ALFRED YOUNG S CONSTRUCTION OF THE IRREDUCIBLE REPRESENTATIONS OF S n

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1 A. Garsia Young s Natural Representations 202B March 3, 204 ALFRED YOUNG S CONSTRUCTION OF THE IRREDUCIBLE REPRESENTATIONS OF S n. The representation matrices If is a partition of n (in symbols ` n), a filling of the Ferrers diagram of by the integers, 2,...,n will be referred to as an injective tableau of shape. The collection of all these tableaux will be denoted by INJ( ). If T 2 INJ( ) and the entries of T increase from left to right in the rows and from bottom to top on the columns then T is said to be standard. The collection of all these tableaux will be denoted by ST( ). Given an injective tableau T with n squares, a permutation =(, 2,..., n) is made to act on T by replacing the entry i by i. The resulting tableau is denoted by T. We say that is in the Row Group of T if the rows of T and T di er only by the order of their entries. The Column Group of T is analogously defined. These two groups will be denoted by R(T ) and C(T )respectively. write Given T,T 2 2 ST( ) we shall say that T precedes T 2 in the Young first letter order and T < YFLO T 2 if the first entry of disagreement between T and T 2 is higher in T than in T 2. For instance we have Here T and T 2 agree in the positions of, 2, 3, 4, 5. The first letter of disagreement is 6 and it is higher in T than in T 2. The positions of the remaining letters do not matter. Given two tableaux T,T 2, not necessarily of the same shape, we let T ^ T 2 be the diagram obtained by placing in the cell (i, j) the intersection of row i of T with column j of T 2. The diagram

2 A. Garsia Young s Natural Representations 202B March 3, below gives an example of this construction Note that in this case there is a cell which contains more than one entry. When this happens we say that the pair T ^ T 2 is bad. Note that in this next example each cell has at most one entry. When this happens we say that T ^T 2 is good. Note that, in the good case, if =(, 2,..., h) is the shape of T and =(, 2,..., k ) is the shape of T 2, then for each i we must have i apple i. This is because the left hand side gives the number of entries in the first i rows of T ^ T 2 and the right hand side gives the number of entries in the tableau (necessarily of shape ) obtained by letting the entries of T ^ T 2 drop down their column until they are packed tight. This means that when T and T 2 have di erent shapes, we must have some motion of cells and a strict inequality in., for at least one i. On the other hand when T and T 2 have the same shape then =, and there can be no motion of cells. In this case T ^ T 2 is also a tableau of shape =, whose rows are a rearrangement of the rows of T and whose columns are a rearrangement of the columns of T 2. In summary when = and T ^ T 2 is good there are two permutations 2 R(T ) and 2 2 C(T 2 ) such that T ^ T 2 = T ; T 2 = 2 T ^ T 2.2 Conversely, it is easily seen, that the existence of two such permutations giving.2 forces T ^ T 2 to be good. This permits us to introduce an important function on pairs of tableau:

3 A. Garsia Young s Natural Representations 202B March 3, Definition. For T,T 2 2 INJ( )welet C(T,T 2 ) = 0 if T ^ T 2 is bad sign( 2 ) if T ^ T 2 is good Here and after, it will be convenient to let S,S 2,...,S f.3 denote the standard tableaux of shape in Young s first letter order. This given, we set C ( ) = kc(s i, S j )k f i 0 j=..4 We have the following basic property Proposition. The matrix C ( ) is upper unitriangular and is therefore invertible over the integers. Note that if T < YFLO T 2 and a is the first letter of disagreement, then there is a letter b such that a and b are in the same column of T and in the same row of T 2. Note that b = 4 for the example given in Fig. In general b is the letter which lies at the intersection of the column of the shape in which a lies in T with the row in which a lies in T 2. This means that T 2 ^ T is bad. So T < YFLO T 2 implies C(T 2,T ) = 0. Thus from the definition of first letter order we derive that C(S i,s j ) = 0 8 i>j. This proves the proposition since C(S i,s i ) = holds true trivially. This allows us to define A ( ) = C ( ) C ( ).5 Our goal in these notes is to show that the collection of matrix functions form a complete set of irreducible representations of S n. {A ( )} `n.6 Remark. The simplicity of the definition in.5 should be compared with the pages and pages of intricate constructions that characterize the treatments of the representations of the symmetric groups given in recent and past litterature following the work of A. Young. Because of peculiarities of Young s style of writing many of his successors never bothered to read, let alone tried to understand, Young s beautiful constructions. As a result you will often see the name Specht module associated

4 A. Garsia Young s Natural Representations 202B March 3, with representations which are in fact none other than those defined in.5. What is more amusing is that Young himself in QSA IV ( ), with surprising premonition, described some alternate ways of constructing representations of S n which include as particular cases all the constructions that followed. Another justification often given to assign some significance to the work of Specht and assorted Jonny-come-latelies is that later developments are Characteristic Free but this is only based on a naive missunderstanding of Young s purely combinatorial arguments. Problems Problem. Show that for any injective tableau T we have = with 2 R(T ) and 2C(T ) if and only if = =. Problem 2. Let T 2 INJ( )with ` n. Show that T ^ T is good if and only if = 2 with 2 2C( T) and 2 R(T ). Show that this factorization is unique. That is show that if = 2 0, 0 with 2 0 2C( T) and 0 2 R(T )thenthen 0 = and 0 =. Problem 3. Construct all standard tableaux apple of shape (3, 2, ) in the Young first letter order construct the matrix A (3,2) ( ) for = Then Problem 4. Show that when is a hook that is =(k, n k ) then the matrix C ( ) reducestothe identity matrix. 2. Young s Tableau idempotents. For a given tableau T here and after we set P (T )=, N(T )= 2C(T ) sign( ) 2. 2R(T ) If T,T 2 2 INJ( )by TT 2 we denote the permutation that sends T 2 into T that is T = TT 2 T 2 This given, it is easy to see that we have a) P ( T) = P(T ) ; N( T) = N(T ) 8 2 S n b) T T 2 P (T 2 ) = P (T ) TT c) T T 2 N(T 2 ) = N(T ) TT 2 ( ) (see The Collected Papers of A. Young, , Math. Expositions V. #2 Univ. of Toronto Press 977

5 A. Garsia Young s Natural Representations 202B March 3, We shall also use the shorthand notations 8 < a) E TT 2 = N(T ) T T 2 P (T 2 ), E T = E TT = N(T )P (T ) : b) F TT 2 = P (T ) T T 2 N(T 2 ), F T = F TT = P (T )N(T ) 2.3 Note that from 2.2 b) and c) we derive that a) ETT 2 = E T T T 2 = TT 2 E T2 b) F TT 2 = F T T T 2 = TT 2 F T2 2.4 Recalling that denote the standard tableaux of shape S,S 2,...,S f 2.5 in Young s first letter order we set ( Ei = E Si ; E ij = E Si S j F i = F Si ; F ij = F Si S j ; ij = Si S j 2.6 When dealing with a fixed shape, to lighten our formulas, sometime we shall omit the superscript from our symbols. We should note that many identities and results involving the elements E ij hold also for the F ij 0 s, with only minor monifications. In each case, to avoid unnecessary repetitions, we shall prove them for one set and leave it to the reader to derive the analogous results for the other set. The group algebra elements P (T ), N(T ) have truly remarkable properties. To prove them we need some auxiliary results. Proposition 2. For a pair of tableaux T and T 2 of the same shape the following properties are equivalent a) T ^ T 2 is good b) 9 2 R(T ) and 2 2 C(T 2 ) such that T 2 = 2 T c) 9 2 R(T ) and 2 C(T ) such that T 2 = T d) R(T 2 ) and 2 2 C(T 2 ) such that T 2 = 2 2 T Moreover we have sign( ) = sign( 2 ) 2.7 We have seen that a) and b) are equivalent. From 2.2 a) we then get that b) implies N(T ) = 2 N(T 2 ) 2 = N(T 2 ) Thus there is a 2 N(T ) such that = 2 2.8

6 A. Garsia Young s Natural Representations 202B March 3, This gives that T 2 = 2 T = ( ) T = T which is c). Note that 2.7 now follows from 2.8. Conversely suppose we have c). Then we may write N(T 2 ) = N(T ) = N(T ) and we must necessarily have a 2 2 C(T 2 ) such that 2 =, giving T 2T = = 2. This shows the equivalence of b) and c). We are left to show that d) is equivalent to b). It turns out that, in this case, the P 0 s play the role the N 0 s played in the previous argument. More precisely from b) and 2.2 a) we derive that P (T 2 ) = 2 P (T ) 2 = 2 P (T ) 2. Thus there must be an 2 2 R(T 2 ) such that 2 = 2 2 and this gives T 2T = 2 = 2 2. Thus b) implies d). We leave it to the reader to show that d) implies b) and complete the proof of the proposition. Remark 2. It is important to note that the permutations, 2,, 2 of Proposition 2. are uniquely determined by T and T 2. In fact for a given Tableau T we cannot have two pairs 0, 2 R(T ) and 0, 2 C(T ) such that 0 0 = Indeed this is equivalent to the existence of an 2 R(T ) and a 2 C(T ) such that = and this is easily seen to be impossible. Let us keep this fact in mind since we are going to make use of it several times in the future. Before we proceed with the next result we need to make some conventions. Young used a very e cient notation to represent some elements of the group algebra of the symmetric group S n. For a given subset S = { apple i <i 2 < <i k apple n} {, 2,...,n}

7 A. Garsia Young s Natural Representations 202B March 3, he let the symbol [S] represent the sum of all elements of S n that permute the elements of S and leave fixed all the elements of the complement of S in {, 2,...,n}. He also used [S] 0 to denote the sum of the same permutations where each is multiplied by its sign. For instance, with this convention we see that for the tableau T = we may write P (T )=[, 4, 8, ] [6, 9, 0] [5, 7] [2, 3] and N(T )=[2, 5, 6, ] 0 [3, 4, 7, 9] 0 [8, 0] 0 Remark 2.2 For a given f 2A(S n ), it will also be convenient to denote by f 0 the element of the group algebra obtained by replacing each permutation by the same permutation preceded by its sign. More precisely, for f = P f( ) we shall set f 0 = sign( )f( ) and refer to the operation f!f 0 as the priming operator. We should note that any identity involving elements of the group algebra of S n generates a companion identity when we apply the priming operator to both sides. In this manner identities for the E 0 ij s may be transformed into identities for the F 0 ij s. In fact, it easy to see that E T,T 2 0 = sign( T,T 2 ) F T >,T > where T > and T > 2 denote the corresponding transposed tableaux ( ). In particular we have E ij 0 = sign( ij ) F 0 ij, 2.0 where 0 denotes the partition conjugate to. Remark 2.3 For two partitions =(, 2,..., h), =(, 2,..., k ) we say that dominates and write apple D ( ) Transposing a tableau simply means reflecting it across the 45 o diagonal emanating from its south-west corner

8 A. Garsia Young s Natural Representations 202B March 3, if and only if i apple i 8 apple i apple min(h, k). Thus our remarks at the beginning of section may be restated by writing T ^ T 2 good! shape(t ) apple D shape(t 2 ). 2. Now it develops that the same implication holds true if we have a) P (T )N(T 2 ) 6= 0 or b) N(T 2 )P (T ) 6= To see this note that, by definition, T ^ T 2 bad implies that there are two elements a, b that are in the same row of T and the same column of T 2. This means that the transposition (a, b) is at the same time in R(T ) and C(T 2 ). We thus have P (T )N(T 2 )= this contradicts 2.2 a). Thus P (T )(a, b) N(T 2 )=P(T ) (a, b)n(t 2 ) = P (T )N(T 2 ). P (T )N(T 2 ) 6= 0! T ^ T 2 good! shape(t ) apple D shape(t 2 ). 2.3 We can proceed similarly and show that N(T 2 )P (T ) 6= 0! T ^ T 2 good! shape(t ) apple D shape(t 2 ) Semi-simplicity of Algebras We have seen that the Group algebra A[ ] of a finite group has two binary operations + (addition) and (multiplication), which are associative and satisfy the left and right distributivity laws. That is for all f,h,g 2A[ ] we have (left) h (f + g) = h f + h g (right) (f + g) h = f h + g h Moreover, since its elements are formal linear combinations of group elements, such as f = 2 f( ) with f a complex valued function on, we also have a multiplication by a scalar operation, defined by setting for any complex number c cf = 2 cf( ) this operation is clearly associative and distributive with respect to addition.

9 A. Garsia Young s Natural Representations 202B March 3, The group algebra has also an identity which we denote which is simply given by the identity element of the group. A structure with these properties with scalars in a field K is usually referred to as a K-Algebra. The collection of n n matrices with entries in K, which we denote M n [K], is the simplest example of a K-Algebra. It easily seen M n [K] isan 2 -dimensional vector space with a natural basis consisting of the n n matrices E i,j with i, j entry equal to and all other entries equal to zero. With this notation, for any matrix A = ka i,j k n i,j= we can write A = n n a i,j E i,j 3. i= j= The E i,j, which (using Young s terminology) will refer as matrix units, are easily seen to satisfy the identities 0n if j 6= r E i,j E r,s = 3.2 if j = r where 0 n here means the n n matrix with all elements equal to 0. The the next simplest example is obtained by taking a direct sum of a finite number of simple matrix algebras. More precisely we will denote E i,s km M nl [K] l= the vector space of M M block matrices, (with M = P k s= n s) of block diagonal form 2 A () A A = 6 (2) 0 km = A (l) with A (l) 2M nl [K] A (k) l= It will be convenient to denote by f (s) i,j the block matrix in 3.3 with A (l) = 0nl E (ns) i,j if l 6= s if i = s 3.4 where here E (ns) i,j denotes the n s n s matrix with i, j element and all other elements 0. It is n s is a basis of L k i,j= i= M n i [K], since it is clearly an easily seen that the collection S k s= f (s) i,j independent set and every element A = L k l= A(l) has an expansion of the form A = k n s n s a (s) i,j f (s) i,j s= i= j= In particular we see that dim km M ni [K] = i= k s= n 2 s

10 A. Garsia Young s Natural Representations 202B March 3, Moreover, from 3.2 and 3.4 it follows that 8 < 0 M if l 6= m fi,j l fr,s m = 0 M if l = m but j 6= r : l = m and j 6= r f l i,s M = P k s= n s 3.5 It is easily shown that only multiples of the identity matrix I 2 M[K] commute with all the matrices in M[K]. In particular, it follows that the center of L k i= M n i [K] (that is the subspace of the elements that commute with all the other elements) is spanned by the elements f s = n s i= f (ns) i,i ( apple s apple k) 3.6 Which are none other than the block matrices in 3.3 with A (l) = n 0nl I (ns) if l 6= s if i = s where I (ns) denotes the n s n s identity matrix. This gives km dim Center M ni [K] i= = k 3.7 It is customary to call the algebras L k i= M n i [K] semi-simple. Frobenius Fundamental Theorem of Representation Theory states that the group algebra of every finite group is isomorphic to a semi-simple algebra. More precisely Frobenius result assures that given a finite group of order N = there are integers {n s } k s= such that k N = n 2 s 3.8 and a bijective map : A( )! s= km M ni [C] 3.9 that respects addition, multiplication and multiplication by a complex number. In particular we deduce from 3.7 that the integer k must be equal to the dimension of the center C( ) of the group group algebra of. But we have seen that C( ) is none other that the subspace of class functions. Thus in particular we derive that k in 3.8 and 3.9 must be none other than the number of conjugacy classes of. Of course in view of 3.5 the Frobenius result amounts to showing the existence in A( ) of a basis S k s= e (s) n s i,j satisfying the identities i,j= 8 < 0 if l 6= m e l i,j e m r,s = 0 if l = m but j 6= r 3.0 : l = m and j = r e l i,s i=

11 A. Garsia Young s Natural Representations 202B March 3, 204 This given, the desired map is simply obtained by setting e l i,j = f l i,j and extending it by linearity to the rest of A( ). Remarkably, Alfred Young, initially unaware of Frobenius work, set himself the task, in a series of papers written in a span of 7 years, to produce a purely combinatorial construction of such matrix units for several reflection groups including all the symmetric groups. While doing so he completely upstaged Frobenius who only succeeded in constructing the k elements f s in 3.6 for S n. In the next few section we will present Young s construction for the symmetric groups and derive some of its most significant properties. 4. Young s matrix units for A(S n ). This section is dedicated to the proof of the following remarkable result Theorem 4.(A.Young) With S,S 2,...,S n the standard tableaux of shape ` n in Young s first letter order, let h = n!/n and set i = h N(S i )P (S i ) 4. This given, for S i,s j = ij, the group algebra elements do not vanish and satisfy the identities e ij = i ij ( j+)( j+2) ( f ) 4.2 e ij e rs = ( eis if = and j = r 0 otherwise 4.3 We will obtain our proof by combining a few, very simple. purely combinatorial properties of the group algebra elements i which we will state as separate Propositions. We will start with the following very beautiful fact which considerably simplifies Young s original argument. Proposition 4. (Von Neuman Sandwich Lemma) For any element f 2A(S n ) and any T 2 INJ( ) we have N(T ) fp(t ) = c T (f) N(T )P (T ), 4.4 with c T (f) = fp(t )N(T ). 4.5

12 A. Garsia Young s Natural Representations 202B March 3, Expanding f in the left hand side of 4.4 we get N(T ) fp(t ) = f( ) N(T ) P(T ). 4.6 Now note that using 2.2 a) we can write N(T ) P(T ) = N(T ) P ( T). That means that the only terms that survive in 4.6 are those for which N(T ) P ( T) 6= 0. However, as we have seen in Remark 2.3, this can happen only if T ^ T is good. But part d) of Proposition 2. gives that we must have T = T with 2 R(T ) and from permutations 2 C(T ). In other words the only terms that survive in 4.6 are those coming of the form = (with 2 R(T ), 2 C(T )) Since any permutation that can be so expressed has a unique expression of this form we can rewrite 4.6 as N(T ) fp(t ) = f( ) N(T ) P (T ) R(T ) 2C(T ) But then the simple identity N(T ) P (T ) = sign( ) N(T )P (T ) reduces 4.7 to N(T ) fp(t ) = sign( )f( ) N(T )P (T ), 2R(T ) 2C(T ) which is 4.4 with c T (f) = sign( )f( ) 2R(T ) 2C(T ) However the latter is but another way of writing 4.5. Q.E.D. Proposition 4. yields us the following fundamental fact Proposition 4.2 For each ` n we have a non vanishing constant h depending only on such that for all tableaux T 2 INJ( ) we have, E T = N(T )P (T ) E 2 T = h E T 4.8

13 A. Garsia Young s Natural Representations 202B March 3, We simply use 4.4 with f = P (T )N(T ) and get N(T ) P (T )N(T ) P (T ) = h N(T )P (T ), with h = P (T )N(T ) P (T )N(T ). Next we need to show that h is the same for all injective tableaux of shape. However this is a simple consequence of the fact that if T = T then P (T )= P(T ) and N(T )= N(T ), thus P (T )N(T ) P (T )N(T ) = P(T )N(T ) P (T )N(T ) = P (T )N(T ) P (T )N(T ) Finally we must show that h 6= 0. But if h =0theelementE T would be nilpotent. This in turn would imply that its image L(E T ) by the left regular representation would be a nilpotent matrix. Since nilpotent matrices have trace zero and for any group algebra element f = P f( ) we have tracel(f) = f( ) tracel( ) = f( )n! the vanishing of h would imply that E T = 0. Now this is absurd since, from Remark 2. we immediately derive that E T =. This completes our proof. Remark 4. We should note that one of the simplifications to Young s arguments due to the Von Neuman Lemma is to avoid a direct identification of the constant h as n!/n. This is obtained in Young s work as the final result of gruesome brute force proof of the idempotency of the group algebra elements E T /h. In the present development the identification of h will be carried out at the very end. This given, until that time we will assume that h is the mysterious constant appearing in Von Neuman s lemma. In particular, this assumption yields us that all the elements i in are idempotent. Proposition 4.3 j i = ( 0 if 6= 0 if = but j>i i if = and j = i We have seen in the proof of Proposition. that if T < YFLO T 2 then there is a pair of entries r, s that are in the same column of T and same row of T 2. This gives P (T 2 )N(T ) = P (T 2 )(r, s)n(t ) = P (T 2 )N(T ) and thus E T2 E T must necessarily vanish. Since S,S 2,...,S n are in Young s first letter order it follows that E Sj E Si = 0 when j>i 4.9

14 A. Garsia Young s Natural Representations 202B March 3, in view of 4. this proves the second case of 4.9. To prove the first case of 4.9 we need only show that if T and T 2 are injective tableaux of shapes and respectively then To this end note that 2.3 immediately gives that 6=! E T E T2 = 0 E T E T2 = N(T )P (T )N(T 2 )P (T 2 ) 6= 0! apple 4.0 On the other hand we expand the mid element P (T )N(T 2 )intheproducte T E T2 we will obtain a sum of terms of the form N(T ) P(T 2 ) = N(T )P ( T 2 ) If E T E T2 does not vanish then at least one of these terms must not vanish. But then from 2.4 we derive that apple which together with 4.0 contradicts 6=. This completes our proof of the first case of 4.9 and we are done since the last case, under our assumption that h is the Von Neuman constant, is as we have seen an immediate consequence of Proposition 4.2. We are now finally in a position to prove Theorem 4., except for the identification of the constant h. We will start with a of the identities in 4.3 Note first that if 6= then the first case of 4.9 gives Thus ( i ) j = j (for all i, j) 4. e ij e rs = ij j ( j+) ( n ) r rs( = ij j r rs ( s+ ) ( n ) = 0 (again by the first case of 4.9 ) We are thus reduced to proving the identities in 4.3 when the second case of 4.9 gives s+ ) ( n ) = in all our elements. Note then that a) j i = 0 as well as b) ( j ) i = i when j>i 4.2 Now note that we can write, for r<j and for r = j e ij e rs = ij j ( j+) ( n ) r rs ( s+) ( n ) ( by 4.2 b) ) = ij j r rs ( ( by 4.2 a) ) = 0 s+ ) ( n ) e ij e rs = ij j ( j+) ( n ) j js ( s+) ( n ) ( by 4.2 a) = ij j j js ( s+) ( n ) ( by the last case of 4.9) ) = ij j js ( s+) ( n ) = i ij js ( s+) ( n ) = e is

15 A. Garsia Young s Natural Representations 202B March 3, Finally for r>j, after successive uses of 4.2 will necessarily get to a point where the resulting expression for e ij e rs contains the factor ( r ) r which of course vanishes by the last case of 4.9. This completes our argument. From these relations we can now establish that Proposition 4.4 The group algebra elements e ij cannot vanish Note first that the identity e i,j e j,i = e i,i reduces us to showing that the e i,i themselves cannot vanish. Now this result is an immediate consequence of the very useful fact (as we shall see) that for any two group algebra elements f,g we have fg = gf 4.3 the reader is urged to work out a proof of this identity in full generality. This given, note first that from the definition in 4.2 and 4.3 it follows that e i,i = i ( i+) ( n ) = ( i+) ( n ) i (by 4.2 b)) = i = /h 6= 0 As a first step in proving that Young s matrix units are a basis we can now establish that Proposition 4.5 The group algebra elements [`n e ij n are independent i,j= Assume if possible that for some constants c ij we have `n n i,j= c ij e ij = 0 then for all possible choices of, r and s we derive from the identities in 4.3 that c rs e rs = e rr `n n i,j= c ij e ij e ss = 0 and since, as we have seen, the e rs 0 s do not vanish we must necessarily have c rs = 0 as desired. To prove that an independent set in a vector space V is a basis we need only show that it spans V or that its cardinality equals the dimension of V. Alfred Young proved that his matrix units are a basis, both ways. We will follow the latter approach first since it it based on two beautiful combinatorial identities.

16 A. Garsia Young s Natural Representations 202B March 3, But before we can state them we need a few preliminary observations. In the figure below we have depicted the Ferrers diagram of the partition (3,, ) and its immediate neighbors in the Young lattice. (the lattice of Ferrers diagrams ordered by inclusion). We have at the center the Ferrers diagram of (3,, ) with the circles indicating its removable (inner) corner cells and the x 0 s showing its addable (outer) corner cells. In the row below we have the partitions obtained by removing one of its inner corner cells and in the row above we have the partitions obtained by adding one of its addable outer corner cells. We can clearly see that every Ferrers diagram has one more addable cells than removable ones. In the same row as (3,, ) we have depicted the partitions that are obtained by going up a row by adding an addable cell then down a row by removing a removable cell. The fundamental property that yields our desired basis result is that the same collection of partitions can be obtained by first removing a removable cell then adding one of the addable cells. Using the symbol! to express that is obtained from by adding an addable cell of, or equivalently, is obtained from by removing one of the removable cells of, we can derive the following basic recursions for the number of standard tableaux. Proposition 4.5 Denoting by n the number of standard tableaux of shape we have a) `n n (!) =n, b) `n+ n (! )=(n + )n 4.4 The identity in 4.4 a) is immediate. In fact, if is a partition of n we obtain a bijection between the standard tableaux enumerated by the right hand side and the collection of standard tableau enumerated by the left hand side by the removal of n and the cell containing it. For 4.4 b) we will proceed by induction on n. We can take as base case = () since we trivially see there is only one standard tableau of shape () and exactly one each of shapes (, ) and (2). So let us assume that 4.4 b) is true for n and note that multiple applicationc of 4.4 a) give `n+ n (! ) = `n+ n (! ) 4.5 `n

17 A. Garsia Young s Natural Representations 202B March 3, Since in the sums on the right hand side we are allowed to remove the cell we are adding, we can see that, if has r inner corner cells (and therefore r + outer corner sells), then by separating out the terms with = we can rewrite 4.5 in the form `n+ n (! ) = `n+ n (! ) + (r + )n 4.6 6= But the observation, that for any given, adding and addable and then removing a removable yields the same collection we obtain by the reverse process of removing a removable and then adding an addable, can be simply translated into the following beautiful identity `n+ n (! ) = 6= Thus, using 4.7 in 4.6 we derive that `n+ n (! ) = `n `n n (! ) 4.7 6= n (! ) + (r + )n 4.8 But now, since has r removable corners we see that the sum P `n n exactly r terms where =. This gives `n Using this 4.8 becomes 6= n (! ) = `n `n+ n (! ) = `n = `n (by 4.4 b) for n ) = completing the induction and the proof. `n `n n (! ) + rn 6= n (! ) + n `n (!) `n n (! ) + n (!) nn + n (by 4.4 a) ) = nn + n = (n + )n These two recursions combined have the following remarkable corollary. (! ) will contain Proposition 4.6 n 2 = n! 4.9 `n In particular it follows that the group algebra elements [`n e ij n i,j= yield a basis for A(S n).

18 A. Garsia Young s Natural Representations 202B March 3, Note that for the base case n =2wehaven 2 (,) + n2 (2) = + = 2!. Thus we can again proceed by induction on n. We will suppose then that 4.9 is true for n. Now, using 4.4 a), the left hand side of 4.9 for n + becomes Completing the induction and the proof. n (! ) n 2 = `n+ n `n+ `n = n n (! ) `n `n+ by 4.4 b) = `n n (n + ) n by induction =(n + ) n! = (n + )! It is good to keep in mind that Young s matrix units satisfy the following useful identities Proposition 4.7 For all we have e i,j = ( 0 if i 6= j if i = j i = h These identities are immediate consequences of 4.3. In fact, to begin with we obtain 4.9 e i,j = e i, e,j = e,j e i, = ( 0 if i 6= j if i = j e But on the other hand we have e = ( 2 ) ( n ) = ( 2 ) ( n ) = = h E S = h as desired. This given, it develops that there is a very simple explicit formula yielding the expansion of any element of the group algebra of S n in terms of Young s matrix units. Theorem 4. For any f 2A(S n ) we have Since the collection [`n f = `n h e ij n i,j= n f e j,i e i,j 4.20 i,j= is a basis there will be coe cients f = `n n c i,j (f) e i,j i,j= c ij (f) yielding

19 A. Garsia Young s Natural Representations 202B March 3, Thus and 4.9 gives f e rs = n i,j= c i,j (f) e i,j e rs = f e rs = c s,r(f) h n c i,r (f) e i,s This proves Note that, if we interpret every permutation 2 S n as the element of A(S n ) which is on and 0 elsewhere, we will have coe cients a i,j ( ) yielding the expansion = `n i,j= Now we have the following remarkable fact i= n a i,j ( ) e i,j 4.2 Theorem 4.2 For each ` n the matrices {A ( )=ka i,j ( )k n i,j= } yield an irreducible representation of S n For any, 2 S n we have, using 4.2 and the identities in 4.2 = `n n i,j= r,s= n = `n i,j= n a i,j ( )a r,s ( ) e i,j e r,s n s= a i,j ( )a j,r ( ) e i,s = `n n n n 4.22 a i,j ( )a j,r ( ) e i,s i= s= j= But on the other hand a direct use of 4.2 for gives = `n n a i,j ( ) e i,j i,j= Comparing with 4.22 yields the desired product identity A ( ) = A ( )A ( ) Note further that an application of 4.20 yields (by 4.9) a i,j ( ) = h e j,i = 0 if i 6= j if i = j In other words we also have A ( ) = I n (the n n identity matrix!).

20 A. Garsia Young s Natural Representations 202B March 3, Next note that using 4.2 we derive that e r,s = n i,j= a i,j ( ) e i,j e r,s = n a i,r ( ) e i,s 4.24 i= In matrix form, this can be written as e,s,e 2,s,...,e n,s = e,s,e 2,s,...,e n,s A ( ) 4.25 In other words the have shown that the subspace L[e,s,e 2,s,...,e n,s] spanned by the independent set {e,s,e 2,s,...,e n,s} is invariant under the action of S n and the matrix corresponding to the action of on the basis {e,s,e 2,s,...,e n,s} is precisely A ( ). To show that A ( ) is an irreducible representation we need only show that there is no nontrivial invariant subspace or, equivalently that any non vanishing element g 2L[e,s,e 2,s,...,e n,s], has any one of the basis elements {e,s,e 2,s,...,e n,s} as one of its images. To this end suppose that n g = c i e i,s with c u 6= 0, (there clearly must be at least one such u). Then note that we must have c u e r,u g = n i= i= c i c u e rue i,s = e r,s and the arbitrariness of r proves the irreducibility, completing our proof. 5. Properties of the representations {A } and their characters To identify Young s matrices A ( ) and their traces we will need an auxiliary fact which reveals a beautiful property of the tableaux function C(T,T 2 ). Proposition 5. For any T,T 2 2 INJ( ) and 2 S n E T2T = C( T,T 2 ) 5. E T2T = sign( 2 ) 2 2 T2T = sign( 2 ) 2 2 T2,T = 22C(T 2) 22R(T 2) 22C(T 2) 22R(T 2) now there may not be a single pair 2, 2 for which 2 2 T2,T =, in which case E T2T =0

21 A. Garsia Young s Natural Representations 202B March 3, But if there is such a pair, there will be one and only one ( ), and then we must have or 2 2 T2T =! 2 2 = T T 2 = T,T 2 T 2, T = 2 which combined with part d) of Proposition 2. gives that T ^ T 2 is good. Thus from 2.7 we derive that 8 < 0 if T ^ T 2 is bad E T2T = : sign( 2 ) if T ^ T 2 is good or 2 E T2T = C( T,T 2 ) as desired. We should note that in particular we must have Q.E.D. E T2T = sign( 2 ) = C(T,T 2 ) 5.2 This brings us in a position to prove Theorem 5. For any injective tableau T of shape ` n we have coe cients a i (T ) giving In fact these coe. E T = n i= a i (T )E S i,t 5.3 cients may be directly obtained from the vector identity 0 a (T ) 0 C(S,T) a 2 (T ) B C(S 2,T). a n (T ) C A = C ( ) C(S n,t) C A 5.4 Assume first that these coe cients do exist. Multiplying 5.3 on the right by TS j gives Equating coe E TS j = n i= a i (T ) E S i S j cients of the identity, and using 5.2, we get. n C(S j,t) = i= a i (T ) C(S j,s i ) ( ) Recall Remark 2.

22 A. Garsia Young s Natural Representations 202B March 3, which, in matrix notation, may be expressed as 0 C(S,T) C(S 2,T). C(S n,t) C A = C ( ) 0 a (T ) a 2 (T ). a n (T ) C A, and formula 5.4 follows upon left multiplication of both sides of this identity by C ( ). It turns out that existence is an immediate consequence of 4.2, namely the general expansion identity n a i,j ( ) e i,j 5.5 = `n i,j= In fact, for any given and apple r apple n the identities in 4.3 give (using 5.5) r = = n i,j= n i= a i,j ( ) e i,j a i,r ( ) i r = i,r r = n i,j= n i= a i,j ( ) i a i,r ( ) i i,j ( j ) ( i,r n ) r Recalling the definition in 4., we derive from this that E S r = n a i,r ( ) E S i,s r i= Now this can be rewritten as E S r = n a i,r ( ) E S i,s r i= and for = T,S r we get E T = n a i,r ( T,S r ) E S i,s r S r,t = i= n a i,r ( T,S r ) E S i,t i= This proves 5.3 with a i (T ) = a i,r ( T,S r ) 5.6 and completes our argument. This proof has a most desirable by-product. Theorem 5.2 For all ` n and all 2 S n we have A ( ) = C ( ) C ( ) 5.7

23 A. Garsia Young s Natural Representations 202B March 3, From 5.6 and 5.4 it follows for 0 a,r ( ) a 2,r ( ). a n,r( ) S r = T that 0 C A = C ( ) C(S, S r ) C(S 2, S r ). C(S n, S r ) C A But this simply says that the r th column of the matrix on the left of 5.7 equals the r th of the matrix on the right. Finally yielding the result identity which we have anticipated since the beginning of these notes. We are now in a position to obtain Young s formula giving the character s of his irreducible representations. This result may be stated as follows. Theorem 5.3 For a given ` n set ( )=trace A ( ) and = 2Sn ( ) 5.8 then = T 2INJ() Using 5.7 and dropping some of the superscripts we can write P (T )(N(T ) n!/n 5.9 = 2Sn tracec( ) C( ) = trace C( ) F 5.0 where F = kf ij k n i= is a matrix with group algebra entries F ij = C(S i, S j ). Now Proposition 5. gives that C(S i, S j ) = C( S i,s j ) = E SjS i Thus F ij = E SjS i = E SjS i = P (S i ) ij N(S j ). Note further that since is a class function it follows that we also have = n!

24 A. Garsia Young s Natural Representations 202B March 3, and we can thus write, using 5.0 = n! trace C( ) F = n! trace C( ) F. But we see that F ij = P(S i ) ij N(S j ) = N(S j )P (S i ) ij = 0 at least when i>j. This means that the matrix F is upper triangular and since C( ) is upper triangular with unit diagonal elements we must conclude that F = n F i,i n! = n! trace 2Sn = n! n i= which is 5.9 precisely as asserted. P(S i )N(S i ) = n n! i= T 2INJ( ) P (T )N(T ) The character Proposition 5.2 Setting we have is in fact very closely related to the matrix units. More precisely e = n e i,i 5. i= = h e 5.2 in particular it follows from the identities in 4.3 that = h 5.3 Note that using 4.2 we get e i,i = e i,i = `n = n r= n a r,s ( ) e r,s e i,i = r,s= n r,s= a r,i ( ) e r,i = a i,i ( ) h (by 4.9) a r,s( ) e r,s e i,i

25 A. Garsia Young s Natural Representations 202B March 3, and summing for i e = n i= e i,i Now note that since e is a central element of A(S n ) it follows that Thus multiplying both sides of 5.4 by e = e. = h ( ) 5.4 and summing over S n gives 5.2 and completes our proof. Now a result parallel to 5.3 can be obtained by direct manipulations of Young idempotents. Proposition 5.3 T 2INJ() N(T )P (T ) 2 = h 2 T 2INJ() N(T )P (T ) 5.5 Note we may rewrite the left hand aside of 5.5 as LHS = N(T )P (T )N( T)P ( T) 5.6 T 2INJ() But P (T )N( T) 6= 0impliesT ^ T is good. Thus from Proposition 2. we derive that the only non vanishing summands in are obtained for = with 2 R(T ) and 2 C(T ). Since we have N(T )P (T )N( T )P ( T ) = N(T )P (T ) N(T )P (T ) the identity in 5.6 becomes LHS = h = N(T )P (T )N(T )P (T ) sign( ) (by 4.8) = h N(T )P (T ) sign( ) T 2INJ() = h T 2INJ() and a further use of 4.8 gives 5.5 as desired. Theorem 5.4 N(T )P (T ) 2R(T ) 2C(T ) N(T )P (T )N(T )P (T ) sign( ) The result we have anticipated since Von Neuman Lemma is now finally within reach. Squaring both sides of 5.9 gives, using 5.3 and 5.5 h h 2 = (n!/n ) 2 P (T )N(T ) h = n! n 5.7 T 2INJ() canceling the common factor h and using 5.9 again gives = and 5.7 follows from the non-vanishing of. h n!/n

26 A. Garsia Young s Natural Representations 202B March 3, The Frobenius map The group algebra A(S n ) has a natural scalar product obtained by setting for any f,g 2 A(S n ) <f,g>= f( )g( ). 6. n! Note that since for all 2 S n f = f( ) = f( ). we have < f,g>= n! = n! f( )g( ) f( )g( )=<f, g>. 6.2 Similarly we show that <f,g>= <f,g >. 6.3 Let us denote by C the element of the group algebra of S n that is obtained by summing all the permutatiions with cycle structure. Now we have shown that the cardinality of this collection of permutations is given by the ratio C = n! z 6.4 where for a with i parts equal to i it is customary to set z = 2 2 n n! 2! n! 6.5 Using this and the fact that conjugacy classes are disjoint it follows that 8 C,C < n! = if = = n! z z : 0 otherwise There is also a natural scalar product in the space =n of symmetric polynomial homogeneous of degree n which is obtained by setting for two power basis elements p,p 6.6 This given note that if we set p,p = ( z if = 0 otherwise 6.7 FC = p /z 6.8 and use the fact that =n and the center C(S n ) have the same dimension, we can define an invertible linear map F : C(S n )! =n

27 A. Garsia Young s Natural Representations 202B March 3, which is also an isometry. In fact, note that from 6.6 and 6.7 it follows that FC,FC = z 2 p,p = z ( = )] = C,C Now Frobenius introduced this map precisely for the purpose of constructing all the irreducible characters of S n. Frobenius result can be stated as follows Theorem 6. The class functions = F s 6.9 form a complete set of the irreducible characters of S n. The surprising fact is that Young s and Frobenius partition indexing of the irreducible characters turn out to be identical. This result, established by Young himself, may be simply be stated as follows Theorem 6.2 (A. Young) F s = n n! T 2INJ( ) N(T )P (T ) 6.0 We should mention that to the best our knowledge there is at the moment no direct proof of this fact. Moreover, even Young s proof is based on a very dubious use of his quite heuristic (hair) raising operators. Our proof as we will later see is very indirect. This given, it is worthwhile taking a close look at what needs to be proved. To begin we need the following very convenient form of the Frobenius map Proposition 6. For each 2 S n set ( ) = p ( ) 6. where ( ) is the partition giving the cycle structure of. This given, for any class function g 2C(S n ) we have Fg = g, 6.2 The proof is immediate. We need only verify 6.2 for the conjugacy class elements C with ` n. In this case the definition in 6. gives FC = C, = n! C p ( ) = n! p n! = p z z which is precisely the definition in 6.8. As a corollary we obtain

28 A. Garsia Young s Natural Representations 202B March 3, Proposition 6.2 The equality in 6.0 amounts to showing that for any injective tableau T of shape s = h 2R(T ) 2C(T ) we have sign( ) p ( ) 6.3 Note that 6.0 may also be written in the form s = n n! F N( T)P ( T) = n n! F N(T )P (T ) Now, using 6.2 we get s = n n! N(T )P (TY), (by 6.2 and 6.3) = n n! (since is central) = n n! N(T )P (TY), N(T )P (T ), = n N(T )P (T ), = n n! 2R(T ) 2C(T ) sign( ) p ( ) This proves 6.3.

29 A. Garsia Young s Natural Representations 202B March 3, Theorem 3.6 For any given tableau T o of shape ` n, the characters of the action of S n on the left ideals a) A(S n )P (T o ) and b) A(S n )N(T o ) are respectively given by the group algebra elements a) P = T 2INJ() P (T )! and b) N = T 2INJ() N(T ) 0! 3.9 Since both P (T o )! and N(T o ) 0! are idempotents, formula 8.6 (R) yields that these characters are P = P (T o )! and N = N(T o ) 0! Now these exressions can be rewritten in the form P = P ( T o )! and N = N( T o ) 0! and the identities in 3.9 follow since, as varies in S n, the tableau T o describes all the injective tableaux of shape. It develops that we have the tools to derive the following basic facts about these these two characters. Theorem 3.7 For ` n and T o 2 INJ(), letk denote the multiplicity of Young s representation A in the action of S n on A(S n )P (T o ) then It then follows that we have the expansions a) K > 0 =) D and b) K = a) P = + > D K and b) N = + < D K In particular we derive that P N = h 3.22 The orthonormality of the irreducible character basis immediately gives the expansions a) P = `n P, and b) N = `n N, 3.23

30 N = `n N = `n A. Garsia Young s Natural Representations 202B March 3, Now note that our definition of priming introduced in section 2 applied to formula 3.9 gives that ( ) 0 = Thus priming 3.23 a), written for replaced by 0, gives 0 P 0,, and this may be rewritten as P 0, 0. Thus we see that 3.2 b) follows from 3.2 a). To prove 3.20 note that we may write K = P, =!h T 2INJ() T 22INJ( ) P (T ),N(T 2 )P (T 2 ) But since P (T ),N(T 2 )P (T 2 ) = N(T 2 )P (T ),P(T 2 ), we see from 2.28 that, if a single summand in 3.25 fails to vanish, we must necessarily have D. This proves 3.20 a). To prove 3.2 a) assume next that P =! P( T o ) and =. We then simultaneously have = h N( T o )P ( T o ) Thus K = P, =! P(To ), =! P (To ), But we see that we have = n!! = n!!h P (To ), P (To ),N( T o )P ( T o ) P (To ),N( T o )P ( T o ) = N( T o )P (T o ),P( T o ) and so the only case in which this summand fails to vanish is when T o = T o with 2 R(T o ) and 2 C(T o )

31 A. Garsia Young s Natural Representations 202B March 3, This reduces 3.26 to K = n!!h = n!!h = n! h 2R(T o) 2R(T o) 2C(T o) 2C(T o) 2C(T o) P (To ), N(T o )P (T o ) sign( ) P (T o ), N(T o )P (T o ) sign( ) P (T o ),N(T o )P (T o ) = n! h = n! h id, N(To )P (T o )N(T o )P (T o ) = n! id, N(T o )P (T o ) = P (To ),N(T o )P (T o )N(T o ) as desired. This completes the proof of our theorem since 3.22 follows immediately from 3.2 because of the relations in 5.0 (R)