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Apr 25, 2020 at 0:08 comment added Sophie M The connection that @PhilTosteson mentions is described in more detail in "Laplacian eigenvectors of graphs: Perron-Frobenius and Faber-Krahn type theorems" by Biyikoglu, Leydold, and Stadler, section 2.7.
Apr 25, 2020 at 0:05 comment added Phil Tosteson Also, if your graph has a symmetry group, then the eigenfunction decomposition will be compatible with the decomposition of functions on the graph into irreducibles. So you may be interested in the laplacian of Cayley graphs: en.wikipedia.org/wiki/Cayley_graph.
Apr 24, 2020 at 22:28 comment added Phil Tosteson Maybe this is too simple, but in the graph case you have a representation of the group $\mathbb Z$ on the functions of the graph, and you are decomposing this representation into irreducibles.
Apr 24, 2020 at 20:42 review First posts
Apr 24, 2020 at 22:49
Apr 24, 2020 at 20:37 history asked Jackson Abascal CC BY-SA 4.0