The Geometry Junkyard

Euler's Formula, Proof 21: Schlegel Projection

This proof comes from Hliněný, and is an induction proving that for all convex polytopes \(P\), \[ \chi(P)=\sum_{i=0}^d (-1)^i f_i=1, \] where \(f_i\) is the number of \(i\)-dimensional faces of \(P\). We include \(f_d=1\) for the whole polytope but do not consider the empty set as a face.

Consider a \((d-1)\)-dimensional Schlegel diagram for the given \(d\)-dimensional polytope, and choose a projection \(\pi\) from \(\mathbb{R}^{d-1}\) to \(\mathbb{R}^{d-2}\) whose projection direction is in general position, so that all faces of the polytope of dimension up to \(d-2\) are projected to a convex set of the same dimension. For a facet \(f\) of the polytope and its Schlegel diagram, we call the faces of \(f\) that project to faces of \(\pi(f)\) the "silhouette" of \(f\).

As in the proof by electrical charge and proof by dual charge, we assign a charge of \(\pm 1\) to each face of the polytope, equal to its contribution to the Euler characteristic, so the charge is \((-1)^i\) for a face of dimension \(i\) and the total of all the charges is just \(\chi(P)\). We then redistribute the charge to facets and sum over the facets. The charge is distributed by choosing a base point in the interior of the face in the Schlegel diagram (it doesn't matter where in the interior), and then redistributing half of it upwards and half downwards to the two facets of the polytope immediately above and below the base point according to the projection direction. This determines what to do with all charges but the one for the whole polytope, which is redistributed to the outer facet of the Schlegel diagram.

redistribution in a Schlegel diagram

Then, if \(f\) is an interior facet of the Schlegel diagram, it gets half the charge from each of its faces, except for the faces of its silhouette, from which it gets nothing. It also gets an extra half-charge from its own base point, since both half-charges go to \(f\) rather than some other facet. Therefore, its contribution to the total redistributed charge is \[ \frac12\Bigl(\chi(f)\pm 1\Bigr) - \frac12\Bigl(\chi\bigl(\pi(f)\bigr)\pm 1\Bigr) = \frac12(1\pm 1) - \frac12(1\pm 1) = 0, \] where the extra charge of \(\pm\tfrac12\) in the part of this formula coming from the silhouette corresponds to the fact that \(\pi(f)\) does not come from a silhouette face but is nevertheless counted in \(\chi\bigl(\pi(f)\bigr)\). The evaluation of both Euler characteristics as \(1\) comes from the induction hypothesis of the induction on dimension. When \(f\) is the outer facet of the Schlegel diagram, we have a similar calculation for its redistributed charge, but with each silhouette face contributing two half-charges rather than none, and with an extra \(\mp 1\) charge redistributed from the whole polytope: \[ \frac12\Bigl(\chi(f)\pm 1\Bigr) + \frac12\Bigl(\chi\bigl(\pi(f)\bigr)\pm 1\Bigr) \mp 1 = \frac12 (1\pm 1) + \frac12 (1\pm 1) \mp 1 = 1. \] Since the redistributed charge is zero for the interior facets and one for the outer facet, the Euler characteristic, which equals the total redistributed charge, is one.

Hliněný goes on to describe an equivalent view of the same proof using projections from two distinguished points in \(\mathbb{R}^d\) rather than a projection of a Schlegel diagram in \(\mathbb{R}^{d+1}\).

Proofs of Euler's Formula.
From the Geometry Junkyard, computational and recreational geometry pointers.
David Eppstein, Theory Group, ICS, UC Irvine.