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I'm trying to understand the Bruhat decomposition of $GL_n$ so I've been doing a simple example with $GL_3$.

We have the Weyl group $W$, which is isomorphic to $S_3$, and a simple root system namely the transpositions (i, i+1). We get cells of the form $BwB$ with $w\in W$, where $B$ denotes the subgroup of upper triangular matrices. I have read a bit in 'Reflection groups and coxeter groups' by Humpreys, and he explains a length function on $W$. The length of $w\in W$ is the number $n$ of simple reflections $r_i$ we need to write $w=r_1\dots r_n$, which is equal to the number of inversions of $w$. On Wikipedia it says that the dimension of the cells corresponds to the length of $w$. I don't really understand how that works. If you know the length how do you calculate the dimension? Why does that work?

Now I've tried to see what happens in one of the $BwB$. Let's take $w=(123)$. Then we can write $(123)=(12)(23)$ and not in less transpositions so the length is 2. I'm trying to see what the dimension of $BwB$ is in this case. During the search I realized that I don't really know what kind of dimension we're even looking for. Am I correct in saying that the dimension of $B(1)B=B$ is 6? I thought perhaps that the length of $w$ gives the dimension that you get extra on top of your original $B$. That would mean that the dimension of $B(123)B$ is 8. How can I see in another way what the dimension is?

I took the example of $(123)$ since that was the only one that did not give me any zeroes when I calculated $bwb'$ for random $b,b'\in B$, but which (might) have a dimension less than 9. The others all gave exactly the amount of zeroes that the (supposed) dimension was less than 9.

Edit: For $B(123)B$ I looked at what happened when you take two invertible upper triangular matrices $a, b$ and just calculated $aP_{123}b$ where $P_\sigma$ denotes the permutation matrix of $\sigma$. You get: $$C=\begin{pmatrix} b_{1,1}a_{1,3} & b_{1,2}a_{1,3}+b_{2,2}a_{1,1} & b_{1,3}a_{3,1}+b_{2,3}a_{1,1}+b_{3,3}a_{1,2}\\ b_{1,1}a_{2,3} & b_{1,2}a_{2,3} & b_{1,3}a_{2,3}+b_{3,3}a_{2,2}\\ b_{1,1}a_{3,3} & b_{1,2}a_{3,3} & b_{1,3}a_{3,3} \end{pmatrix}.$$ You know that $b_{1,1}a_{3,3}\neq 0.$ I can see that for instance you can calculate $c_{2,2}$ when you know $c_{2,1}$, $c_{3,1}$ and $c_{3,2}$, so that brings the dimension down. But how can I see this more obviously since this is just a bit of messing around? Also this cell and $B(13)B$ should be disjoint but $B(13)B$ also gives matrices with $c_{3,1}\neq 0$.

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Can you describe the set you found when computing $B(123)B$? –  Mariano Suárez-Alvarez Mar 15 '12 at 23:38
    
I usually prefer to think in terms of maximal flags $(0=V_0<V_1<V_2<V_3=V)$ of subspaces $V_i$ of $V$. $GL_3$ acts transitively on the set $\mathcal{F}$ of maximal flags and $B$ is the stabilizer subgroup of some maximal flag. The double cosets $BwB$ correspond then to orbits on pairs of flags or - if you prefer - to the orbits of $B$ on $\mathcal{F}$. See if you can find some useful meaning of dimension for subsets/subspaces of $\mathcal{F}$. –  j.p. Mar 16 '12 at 9:15
    
Another thing you should understand is the Jordan-Hölder permutation belonging to a pair of maximal flags and - given a maximal flag $f$ and an element $g$ of $GL_3$ - the Jordan-Hölder permutation belonging to the pair $f$, $f^g$. –  j.p. Mar 16 '12 at 9:17
    
@jug I don't really know that much about flags, this is a sort of introduction to those kinds of things, so I'm not sure how it works with the set of maximal flags and such. How does $GL_3$ act on it? –  dropfruitduo Mar 16 '12 at 13:09
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The action of $GL_3$ on $V$ induces an action on all subsets of $V$, which maps subspaces to subspaces, preserves dimensions and inclusion. –  j.p. Mar 16 '12 at 19:35

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