# Sufficient condition for two matrices to satisfy braid relation

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.
• Does the braid relation necessarily involve an even number of terms (so that $\{A,B\} \in \mbox{ braidPairs } \wedge s\in \Bbb R \implies \{sA,B\} \in \mbox{ braidPairs }$)? – Mark Fischler Jun 27 '19 at 16:05
• The braid relations can be any length. – mawir Jun 28 '19 at 8:29

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.