# Problem 106

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Is the multiplicity of $1$ as an eigenvalue of $A$ a flow conjugacy invariant of $\Sigma _A$? How about $\Pi_{\lambda_i \not = 1} (1- \lambda_i)$?

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### 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) := Zn/(I-A)Ztogether with the integer det(I-A)  (see  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  based on results of Parry and Sullivan.

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

### References

1. [MR758893] Franks J.  1984.  Ergodic Theory Dynam. Systems. 4:53–66.
2. [bowen1977homology] Bowen R, Franks J.  1977.  Annals of Mathematics. 106:73–92.
3. [MR1907894] Boyle M.  2002.  Pacific J. Math.. 204:273–317.

### 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.