proof of Dehn’s theorem

We define the Dehn’s invariant, which is a number given to any polyhedron which does not change under scissor-equivalence.

Choose an additive function $f:ℝ\to ℝ$such that $f(\pi )=f(0)=0$and define for any polyhedron $P$the number (Dehn’s invariant)

$$D(P)=\sum _{e\in \{\text{edges of }P\}}f({\theta }_e)\mathrm{\ell }(e)$$

where ${\theta }_e$ is the angle between the two faces of $P$ joining in $e$, and $\mathrm{\ell }(e)$ is the length of the edge $e$.

We want to prove that if we decompose $P$ into smaller polyhedra $P_1,\mathrm{\dots },P_N$ as in the definition of scissor-equivalence, we have

$$D(P)=\sum _{k=1}^{N}D(P_k)$$(1)

which means that if $P$ is scissor equivalent to $Q$ then $D(P)=D(Q)$.

Let $P_1,\mathrm{\dots },P_N$ be such a decomposition of $P$. Given any edge $e$ of a piece $P_k$ the following cases arise:

1. $e$ is contained in the interior of $P$. Since an entire neighbourhood of $e$ is contained in $P$ the angles of the pieces which have $e$ as an edge (or part of an edge) must have sum $2\pi$. So in the right hand side of (1) the edge $e$ gives a contribution of $f(2\pi )\mathrm{\ell }(e)$ (recallthat $f$ is additive).

2. $e$ is contained in a facet of $P$. The same argument as before is valid, only we find that the total contribution is $f(\pi )\mathrm{\ell }(e)$.

3. $e$ is contained in an edge $e^{\prime }$ of $P$. In this case the total contribution given by $e$ to the right hand side of (1) is given by $f({\theta }_{e^{\prime }})\mathrm{\ell }(e)$.

Since we have choosen $f$ so that $f(\pi )=0$ and hence also $f(2\pi )=0$ (since $f$ is additive) we conclude that the equivalence (1) is valid.

Now we are able to prove Dehn’s Theorem. Choose $T$ to be a regular tetrahedron with edges of length $1$. Then $D(T)=6f(\theta )$ where $\theta$ is the angle between two faces of $T$. We know that $\theta /\pi$ is irrational, hence there exists an additive function $f$ such that $f(\theta )=1$ while $f(\pi /2)=0$ (as there exist additive functions which are not linear).

So if $P$ is any parallelepiped we find that $D(P)=0$ (since each angle between facets of $P$ is $\pi /2$ and $f(\pi /2)=0$) while $D(T)=6$. This means that $P$ and $T$ cannot be scissor-equivalent.