Timeline for Number of periodic points of subshift of finite type
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Oct 30 at 7:01 | comment | added | Ville Salo | Oh, you seem to be restricting to vertex shifts. Then it might work but it's not the same question. | |
| Oct 30 at 6:54 | comment | added | Ville Salo | For example by a result of Boyle, Lind and Rudolph, we can find a subshift of finite type with the correct period counts whose language agrees with that of the 4-shift for the first 100 steps. | |
| Oct 30 at 6:47 | comment | added | Ville Salo | Quas are you saying there are just a few matrices with the correct number of periodic points? The 2-shift has a pretty large orbit under the automorphism group of the 4-shift, so I don't get how such a naive approach can work. | |
| Oct 30 at 2:56 | comment | added | Anthony Quas | How to see all of this? The matrix is supposed to have an eigenvalue 2; and an eigenvalue 0 with algebraic multiplicity 3. This means the Jordan form of the matrix has a one-dimensional block with eigenvalue 2; and one of {three one-dimensional blocks with eigenvalue 0; a two-dimensional block with eigenvalue 0 and a one-dimensional block with eigenvalue 0; a three-dimensional block with eigenvalue 0}. Going through the cases is a bit painful, and I have to admit I have not written out full details (although I am quite confident in the outcome) | |
| Oct 30 at 2:49 | comment | added | Anthony Quas | What I mean is that in the two-sided version, even though the symbols 2 and 3 are in the alphabet, they do not appear in the actual 2-sided points because 2 and 3 do not follow any other symbol in the alphabet. In this case, the 2-sided shift with the $4\times 4$ matrix is equal to the original 2-sided shift. The same will be true for any SFT with these properties: there will be exactly two symbols that appear in the 2-sided shift; and they can occur in arbitrary concatenations. | |
| Oct 29 at 23:50 | comment | added | user119197 | Thanks a lot. Could you explain a little more for the last paragraph? Do you mean if the SFT is supposed to be two-sided, then it is topological conjugate to full shift? How to show that " any example is essentially of this type "? | |
| Oct 29 at 1:29 | history | answered | Anthony Quas | CC BY-SA 4.0 |