.. _euclidean_moebius:

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

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

.. math::

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

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

.. math::

    \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}

where we used that :math:`\langle \tilde{x}, e_\infty\rangle = \langle
\tilde{x}, b\rangle = \langle \tilde{x}, O\rangle = 0`.

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

.. math::

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

and finally:

.. math::

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

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

.. math::

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


.. rubric:: Footnotes

.. [#]

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