LECTURE 14: YANG-BAXTER EQUATION AND DOUBLESCHUBERT POLYNOMIALSCARLOS BARRERA-RODRIGUEZ1. The Yang-Baxter Equation (Continued). Double SchubertPolynomials.Last time we talked about the nil-Coxeter algebra, and we saw that the nil-Coxeter relations for u1, u2, . . . un−1are given byuiuj= ujuifor |i − j| > 1uiui+1ui= ui+1uiui+1ui2= 0We showed the following result for hi(x) = 1 + xui(x).Lemma 1.1.hi(x)hi(y) = hi(x + y)hi(x)hj(y) = hj(y)hi(x) |i − j| > 1hi(x)hj(x + y)hi(y) = hj(y)hi(x + y)hj(x) |i − j| = 1The last relation is ca lled the Yang-Baxter equation.We also learned that we can associate to a strand configur ation C a polynomialx1x2xns1s2Figure 1. Strand repre sentationΦ(C) in H[x]. In the above exa mple Φ(C) = hs2(x3− x2)hs1(x3− x1).Next we consider a particular configuration, as shown in the following figure.Date: February 6th, 2009.1yn−1y2y1x1x2xn−1Figure 2. Particular configurationΦ(Csp) =n−2Yd=2−nYi−j=di+j≤nhi+j−1(xi− yj).Note that the order of the factors in the pro duct is important! Deforming Fig. 2we obtain Fig. 3 (using only braid and commutation relations which we showedlast time do not change Φ(Csp)), which simplifies the calculation of Φ.yn−1y2y1x1x2xn−1Figure 3. Simplified particular configuration2By this simple observation it is easy to see that we can rewrite Φ as:Φ(Csp) =n−1Yi=11Yj=n−ihi+j−1(xi− yj)where recall hi(x) = 1 + xui.Theorem 1.2. If one decomposes Φ(Csp) in H[x, y] asΦ(Csp) =Xw∈SnΦw(Csp)wthenΦw(Csp) = σw(x, y).Proof. Let us first look atΦw0(Csp) =Yi+j≤n(xi− yj) = ∆(x, y) = σw0(x, y)Recall ∂iσw= σwsiif ℓ(wsi) = ℓ(w) − 1. Hence it rema ins to show that the samerecursion holds for the coefficient polynomials in Φ(Csp). But we have that∂iΦwsi(Csp) = Φw(Csp) for ℓ(wsi) = ℓ(w) − 1if and only if∂iΦ(Csp) = Φ(Csp)ui.SetHi(x) = hn−1(x) · · · hi+1(x)hi(x),Then no te thatHi(x) = Hi+1(x)hi(x),hi(x)Hj(y) = Hj(y)hi(x) if j > i + 1,hi(x)hi(−x) = 1.Lemma 1.3.(a) Hi(x)Hi(y) = Hi(y)Hi(x)(b) Hi(x)Hi+1(y) − Hi(y)Hi+1(x) = (x − y)Hi(x)Hi+1(y)ui.Proof.(a) This follows be descending induction on i:Hi(x)Hi(y) = Hi+1hi(x)Hi+2(y)hi+1(y)hi(y)= Hi+1(x)Hi+2(y)hi(x)hi+1(y)hi(y − x)hi(x)= Hi+1(x)Hi+2(y)hi+1(y − x)hi(y)hi+1(x)hi(x)(this latter is theY-B eq.)= Hi+1(x)Hi+1(y)hi+1(−x)hi(y)hi+1(x)hi(x)= Hi+1(y)Hi+1(x)hi+1(−x)hi(y)hi+1(x)hi(x)(but Hi+2(x) = Hi+1(x)hi+1(−x))= Hi+1(y)hi(y)Hi+2(x)hi+1(x)hi(x)= Hi(y)Hi(x)3(b)Hi(x)Hi+1(y) − Hi(y)Hi+1(x) = Hi(x)Hi(y)hi(−y) − Hi(y)Hi(x)hi(−x)(and hi(−y) = 1 − yui, hi(−x) = 1 − xui)= Hi(x)Hi(y)(−yui) + Hi(x)Hi(y)xui= (x − y)Hi(x)Hi(y)ui= (x − y)Hi(x)Hi+1(y)(1 + yui)ui(but 1 + yui= 0 )= (x − y)Hi(x)Hi+1(y)ui Lemma 1.4.(a) hi(x − y) = Hi+1−1(x)Hi−1(y)Hi(x)Hi+1(y)(b) hn−1(x−yn−1) · · · hi(x−yi) = Hn−1−1(yn−1) · · · Hi−1(yi)Hi(x)Hi+1(yi) · · · Hn(yn−1)Proof.(a) Observe that the equality is equivalent toHi(y)Hi+1(x)hi(x)hi(−y) = Hi(x)Hi+1(y)but this latter is equivalent to Hi(y)Hi(x) = Hi(x)Hi(y), which corre sponds pre-cisely to part (a) of the previous lemma.(b) This part can be proved by descending induction on i a nd the previous lemma.We leave the details to the reader. We now complete the proof of Theorem 1 .2. Using Lemma 1.4 (b) we find thatΦ(Csp) =n−1Yi=11Yj=n−ihi+j−1(xi− yj)= σ−1(y)σ(x)where σ(x) = H1(x1)H2(x2) · · · Hn−1(xn−1). Hence it remains to show that∂iσ(x) = σ(x)ui.But we can see that∂iσ(x) =H1(x1) · · · Hn−1(xn−1) − H1(x1) · · · Hi(xi+1)Hi+1(xi) · · · Hn−1(xn−1)(xi− xi+1)= H1(x1) · · · Hn−1(xn−1)ui= σ(x)uiby Lemma 1.3 (b).