While looking at a preprint I've just bumped into a question about the longest element $w\_0$ of a Weyl group $W$ (say irreducible of a Lie type $A$ - $G$ and of rank $n>1$, to simplify). Suppose this element is written as a not necessarily reduced product of the form $w s\_n w'$, where both $w$ and $w'$ lie in the proper parabolic subgroup of $W$ generated by simple reflections other than $s\_n$. My first impression is that this can occur only for type $A\_n$, where for example when $n=2$ we get $w\_0 = s\_1 s\_2 s\_1 = s\_2 s\_1 s\_2$. There is some relevant discussion of *reduced* expressions for $w\_0$ in an earlier post [here](https://mathoverflow.net/questions/54926/longest-element-of-weyl-groups). But getting to a reduced expression from an arbitrary expression takes some work, as shown in the standard Tits algorithm. So I may be overlooking something. 



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> Is my impression stated above correct, and if so how can it best be made rigorous without case-by-case arguments?
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EDIT: As Ben points out, I need to avoid type A here, so my question is really about the remaining simple types. (I've edited the language above.)



Occurrences of a simple reflection in the longest element of a Weyl group?
First, I'll show that you can assume that $ws\_nw'$ is a reduced product.


To see this, first note that if we multiply $s\_n$ by a reduced expression for $w'$, the result will still be reduced, and this can then be extended to a reduced expression for the longest element by multiplying some reduced expression $s\_{i\_1} \cdots s\_{i\_k}$ on the left, where $k$ is $1 + l(w')$ less than the length of the longest element...although, as of yet, we don't know that none of the $i\_p$ are equal to $n$.


But then $s\_{i\_1} \cdots s\_{i\_k} s\_nw' = ws\_nw'$ so that $s\_{i\_1} \cdots s\_{i\_k} = w$. By the nature of the braid relations, any simple reflection that appears in $s\_{i\_1} \cdots s\_{i\_k}$ must appear in any expression for $w$, so if $s\_n$ appears in $s\_{i\_1} \cdots s\_{i\_k}$, then it must also appear in $w$, which is a contradiction.


So, if we have your expression $ws\_nw'$, we then in fact have a reduced expression for the longest element that has only a single occurrence of $s\_n$.


Now consider any positive root $\alpha$ that has a nonzero coefficient on the simple root $\alpha\_n$ corresponding to $s\_n$ and also satisfies $w\_0 \alpha = -\alpha$.


Claim: $\alpha$ must go negative at the location of $s\_n$ in our reduced expression.


Proof: As you apply the simple reflections in our reduced expression one by one to $\alpha$, the only simple reflection that can affect the coefficient of $\alpha\_n$ is $s\_n$, and so when you get to $s\_n$, the simple reflection $s\_n$ must negate the coefficient of $\alpha\_n$ (for otherwise $w\_0 \alpha$ could not be equal to $-\alpha$). (That is, if $s\_n$ just nullified the coefficient on $\alpha\_n$, because $s\_n$ doesn't occur again, one could not end up at $-\alpha$ after all the simple reflections in the reduced expression for $w\_0$ have been applied.)


In every case except for type A, there are at least two such $\alpha$...for instance, you can take the sum of the simple roots and the highest root. But they cannot both go negative at the occurrence of $s\_n$ in our reduced expression for $w\_0$!


Therefore, no such expression exists except in type A.


And then in type A, there's the expression 1 21 321 4321 ... n (n-1) (n-2) ... 321.


