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Irreducible Representations of the Symmetric Group

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Irreducible Representations of the Symmetric Group IRREDUCIBLE REPRESENTATIONS OF THE SYMMETRIC GROUP REDMOND MCNAMARA Abstract. We construct the Specht modules and prove that they completely characterize the irreducible representations of the symmetric group. We will prove certain properties of these representations using combinatorial tools (such as calculating the dimension using Hook's length formula). Only an introductory knowledge of group theory and linear algebra will be assumed and representation theory concepts will be introduced as necessary. Contents 1. Introduction 1 2. Basics of Representation Theory 2 3. Characters and Conjugacy Classes 5 4. Specht Modules 8 5. Irreducibility of the Specht Modules 11 6. Dimensions of the Irreducible Representations 13 Acknowledgments 16 References 16 1. Introduction Representation theory studies homomorphisms of a group G into the group of linear transformations. Generally, it is very difficult to find all possible homomor- phisms of a group. These problems occupy professional representation theorists, day in and day out. In the business of classifying every possible representation of a group, perhaps the simplest case is the cyclic group. This paper will explore what may be the second most accessible example, the symmetric group. We will begin by establishing the basics of representation theory by addressing questions such as: what is a representation? what is a module? how many irre- ducible representations are there? etc. We then apply these basic components to the symmetric group. We define our terms, building the basics of representation theory from introductory knowledge of linear algebra, group theory and the sym- metric group. We find an upper bound on the number of inequivalent, irreducible representations. In order to prove that upper bound exists, and it is just the num- ber of conjugacy classes, we will introduce the machinery of group characters and character inner products; similarly, in order to define the Specht modules we will introduce Young diagrams, Young tableaux, tabloids and polytabloids; in order to Date: August, 2013. 1 2 REDMOND MCNAMARA prove that the polytabloids associated with the standard Young tableaux form a basis for the Specht modules, we introduce both an ordering of tabloids and Garnir elements. We will work through several examples to build intuition and illuminate concepts. The strategy is as follows: in the second section, we define the basic components of representation theory. In the third section, we prove that there is at most one irreducible representation for each conjugacy class. In the fourth section, we will start constructing modules called Specht modules. In the fifth section, we will prove that the Specht moduels are irreducible, which allows us to calculate the number of irreducible representations. In the sixth section, we will find a basis for the Specht modules and use it to calculate the dimensions of the irreducible representations. 2. Basics of Representation Theory We want to study nice homomorphisms from one group G to another G0. A case when these homomorphisms are useful and workable is when we restrict G0 to be the general linear group of size n{the set of all linear transformations from an n dimensional space to itself, or equivalently the set of all n-by-n invertible matrices. Definition 2.1. A representation ' of a group G is a homomorphism ' : G ! GL(n; C), where GL(n; C) is the complex general linear group. Next, we define a module. Modules allow us to take problems about linear maps and turn them into problems about vector spaces. Working with vector spaces can be extremely convenient, making some results immediately clear. Definition 2.2. A G − module, call it V , is a vector space for which there exists a homomorphism ' : G ! GL(V ), where GL(V ) is the set of linear maps V ! V . If the group is clear, we will usually just refer to V as a module. The representation ' induces a multiplication of G in V . Given g 2 G and v 2 V , we define the operation g ? v as '(g)v, which is a vector.1 This operation is linear. If g; h; e 2 G, where e is the identity element, α 2 C and v; w 2 V , then, (1) g ? (v + w) = '(g)(v + w) = '(g)v + '(g)w = g ? v + g ? w, (2) α(g ? v) = α('(g)v) = '(g)(αv) = g ? (αv), (3) (gh) ? v = '(gh)v = '(g)'(h)v = '(g)(h ? v) = g ? h ? v, (4) e ? v = '(e)v = Iv = v. A natural question arises: can we define a vector space built out of elements of G under some representation '? It turns out, we can. Definition 2.3. The group algebra, denoted C[G], is the set of all linear com- binations of elements of G. If G has n elements, then C[G] = fc1g1 + ··· + cngn j c1; : : : ; cn 2 Cg and h ? (c1g1 + ··· + cngn) = c1(hg1) + ··· + cn(hgn) for all h 2 G. A few examples will build our intuition for these objects. Example 2.4. Consider the group Z=2Z and the following three representations: 1The notation ? is specific to this paper. It is introduced simply to distinguish between the different types of multiplication. We use · for scalar, vector and matrix multiplication, and ◦ for composition. IRREDUCIBLE REPRESENTATIONS OF THE SYMMETRIC GROUP 3 (1) The trivial representation, denoted '(1), which sends both 0 and 1 to 1, (2) The representation '(2) which similarly sends 0 7! 1 but sends 1 7! −1 (3) The representation '0 which sends elements 0 and 1 to the set of two-by-two 1 0 0 1 matrices: 0 7! and 1 7! . 0 1 1 0 Next, we compute the corresponding modules. (1) The module V (1) is the one-dimensional vector space C where multiplica- tion by elements of Z=2Z is trivial; for any g 2 Z=2Z and any c 2 C, g?c = c. (2) The module V (2) is the one-dimensional vector space C where multiplica- tion by 1 simply switches the sign of a complex number; for any complex number c, 0 ? c = c and 1 ? c = −c. (3) The module V 0 is the two-dimensional vector space C2 where multiplication x x by 1 just switches the x and y components; for any in 2, 1 ? = y C y 0 1 x y = ; of course, 0 still leaves any vector unchanged.2 1 0 y x For the third example, we ask if there is a subspace of V 0 that is invariant under 1 any action by G. A trivial example is 0. Another example is W = span , 1 x which has vectors of the form w = for some x in C. Therefore 1 ? w = x 0 1 x x = = w, so 1 ? w = w, which is in the subspace W . 1 0 x x If a module has a non-trivial, invariant proper subspace, then it is said to be reducible. A module that is not reducible is said to be irreducible. (" p1 #) We normalize the basis of the invariant subspace so that it is 2 . We then p1 2 (" p1 # " p−1 #) complete the basis of V 0, B0 = 2 ; 2 . p1 p1 2 2 We want to look at '0 in this new basis. Under '0 each element of Z=2Z can be treated like a linear transformation. Thus, using the change in basis formula, we −1 note that there exists a change of coordinates matrix such that TB0 = PTBP . " p1 p−1 # " p1 p1 # In this case, P = 2 2 . Similarly, P −1 = 2 2 . The reader can p1 p1 p−1 p1 2 2 2 2 −1 0 check that 0 7! I , but 1 7! . Therefore, in our new basis, any element 2 0 1 '(2)(g) 0 g 7! . In this form, the representation is clearly reducible. 0 '(1)(g) 2This example, 2.4.3, and subsequent work on it come from [4]. 4 REDMOND MCNAMARA Proposition 2.5. 3 Suppose '0 is a reducible representation. There exists an A(g) B invertible matrix P for which P'0(g)P −1 = where A and C are 0 C(g) representations themselves. Exercise 2.6. 4 Let ' be a representation and P be an invertible matrix. Then '0(g) = P'(g)P −1 is itself a representation. In fact, every representation of a group can be decomposed into a direct sum of irreducible ones; thus, knowing the irreducible representations of a group is of paramount importance. Let us work out another example for the symmetric group. Example 2.7. Consider the following representations of S3: (1) The trivial representation '(1) that sends every element to 1, (2) The sign representation '(2) that sends every element to its sign. For ex- ample, (12) 7! −1 and (132) 7! 1 (later we use sgn: Sn ! C), (3) The defining representation, '0, which just permutes the columns of the 20 1 03 20 0 13 3-by-3 identity matrix: (12) 7! 41 0 05, (132) 7! 41 0 05, etc. 0 0 1 0 1 0 The module corresponding to '(1) is just the one-dimensional vector space C. For all c 2 C, g ? c = c. The module corresponding to '(2) is also C. Here, for all c 2 C, g?c = c if g is an even permutation and g?c = −c if g is an odd permutation. Both of these modules are irreducible since they are one-dimensional. The module V 0 corresponding to '0 is the three-dimensional vector space C3. An action by a group element just permutes the coordinates of the vector. For 223 233 02131 example, (12)?435 = 425. As above, W = span @415A is an invariant subspace.
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