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. 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.

Is my impression stated above correct, and if so how can it best be made rigorous without case-by-case arguments?

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.)

  • 2
    $\begingroup$ In $A_3$ one can write $w_0$ as $321232$. $\endgroup$ Commented Mar 21, 2011 at 0:00

3 Answers 3


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.

  • $\begingroup$ This looks optimal for the purpose. At first I was just trying to understand the argument in the preprint more straightforwardly, but I oversimplified the role of type A as an exception (which in turn forces some kind of case distinction along the way). $\endgroup$ Commented Mar 21, 2011 at 13:55

For all $S_n$, there are exactly two two double cosets for the action of $S_{n-1}$ (the one that fixes $n$ and the one that doesn't). Obviously $s_n$ and $w_0$ are in the same coset.

You can check that this doesn't work for any reflection in type A other than the ones at the end of the diagram (since cosets count the number of elements switched past the point where you take the reflection out).

In general, the double coset of $s_n$ is the shortest non-identity one and $w_0$ the longest. The coincide if and only if there are two double cosets. I think maybe this also happens with one of the D_n inclusions (even into odd, I think).

  • $\begingroup$ What do you man exactly by "cosets count the number of elements switched past the point where you take the reflection out"? $\endgroup$ Commented Mar 21, 2011 at 3:24
  • $\begingroup$ Let $G = S_n$ act on $[1,n]$. We're considering the parabolic subgroup $H = S_k \times S_{n-k}$, wlog $k \leq \frac{n}{2}$, where the first summand acts on $[1,k]$ and the second on $[k+1,n]$ say. Now given any element $g \in G$, I claim that the $(H,H)$-double coset which $g$ lies in is completely determined by how many numbers in $[1,k]$ it sends to the $[k+1,n]$ range. ... $\endgroup$
    – ndkrempel
    Commented Mar 21, 2011 at 14:54
  • $\begingroup$ Certainly, this number is invariant within each double coset, and each number from $0$ to $k$ occurs (this is the bit that requires $k \leq n-k$). In the other direction, we can pick $e, (1,k+1), (1,k+1)(2,k+2), \ldots, (1,k+1)(2,k+2)(3,k+3)\ldots(k,2k)$ as our $k+1$ representatives, since given any $g \in G$ we can first permute $[1,k]$ such that the numbers sent to the $[k+1,n]$ range occur first, then we can permute the $[k+1,n]$ range so that the numbers coming from $[1,k]$ occur first and in the same order, then we can apply one of our representatives above to get an element lying in $H$. $\endgroup$
    – ndkrempel
    Commented Mar 21, 2011 at 14:59
  • $\begingroup$ Right, what ndkrempel said. $\endgroup$
    – Ben Webster
    Commented Mar 21, 2011 at 16:44

I can give a general solution under the additional assumption that the long word acts by -1 on the root system.

Let $\alpha_i$ denote the simple roots and $\beta_i$ be a vector orthogonal to all $\alpha_j$ with j different from i. Let (,) be a W-invariant inner product. The point: For any vector v and any w in the maximal proper parabolic corresponding to i, we have $(v,\beta_i)=(wv,\beta_i)$.

A simple Lemma: If $(\gamma,\beta_i)=(\alpha_i,\beta_i)$, $(s_i\gamma,\beta_i)=-(\alpha_i,\beta_i)$ and $||\gamma||=||\alpha_i||$, then $\gamma=\alpha_i$.

Since rank at least two, there exists a root $\alpha'_i$ with $(\alpha_i,\beta_i)=(\alpha'_i,\beta_i)$. and $||\alpha'_i||=||\alpha_i||$

Now if $w_0=w_1s_iw_2$ with $w_1,w_2$ in the max. parabolic, lets act on $\alpha_i$ and $\alpha_i$. Since $w_1,w_2$ can't change value of inner product with $\beta_i$, and applying $w_0$ must multiply this product by -1, by our lemma we must have $w_2\alpha_i=w_2\alpha'_i=\alpha_i$, a contracition.


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