PermalinkSubmitted by johnmfranks on Mon, 07/03/2017 - 11:02

I suspect that "flow conjugacy" here is intended to mean "flow equivalence" since otherwise it would be the same as topological conjugacy and det(I -At) is an invariant.

For flow equivalence of subshifts of finite type, complete invariants are known. If the shift is irreducible and non-trivial the invariant for flow equivalence is the group BF(A) := Z^{n}/(I-A)Z^{n }together with the integer det(I-A) (see [1] for the final part of the proof and references). The multiplicity of 1 as an eigenvalue of A is the rank of the free part of the group BF(A). So techically the answer to the first question is yes the multiplicity of 1 is an invariant for irreducible SFTs. The fact that BF(A) is an invariant is proved in [2] based on results of Parry and Sullivan.

Complete invariants are also known for non-irreducible SFTs, but the answer is more complicated, see [3] for details and further references.

PermalinkSubmitted by Mike Boyle on Thu, 07/06/2017 - 17:39

The answer to the second question is indeed no in the case that A has 1 as an eigenvalue. For example, compare $\begin{pmatrix} 2&1\\2&3 \end{pmatrix}$ and $\begin{pmatrix} 2&1\\1&2 \end{pmatrix}$.
For the first question, note: the multiplicity of 1 as an eigenvalue of A means the dimension of the eigenspace, which can be smaller than the mutliplicity of 1 as a root of the characteristic polynomial of A.

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## Comments

## FL

Someone (Caroline Series?) wrote "no" near this second question.

## It certainly wasn't me. It

It certainly wasn't me. It looks to me like Rufus' hand and in the same pen as the rest of the problem.

Caroline Series

## I suspect that "flow

I suspect that "flow conjugacy" here is intended to mean "flow equivalence" since otherwise it would be the same as topological conjugacy and det(I -At) is an invariant.

For flow equivalence of subshifts of finite type, complete invariants are known. If the shift is irreducible and non-trivial the invariant for flow equivalence is the group BF(A) := Z

^{n}/(I-A)Z^{n }together with the integer det(I-A) (see [1] for the final part of the proof and references). The multiplicity of 1 as an eigenvalue of A is the rank of the free part of the group BF(A). So techically the answer to the first question is yes the multiplicity of 1 is an invariant for irreducible SFTs. The fact that BF(A) is an invariant is proved in [2] based on results of Parry and Sullivan.Complete invariants are also known for non-irreducible SFTs, but the answer is more complicated, see [3] for details and further references.

## References

## The answer to the second

The answer to the second question is indeed no in the case that A has 1 as an eigenvalue. For example, compare $\begin{pmatrix} 2&1\\2&3 \end{pmatrix}$ and $\begin{pmatrix} 2&1\\1&2 \end{pmatrix}$.

For the first question, note: the multiplicity of 1 as an eigenvalue of A means the dimension of the eigenspace, which can be smaller than the mutliplicity of 1 as a root of the characteristic polynomial of A.

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