By I. G. Macdonald

This new and lots more and plenty elevated variation of a well-received publication continues to be the one textual content to be had with regards to symmetric features and corridor polynomials. There are new sections in virtually each bankruptcy, and plenty of new examples were incorporated all through.

Read or Download Symmetric Functions and Hall Polynomials PDF

Similar combinatorics books

Proofs from THE BOOK

This revised and enlarged 5th version beneficial properties 4 new chapters, which comprise hugely unique and pleasant proofs for classics corresponding to the spectral theorem from linear algebra, a few newer jewels just like the non-existence of the Borromean jewelry and different surprises. From the Reviews". .. inside of PFTB (Proofs from The publication) is certainly a glimpse of mathematical heaven, the place smart insights and lovely rules mix in brilliant and wonderful methods.

Combinatorial Algebraic Geometry: Levico Terme, Italy 2013, Editors: Sandra Di Rocco, Bernd Sturmfels

Combinatorics and Algebraic Geometry have loved a fruitful interaction because the 19th century. Classical interactions comprise invariant conception, theta features and enumerative geometry. the purpose of this quantity is to introduce fresh advancements in combinatorial algebraic geometry and to technique algebraic geometry with a view in the direction of functions, equivalent to tensor calculus and algebraic records.

Finite Geometry and Combinatorial Applications

The projective and polar geometries that come up from a vector house over a finite box are quite invaluable within the development of combinatorial items, resembling latin squares, designs, codes and graphs. This publication offers an creation to those geometries and their many purposes to different components of combinatorics.

Extra resources for Symmetric Functions and Hall Polynomials

Example text

Then for each integer r > 0 we have n-1 Mn+r= r- ( -1)"S(rIp)(X1.... ,Xn)M"-p-1 P-0 (If Mntr=EapMn-p-1, we have these equations for a0, . )) 16. Let A, µ be partitions of length n, and let PP(A, FA,) = det(PA;+µj+2n-i-j)lti,jtn' 50 SYMMETRIC FUNCTIONS I with the understanding that po = n. Then in A we have sAS, = P (A, µ) /P (0, 0). ) (Observe that 17. (a) Let p be an integer > 2 and let (o = e2"`1P. , wp-1) = ± 1 if A 0 (§1, Example 8(f)) and is zero otherwise. , WP-') = vP(A), where op(A) is the sign e(w) of the unique permutation w E SP such that A + SP w8P(mod.

X,,, it is therefore divisible by AS(x I a) in R[x1, ... , x,, ]. As in the text, we may assume that a1 > a2 > ... e. that a = A + 8 where A is a partition of length < n.

X. 6) For any a = (a1, ... , an) E Nn, let Aa = (x70, Ha = (ha;-n+j) (n x n matrices). Then Aa = Ha M. Proof Let n-1 e(k)tr = Fj (1 + xit). E(k)(t) _ r-O i,'k Then H(t)E(k)( -t) = (1-Xkt)-1, 42 SYMMETRIC FUNCTIONS I By picking out the coefficient of t" on either side, we obtain n F, ha;-n+j . ( -1) n-j (k) _ a, j-1 en-j -xk and hence HaM=A,,. 6): we obtain as = det(Aa) = det(H,,)det(M) for any a e Nn, and in particular det M = as, since det(H8) = 1. 8) W(SA) = SA. for all partitions A. t t It should be remarked that if R, R' are raising operators, RR'hA = hRR'A is not necessarily equal to R(R'hA).

Download PDF sample

Rated 4.06 of 5 – based on 41 votes