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# Class Note for MATH 796 at KU 5

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This 5 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Kansas taught by a professor in Fall. Since its upload, it has received 24 views.

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Date Created: 02/06/15
Friday 425 Symmetric Functions We continue our catalog of important symmetric functions We have already seen 1 the monomial sym metric functions A A A 7m E Iaila quot391af7 a1azClP 2 the elementary symmetric functions 6k E film rm 6A A1quot39 Ai 0lti1lti2ltltic and 3 the complete homogeneous symmetric functions hk E zilziznzik h h1hz 0lti1 i2 miik 4 Power sums These are de ned by PkrfII mk7 PA PA1quot39PAZA For example in degree 2 P2 77127 P11 11I2quot392 m22m11 While p2p11 is a Qvector space basis for AQ it is not a Z module basis for A2 To put this in a more elementary way not every symmetric function with integer coef cients can be expressed as an integer combination of the powersums for example mu p11 7 p22 5 SChur functions The de nition of these power series is very different from the preceding ones and it looks quite weird at rst However the Schur functions turn out to be essential in the study of symmetric functions De nition 1 A columnstrict tableau T of shape A or ACST for short is a labeling of the boxes of a Ferrers diagram with integers not necessarily distinct that is o weakly increasing across every row and o strictly increasing down every column The partition A is called the shape of T and the set of all columnstrict tableaux of shape A is denoted CSTA The content of a CST is the sequence a 011012 where ai is the number of boxes labelled i and the weight of T is the monomial IT I Ilz 2 For example here are two CST s and one tableau that is not an CST of shape A 32 13 111 123 23 48 1 2 2 X1 X2 X3 Xi X4Xs Not a SST De nition 2 The Schur function corresponding to a partition A is 1 2 IT TECSTA It is far from obvious that 81 is symmetric but in fact it is We will prove this shortly Example 1 Suppose that A is the partition with one part so that the corresponding Ferrers diagram has a single row Each multiset of n positive integers with repeats allowed corresponds to exactly one CST in which the numbers occur left to right in increasing order Therefore 1 8n hn Em Ahn At the other extreme suppose that A 11 1 is the partition with n singleton parts so that the corresponding Ferrers diagram has a single column To construct a CST of this shape we need n distinct labels which can be arbitrary Therefore 2 S111 5n m111 Let A 21 We will express 81 as a sum of monomial symmetric functions No tableau in CSTA can have three equal entries so the coef cient of mg is zero For weight zazbzc with a lt b lt 5 there are two possibilities shown below Therefore the coef cient of mul is 1 Finally for every a f b E N there is one tableau of shape A and weight 1211 7 either the one on the left if a lt b or the one on the right if a gt b Therefore 821 2771111 m21 Proposition 1 51 is a symmetric function for all A Proof First observe that the number 3 CAa T e CSTA IT may depends only on the ordered sequence of nonzero exponents in a For instance for any A h 8 there are the same number of ACST s with weights zlzgz z and zlzgz z because there is an obvious bijection between them given by changing all 37s to 77s or vice versa To complete the proofthat 81 is symmetric it suf ces to show that swapping the powers of adjacent variables does not change CAa That will imply that 81 is invariant under every adjacent transposition k k 1 and these transpositions generate the group 600 This is precisely the statement that s is a quasisymmet c function We will prove this by a bijection which is easiest to show by example Let A 9 74 32 We would like to show that there are the same number of A CST s with weights 3233473 3233743 1112131456z7 and 11121314175617i Let T be the following A CST 111235666 2345677 3456 466 57 Observe that the occurrences of 5 and of 6 each form snakes from southwest to northeasti munJab ax kIIJBWNt I lmJBWt I To construct a new tableau in which the numbers of 57s and of 67s are switched we ignore all the columns containing both a 5 and a 6 and then group together all the other strings of 57s and 67s in the same rowi 7 2 3 5 5 6 7 6 Then we swap the numbers of 57s and 67s in each of those contiguous blocksi This construction allows us to swap the exponents on 1k and zk1 for any h concluding the proof D Theorem 2 For each n 2 1 the sets mAlAbn eAlAbn hAlAbn and sAlAbn are all Zhases for A 239e bases for AZ as a free Zmodule and pAlAbn is a Qhasz39s for A 239e a basis for AQW as a vector space Moreover e1e2 and h1h2 generate A as a polynomial algebra over R Sketch of proof It is more or less obvious that the m are a Z basis To show that the Schur functions are a Z basis we show that they can be obtained from m by a unitrz39ahgular change of basis Speci cally we write each Schur function as an integer linear combination of monomial symmetric functions as SA E KAHmH Abn and then show that the matrix XML is triangular with is on the main diagonal therefore it is invertible and its inverse has integer entries Note that by the de nition of Schur functions the coef cient XML is the number of columnstrict tableaux with shape A and content h these are the socalled Kostka numbers Of course to do this we have to specify an ordering on the partitions Rather than the lexicographic total order we have worked with before it turns out to be convenient to work with a partial order as follows De nition 3 Let A A1 A1 and h M1 hm be partitions of n We say that A dominates h written A E h if E S m and A12M17 A1A2 2M1M27 A1mMZM1mMo Proposition 3 KM 1 for all A Moreover KM 0 unless A E h The proof of this fact is a homework problemi As a corollary the matrix of Kostka numbers is unitriangular for any total order such as the lexicographic order which re nes dominance A similar result holds for the elementary symmetric functions If we write EA ZBk mH M then the coef cients BAH have a nice combinatorial interpretation and it turns out that 1 if X BAH M 7 0 1f M EX where X denotes the conjugate or transpose of A The proof that the p form a Q basis is analogous although in this case the change of basis has non17s on the diagonal and so is not invertible over Z but it is invertible over The h s are different They have lots and lots of terms so the coef cients of the transition matrix are all nonzero and we can7t use triangularity to prove that they are a basis However we can do something else cleveri

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