The Specht modules are irreducible

We …rst consider a few examples of Specht modules as irreducible representations of Sn. Consider a tableau t of shape = (n):

t = 1 2 ::: n : The associated polytabloid according to (4.3) is

e_t= 1 2 ::: n :

Evidently e_t= e_t so that e_t forms the trivial one-dimensional irreducible rep-resentation.

Consider now a tableau t of shape = (1; 1; :::; 1):

t = 1 2 :::

n

Since there is only one element in each row, the associated polytabloid according to (4.3) is a combination of all possible permutations of the elements along the column with the appropriate permutation sign. Operating an arbitrary permutation on e_t gives

e_t= e _t= X

2C t

sign( ) ~t:

De…ning = , noting that C _t = C_t comprises all permutation elements, and utilising the Rearrangment Theorem yields

e_t= X

2Ct

sign( ¹) ~t

= sign( )et:

Thus, e_t forms a basis for the one-dimensional sign representation, which is di¤erent from the identity representation.

Yet another example of an irreducible representation that can be readily constructed is the standard representation. Consider a tableau

t = a i j :::

b :

From (4.3) the associated polytabloid is e_t= a i j :::

b

b i j :::

a :

The tabloids are labelled by the number in the lower box and there are n inde-pendent of them. Due to the form of the polytabloids with a pair of tabloids of opposite sign, the number of independent polytabloids, however, is n 1. This provides a basis for the (n 1)-dimensional irreducible standard representation.

To prove that Specht modules are irreducible representations of Sn, we …rst de…ne a projection operator for a given tableau t of shape

t= X

2C^t

sign( ) ;

which when acts on the tabloid ~t reproduces e_t. Let u =P

ic_i~t_i2 U, where U is a sub-module of M and ~t_i are tabloids of shape . Operating _ton u gives

4.6. SPECHT MODULES 69

tu =X

i

c_i X

2Ct

sign( ) ~t_i:

We will show that if tu is not zero, it must be proportional to et, i.e., t

projects out the components of u belonging to et. First, we note that

~

eti = X

2Ct

sign( ) ~ti= X

2Ct

sign( ) r~ti;

where ris a permutation that shu- es elements in the same rows of tibut does not permute elements belonging to di¤erent rows of ti. The above equation is true because ~eti consists of tabloids. Assume that ti is such that each column of rti has the same elements as those in the corresponding column of t, but not necessarily in the same order. Now let c permute elements in the columns of t such that rti= ct. We then have

~

e_ti = X

2Ct

sign( ) _r~t_i= X

2Ct

sign( ) _c~t:

Since consists of column permutations, we …nd

~

e_ti = X

0 1

c 2Ct

sign( ⁰ _c¹) ⁰~t

= sign( _c¹) X

02C^t

sign( ⁰) ⁰~t = sign( c)et:

The second step utilises the Rearrangment Theorem. If for a given ti there is no

r such that each column of rti and the corresponding column of t share the same elements then ~e_ti= 0. The reason is that for every 2 Ctthere is another element belonging to C_tthat produces the same tabloid but with opposite sign.

This does not happen in the case of t because every 2 Ct acting on t yields a new tabloid as can be seen in an example in (4.4). Some examples should clarify the point. Consider the following tableaux:

t = 1 2 3

4 5 ; t1= 2 3 4

1 5 ; t2= 1 5 2

4 3 :

In this case C_t= e; (14); (25) and (14)(25). Applying _ton t₁ yields

tt₁= 2 3 4 1 5

2 3 1

4 5

5 3 4

1 2 + 5 3 1

4 2 = e_t; where et is given in (4.4). On the other hand

tt2= 1 5 2 4 3

4 5 2

1 3

1 2 5

4 3 + 4 2 5

1 3 = 0:

In the case of t_{1, r}= (234) will bring _rt₁to a tableau with the same elements as t in each column but for t₂there is no such _r.

To formally prove that tti = 0 for any tableau ti that cannot be brought into a tableau that is column-equivalent to t by a row-permutation r, we note that there must be numbers a and b in the same column of t and the same row of ti. The tableau t2 above is such an example, with a = 2 and b = 5. We can then write Ct in the form of cosets: Ct= 1; :::; r; 1(ab); :::; r(ab). However,

k and k(ab) have opposite signs but otherwise yield the same tabloid when acting on ti because a and b are in the same row. When applying ton ti the terms kti and k(ab)ti cancel out and hence tti= 0.

We conclude that if tu 6= 0 then it must be proportional to e^t, i.e., t

projects out the e_t-component of u. Supposing that _tu is a non-zero multiple of e_t, then e_t2 U and it follows that S U . On the other hand, suppose that for all u 2 U, tu = 0. Then

(u; et) = u;X

2C^t

sign( ) ~t

!

= X

2C^t

sign( ) u; ~t

= X

2C^t

sign( ¹) ¹u; ~t

!

= tu; ~t = 0;

and it follows that U S ^?. Thus, if U is a sub-representation of S , then either S U or U S ^? but since S is non-zero then U = S . In other words, S is an irreducible representation of Sn.

Standard Young tableaux as basis forS

Standard Young tableaux of a given shape can be used to construct a basis for the irreducible representation S . The set fe^tg where t is a standard Young tableau of shape form a basis for S . Unlike the tabloids, the number of standard Young tableaux corresponds exactly to the dimensionality of the irre-ducible representation, which can be calculated using the hook-length formula described below. As an example, consider a tableau of shape = (4; 1). There are 5!=(4!1!) = 5 corresponding tabloids labeled by the number in the second row:

1 2 3 4

5 ; 1 2 3 5

4 ; 1 2 4 5

3 ;

1 3 4 5

2 ; 2 3 4 5

1 ;

whereas there are 4 standard Young tableaux:

1 2 3 4

5 , 1 2 3 5

4 ; 1 2 4 5

3 ; 1 3 4 5

2 :

4.6. SPECHT MODULES 71 The polytabloids generated from the fourth and …fth tabloids are equivalent, except for the sign, so that the number of independent polytabloids e_t is four, which is what we get if we start from the standard Young tableaux. In general, the standard Young tableaux remove double-counting present in the tabloids arising from permutations in the columns by imposing the rule that numbers along any given row (from left to right) or any given column (from top to bottom) is increasing.

The dimension of S : the hook-length formula

The number of independent polytabloids e_tcorresponding to a given shape or the dimension of the Specht module S , which is the irreducible representation of S_n, is given by the hook-length formula. In each box draw an arrow to the right and an arrow to the bottom. Fill in the box with the number of boxes covered by the arrows. The dimension of the irreducible representation S is given by

n!

m1m2:::mn

where m_i is the number in box i. For example, for a given shape = (3; 2)

! 4 3 1

2 1

the dimension is 5!=(4:3:1:2:1) = 5. The proof of the hook-length formula is rather complicated and it is not included here.