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Let $P \subset \mathbf{R}^3$ be a convex polyhedron. Let $rP$ be the dilation of $P$ by the positive real number $r$.

The Dehn invariant of $P$ is an element of the weird real vector space $\mathbf{R} \otimes_{\mathbf{Z}} \mathbf{R}/(\pi\mathbf{Z})$, given by the formula $$ D(P) = \sum_e (\text{length of }e) \otimes (\text{angle between the two faces containing }e) $$

The mean width $W(P)$ is a real scalar, given by the formula $$ W(P) = \int_{\vec{u} \in S^2} \left(\text{length of the interval }\left\{\vec{v} \cdot \vec{u} \mid \vec{v} \in P\right\}\right) $$

Both $D$ and $W$ scale linearly with dilations: $D(rP) = rD(P)$ and $W(rP) = W(P)$. And they are both invariant under the scissors congruence relation.

Is there a formula for $W$ in terms of $D$?

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    $\begingroup$ I do not think $W$ is invariant under scissor congruence. Compare the orders of magnitude of mean widths of $1\times 1\times N^3$ amd $N\times N\times N$ boxes. $\endgroup$ Jan 23, 2015 at 5:11
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    $\begingroup$ As Ilya points out, $W$ is not invariant under scissors congruence (if it were, then Hilbert's third problem would not have made much sense, unless Hilbert didn't know about the mean width). However, there is a formula for $W$ that is similar to the formula for $D$, namely: $W = \sum_e |e|\theta_e$ (up to correct normalization), where $\theta_e$ is the deficiency angle from the edge being flat. This is just an instance of the fact that the mean width is the total integrated mean curvature over the surface, but for a polyhedron, the mean curvature is concentrated singularly at the edges. $\endgroup$ Jan 23, 2015 at 6:05
  • $\begingroup$ Hi Ilya and Yoav. Have I misunderstood the definition of scissors congruence, or the discussion of mean width on this wikipedia page? en.wikipedia.org/wiki/Hadwiger_theorem $\endgroup$ Jan 23, 2015 at 6:16
  • $\begingroup$ Ilya, what is the mean width of the 1x1xN^3 box? $\endgroup$ Jan 23, 2015 at 6:21

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Additivity in the sense of scissors congruence means that if $A$ and $B$ have disjoint interiors, then $$D(A \cup B) = D(A) + D(B).$$

Valuations such as the mean width satisfy $$W(A \cup B) = W(A) + W(B) - W(A \cap B).$$

That last term is not necessarily $0$ when $A$ and $B$ have disjoint interiors. For example, in two dimensions, let $A$ and $B$ be (closures of) rectangular halves of a unit square. $A\cap B$ is a unit line segment. In two dimensions, the mean width is the perimeter of the convex hull divided by $\pi$. $W(A) = W(B) = 3/\pi$. $W(A \cup B) = 4/\pi$. $W(A\cap B) = 2/\pi$. So, the mean width is not additive in the sense of scissors congruence, as was pointed out in the comments. The last term is necessary in $4/\pi = 3/\pi + 3/\pi - 2/\pi$.

In three dimensions, the mean width is not determined by the Dehn invariant together with volume, since there are figures with equal Dehn invariants and volumes but different mean widths, such as the rectangular solids given by Ilya Bogdanov.

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