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EDIT: as pointed-out below, this has been posted on math.stackexchange. I'll leave it up to the community whether or not to delete this question, but I do think there is room for a more technical answer than the one posted on math.stackexchange.


A famous question related to Hilbert's third problem:

Given any two polyhedra of equal volume, is it always possible to cut the first into finitely many polyhedral pieces which can be reassembled to yield the second?

This was answered in the negative by Max Dehn, who showed that it was impossible to cut a cube into a finite number of polyhedral pieces that can be reassembled into a tetrahedron. This was accomplished by associating to each polyhedra $P$, an element $D(P)$ of the group $\mathbb{R}\otimes_{\mathbb{Q}}(\mathbb{R}/\mathbb{Q}\pi)$ that is invariant under cutting and pasteing. To obtain his result, Dehn then proved that the cube had value 0 in the group above, and the tetrahedron had a non-zero value.

My question is this: is there a more conceptual explanation for why this question should be answered in the negative? In particular, the corresponding statement about polygons is true: given two polygons with the same area, there is a decomposition of the first polygon into polygonal pieces that may be reassembled into the second (see the Wallace-Bolya-Gerwien Theorem). Is there a specific property of polygons that we are exploiting to prove WBG, that fails when we go to three dimensions?

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    $\begingroup$ math.stackexchange.com/a/95526/87355 $\endgroup$ – Carlo Beenakker Jan 3 '17 at 11:20
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    $\begingroup$ I remember I have once attended a lecture where Dehn's invariant was explained in an intuitive way. Alas I don't remember the details, but it came down to the fact that the dihedral angles were irrational, so one could then create a nontrivial function satisfying equation f(x+y)=f(x)+f(y) which is 0 on rational multiples of pi. I believe that f(sum of all dihedral angles in a dissection) is invariant and can be made different for a cube and a tetrahendron. $\endgroup$ – Wojowu Jan 3 '17 at 12:05
  • $\begingroup$ Additionally, we don't need the axiom of choice above, since f only needs to be defined on a finite dimensional vector space over Q. (sorry for no LaTeX, on mobile right now) $\endgroup$ – Wojowu Jan 3 '17 at 12:07
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    $\begingroup$ From the link Carlo provided: "There is intuition behind the Dehn invariant, essentially relying on a distinction between dihedral angles that are rational multiples of $\pi$ and those that are not. The cube is not scissors congruent to the regular tetrahedron basically because the latter has irrational (multiples of $\pi$) dihedral angles while the cube has rational dihedral angles, $\frac{1}{2}\pi$. Of course, there is much more, but this is somehow the essence." $\endgroup$ – Joseph O'Rourke Jan 3 '17 at 13:26
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    $\begingroup$ I am going to self promote here, sorry. I tried to explain this in my book - see sections 15-17. You might enjoy the reading... math.ucla.edu/~pak/book.htm $\endgroup$ – Igor Pak Jan 5 '17 at 11:10
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The Dehn functional of a polyhedron $P$ with edge lengths $\ell_i$ and exterior dihedral angles $\theta_i$ is $D(P) = \sum_i \ell_i \otimes_{\mathbb Q} (\theta_i\, \mathrm{mod}\, \mathbb{Q}\pi)$. It looks similar to the (twice) the total mean curvature $S(P) = \sum_i \ell_i \theta_i$. Both of them satisfy the "valuation property": $$ F(P \cup Q) = F(P) + F(Q) - F(P \cap Q) $$ for any two convex polyhedra whose union is convex. In particular, if $P \cap Q$ is a polygon, then $S(P \cap Q)$ is $\sum_i \ell_i \cdot \pi$. Dehn's idea is to take $\theta\, (\mathrm{mod}\, \pi)$ to make this vanish. Since we need linearity and distributivity, the product of $\ell$ with $\theta$ becomes the tensor product over $\mathbb{Q}$. Because of $D(P \cap Q)=0$, Dehn's functional is invariant with respect to cutting a polyhedron along a plane, so on equidecomposable polyhedra the functional has the same value.

Dehn almost surely knew about the total mean curvature (Minkowski studied more general functionals shortly before), so this might be the way he invented his functional.

As for the polygons, a similar argument leads to a functional that is the angle sum modulo $\pi$, so always vanishes.

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  • $\begingroup$ Very interesting! Do you know if one can actually obtain something like curvature by passing to limit wrt more and more thin polyhedral subdivisions of a surface? $\endgroup$ – მამუკა ჯიბლაძე Mar 17 '17 at 10:30
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    $\begingroup$ If you approximate a smooth convex surface by convex polyhedra, then $1/2 \sum \ell_i \theta_i$ converges to the integral of the mean curvature. This is the continuity of the mixed volumes (more specifically, of intrinsic volumes). $\endgroup$ – Ivan Izmestiev Mar 17 '17 at 10:36
  • $\begingroup$ Great! Do you know where to read about such things? $\endgroup$ – მამუკა ჯიბლაძე Mar 17 '17 at 10:47
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    $\begingroup$ "Theory of convex bodies" of Bonnesen and Fenchel is quite good. $\endgroup$ – Ivan Izmestiev Mar 17 '17 at 11:04

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