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# Class Note for MATH 796 at KU 7

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This 3 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 14 views.

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Date Created: 02/06/15

Friday 52 The Frobenius Characteristic Let R be a ring Denote by CZRGn the vector space of RValued class functions on the symmetric group Gni If no R is speci ed7 we assume R C De ne Cos g9 Chen n20 We make CZG into a graded ring as follows For f1 6 CZGn1 and f2 6 CZGM7 we can de ne a function f1 f2 6 Gm X Gm by f1 130017102 f1w1f2w2 There is a natural inclusion of groups Gm gtlt Gm Gn1n27 so we can de ne f1 f2 6 CZGMM by means of the induced character 6 n f1f2 Ind6n11gmm f2 since the formula for induced characters can be applied to arbitrary class functions This product makes CZG into a graded Calgebrai We won7t prove this For a partition A b n7 let 1A be the indicator function on the conjugacy class CA C Gm and let GA G11 gtlt G1112 gtlt gtlt Gn741n C 5w For w E Gm denote by Aw the cycleshape of w7 expressed as a partition De nition 1 The Frobenius Characteristic is the map Ch I Czc6 A AC de ned on f 6 C465 by 1 7 Chf 5 Z fwpw we6n Equivalently7 lt gt ch f M 6 where 1 is the class function 6n A A de ned by 1 WW pAw Theorem 1 1 ch is a ring isomorphism 2 ch is an isometry ie it preserves inner products ltf79gt6n ltChf70h9gtA 3 ch restricts to an isomorphism CZZG A A2 4 1A gt gt pAZ 5 Inng XmV H hA 6 lndg Xsign gt gt e 7 The irreducible characters of Sn are the ch 1s 8 For all characters x we have chX XmV wchx llll prove a few of these assertionsi Recall that 2 lCAl TLlZ Therefore 1 Ch x a 2 PA IDA2A 10er which proves assertion It follows that Ch is at least a graded Cvector space isomorphism since 1 and pk2A are graded Cbases for CZG and A respectively To show assertion 2 it suf ces to check it on these basesi Let A M F n then 1 7 1 lt1A71ugt6n 5 Z 1Aw1uw glCAl SM SM2A7 39LUEGn ltampampgt1lt Pugt5k5kzk ZklzuA m mA m Next we check that Ch is a ring homomorphism hence an isomorphism Let f E G g E Gk and n j kl Then chf y Indgfxmfmt wgt6 where 1 is as in f g Resggtlt6k gt 67 X 6k by Frobenius reciprocity 1 Z f 9ltw7 mam 1 101E67X6k 2 WW G 2 lm wee meGk chf 0119 More Fundamental Results 1 The Murnaghan Nakayama Rule We now know that the irreducible characters of 6 are X Ch 13 for A F n The Murnaghan Nakayama Rule gives a formula for the value of the character X on the conjugacy class OH in terms of m hook tableaux Here is an example of a rimhook tableau of shape A 543 3 l and content M 6332 11 13 1 4l4 1 2 3i 3 6 5 Note that the columns and row are weakly increasing and for each i the set Hi T of cells containing an i is contiguousi Theorem 2 Murnaghan Nakayama Rule 1937 XACM Z lt711htHT rimehook tableaux T 11 of shape A and content it For example the heights of H1 H5 in the rimhook tableau above are 4 3 2 1 1 1 There are an even number of even heights so this rimhook tableau contributes 1 to xACH An important special case is when M 11 1 ie since then X CH X 16 ie the dimension of the irreducible representation S of Sn indexed by A On the other hand a rimhook tableau of content 1 is just a standard tableau So the Murnaghan Nakayama Rule implies the following Corollary 3 dim S f This begs the question of how to calculate f which you may have been wondering anyway There is a beautiful formula in terms of hooks For each cell 1 in the Ferrers diagram of A let hz denote its hook length the number of cells due east of due south of or equal to 1 1n the following example hz 6 Theorem 4 Hook Formula of Frame Robinson and Thrall 1954 Let A k n Then nl F 7 H 11 EEA Example 1 For A 5 433 1 k 16 as above here are the hook lengths 9 7 6 3 l 7 5 4 l 5 3 2 4 2 l 1 Therefore fA 1 4 2288 97265242322214 Example 2 For A nn b 2n the hook lengths are n1nn71 2 top row n n71 n72 1 bottom row Therefore f 7 2n 7 1 2n 7 n 1 ml 7 n 1 n which is the nth Catalan number as we already know

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