6
$\begingroup$

Let $a$ and $b$ be two generators in a Coxeter group which do not commute. Is it possible for $ab$ to be equal to a product of generators where all instances of $b$ come before all instances of $a$?

I've tried coming up with invariants that are preserved after applying the conditions of a Coxeter group to a string of generators, but the fact that one can insert either $aa$ or $bb$ at any point in the string has made this complicated. Particularly, the more general claim that a string with all $a$s before $b$s can't be turned into one with all $b$s before $a$s turns out to be false, since for instance $aab=baa$.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
12
$\begingroup$

The answer is no. The Deletion Condition says that any expression in the generators of a Coxeter group contains a reduced expression for the same element as a subexpression. Since $a$ and $b$ don't commute, the unique reduced expression for $ab$ is $ab$, which must therefore occur as a subexpression of any longer expression that evaluates to $ab$, i.e. any such expression must contain an $a$ before a $b$.

Improve this answer
$\endgroup$
4
  • 3
    $\begingroup$ Isn't the "deletion condition" a theorem rather than a "condition"? $\endgroup$
    – YCor
    Commented May 13, 2021 at 7:27
  • 2
    $\begingroup$ @YCor: I agree that the terminology is perhaps strange, but that's how it is in the book by Humphreys. $\endgroup$
    – Matt Fayers
    Commented May 13, 2021 at 9:03
  • 1
    $\begingroup$ Is there any standard reading for this topic? $\endgroup$
    – ViHdzP
    Commented May 13, 2021 at 15:19
  • 1
    $\begingroup$ @URL: Björner and Brenti's "Combinatorics of Coxeter Groups" talks about the Deletion Property. Look at Section 1.4. (By the way, Björner and Brenti call it a "property", not a "condition".) $\endgroup$
    – Nathan Reading
    Commented May 14, 2021 at 0:58

Not the answer you're looking for? Browse other questions tagged or ask your own question.