{R}^{3,1}:=(\mathbb{R}^4,\phi)$ where $\phi:\mathbb{R}^4\rightarrow\mathbb{R}$ is a quadratic form of signature $(3,1)$. Lying within this is a hyperboloid model for real hyperbolic 3-space $\mathbb{I}^3:=\{p=(w,x,y,z)\in\mathbb{R}^{3,1}\mid\phi(p)=-1\}/\{\pm 1\}$.

Consider replacing $\phi$ with a complex quadratic form $\psi:\mathbb{R}^{3,1}\rightarrow\mathbb{C}$, and consider a set of the form $S:=\{p=(w,x,y,z)\in\mathbb{R}^{3,1}\mid\psi(p)=-1, w>0\}$. What sort of shape does $S$ have? Are there choices for $a, b,c,d$ (besides $-1, 1, 1, 1$) that make $S/\{\pm1\}$ a hyperboloid model?

EDIT: I've posted the answer to the above question, so here is a follow-up question. Is there some insight to be gained from the cases discovered in the answer? Would something more enlightening happen if this were carried out in higher dimensions, i.e. taking a level set of a complex quadratic form on $\mathbb{R}^{n,1}$? That may be an imprecise question, but I wouldn't mind hearing opinions...

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$S$ cannot be a 3-dimensional real hyperboloid unless $\psi$ is of the same form as $\phi$, though it can be a 2-dimensional real hyperboloid*. Excluding the possibilities achievable when $\psi$ is a real quadratic form, and working up to similarity of $\psi$ over $\mathcal{R}$, $S$ is one of the following:

$\quad\bullet\quad\emptyset$

$\quad\bullet\quad$ a point

$\quad\bullet\quad$ a 2-dimensional quadric surface in the form of a cone, an ellipsoid, a hyperboloid of two sheets*, or a hyperboloid of one sheet.

$\quad\bullet\quad$ the Cartesian product of a circle with two intersecting lines

$\quad\bullet\quad$ the Cartesian product of a hyperbola with two intersecting lines

PROOF:

By the inertia theorem, up to similarity over $\mathbb{R}$, there exist $a,b,c,d\in\{\pm 1,\pm i\}$ such that $\psi(w,x,y,z)=aw^2+bx^2+cy^2+dz^2$. In the case where $a,b,c,d\in\{\pm 1\}$, we have the usual quadrics obtainable by real quadratic forms on $\mathbb{R}^4$, and in this case $S$ is a hyperboloid model if and only if $\psi$ takes the form of $\phi$, contributing nothing new. So assume that not all of $a,b,c,d$ are $\pm 1$, and we will work up to permutation of coordinates.

Suppose $a,b,c,d\in\{\pm i\}$. Then $\psi=i\phi'$ where $\phi'$ is a real quadratic form, therefore $i\phi'(w,x,y,z))$ is purely imaginary, giving $\forall w,x,y,z\in\mathbb{R}:\Re(\psi(w,x,y,z))=0\neq-1$. So $S=\emptyset$.

Suppose $a\in\{\pm1\}$ and $b,c,d\in\{\pm i\}$. Then $\Re(\psi(w,x,y,z))=\pm w$ and $\Im(\psi)=i\phi'$ where $\phi'$ now is a real quadratic form on $\mathbb{R}^3$. So we require $w=\pm 1$, and $\phi'(x,y,z)=b'x^2+c'y^2+d'z^2=0$ (where $b'$ is the real number $\Im(b)$, etc.) which gives either $x=y=z=0$ (when the signs agree) or $(x,y,z)$ lie on a two-dimensional cone (when the signs do not agree). So $S$ is either the point $(\pm1,0,0,0)$, or a cone lying in the copy of $\mathbb{R}^3$ at $w=1$ or $w=-1$.

Suppose $a,b\in\{\pm1\}$ and $c,d\in\{\pm i\}$. If $a,b=1$ then $\Re(\psi(w,x,y,z))=w^2+x^2\neq-1$, making $S=\emptyset$. If $a,b=-1$ then $\Re(\psi(w,x,y,z))=-w^2-x^2=-1$ $\Longrightarrow(w,x)$ lie on a unit circle. If $a=1, b=-1$ then $\Re(\psi(w,x,y,z))=w^2-x^2=-1\Longrightarrow(w,x)$ lie on a regular hyperbola. But for $\phi(w,x,y,z)$ to be $-1$ we need $\Im(\phi(w,x,y,z))=0$. If $c=d=\pm i$ then this forces $y=z=0$. If $c=i$ and $d=-i$ then this forces $y^2=z^2$, i.e. $(y,z)$ lie on two perpendicular lines. Combining this with the observations about the real part, we have that $S$ is either $\emptyset$, or a cartesian product of a curve with a pair of perpendicular lines, where the curve is either a circle or a hyperbola.

Lastly, suppose $a,b,c\in\{\pm1\}$ and $d\in\{\pm i\}$. Then to get $\phi(w,x,y,z))=-1$ we require that $z=0$, and that the solution for $(w,x,y)$ to $aw^2+bx^2+cy^2=-1$ is either empty, an ellipsoid, a hyperboloid of one sheet, or a hyperboloid of two sheets, as determined by the signs.

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