From symmetry-labeled quotient graphs of crystal nets to coordination sequences

Abstract

The combinatorial topology of crystal structures may be described by finite graphs, called symmetry-labeled quotient graphs or voltage graphs, with edges labeled by symmetry operations from their space group. These symmetry operations themselves generate a space group which is generally a non-trivial subgroup of the crystal space group. The method is an extension of the so-called vector method, where translation symmetries are used as vector labels (voltages) for the edges of the graph. Non-translational symmetry operations may be used as voltages if they act freely on the net underlying the crystal structure. This extension provides a significant reduction of the size of the quotient graph. A few uninodal and binodal nets are examined as illustrations. In particular, various uninodal nets appear to be isomorphic to Cayley color graphs of space group. As an application, the full coordination sequence of the diamond net is determined.

Appendix

We say that a triangular face of the cycle-figure is of type (p, q, r) if its vertices correspond to cycles of the quotient graph of lengths p, q and r. The topological density of a three-periodic net is given [4] after triangulation of the cycle-figure by

$$ \rho = Z\sum\limits_{\sigma } {(p.q.r)}^{ - 1} /6 $$

where Z is the order (number of vertices) of the quotient graph and the summation is over the triangular faces σ of the cycle-figure.

Every cycle in the quotient graph of dia has length 2 and, after triangulation, there are 20 faces in the cycle-figure, hence

$$ \rho \left( {{\mathbf{dia}}} \right) \, = { 2}\left[ { 20.\left( { 2. 2. 2} \right)^{ - 1} } \right]/ 6 { } = { 5}/ 6. $$

After triangulation, the cycle-figure of ths contains 8 faces of type (2,2,4), 8 faces of type (2,4,4) and 4 faces of type (4,4,4), hence

$$ \rho \left( {{\mathbf{ths}}} \right) \, = { 4}\left[ { 8.\left( { 2. 2. 4} \right)^{ - 1} + 8.\left( { 2. 4. 4} \right)^{ - 1} + 4.\left( { 4. 4. 4} \right)^{ - 1} } \right]/ 6= 1 3/ 2 4. $$