Timeline for Closed geodesics and eigenvalues in a non-regular graph
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Jan 11, 2021 at 14:46 | comment | added | H A Helfgott | A non-backtracking closed path that is non-backtracking even if you change the vertex you consider as the origin (i.e., the last edge is not the inverse of the first edge). You can assume that the number of non-backtracking closed paths is also $\geq \gamma^{2 k}$. | |
| Jan 11, 2021 at 13:03 | comment | added | Fedor Petrov | what is a closed geodesic? | |
| Jan 11, 2021 at 10:33 | comment | added | markvs | If the graph is undirected what does it mean that an edge flows into another edge? Def on p 4. | |
| Jan 11, 2021 at 10:22 | history | edited | H A Helfgott | CC BY-SA 4.0 |
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| Jan 11, 2021 at 10:16 | comment | added | H A Helfgott | "edge adjacency matrix". And it's Hashimoto. | |
| Jan 11, 2021 at 10:15 | comment | added | markvs | On page 4 the word Hashimoto does not appear. | |
| Jan 11, 2021 at 9:18 | comment | added | H A Helfgott | @dodd See page 4. | |
| Jan 11, 2021 at 9:10 | comment | added | markvs | For the sake of correctness the paper you linked to does not have a definition of Hashimoto matrix. | |
| Jan 11, 2021 at 8:15 | comment | added | H A Helfgott | For the definition of Hashimoto's edge-adjacency operator, see, e.g., ftp.gwdg.de/pub/misc/EMIS/journals/EJC/Volume_16/PDF/… | |
| Jan 11, 2021 at 8:12 | comment | added | H A Helfgott | Well, if you know the number of closed walks of length $2 k$ for some $k$, then that gives you an upper bound on the largest eigenvalue of the adjacency matrix almost immediately, without needing to go through Perron-Frobenius. The eigenvalues of $A^{2 k}$ are all non-negative. (That is still the case if you deal not with the adjacency matrix $A$, but with other, related symmetric operators.) | |
| Jan 11, 2021 at 7:46 | comment | added | markvs | I do not know what Hashimoto operator is. I was talking about the adjacency matrix and its largest eigenvalue assuming that the graph is connected. | |
| Jan 11, 2021 at 7:06 | comment | added | H A Helfgott | Again : an upper bound on what? On the largest eigenvalue of the Hashimoto operator? Sure. | |
| Jan 11, 2021 at 4:46 | comment | added | markvs | I thought that the way to prove that the largest eigenvalue is unique also gives an upper bound. | |
| Jan 11, 2021 at 0:29 | comment | added | H A Helfgott | Alternatively: you can use Perron-Frobenius to bound the eigenvalues of Hashimito's edge-adjacency operator, but how do you go from there to the eigenvalues of A when the degree is non-constant? | |
| Jan 11, 2021 at 0:16 | comment | added | H A Helfgott | Is that enough? If we had a bound on the number of closed walks of length 2k, the problem would be easy (you wouldn't need Perron-Frobenius). What we are actually given is a bound on the number of closed geodesics, i.e., closed, non-backtracking walks. | |
| Jan 11, 2021 at 0:13 | history | edited | H A Helfgott | CC BY-SA 4.0 |
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| Jan 10, 2021 at 23:46 | comment | added | markvs | Perron-Frobenius? | |
| Jan 10, 2021 at 23:43 | history | asked | H A Helfgott | CC BY-SA 4.0 |