Timeline for Conceptual explanation for the gap in the spectrum of biregular trees
Current License: CC BY-SA 4.0
6 events
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| Jun 30, 2022 at 7:01 | comment | added | Amitay | You can look at Hashimoto's paper "Zeta functions of finite graphs and Representations of p-adic groups" for the action of the operator on finite graphs, but there are also many other references. The conceptual reason this operator has square root cancellation is that it is "collision-free" on the tree (its powers are simple and sum different edges), unlike the usual adjacency operator whose powers are more complicated, but the actual proof requires some more. I can't point to a specific reference, but see arxiv.org/abs/1702.05452 or arxiv.org/abs/1609.04433, section 3.5. | |
| Jun 29, 2022 at 15:02 | comment | added | Antoine Labelle | @Amitay Do you have a reference for this operator? | |
| Jun 29, 2022 at 9:37 | comment | added | Amitay | A somewhat conceptual reason is to look at Hashimoto'a non-backtracking operator on directed edges or its square to make it slightly more symmetric. This is a non symmetric operator summing $q_1q_2$ elements, and there is a "simple" reason that its spectrum on the tree is bounded in absolute value by $\sqrt{q_1q_2}$ (and is actually all $|\lambda|=\sqrt{q_1q_2}$). Then you translate to usual adjacency operator (as in the zeta function of a finite biregular graph) and get the result. | |
| Jun 28, 2022 at 17:00 | comment | added | Will Sawin | If $q_2=1$ you can interpret it as the edge-subdivision of the regular tree. One can relate the eigenvectors and eigenvalues of the regular tree and its subdivision in a direct manner, recovering the formula, which doesn't pass through walk counting (except on the regular tree where it's maybe more intuitive). | |
| Jun 28, 2022 at 16:57 | comment | added | Will Sawin | If $q_2=1$ you can interpret it as the edge-subdivision of the regular tree. One can relate the eigenvectors and eigenvalues of the regular tree and its subdivision in a direct manner, recovering the formula, which doesn't pass through walk counting (except on the regular tree where it's maybe more intuitive). | |
| Jun 28, 2022 at 14:08 | history | asked | Antoine Labelle | CC BY-SA 4.0 |