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In Thurston's Three-Dimensional Geometry and Topology, he gives a recipe for a non-standard model for hyperbolic space which he calls the paraboloid model. I'd like to use the model to try out certain parametrizations of hyperbolic isometries that do not work out in the standard models.

The recipe occurs as Problem 2.3.13, which guides one through the construction of a non-standard paraboloid model for hyperbolic space, wherein he instructs "Write down a projective transformation that maps the unit sphere of $\mathbb{R}^n$ to the paraboloid $x_n=x_1^2+\dots+x_{n-1}^2$." On the previous page, he defines "A projective transformation is a self-map of $\mathbb{RP}^n$ obtained from an invertible linear map of $\mathbb{R}^{n+1}$ by passing to the quotient." I am able to write maps from the unit sphere to the paraboloid, but I cannot think of one that meets his definition of a projective transformation.

Edit: It is not necessary to read the rest of the question to understand the answer. The remainder is my attempt to interpret the problem without the proper framework of projective geometry.

First of all, what does he mean by "linear?" An invertible linear transformation can't send a point on the unit sphere (non-zero) to the paraboloid's vertex (zero), for instance. And a Mobius transfromation can't send a circle to a parabola (the image of a parabola under Mobius transformation is a curve with a cusp, for instance a cardioid) and so a conformal map can't send a sphere to a paraboloid (I don't think).

We could stereographically project the sphere to a plane union $\infty$, then push the third coordinate up appropriately, e.g. $(x_1,\dots,x_n)\mapsto\big(\frac{x_1}{1-x_n},\dots,\frac{x_{n-1}}{1-x_n},\frac{x_1^2+\dots+x_{n-1}^2}{(1-x_n)^2}\big)$ (sending the north pole to $\infty$), but I don't see how maps of this nature could have been induced by an invertible linear map on $\mathbb{R}^{n+1}$.

It seems better to write the homogeneous $(x_0,\dots,x_n)\mapsto[x_nx_1:\dots:x_nx_{n-1}:x_1^2+\dots+x_{n-1}^2]\in\mathbb{RP}^n$, which arises from projection followed by pushing up the last coordinate, but again, what would be the corresponding invertible linear map on $\mathbb{R}^{n+1}$?

Lastly, imagine the unit sphere as the cross-section in $\mathbb{R}^n$ of a cone in $\mathbb{R}^{n+1}$, and tilt the cone so that the cross-section being taken is a parabola. For instance, the cross section at $x_3=1$ of the cone $x_1^2+x_2^2=x_3^2$ in $\mathbb{R}^3$ is $S^1$, but if we rotate the cone by $\pi/4$ about an axis tangent to the circle (toward the circle's center) then the plane $x_3=1$ would intersect the cone in a parabola. This interpretation feels like the best analogy to the Klein model, but does not map the points of the sphere to the points of the paraboloid. To achieve that would be like thinking of the sphere as an ellipsoid with both foci in the same place, and the parabola as an ellipsoid with a focus at $\infty$, so we want a map moving that focus from the center of the sphere to $\infty$, but I don't know how to write that down as a map on $\mathbb{R}^n$, nevermind one induced by an invertible linear map on $\mathbb{R}^{n+1}$.

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  • $\begingroup$ Regarding the vote to close this question: the paraboloid model has potential applications, and as far as I can tell an explicit description of its construction does not exist - any of the rare references to it I can find only refer to this problem, in which said construction is left to the reader. $\endgroup$
    – j0equ1nn
    Commented Nov 5, 2014 at 8:53
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    $\begingroup$ Your trouble arises because Thurston assumes some familiarity with classical projective geometry. Put $n=3$ for example. Consider $\mathbb{R}^3\subset\mathbb{RP}^3$ as an affine chart, say, the chart $\{[1:x_1:x_2:x_3]|x_i\in\mathbb{R}\}$. The paraboloid is $P=\{[1:x_1:x_2:x_3]| x_3=x_1^2+x_2^2\}=\{[x_0:x_1:x_2:x_3]| x_0x_3=x_1^2+x_2^2,\ x_0\neq 0\}$. $P$ has a "point at infinity" $[1:0:0:0]$. What Thurston really means by paraboloid when saying "maps the unit sphere to paraboloid" is $P$ with this point at infinity attached, i.e. $\{[x_0:x_1:x_2:x_3]| x_0x_3=x_1^2+x_2^2\}$ $\endgroup$
    – Xin Nie
    Commented Nov 5, 2014 at 9:14
  • $\begingroup$ Okay yeah that's pretty simple, guess I got hung up on technicalities there. Thanks for that, and do you happen to know of any reference where this model is used? $\endgroup$
    – j0equ1nn
    Commented Nov 5, 2014 at 9:14
  • $\begingroup$ Incidentally, when I go and embarrass myself like this I'm always reassured by Thurston's ghost in the form of his Mathoverflow bio. $\endgroup$
    – j0equ1nn
    Commented Nov 5, 2014 at 9:19
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    $\begingroup$ No, but the point is to understand that all conic hypersurfaces in $\mathbb{RP}^n$ are projectively equivalent (in other words, in projective geometry there is no distinction among ellipsoid, paraboloid and hyperboloid), and such a hypersurface delimits a convex body which serves as a model of $\mathbb{H}^n$, where geodesics are straight lines $\endgroup$
    – Xin Nie
    Commented Nov 5, 2014 at 9:22

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Your trouble arises because Thurston assumes some familiarity with classical projective geometry. Put n=3 for example. Consider $\mathbb{R}^3\subset\mathbb{RP}^3$ as an affine chart, say, the chart $\{[1:x_1:x_2:x_3]\mid x_i\in\mathbb{R}\}$. The unit sphere is $$ S=\{[1:x_1:x_2:x_3]\mid x_1^2+x_2^2+x_3^2=1\}=\{[x_0:x_1:x_2:x_3]\mid x_1^2+x_2^2+x_3^2=x_0^2\}, $$ while the paraboloid is $$P=\{[1:x_1:x_2:x_3]\mid x_3=x_1^2+x_2^2\}=\{[x_0:x_1:x_2:x_3]\mid x_0x_3=x_1^2+x_2^2,\ x_0\neq 0\}.$$ $P$ has a "point at infinity" $[0:0:0:1]$. What Thurston really means by "paraboloid" when saying "maps the unit sphere to paraboloid" is P with this point at infinity attached, i.e. $\{[x_0:x_1:x_2:x_3]\mid x_0x_3=x_1^2+x_2^2\}$. To find the required map, try something like $[x_0:x_1:x_2:x_3]\mapsto [x_0+x_3:x_1:x_2:x_0-x_3]$.

The point is to understand that all conic hypersurfaces in ℝℙn are projectively equivalent (in other words, in projective geometry there is no distinction among ellipsoid, paraboloid and hyperboloid), and such a hypersurface delimits a convex body which serves as a model of $\mathbb{H}^n$, where geodesics are straight lines.

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    $\begingroup$ I have edited to indicate an example of the map. Note that the map is not unique since precomposing with an automorphism of the Klein model would give another one. $\endgroup$
    – Xin Nie
    Commented Nov 5, 2014 at 9:57
  • $\begingroup$ Much appreciated. For future reference I will read up on my classical projective geometry. But at least now the details of this construction exist online. $\endgroup$
    – j0equ1nn
    Commented Nov 5, 2014 at 10:15

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