16
$\begingroup$

The Dehn invariant of a polyhedron is a vector in $\mathbb{R}\otimes_{\mathbb{Z}}\mathbb{R}/2\pi\mathbb{Z}$ defined as the sum over the edges of the polyhedron of the terms $\sum\ell_i\otimes\theta_i$ where $\ell_i$ is the length of edge $i$ and $\theta_i$ is its dihedral angle.

Are all vectors in this space realizable as the Dehn invariants of polyhedra?

(To be very specific let's define a polyhedron to be a bounded closed manifold embedded into Euclidean or hyperbolic space as a subset of the union of finitely many planes.)

You can add two representable Dehn invariants by taking the disjoint unions of their polyhedra (or gluing them together on any pair of faces, even with mismatched face shapes, if you prefer a single connected polyhedron). For Euclidean polyhedra, you can multiply the Dehn invariant by any scalar by scaling the polyhedron by the same factor. So the left sides of the products in the sum are controllable but the right sides are not, and getting a nonzero Dehn invariant that represents only a single angle $\theta$ and its rational multiples seems difficult except for some very special cases.

The Dehn invariant also makes sense in hyperbolic space but there you have the opposite problem. You can get a polyhedron whose dihedrals are all rational multiples of the same angle $\theta$, for your favorite $\theta$, by scaling one of the Platonic solids, but scaling the lengths by arbitrary real factors (while keeping the angles fixed) becomes difficult. So I'd be interested in answers both for the Euclidean and for the hyperbolic cases, especially if they require different arguments or have different answers.

If an answer can be found in the published literature, references would be helpful.

$\endgroup$
3
$\begingroup$

In hyperbolic, spherical and Euclidean geometry the answer is no.

For Euclidean polytopes it is a result of B. Jessen (proved here). Namely, let $\Omega^1_{\mathbb{R}/\mathbb{Q}}$ be a $\mathbb{Q}-$vector space of Kähler differentials. Then the image of Dehn invariant equals to the kernel of a surjective map $$ \mathbb{R}\otimes \mathbb{R}/2 \pi \mathbb{Z} \longrightarrow \Omega^1_{\mathbb{R}/\mathbb{Q}} $$ sending $l \otimes \alpha$ to $l \dfrac{d \sin(\alpha)}{\cos(\alpha)}.$

For hyperbolic and spherical polytopes similar results were proven by Dupont and Sah. The role of Kähler differentials is played here by Milnor $K-$group $K_2(\mathbb{C})$. For instance for hyperbolic case the image of Dehn invariant equals to the kernel of a map $$ \mathbb{R}\otimes \mathbb{R}/2 \pi \mathbb{Z} \longrightarrow K_2(\mathbb{C}) $$ sending $l \otimes \alpha$ to $\{e^{l},e^{i \alpha}\}.$

A. Goncharov proposed a conjectural higher dimensional generalization of these results, see here.

$\endgroup$
  • 1
    $\begingroup$ As Douglas Zare never posted his 2017 comment (which pointed to roughly the same material) as an answer, I am accepting this one. $\endgroup$ – David Eppstein Jul 20 at 16:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.