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Let $\underline{w} = [s_1, s_2, \dots ,s_n]$ be a reduced expression in a Coxeter group $W$. Given $x$ in $W$ one can consider the set $\Pi(\underline{w},x)$ consisting of all subexpressions of $\underline{w}$ with product $x$. (A subexpression of $\underline{w}$ is a sequence $[t_1, t_2, \dots, t_n]$ such that $t_i$ is either the identity or $s_i$ for all $i$. Given a subexpression $[t_1, \dots, t_n]$ its product is (obviously) $t_1 \dots t_n$.

There are lots ways of thinking about subepxpressions. Probably one of the nicest is as a path in the Coxeter complex, where at time $= i$ one either chooses to either stay put or to cross a wall coloured by $s_i$. In this interpretation the product is the end-point of the path.

My question is the following:

How should one go about calculating the set $\Pi(\underline{w},x)$?

Of course there is the obvious answer: enumerate all $2^n$-subsequences and calculate all products (or perhaps take some obvious subset given, for example by length restrictions). However, this obviously gets difficult quickly.

I am wondering if anybody has studied this problem from a computational point of view. In small examples it seems like there are a lot of tricks that give an answer much faster than the brute-force method above.

(My motivation is that I have recently been writing software to do calculations in the Hecke algebra, where this problem is the bottleneck. Often I have a long reduced expression (length 40 or so) and think that one should be able to calculate $\Pi(\underline{w},x)$ for elements which are not too much smaller than w (say length 30). However either it's hard or I'm missing something!)

Myself and my computer would like to thank you in advance for any labour saving tips!

share|cite|improve this question might be helpful. – David Speyer Oct 6 '11 at 11:57
Moreso, might be; our paper David references is about subwords whose Demazure product is $x$, instead of the actual product, as in the question. – Allen Knutson Oct 6 '11 at 13:52
Maybe use dynamical programming? Figure out values of subsequences recursively on length building a table. Then compute longer subsequences by taking products of two shorter subsequences by breaking it roughly in half. This way you need only have computed subsequences of length 4 to compute your set for a word of length 8. I am not sure of the complexity of this. – Benjamin Steinberg Oct 6 '11 at 14:42
@Ben: this is a nice idea. It seems believable that this would be faster. I will add another comment if I get this working. – Geordie Williamson Oct 10 '11 at 8:49

I have found a much more efficient way of solving this problem on computer. Having asked the question I guess I should provide a brief account. However I feel like the algorithm is technical and not very enlightening, and so I'll be brief. Please leave comments if you would like more detail.

First we consider an important special case: Is there an efficient algorithm to find all subexpressions of a fixed (reduced) expression $\underline{w} = [s_1, \dots, s_m]$ whose product is the identity? In the language of the question, how does one calculate $\Pi(\underline{w}, id)$?

Of course, there is one canonical such subexpression: $(id,id,…,id)$. Moreover, given any subexpression for the identity we can apply "cancellation moves" to get to this expression.

By a "cancellation move" I mean the following: write $\pi[i,j]$ for the product $t_it_{i+1}…t_j$. Suppose that

$s_i \pi[i+1,j-1] = \pi[i+1,j-1] s_j$

then a "cancellation move" is one of the following:

$\pi = [\dots ,s_i, \dots, id, \dots] \mapsto \pi ' = [\dots, id,…,s_j, \dots]$

$\pi = [\dots ,s_i, \dots, s_j, \dots ]\mapsto \pi' = [\dots ,id, \dots, id, \dots]$

where $\pi$ remains unchanged except at the $i^{th}$ and $j^{th}$ place. Obviously, applying a cancellation move to a subexpression does not change the product. Also, cancellation moves either increase the number of $id$'s or increase the number of $id$'s to the left. It follows that by repeatedly applying cancellation moves to any subexpression for the identity we end up at the canonical subexpression.

For example, consider the Coxeter group with simple reflections $s$, $t$ and $u$ and only braid relation $(st)^3 = id$ (so that $su$ and $tu$ have infinite order). Let $\underline{w} = [s,t,s,u,t,s,t]$. Then the following is a sequence of cancellation moves:

$[s,t,s,id,t,s,t] \mapsto [id,t,s,id,id,s,t] \mapsto [id,t,id,id,id,id,t] \mapsto [id, id, id, id, id, id, id]$.

Now one can reverse this, and define "reverse cancellation moves". Applying all possible reverse cancellation moves to the canonical subexpression yields all subexpressions for the identity. (This is easily programmed.)

We now turn to the general case. The above "cancellation moves" make sense for any subexpression and one can show that again there is a canonical subexpression. (It is characterised by the fact that the subexpression is reduced, and the $t_i \ne id$ occur as far to the right as possible.) Again, once one has this canonical subexpression then it is a simple matter to reconstruct all subexpressions by reverse cancellation moves.

Of course this begs the question: can one find this canonical subexpression efficiently? Yes. Firstly, consider $R(x)$ the right descent set of $x$, and choose $s \in R(x)$ such that s occurs as far as possible to the right in $\underline{w}$. Now, delete everything to the right and including the right-most occurrence of $s$ and repeat with $x$ replaced by $xs$. (This also gives a reasonably efficient algorithm to decide whether $x \le y$ in the Bruhat order.)

(One can ask what this canonical expression "means". Here is one possible explanation: because Schubert varieties are normal the fibre of the Bott-Samelson resolution over point corresponding to $x \le w$ is connected and hence its homology is one-dimensional in degree zero. Now subexpressions also index BB-cells, and hence a basis for the cohomology of the fibre. This "canonical subexpression" corresponds to a generator of the one-dimensional degree zero part.)

For example, take $W = S_7$ (with simple reflections = simple transpositions) and take the following reduced expression for the longest element:

$w_0 = 121321432154321654321$

then there are 6408 subexpressions with product the identity. On my laptop the above algorithm takes 3.33 seconds to find them. A brute force attack takes 93 seconds.

Here is another example which I actually cared about in my calculations. Take the reduced expression

$\underline{w} = 13572613574352461357$

in $S_8$. There are 80 subexpressions with product the longest element in the standard parabolic subgroup generated by all simple transpositions except 4. The above algorithm takes 0.03 seconds to find all of them, whereas the naive approach takes roughly 15 seconds. (So here the above algorithm is 500 times as fast.)

I imagine that the differences become (even) more pronounced with longer words.

share|cite|improve this answer
(Thanks to Jean Michel for pointing out that $W$ should be of type $A_6$ (rather than $A_5$) in the first example above.) – Geordie Williamson Oct 13 '11 at 18:00

To replace my somewhat fuzzy comment, maybe I can formulate a skeptical semi-answer. At any rate your question probably doesn't have a clearcut answer unless you impose strong enough restrictive conditions.

First, I'm assuming you want to work with an infinite Coxeter group. The finite ones are more often studied concretely as finite (real) reflection groups. For them one has the advantage of being able to specify upper bounds on most computational problems; even then the problems may get arbitrarily difficult to manage computationally as the rank grows.

In an infinite group of rank $r$, probably the best organized situation of the type you outline involves a fixed choice of generators and resulting Coxeter element $x=s_1 \cdots s_r$. By a theorem of David Speyer, each power $w=x^t$ has an obvious reduced expression of length $n=tr$. By increasing $t$, you quickly get complicated sets $\Pi$ of the type you want to compute, and even for relatively small values of $r$ the choice of methods doesn't look promising to me.

I have too little computational experience to offer practical advice, but it's worth looking at what John Stembridge and others have been able to compute for general Coxeter groups.

share|cite|improve this answer
Actually, I already don't know how to do this efficiently for finite Coxeter groups, or even the symmetric group. However, when one does examples in the symmetric group is seems convincing that there should be some easier way to do it. – Geordie Williamson Oct 10 '11 at 8:47

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