Gruber P. Convex and Discrete Geometry

288 Convex Polytopes
An Alternative Approach to the Existence of the Elementary Volume
on Convex Polytopes
Using a well known determinant formula from analytic geometry, define the elementary
volume of simplices. It is easy to show that this is a valuation on the space of
all simplices in E d . The latter can be extended, by the extension result of Ludwig
and Reitzner [667] mentioned in Sect. 7.1, to a unique, simple, translation invariant,
monotone valuation on the space P of all convex polytopes in E d . Since it is easy to
see that this valuation is 1 for the unit cube, it coincides by Theorem 16.1 with the
earlier notion of elementary volume.
Why not Simply use Lebesgue Measure?
There is general agreement, that a notion of volume or measure on the space of
convex polytopes should be at least a translation invariant, simple and monotone
valuation which assumes the value 1 for the unit cube. Lebesgue measure has these
properties and is unique (on the space of Lebesgue measurable sets). So, why not
simply use Lebesgue measure instead of the elementary volume, the introduction
and uniqueness of which are so complicated to show. The reason is that it is by
no means clear that the restriction of Lebesgue measure to the small subspace of
convex polytopes is the only valuation having the mentioned properties. (Note that
the requirements of being translation invariant, etc. on the space of convex polytopes
is a much weaker property than the analogous requirement on the large space of
measurable sets.) A similar remark applies to Jordan measure.
16.2 Hilbert’s Third Problem
Let G be a group of rigid motions in E d . Two proper convex polytopes P, Q are
G-equidissectable if there are dissections {P1,...,Pm} of P and {Q1,...,Qm} of
Q such that
Pi = mi Qi with suitable mi ∈ G for i = 1,...,m.
By equidissectability we mean G-equidissectability, where G is the group of all rigid
motions.
If two convex polytopes are equidissectable, then they have equal volume. Does
the converse hold? For d = 2 the answer is yes. While this can easily be shown with
the geometric tools already known in antiquity, the first rigorous proofs are due to
Bolyai [146] and Gerwien [372]. Farkas Bolyai published his proof in a book for
high schools. In an appendix of this book Farkas’s son János published his famous
result on non-Euclidean geometry. Gerwien was a Prussian officer and amateur mathematician.
Gauss, perhaps, was in doubt whether for d = 3 the answer still is yes,
see his letters [365] to Gerling. In the late nineteenth century there were several attempts
to prove that the answer was no, for example by Bricard [166], unfortunately
with a gap. This seems to have been the motive for Hilbert [501] to state, in the third
problem of his famous list of 23 problems, the following.