Timeline for Effect on finite transformation semigroup under a particular modification of the generators
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Aug 1, 2021 at 14:13 | comment | added | Benjamin Steinberg | A more matrix theoretic proof would be nice. I haven't been following recent work but last I looked all proofs are improvements on Trahtman's method but use his main idea so a genuinely new proof would be interesting | |
| Jul 31, 2021 at 19:41 | comment | added | Sophie M | @BenjaminSteinberg Thanks for these comments! You're right about the transitivity. I have removed the other condition that I stated to prevent confusion. I've been reading your book, but my knowledge of modules is remedial at best, so I thought that while I work on that remediation, I'd see if phrasing the question in a slightly different way could dredge up any ideas that might be latent in the community. What I'm ultimately interested in is a potentially different proof of road colouring that might generalize to other, related problems more readily. | |
| Jul 31, 2021 at 16:49 | history | edited | Sophie M | CC BY-SA 4.0 |
the deleted condition was not equivalent to transitivity
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| Jul 30, 2021 at 22:38 | comment | added | Benjamin Steinberg | In some sense a similar problem arises in markov chains. Any stochastic matrix is a convex combination of function matrices but in a nonunique way. How do you choose the convex combination so that the semigroup they generate is the most useful to analyze the chain | |
| Jul 30, 2021 at 22:36 | comment | added | Benjamin Steinberg | I think it is hard to say what happens without road coloring which is already quite hard in some sense but I'd be happy to see an answer that says something concrete | |
| Jul 30, 2021 at 22:31 | comment | added | Benjamin Steinberg | To make sure I understand you start with a strongly connected two letter automaton and you allow yourself to relabel the edges and see what semigroups you can get and what you can say about them | |
| Jul 30, 2021 at 22:24 | comment | added | Benjamin Steinberg | Your condition doesn't seem the same as transitive to me. For insurance if f_1 is the idempotent sending n to n-1 and fixing all other inputs and f_2 is the constant map to n you meet your condition but it is not transitive. You can move any element to n and n to n-1 but you can't move n to anything else | |
| Jul 30, 2021 at 19:24 | history | asked | Sophie M | CC BY-SA 4.0 |