Euclidean geometry as a subgeometry of Möbius geometry

Constructing a subgeometry of Möbius geometry that is isometric to Euclidean geometry is fairly commonly done, but often only a special case is treated. This is an attempt at generalization.

Let \(Q \subset \RP^n\) be a quadric of signature \((n+1, 1)\), \(\langle\cdot,\cdot\rangle\) its bilinear form, \(\mathbf{e_\infty}\) a point in \(Q\) and \(B = \mathbf{b}^\perp\) a hyperplane not containing \(\mathbf{e_\infty}.\) Also fix a representative vector \(e_\infty\) of \(\mathbf{e_\infty}\). This choice will scale the metric we get in the end.

Set \(\ell_\infty \coloneqq B \cap \mathbf{e_\infty}^\perp\). In these terms, for inverse stereographic projection \(\sigma^{-1}\colon B \setminus \ell_\infty \to Q \setminus\{\mathbf{e_\infty}\}\) via \(\mathbf{e_\infty}\) we have the formula

\[\sigma^{-1}\colon [x] \mapsto [-2\langle x, e_\infty\rangle x + \langle x, x \rangle e_\infty].\]

Next, set \(\mathbf{O} \coloneqq \operatorname{join}(\mathbf{b}, \mathbf{e_\infty}) \cap B\). If \(O\) is any representative vector (we will choose a good one later), we can now decompose \(B \setminus \ell_\infty \ni [x] = [O + \tilde{x}]\) with \([\tilde{x}] \in \ell_\infty.\) \(\tilde{x}\) is unique: it is the intersection of the line \(-O + tx\) with the vector subspace \(\ell_\infty\) in \(\R^{n+1}.\) Scaling \(O\) will scale \(\tilde{x}\) in the same way. Using such a decomposition, we get

\[\begin{split}\begin{aligned} \sigma^{-1}([\tilde{x} + O]) &= [-2\langle \tilde{x} + O, e_\infty\rangle (\tilde{x} + O) + \langle \tilde{x} + O, \tilde{x} + O \rangle e_\infty] \\ &= [-2\langle O, e_\infty\rangle O + \langle O, O\rangle e_\infty -2\langle O, e_\infty\rangle \tilde{x} + \langle \tilde{x},\tilde{x}\rangle e_\infty], \end{aligned}\end{split}\]

where we used that \(\langle \tilde{x}, e_\infty\rangle = \langle \tilde{x}, b\rangle = \langle \tilde{x}, O\rangle = 0\).

Now choose the representative \(O\) such that \(\langle O, e_\infty\rangle = \pm\frac{1}{2}\) (sign doesn’t matter). [1] Then set \(\mathbf{e_0} \coloneqq \sigma^{-1}(\mathbf{O})\) and fix the representative \(e_0 \coloneqq -2\langle O, e_\infty\rangle O + \langle O, O\rangle e_\infty.\) Note that \(\langle e_\infty, e_0\rangle = - \frac{1}{2}\) automatically and that \(\langle \tilde{x}, e_0\rangle = 0.\) We get

\[\sigma^{-1}([\tilde{x} + O]) = [e_0 \pm \tilde{x} + \langle \tilde{x},\tilde{x}\rangle e_\infty]\]

and finally:

\[\Bigl\langle e_0 \pm \tilde{x} + \langle \tilde{x},\tilde{x}\rangle e_\infty,\; e_0 \pm \tilde{y} + \langle \tilde{y},\tilde{y}\rangle e_\infty \Bigr\rangle = - \frac{1}{2} \langle \tilde{x} - \tilde{y}, \tilde{x} - \tilde{y} \rangle.\]

Since \(\langle\cdot,\cdot\rangle\bigr|_{\ell_\infty}\) is positive definite, this is a transformed version of the Euclidean metric.

Every \([a] \in Q \setminus \{\mathbf{e_\infty}\}\) has a representative of the form \(e_0 \pm \tilde{x} + \langle \tilde{x},\tilde{x}\rangle e_\infty\) which can be found by normalizing, namely by dividing \(a\) by \(-2\langle a, e_\infty \rangle.\) We get

\[-\frac{1}{2} \frac{\langle a, b \rangle} {\langle a, e_\infty\rangle \langle b, e_\infty\rangle} =\langle \tilde{x} - \tilde{y}, \tilde{x} - \tilde{y} \rangle\]

for any \([a] = \sigma^{-1}([x]) \in Q \setminus \{\mathbf{e_\infty}\}\) and the same for \(b\) and \(y.\)

Footnotes

[1]

It is also possible to go backward and choose \(O\) first and then choose the representative \(e_\infty\) so that this relation is fulfilled. If, for example, we wanted to match the Euclidean metric in affine coordinates, we would dehomogenize \(O\).