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Let $A=\Delta^m A_1 A_2...A_k$ be the Garside's normal form of a braid. Then its canonical length is $k$. I need to prove that for any $A,B \in B_n$ we have ${\rm len}(AB) \leq{\rm len}(A)+{\rm len}(B)$ where ${\rm len}$ is its canonical length.

If $A=\Delta^m A_1 A_2...A_k$ and $B=\Delta^s B_1 B_2...B_r$, I can see how multiplying them may reduce the canonical factors but I am not sure how to go about it. Any hints or a nudge would be highly appreciated.

For definitions and concepts, one may check https://eprint.iacr.org/2018/1142.pdf

Shaun
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    $AB = \Delta^{m+s}A_1'A_2'\cdots A_k'B_1B_2 \cdots B_r$, where $A_i'$ is the conjugate of $A_i$ under $\Delta$ and is also a divisor of $\Delta$. The process of putting the positive braid $A_1'A_2'\cdots A_k'B_1B_2 \cdots B_r$ into right greedy normal form does not increase its canonical length, although it could decrease it. – Derek Holt Nov 18 '19 at 10:36
  • Hi! That's exactly what I want to prove. I know that $\Delta^m A=A \Delta^m$ if m is even, and $A \Delta^m = \Delta^m R(A)$ if m is odd, where R is the reflection automorphism on $B_n$. But, when we multiply,I cannot see how the process of putting AB into left weighted canonical form (or garside's normal form) does not increase its canonical length. Is there a text I could refer to, which gives this proof,if it's too technical to elaborate here? –  Nov 20 '19 at 18:22
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    The critical point is that, for any positive word $w$ in the braid generators, there is a unique longest divisor $u$ of $\Delta$, such that $w$ is equal in the braid group to a positive word $uv$ for some $v$. So if $w = A_1A_2 \cdots A_k$ is a product of divisors of $\Delta$, then $A_1$ must be a left divisor of the unique longest such left divisor $u$. Then an easy inductive argument shows that the left greedy normal form of $w$ has at most $k$ factors. – Derek Holt Nov 22 '19 at 14:59
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    I learnt this theory from Chapter 9 of the book Word Processing in Groups by Epstein et al (which is perhaps not surprising because I am one of the al), but there are probably other standard references. – Derek Holt Nov 22 '19 at 14:59
  • Thanks! I shall look into it. –  Nov 24 '19 at 06:14

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