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Suppose we have two matrices $A$ and $B$. Is there 'simple' condition that implies $A$ and $B$ satisfy a braid relation of some length? i.e.

$ABABA\ldots = BABAB\ldots$

where both sides are of equal length.

  • By 'simple' I was thinking in terms of the eigen-structure of $A$ and $B$. I would like to be able to automate the checking of this condition in e.g. Maple.
  • I'm particularly interested in matrices in $SU(n,1)$, but results for other matrix groups are welcome.
  • A negative condition would also be useful, i.e. if $A$ and $B$ have some property then they do not satisfy a braid relation of any length.
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  • $\begingroup$ Does the braid relation necessarily involve an even number of terms (so that $\{A,B\} \in \mbox{ braidPairs } \wedge s\in \Bbb R \implies \{sA,B\} \in \mbox{ braidPairs }$)? $\endgroup$ – Mark Fischler Jun 27 '19 at 16:05
  • $\begingroup$ The braid relations can be any length. $\endgroup$ – mawir Jun 28 '19 at 8:29
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First, if the length is odd, say, $F:=ABABA=BABAB$ then $ABABABABAB=F^2=BABABABABA$. So if the braid relation is satisfied for some odd length then it is satisfied for some even length. Therefore we can solve for even lengths only.

Now let's say there exists a natural number $m$ such that $C:=(AB)^m=(BA)^m$. We have that if $C:=ABABAB=BABABA$ then $$AABABAB=ABABABA$$ $$AABABAB=BABABAA$$ $$AC=CA$$

Similarly, $$BABABAB=BBABABA$$ $$ABABABB=BBABABA$$ $$CB=BC$$

By this, if $C$ has $n$ distinct eigenvalues then $AB=BA$. Check if $BA=AB$. This is a sufficient condition. It is also necessary if $C$ has $n$ distinct eigenvalues. How can we check that? Look at the eigenvalues of $AB$ (which are the same as the eigenvalues of $BA$). Since $A$ and $B$ are unitary, $AB$ is unitary. So its eigenvalues are on the unit circle. They are equivalent if there exists a natural number $m$ such that $\mathrm{e}^{im\theta}=\mathrm{e}^{im\phi}$. This happens iff there exists an integer $k$ such that $m(\theta-\phi)=2\pi k \iff \theta-\phi=q\pi$ for some rational number $q$. If $AB$ has no equivalent eigenvalues then $(AB)^m$ has distinct eigenvalues (no matter what $m$). In this case, a necessary (and sufficient) condition would be that $AB=BA$ which is easy to check.

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