Timeline for Symmetry preserving graph products
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 21, 2012 at 0:43 | comment | added | Aaron Meyerowitz | I used Google Sketchup which is a pretty silly way to do it but free and easy to use. | |
| Jul 20, 2012 at 22:42 | comment | added | Hans-Peter Stricker | @Aaron: By the way: how do you draw such graphs? | |
| Jul 20, 2012 at 22:12 | vote | accept | Hans-Peter Stricker | ||
| Jul 20, 2012 at 22:11 | comment | added | Hans-Peter Stricker | Thank you very much for this elaborate answer! Now I am convinced. | |
| Jul 20, 2012 at 21:59 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
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| Jul 20, 2012 at 21:18 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
added 142 characters in body
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| Jul 20, 2012 at 14:18 | comment | added | Hans-Peter Stricker | I am thinking about $C_3\square_\pi C_4 = K_3\square_\pi C_4$ and how it could pose a problem. If this is not the smallest counter-example, maybe $C_3\square_\pi C_5 = K_3\square_\pi C_5$? (Looking for counter-examples is appropriate if one doubts a conjecture. I still believe in it - until I see a counter-example. So I should try to prove the conjecture - that $G\square_\pi H$ is vertex-transitive for vertex-transitive $G,H$.) Nevertheless I'd appreciate any worked-out counter-example! | |
| Jul 20, 2012 at 9:30 | comment | added | Hans-Peter Stricker | Could you please give me a hint, how $K_2\square_\pi C_{11}$ could be problematic. | |
| Jul 20, 2012 at 7:55 | comment | added | Aaron Meyerowitz | You are right on both points, I wasn't paying enough attention. My feeling however is simple examples will show problems in all directions. I don't think $C_3 \square_\pi C_m$ would usually be vertex transitive. I might be wrong again. Even $K_2 \square_\pi C_{11}$ seems problematic sometimes. | |
| Jul 20, 2012 at 7:24 | comment | added | Hans-Peter Stricker | $K_6$ minus a matching equals the complement of three copies of $K_2$ and thus is constructable according to my definition. | |
| Jul 20, 2012 at 7:08 | comment | added | Hans-Peter Stricker | Are you sure that your permutation $\pi_2(2)$ is in the normalizer of $\text{Aut}(C_m)$? | |
| Jul 20, 2012 at 6:38 | history | answered | Aaron Meyerowitz | CC BY-SA 3.0 |