Диссертация (1136188), страница 49
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. . , uk−1 ) by the inequalities1 u 1 2 u 2 · · · k − 1 u k −1 k .(C)Namely, we set u 1 = u 1,i 1 , u 2 = u 1,i 1 +i 2 , . . . , uk−1 = u 1,i 1 +···+ik−1 . Observe that all vertices of GZ ( p )project to vertices of the cube C . Thus it suffices to describe the fibers of the projection π over thevertices of the cube C .It will be convenient to label the vertices of the cube C by the monomials in the expansion ofthe polynomial A (x1 · · · xk ). Namely, to fix a vertex of C , one needs to specify, for every j between 1and k − 1, which of the two inequalities j u j or u j j + 1 turns to an equality.
Similarly, to fix amonomial in the polynomial A (x1 · · · xk ), one needs to specify, for every j between 1 and k − 1, whichterm is taken from the factor (x j + x j +1 ), the term x j or the term x j +1 . This description makes thecorrespondence clear.ααLet v be the vertex of the cube C corresponding to a monomial x1 1 · · · xk k . It is not hard to seeπ −1 ( v ) is combinatorially equivalent toGZ 1i 1 −1+α1 · · · kik −1+αk .that the polytopeDefine the coefficients c α1 ···αk so thatA (x1 · · · xk ) =ααc α1 ···αk x1 1 · · · xk k .α1 ,...,αkThen we haveV 1i 1 · · · k i k =c α1 ···αk V 1i 1 −1+α1 · · · kik −1+αk .α1 ,...,αkSince for any k-tuple of indices α1 , . .
. , αk , for which the corresponding coefficient c α1 ,...,αk is nonzero,the Gelfand–Zetlin polytope GZ (1i 1 −1+α1 · · · kik −1+αk ) has smaller dimension than GZ (1i 1 · · · kik ), wecan assume by induction that i −1 +α 1i −1+αk .V 1i 1 −1+α1 · · · kik −1+αk = A ∞ x11· · · xkkHence we haveV 1i 1 · · · k i k =i −1 +α 1c α1 ···αk A ∞ x11α1 ,...,αkThe desired statement follows.2i −1+αk · · · xkk i = A ∞ A x1i 1 · · · xkk .964P. Gusev et al. / Journal of Combinatorial Theory, Series A 120 (2013) 960–9693. Equations on generating functions E k and G kIn this section, we deduce equations on the generating functions E k and G k . In particular, we proveTheorems 1.1 and 1.2.ααFor a multi-index α = (α1 , .
. . , αk ), we let zα denote the monomial z1 1 · · · zk k , and α ! denote theproduct α1 ! · · · αk !. The partial derivation with respect to z will be written as ∂ . The power ∂ α willααmean ∂1 1 · · · ∂k k . We will write I forthe operator of integration with respect to the variable z . This∞noperator acts on the power seriesn=0 an z , where an are power series in the other variables, asfollows:I∞n=an zn =0∞n =0zn+1an .n+1We will use the expansion(x1 + x2 ) · · · (xk−1 + xk ) =c α xα ,αin which the coefficients c α can be computed explicitly.
Let E k∗ be the sum of all terms in E k divisibleby z1 · · · zk . Then we have (i, j, α being multi-indices of dimension k)E k∗ =i >0= zi zic α A ∞ x i −1 +α=i!i!αA ∞ xii >0cα ∂ α I 1 · · · I kαA ∞ x i −1 +αz i −1 +αi >0(i − 1 + α )!=αcα ∂ α I 1 · · · I kj α zj.j!A∞ x jApply the differential operator ∂1 · · · ∂k to both sides of this equation. Note that ∂1 · · · ∂k ( E k∗ ) =∂1 · · · ∂k ( E k ). Thus we have∂1 · · · ∂k ( E k ) =∂αj ααObserve also that, sincecα ∂ α zjA∞ x jj!.α 0 whenever cα = 0, we have zj= ∂ α Ek .j!A∞ x jj αThis implies Theorem 1.1.Example (k = 1 and k = 2).
In the case k = 1, we have E 1 = e z1 . Consider now the case k = 2. SetE = E 2 , x = z1 and y = z2 . By Theorem 1.1, the function E satisfies the following partial differentialequation:E xy = E x + E y .This equation can be simplified by setting E = e x+ y u. Then the function u satisfies the equationu xy = uand the boundary value conditions u (x, 0) = u (0, y ) = 1. We can now look for solutions u that havethe form v (xy ), where v is some smooth function. This function must satisfy the initial conditionv (0) = 1 and the ordinary differential equationt v (t ) + v (t ) − v (t ) = 0.√It is known that the only analytic solution of this initial value problem is I 0 (2 t ), where I 0 is the√modified Bessel function of the first kind.
Thus I 0 (2 xy ) is a partial solution of the boundary valueP. Gusev et al. / Journal of Combinatorial Theory, Series A 120 (2013) 960–969965problem u xy = u, u (x, 0) = u (0, y ) = 1. The solution of this boundary value problem is unique (notethat the boundary values are defined on characteristic curves!).
Therefore, we must conclude that√E (x, y ) = e x+ y I 0 (2 xy ).The proof of Theorem 1.2 is very similar to the proof of Theorem 1.1. Let G k∗ be the sum of allterms in G k that are divisible by y 1 · · · yk , i.e.G k∗ = V 1i 1 · · · k i k y i .i >0Then, similarly to a formula obtained for E k∗ , we haveG k∗ =1 −α 1cα y11−αk· · · yk A∞ x j y j .j ααApplying the operator 1 · · · k to both sides of this equation, we obtain Theorem 1.2. Similarly tothe proof of Theorem 1.1, we need to use that 1 · · · k G k∗ = 1 · · · k (G k )and thatα1−αkα11 · · · k k (G k ) = y −· · · yk1α A∞ x j y j .j αWe will now discuss several examples.Example (k = 1 and k = 2).
For k = 1, we have the following equation: 1 G 1 = G 1 , i.e. G 1 ( y 1 ) −G 1 (0) = y 1 G 1 ( y 1 ). Knowing that G 1 (0) = 1, this givesG 1 ( y 1 ) = 1 + y 1 + y 21 + · · · =11 − y1.Suppose that k = 2. Set G = G 2 , x = y 1 , y = y 2 . The function G satisfies the following equationx y G = x G + y G .Note that G (x, 0) = G 1 (x) and G (0, y ) = G 1 ( y ).
Therefore, the right-hand side can be rewritten as11G − 1−G − 1−yx+.xyThe left-hand side isx1G − 1−xy=1x11−1G − 1−1− yx.−yySolving the linear equation on G thus obtained, we conclude thatG=11−x− y.Example (k = 3). We set G = G 3 , x = y 1 , y = y 2 and z = y 3 . The function G satisfies the followingequation: x y z G = (x + y )( y + z )G. This equation can be rewritten as follows:2y G =G (1 − x − y − z) − 1xyz.Suppose that G = G (x, 0, z) + A (x, z) y + O ( y 2 ). Then we havey 2 2y G = G − G (x, 0, z) − A (x, z) y = G −11−x−z− A (x, z) y .966P. Gusev et al. / Journal of Combinatorial Theory, Series A 120 (2013) 960–969Substituting this into the equation, we can solve the equation for G in terms of A:G=−xz + y (1 − x − z)(1 − A (x, z)xz).(1 − x − z)( y − (x + y )( y + z))Since the power series 1 − x − z is invertible, it follows that G has the forma + byy − (x + y )( y + z),μ be the two solutions of the equationwhere a and b are some power series in x and z.
Let λ andy = (x + y )( y + z), namely,λ, μ =1−x−z±1 − 2(x + z) + (x − z)22.The signs are chosen so that, at the point x = z = 0, we have λ = 1 and1y − (x + y )( y + z)=cy−λ+dy−μμ = 0. Then,where c and d are some power series in x and z. Note that, since ( y − λ)−1 makes sense as a powerseries, c (a + by )/( y − λ) can be represented as a power series in x, y and z. Thus the function d(a +by )/( y − μ) must also be representable as a power series in x, y and z. However, this is only possibleif the numerator is a multiple of the denominator, i.e.
(a + by ) = e ( y − μ), where the coefficient eis a power series of x and z. It follows that G is equal to e ( y − λ)−1 . The coefficient e can be foundfrom the condition G (x, 0, z) = 1−1x−z :G==1λ1−x−z λ− y2xz − y (1 − x − z) − y 1 − 2(x + z) + (x − z)22(1 − x − z)((x + y )( y + z) − y ).4. Proof of Theorem 1.4In this section, we will prove Theorem 1.4, which expresses the numbers V k,,m as coefficients ofcertain polynomials. The numbers V k,,m satisfy the following recurrence relation:V k,,m = V k−1,,m + V k,−1,m + V k,,m−1 + V k−1,+1,m−1provided that k, , m > 0, and the following initial conditions:V 0,,m = V ,m ,V k,0,m = V k,m ,V k,,0 = V k, .Set V ks,m = V k,s−k−m,m . Then we can write the following recurrence relations on the numbers V ks,m :−1−1V ks,m = V ks−+ V ks,−m1−1 + V ks−+ V ks,−m11,m1,m−1provided that k 1, m 1, k + m s − 1, and−1V ks,m = V ks−+ V ks,−m1−11,mprovided that k + m = s.For a fixed s, we can arrange the numbers V ks,m into a triangular table T s of size s as shown inFig.
1. Namely, the number V ks,m is placed into the cell, whose southwest (lower left) corner is atposition (k, m). The next table T s+1 can be obtained from the table T s as follows. First, we add toevery element of T s its south, west and southwest neighbors. Next, we add a line of cells, whosepositions (k, m) satisfy the equality k + m = s. In every cell of this line, we put the sum of the southP.
Gusev et al. / Journal of Combinatorial Theory, Series A 120 (2013) 960–969967Fig. 1. Triangular tables T s containing the numbers V ks,m . Southwest corners of these tables are located at (0, 0).Fig. 2. The skew-symmetric tables T̃ s .and west neighbors. Note that, by construction, the boundary of every table T s consists of binomialcoefficients.Consider the generating function G = G 3 for the numbers V k,,m . The splitting of G into homogeneous components can be obtained by expanding the function G (xy , y , zy ) into powers of y.
WesetG (xy , y , zy ) =∞g s (x, z) y s .s =0Then we haveg s (x, z) =s−ks V ks,m xk zm .k =0 m =0Thus the coefficients of the polynomial g s are precisely elements of the table T s . The recurrencerelations on the numbers V ks,m displayed above imply the following property of the generating functions g s :Proposition 4.1. The polynomials g s satisfy the following recurrence relations:g s+1 = (1 + x + z) g s + τs (xzg s ),where the truncation operator τs acts on a polynomial by removing all terms, whose degrees exceed s.Consider the polynomialsh s (x, z) = g s (x, z) − (xz)s g s z−1 , x−1 .Geometrically, these polynomials can be described as follows.
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