LECTURE 7: BRUHAT ORDERCARLOS BARRERA-RODRIGUEZMOHAMED OMARIn this lecture, as before, we let (W, S) be a Coxeter system and T be the set ofall conjugates:T = {wsw−1|s ∈ S, w ∈ W }.1. Bruhat OrderDefinition 1.1. Let u, v be two elements in W(1) ut−→ w means ut = w for t ∈ T and `(u) < `(w)(2) u → w means ut−→ w for some t ∈ T(3) u ≤ w means that there exist ui∈ W such thatu = u0→ u1→ u2→ · · · → uk= w.The Bruhat graph is a directed graph whose nodes are elements of W and whoseedges are given by condition (2) of the previous definition. Also Bruhat order isthe partial order given by condition (3).Remark 1.2. From the definition it is clear that(1) u < v implies that `(u) < `(w)(2) u < ut if and only if `(u) < `(ut), with u ∈ W and t ∈ T .(3) e ≤ w ∀w ∈ W .The last remark is true because if w = s1s2· · · skis a reduced expression for w,then we have the sequence of stepse → s1→ s1s2→ · · · → s1s2· · · sk= wRemark 1.3. The condition w = ut in definition 1.1 can be replaced by w = t0ufor t0∈ T and t0= utu−1∈ T .[Bruhat order in Sn] Consider the system (Sn, S) whereS = {s1, . . . , sn−1}and si= (i, i + 1). We observe thatxsix−1= (x(i), x(i + 1)),for x ∈ Sn, si∈ S. Thus the reflection set is the set of all transpositionsT = {(a, b)|1 ≤ a < b ≤ n}.Since reflections in Snare transpositions (a, b) and the length of a permutationequals its inversion number, x(a,b)−−−→ y means that one moves from the permutationDate: January 21st, 2009.1Figure 1. Bruhat Order for S3Figure 2. Bruhat Graph for S3x = [x1, . . . , xa, . . . , xb, . . . , xn] to the permutation y = [x1, . . . , xb, . . . , xa, . . . , xn]obtained by switching xaand xb, where a < b and xa< xb. For example21543(2,5)−−−→ 23541.From this construction we naturally get the Bruhat graph for Sn, a directed graphwith edges between x and y if x < y. We also have a Hasse diagram for the Bruhatorder on Snobtained from the Bruhat graph by relaxing directedness and keepingedges corresponding to covering relations. Figure 1 shows the Bruhat order for S3and Figure 2 shows the Bruhat graph for S3.Lemma 1.4. Let x, y ∈ Sn. Then, x is covered by y in the Bruhat order if andonly if y = x(a, b) for some a and b, a < b, such that x(a) < x(b) and there doesnot exist c, a < c < b, such that x(a) < x(c) < x(b).Proof. (⇐) If y = x(a, b) with the stated conditions, then inv(y) = inv(x) + 1, thuswe have a Bruhat covering relation.2(⇒) Conversely, suppose that y = x(a, b), a < b, and inv(y) > inv(x). Thenx(a) < x(b). If x(a) < x(c) < x(b) for some a, b and c, a < c < b, thenx < x(a, c) < y, so x < y is not a Bruhat covering. Recall that Bruhat order is a partial order; two elements not necessarily com-parable. This leads to the natural question: what are the necessary and sufficientconditions for determining when two elements in Snare comparable in Bruhat or-der ? For example, how can we determine if x = 368475912 and y = 694287531 arecomparable in Bruhat order ?Let x ∈ Sn, and consider the collection of points in the square [n] × [n] given by(i, x(i)). We define the number of dots north-west of (i, j) in the diagram as follows(1.1) x[i, j] = |{a ∈ [i] : x(a) ≥ j}|For example, in the following diagram for x = 31524 ∈ S5we observe that thenumber of points north-west of the point (4, 2) is 4 , which corresponds to thenumber of points in the shaded area, thus x[4, 2] = 3.Remark 1.5. Note that for any x ∈ Snx[n, i] = n + 1 − i and x[i, 1] = i.The following lemma is essentially to determining conditions for Bruhat compa-rability.Lemma 1.6. x[i, j] − x[k, j] − x[i, l] + x[k, l] = |{a ∈ [k + 1, i] : j ≤ x(a) < l}|∀1 ≤ k ≤ i ≤ n and ∀1 ≤ j ≤ l ≤ n.The proof is evident by considering the diagram below.3We can now state our theorem characterizing Bruhat comparability.Theorem 1.7. Let x, y ∈ Sn, then the following are equivalent:(i) x ≤ y(ii) x[i, j] ≤ y[i, j] ∀i, j ∈ [n]As an application, let us determine if x = 368475912 and y = 694287531are comparable in Bruhat order. Considering the corresponding overlapped di-agrams we observe (see the figure below) that x[1, 6] = 0 < 1 = y[1, 6], andx[4, 3] = 4 > 3 = y[4, 3]. Consequently, x and y are not comparable in Bruhatorder.Proof. (i) ⇒ (ii). Suppose that x ≤ y, and without loss of generality assume x → y .Then there exist a and b, 1 < a < b < n, such that y = x(a, b) and x(a) < x(b). By4definition (1.1) this implies thaty[i, j] =x[i, j] + 1 , if a ≤ i < b, x(a) < j ≤ x(b)x[i, j] , otherwiseso (ii) follows.(i) ⇐ (ii). see Bj¨orner and Brenti.