AN EXPOSITION ON GROUP CHARACTERS
THESIS
Presented in Partial Fulfillment of the Requirements for the Degree Master of
Mathematical Science in the Graduate School of the Ohio State University
By
Aaron Margraff, BS
Graduate Program in Mathematics
The Ohio State University
2014
Master’s Examination Committee:
Dr. James Cogdell, Advisor
Dr. Warren Sinnott c Copyright by
Aaron Margraff
2014 ABSTRACT
This paper is an educational approach to group characters through examples which introduces the beginner algebraist to representations and characters of finite groups. My hope is that this exploration might help the advanced undergraduate student discover some of the foundational tools of Character Theory. The prerequisite material for this paper includes some elementary Abstract and Linear Algebra. The basic groups used in the examples are intended to excited a student into exploration of groups they understand from their undergraduate studies. Throughout the section of examples there are exercises used to check understanding and give the reader opportunity to explore further. After taking a course in Abstract Algebra one might
find that groups are not concrete objects. Groups model actions, rotations, reflections, movements, and permutations. Group representations turn these abstract sets of objects into sets of n × n matrices with real or complex entries, which can be easily handled by a computer for any number of calculations.
ii ACKNOWLEDGMENTS
The paper that is before you is not only evidence of my hard work and under- standing of characters and representations but also evidence of the patience of many teachers that have poured their time and knowledge into me. It was my honor to work with an advisor as accomplished as Dr. James Cogdell. As much as he instructed me in mathematics he also instructed me with his patience. As a future educator I greatly esteem the self-restraint he showed when I presented him with questions that he had previously answered. His calm approach to education removed the barriers that would otherwise impede learning. Thank you for investing your valuable time into a growing student.
I would also like to thank Dr. Warren Sinnott for his willingness to participate as a member on my thesis committee and spending his time to read and review my work.
I would be amiss not to pour out my gratitude to my family and church for their support of my schooling and education over the past 20 years of my life. I attribute my position spiritually, mentally, bodily, and financially to your faithfulness in caring for me. You all are truly a gift from God.
iii VITA
1989 ...... Born in Columbus, Ohio
2012 ...... Bachelor of Science in Education at Youngstown State University
2012-Present ...... Graduate Teaching Associate, The Ohio State University
FIELDS OF STUDY
Major Field: Mathematics
Specialization: Education
iv TABLE OF CONTENTS
Abstract ...... ii
Acknowledgments ...... iii
Vita...... iv
CHAPTER PAGE
1 Introduction ...... 1
1.1 Introductory Definitions and Facts for Representations and Char- acters of Finite Groups ...... 1 1.2 Helpful Facts for Finding Representations ...... 4
2 Examples ...... 8
2.1 Characters of Abelian Groups ...... 8 2.2 Characters of Non-Abelian Groups ...... 16
3 Proofs ...... 32
Bibliography ...... 52
v CHAPTER 1
INTRODUCTION
1.1 Introductory Definitions and Facts for Representations
and Characters of Finite Groups
The structure of this paper is modeled after the method in which I was taught to
find group characters. I was given the definitions and major results of representations and characters, taking them as truth, and used them to discover the character tables of various groups. In formal mathematical developments it is unsettling to use a result before the result is proven to be true, so this method is perhaps an educational approach to character theory. I begin this paper with an introduction to definitions and results I used in my discovery process. In the second chapter I use the definitions and results to derive the character tables for some abelian and non-abelian groups.
In the last chapter of the paper I give formal definitions and prove the results given in this section.
Let’s look at some of the basic definitions and results to begin the study. Words in italics indicate a new definition. A representation of a group G is a homomorphism
ρ : G → GL(V ) where V is a vector space over the field C and ρ(g) is the image of an element g. No- tice each element ρ(g) is a element of the group of invertible matrices. Also, since ρ is
1 a homomorphism these matrices preserve the group action, that is ρ(gh) = ρ(g)·ρ(h)
[1]. Many groups have a geometric interpretation and have representations that reflect their geometric meaning. One can find such a representation by choosing a suitable basis for the vector space for which the geometric interpretation is understood. The dimension of the representation is given by the dimension of the vector space chosen,
V . In some sense it would be nice to know when a representation captures the com- plete behavior of a group, which leads us to defining the next term. A representation is called faithful when it maps G isomorphically to its image.
Every finite group has a finite number of irreducible representations. These are the building blocks for all other representations of the group. For this section of the paper irreducible representation will mean a representation that cannot be written as the direct sum of two or more representations. In Chapter 3 we will develop a more rigorous definition of irreducible representations. Also in Chapter 3 we find that the number of irreducible representations is the number of conjugacy classes of the group, hence a finite number of them. It is natural at this point to ask, what are the dimensions of the irreducible representations? With the number of irreducible representations known one can specify the dimension of the irreducible representations by using the property that the sum of the squares of the dimensions of the irreducible representations is equal to the order of the group. This result is written as,
2 2 2 |G| = d1 + d2 + ··· + dr
where di is the dimension of the irreducible representation ρi and r is the number of conjugacy classes of G.
Let G be a group and ρ : G → GL(V ) be a representation on a vector space V over the field C. The character of ρ is the mapping
χρ : G → C 2 defined by χρ(g) = tr ρ(g). The degree of the character is the dimension of its cor- responding representation. In general the trace is not a homomorphism, hence the character is not always a homomorphism either. The character χo assumes the value of 1 at each element of G, this is called the trivial character. Sometimes the trivial character is written as 1, 1G or simply as 1. The mathematical term “character” was first introduced by Carl Friedrich Gauss in Disquisitiones Arithmeticae. He used characters to assign numerical information to classes of binary quadratic forms, in order to separate classes of forms with the same determinant [3]. The characters in representation theory are associated with a respective representation where χρ(g) is the trace of ρ(g). An irreducible character is the character of an irreducible represen- tation. Hence there are r irreducible characters and representations. Later we will
find that characters are class functions, which means they are constant over conjugacy classes,
−1 χρ(hgh ) = χρ(g).
A character χ can be thought of as vectors where elements of a group G are listed as
G = {g1, g2, . . . , gn} and the vector representation of χ is
t χ = (χ(g1),..., χ(gn)) , where t denotes the transpose and n is the order of the group [1]. Since characters are constant over conjugacy classes it is natural to list the group elements according to their conjugacy class. The most valuable result in relating group characters to vectors is found by constructing a hermitian dot product as:
1 X hχ, χ0i = χ(g)χ0(g). |G| g∈G
With this inner product we have that the irreducible characters are orthonormal.
This fact is extremely valuable when one needs to determine the reducibility of a
3 representation. If taking the inner product of the character of the representation with itself results in unity then the character came from an irreducible representation, if not the representation is reducible. If we let χ be the trivial character and χ0 is any non-trivial irreducible character then X χ0(g) = 0. g∈G This fact is useful when finding the rows in a character table. A character table of a group is a two dimensional table whose rows correspond to the irreducible characters and columns correspond to conjugacy classes. The sum of the rows of the table weighted by the size of the conjugacy class is zero. A similar statement about the columns can be found in Theorem 3.0.16. Its result is that, X deg(χ0) · χ0(g) = 0, for g 6= 1. χ0 irreducible Another way that the above relation can be stated is that the sum of the irreducible characters, weighted by their dimension, evaluated at a non-trivial conjugacy class is zero. If you sum the irreducible characters, weighted by their dimensions, over the trivial conjugacy class you will receive the order of the group.
1.2 Helpful Facts for Finding Representations
Finding representations of groups can be difficult, and sometimes not necessary to complete the character table. But when it is helpful to find the representation there are some standard methods which can be used to isolate potential representations.
One-dimensional representations can be found by using parity, the tensor product of two other one-dimensional representations, or perhaps the determinant of a higher dimension representation. Here are a few results that are helpful when looking for new representations of a group. Recall the definition of a representation: An n dimensional matrix representation of a group G is a homomorphism ρ : G → GLn(C). 4 Problem 1.2.1. Let ρ be a representation of a group G. The determinant of ρ is a one-dimensional representation.
Proof. Let ρ : G → GLn(C), if g1, g2 ∈ G then ρ(g1g2) = ρ(g1)ρ(g2) since ρ is a homomorphism. From linear algebra it is true that det(AB) = det(A) det(B) for n × n square matrices. Therefore,
det ρ(g1g2) = det(ρ(g1)ρ(g2))
= det(ρ(g1)) det(ρ(g2)) ∈ C
Hence det ρ : G → GL1(C) is a one dimensional representation of G.
This previous result can be helpful when looking for a one-dimensional represen- tation of a group; if a higher dimensional representation is known calculating the determinate may yield a new one-dimensional representation.
Problem 1.2.2. If G is a group with a faithful representation by diagonal matrices then G is abelian.
Proof. Let g1, g2 ∈ G then let ρ be a faithful representation by diagonal matrices, then ρ(g1) and ρ(g2) are the diagonal matrix images of g1 and g2.
ρ(g1g2) = ρ(g1)ρ(g2)
= ρ(g2)ρ(g1) ← diagonal matrices commute
= ρ(g2g1)
Since ρ(g1g2) = ρ(g2g1) and ρ is faithful we have g1g2 = g2g1. Therefore G must be abelian.
Problem 1.2.3. The only one-dimensional representation of the symmetric group Sn are the trivial representation and the sign representation.
5 Proof. Transpositions in Sn have order two. Hence a transposition’s image under a one-dimensional representation has either order one or two.
Case I: ρ is a one-dimensional representation and α ∈ Sn is a transposition such that ρ(α) = 1. Since all transpositions belong to the same conjugacy class we can write
−1 α = γβγ for any other transposition β ∈ Sn and for some γ ∈ Sn. Then we have,
1 = ρ(α) = ρ(γβγ−1)
= ρ(γ)ρ(β)ρ(γ−1)
= ρ(γ)ρ(γ−1)ρ(β) ← one-dimensional matrices commute
= ρ(γγ−1β)
= ρ(β).
Hence any transposition is mapped to 1. Since Sn is generated by the transpositions every element of Sn is mapped to 1. Therefore ρ is the trivial representation.
Case II: ρ is a one-dimensional representation and α ∈ Sn is a transposition such that ρ(α) = −1. This is the only one-dimensional matrix of order two. As in the first case we will use the fact that transpositions are conjugate. For any transposition β there exists γ such that α = γβγ−1. By the same reasoning,
− 1 = ρ(α) = ρ(γβγ−1)
= ρ(γ)ρ(β)ρ(γ−1)
= ρ(γ)ρ(γ−1)ρ(β)
= ρ(γγ−1β)
= ρ(β).
Hence all transpositions have order two and are mapped to −1. Now, since the transpositions generate Sn, elements of Sn that are composed of an even number of
6 transpositions are mapped to 1 and elements composed of an odd number of trans- positions are mapped to −1, this is called the sign representation.
7 CHAPTER 2
EXAMPLES
The purpose of this chapter of the paper is to give the reader some experience relating their knowledge of elementary algebra to representations of groups and then to their characters and character tables. In some sense the easiest groups to understand are abelian groups. Of the abelian groups cyclic groups are the most fundamental. Let us begin this section of examples by considering some basic cyclic groups and their representations.
2.1 Characters of Abelian Groups
Recall the Fundamental Theorem of Finite Abelian Groups which states that every
finite abelian group G can be expressed as the direct sum of cyclic groups of prime power order. So, if we can classify cyclic groups and their representations then our goal is to extend the classification to all abelian groups. We will start with the cyclic group of order three.
2 Problem 2.1.1. Consider the cyclic group C3 = {1, u, u }. Find all irreducible rep- resentations and characters of C3. Once you have found them construct the character table for the group.
First, we must identify the number of irreducible representations in C3. We know that
−1 −1 all cyclic groups are abelian and for all c1, c2 ∈ C3 we have that c1c2c1 = c1c1 c2 = c2. 8 Hence each element of C3 is in its own conjugacy class and there must be three irreducible representations. If we combine this new knowledge with the fact that the sum of the square of the dimensions of the representations is equal to the order of the group we see it is impossible to have an irreducible representation of C3 of order two. Hence all three irreducible representations are one dimensional. This process will be used often throughout the remainder of this chapter to specify the number and dimension of irreducible representations. The trivial representation we will call ρ0 which will send each element of C3 to 1. One may interpret C3 as the set of rotations
2 of an equilateral triangle triangle in R . This interpretation is a real two dimensional 0 0 representation, call it ρ1. Before we find ρ1 we should look at the more common and preferred way of constructing these rotational representations using the complex
2πi representation ρ1 : G → GL1(C) where ρ1(u) = ω = e 3 a primitive cube root of unity. This representation is one dimensional and captures the action of the group completely. One can use the polar representation of ω to realize the above mentioned √ 1 3 real two-dimensional representation, ρ0 . Let a = Re(ω) = − and b = Im(ω) = . 1 2 2 0 We can use this fact to find a real representation of C3, ρ1 : C3 → GL2(R). 1 0 a −b a b ρ0 (1) = , ρ0 (u) = , ρ0 (u2) = 1 1 1 0 1 b a −b a
Currently we have two one-dimensional representations, the trivial and ρ1. The last
2 representation is given by ρ2(u) = ω . Now ρ1 and ρ2 differ in that the generator of
C3 is mapped to a different cube root of unity. In ρ0 the generator is mapped to 1, hence all other elements. In ρ1 then generator is mapped to ω. In ρ2, u is mapped
2 to ω . Each of these produce a new one dimensional representation. Notice ρ1 and
ρ2 are faithful representations on diagonal matrices, this is consistent with Problem
1.2.2 since C3 is abelian.
9 For cyclic groups the trace of the representations is the representation itself be- cause the representation is one-dimensional. In this case the characters are χ0 = ρ0,
χ1 = ρ1, and χ2 = ρ2. This result can be seen in the character table. The columns are given by the conjugacy classes and the rows by the irreducible characters. It is also standard to label the order of the conjugacy class by placing the size in parenthesis above the class. In this case each class is of equal size, one.
1 u u2
χ0 1 1 1
2 χ1 1 ω ω
2 χ2 1 ω ω
Table 2.1: The character table for C3
Recall that the sum of the values of a nontrivial character weighted by the size of the conjugacy classes is zero. This means that the sum of the rows weighted by the size of the conjugacy class is zero. Let us check this fact on χ1 with each conjugacy class of size one,
2 2 1 · χ1(1) + 1 · χ1(u) + 1 · χ1(u ) = 1 + ω + ω √ √ 1 3 1 3 = 1 + (− + i) + (− − )i 2 √ 2 √ 2 2 3 3 = 1 − 1 + i( − ) = 0. 2 2
Exercise 1. Check that the row for χ2 also sums to zero.
10 Recall again the sum irreducible characters over a nontrivial conjugacy class weighted by their degree is zero. We will check this fact over u,
X deg(χi) · χi(u) = 1 · χ0(u) + 1 · χ1(u) + 1 · χ2(u)
χiirreducible = 1 + ω + ω2
= 0.
It is also true that the sum of the irreducible characters over the conjugacy class u2 is zero. Lastly, the sum of the irreducible characters over the trivial conjugacy class is the order of the group, in this case, χ0(1) + χ1(1) + χ2(1) = 1 + 1 + 1 = 3 = |C3|. This method can be extended to all cyclic groups. To see this result in general let
2πi th ωn = e n a primitive n root of unity. When we multiply ωn with another complex 2π number its coordinates rotate in the complex plane. If we use Euler’s formula to n expand ωn, the product of ωn with a complex number z = x − iy is
2πi ωn · z = e n z 2π 2π = (cos( ) + i sin( ))(x − iy) n n 2π 2π 2π 2π = (x cos( ) + y sin( )) + i(x sin( ) − y cos( )). n n n n
Using vector notation for C we have that 2π 2π cos( ) − sin( ) x n n ωn · z = 2π 2π . sin( ) cos( ) −y n n a This gives a real 2 × 2 matrix representing ωn. The the rotation matrix for ωn is therefore 2aπ 2aπ cos( ) − sin( ) Re(ωa) −Im(ωa) n n n n 2aπ 2aπ = . sin( ) cos( ) Im(ωa) Re(ωa) n n n n Consider next a more general problem.
11 Problem 2.1.2. A common number theoretic example to consider is a cyclic group of additive cosets
Zn = Z/nZ = {a + nl|a ∈ Z, l ∈ Z}, where n is an integer. Find the character table of Zn.
Let a represent the residue class of a, so that a ≡ a(mod n). The representations
ka 2πi th and characters of Zn are defined by χk(a) = ω , where ω = e n , a primitive n root of unity. The multiplicative structure of the roots of unity preserves the multiplicative property required for representations and thus also the characters,
ka kb k(a+b) χk(a + b) = χk(a)χk(b) = ω ω = ω .
The characters of the generator of Zn, 1, are mapped to ω to the power of the n distinct residue classes hence yielding n distinct characters of degree one. There are n conjugacy classes and therefore n irreducible representations and characters.
Therefore, these characters must be the n irreducible characters for this cyclic group of order n. This result can be observed in Table 2.2. One may initially suspect that
0 1 2 ··· n − 1
χ0 1 1 1 ··· 1
2 n−1 χ1 1 ω ω ··· ω
2 2·2 2·(n−1) χ2 1 ω ω ··· ω ......
n−1 (n−1)·2 (n−1)·(n−1) χn−1 1 ω ω ··· ω = ω
Table 2.2: The character table for Zn
each of the characters of the generator is a primitive root of unity, this is only the 12 case when the order of the cyclic group is a prime hence all descending orders are co-prime to the order of the group and are therefore a primitive root of unity. Before we continue let us apply the result from above to a composite order cyclic group in order to observe this result.
Problem 2.1.3. Use the above result to construct the character table for Z4.
First recall the representations are the characters since they are all one dimen- sional. We have four characters each with the generator sent to a unique power of a
πi primitive fourth root of unity, e 2 = i.
0 1 2 3 χ0(1) = i = 1, χ1(1) = i = i, χ2(1) = i = −1, χ3(1) = i = −i.
Now construct the table using the mapping of the generator from above. Observe that
0 1 2 3
χ0 1 1 1 1
χ1 1 i −1 −i
χ2 1 −1 1 −1
χ3 1 −i −1 i
Table 2.3: The character table for Z4
χ2 does not take the generator 1 to a primitive fourth root of unity. This occurrence happens because 2 is not relatively prime to the order of the root of unity hence missing i and −i.
Hence character tables of cyclic groups are understood here by mapping the gen- erator to unique powers of a primitive root of unity. Now we want to extend this idea 13 to find the tables for any abelian group. Let us consider a product of cyclic groups say, C2 × C3.
Problem 2.1.4. Find the character table for C2 × C3.
Since the representations are equal to the characters in cyclic groups we will strictly consider the characters of the group. Call the characters of C2, χi for i = 0, 1.
0 Call the characters of C3, χj for j = 0, 1, 2. Since C2 × C3 is abelian there are six conjugacy classes so there are six characters for C2 ×C3 call them χei,j. The characters for C2 ×C3 will be determined by the image of the generators of each of the individual cyclic groups. In general the characters of C2 × C3 will be given by
χ n m χ n χ0 m χ n χ0 m ei,j(a , b ) = i(a ) · j(b ) = i(a) · j(b) ,
where a and b are the generators of C2 and C3 respectively, and n and m are integers.
The characters of C2 and C3 are defined in this way,
0 1 Character for C2: χ0(a) = (−1) = 1, χ1(a) = (−1) = −1
0 0 0 1 0 2 Character for C3: χ0(b) = ω = 1, χ1(b) = ω = ω, χ2(b) = ω , where ω is defined as a primitive cube root of unity as in Problem 2.1.1. The character table is shown in Table 2.4. The group C2 × C3 is isomorphic to C6.
Notice the characters χe1,1 and χe1,2 both map the generator ab to a primitive sixth root of unity. In fact the character tables for C2 × C3 and C6 are identical.
Exercise 2. Construct the character table for C6. Match the conjugacy classes of
C6 with the conjugacy classes of C2 × C3 by matching the generators. Match the characters according to Table 2.4.
The two groups are isomorphic because the order of the individual cyclic groups are relatively prime. Let us look at an example where the order of the groups are 14 1 ab b2 a b ab2 χ χ χ0 e0,0 = 0 0 1 1 1 1 1 1 χ χ χ0 2 2 e0,1 = 0 1 1 ω ω 1 ω ω χ χ χ0 2 2 e0,2 = 0 2 1 ω ω 1 ω ω χ χ χ0 e1,0 = 1 0 1 −1 1 −1 1 −1 χ χ χ0 2 2 e1,1 = 1 1 1 −ω ω −1 ω −ω χ χ χ0 2 2 e1,2 = 1 2 1 −ω ω −1 ω −ω
Table 2.4: The character table for C2 × C3
multiples of one another. We have already observed the character table for Z4 let us
find the character table for another group of order four, the Klein four-group, Z2 ×Z2.
Problem 2.1.5. Find the character table for Z2 × Z2.
Let χ0(1) = 1 and χ1(1) = −1 represent the two characters of Z2, the square roots of one. Let χei,j be the four multiplicative combinations of χ0 and χ1, where i = 0, 1 and j = 0, 1. The character table is then given by these four characters over the four elements of Z2 × Z2. Notice that the table for Z2 × Z2 is not identical
(0, 0) (0, 1) (1, 0) (1, 1)
χe0,0 = χ0χ0 1 1 1 1
χe0,1 = χ0χ1 1 −1 1 −1
χe1,0 = χ1χ0 1 1 −1 −1
χe1,1 = χ1χ1 1 −1 −1 1
Table 2.5: The character table for Z2 × Z2
to the table for Z4. The Klein four group is not isomorphic to Z4 and this can
15 be observed in the character table. At this point representations and characters of
finite abelian groups are relatively trivial. Mapping each product of elements to its respective product of characters gives all possible characters. Of course this is also true for their representations.
2.2 Characters of Non-Abelian Groups
One of the biggest differences between this section and the last section is the effort spent on finding a matrix representation of the group. Many times it is not necessary to find the representations to complete the character table of a group, but we will still explore finding the representations for the occasions in which they are needed.
With abelian groups the representations were all one dimensional and therefore were much easier to pin down. We must now use the fact that the sum of the square of the dimensions of the irreducible representations is equal to the order of the group.
We will also use the fact that the number of irreducible representations is equal to the number of conjugacy classes of the group. These two facts combined will allow us to find the number of irreducible representations along with their dimensions. Lets begin with the symmetric group S3 which is isomorphic to the dihedral group, D3.
Problem 2.2.1. Find all irreducible representations of S3. Use these to find the character table for S3.
Recall the elements of S3 in cycle notation {1, (123), (132), (12), (13), (23)}. I use the convention of feeding cycles from the right such that (123)(12) = (13). Recall, the conjugacy classes of S3 are found by collecting all conjugations of a permutation
−1 by all others so that Ca = {gag |a ∈ S3}. Below I will show the work for the
16 calculation of C(123),
(12)(123)(12)−1 = (12)(123)(12) = (132)
(132)(123)(132)−1 = (132)(123)(123) = (123)
(13)(123)(13)−1 = (13)(123)(13) = (132)
(23)(123)(23)−1 = (23)(123)(23) = (132)
(123)(123)(123)−1 = (123)(123)(132) = (123).
Therefore C(123) = {(123), (132)}. Similar calculations give C(12) = {(12), (13), (23)} and C1 = {1}. It is worth noting here that for symmetric groups the conjugacy classes are determined by cycle type. Once an element is in a conjugacy class there is no point to check its own conjugacy class, the calculation will yield the same conjugacy class as the first. Also, the identity element is always in a class of its own since it commutes will all other elements, i.e. 1g = g1 or g1g−1 = 1. There are three conjugacy classes for S3 so we have three irreducible representations. Call the three representations
ρ0, ρ1, and ρ2, with dimensions d0, d1, and d2 respectively. We know that the trivial representation is always one of the irreducible representations, assign it to ρ0, with dimension, d0 = 1. Recall the property that the sum of the square of the dimensions
2 2 2 of irreducible representations is equal to |S3| = 6. Therefore 1 + d1 + d2 = 6
2 2 or d1 + d2 = 5. The only possibilities for the dimensions are a one-dimensional representation and a two-dimensional representation. By Problem 1.2.3 we found that the only one-dimensional representations of a symmetric group are the trivial representation and the sign representation. Let ρ1 be the sign representation. The dimension of ρ2 is known, d2 = 2, so its character, χ2, of the identity is the trace of the two by two identity matrix, hence χ2(1) = 2. The characters χ0 and χ1 are given by
ρ0 and ρ1 so the partial character table is shown below. The order of each conjugacy class is placed above the class in parenthesis. At this point one can complete the
17 (1) (2) (3) C1 C(123) C(12)
χ0 1 1 1
χ1 1 1 -1
χ2 2
Table 2.6: Partial character table for S3
character table without knowledge of the third matrix representation using the fact that the sum of the irreducible characters, weighted by their dimensions, evaluated at a non-trivial conjugacy class is zero.
X deg(χi) · χi(123) = 1 · χ0 ((123)) + 1 · χ1 ((123)) + 2 · χ2 ((123)) = 0
χiirreducible
1 + 1 + 2 · χ2 ((123)) = 0
2 · χ2 ((123)) = −2
χ2 ((123)) = −1.
Exercise 3. Find χ2((12)) using the method shown above.
Because we are interested in checking our work and the consistency of our table we will find the two-dimenstional representation. It can be found using the geometric interpretation of S3, the symmetries of an equilateral triangle. We will place points 1, 2, and 3 on the vertices of an equilateral triangle centered at the origin with the
2 standard orthonormal basis for R . We will use the following coordinates for the vertices, 1 1 1 − − 1 : , 2 : √2 , 3 : √2 . 3 3 0 − 2 2 We have seen one way to find the matrix representation using the rotation matrix in
Problem 2.1.1. Another common way to determine the matrix representation is to 18 use these coordinate vectors for each of the vertices, take an arbitrary 2 × 2 matrix
(aij) to represent ρ2((123)) and multiply the coordinate vectors to create a system. 1 1 a11 a12 1 − a11 = − ρ ((123))1 = = √2 =⇒ √2 2 3 3 a21 a22 0 a = 2 21 2
This result forces the matrix to keep the group action of S3. We feed the vector associated with vertex 1 and equate it to the vector 2 in keeping with the action of element (123). We will now to the same feeding vector 2 and equating it with vector
3. √ 1 1 1 3 1 a11 a12 − − − a11 a12 = − ρ ((123))2 = √2 = √2 =⇒ 2 2 √ 2 √ 2 3 3 1 3 3 a21 a22 − − a + a = − 2 2 2 21 2 22 2 Solving this system of four equations and four unknowns the matrix representation is, √ 1 3 − − ρ ((123)) = √2 2 . 2 3 1 − 2 2 The remaining representations can be found in a similar way. They are, √ √ 1 3 1 3 1 0 − − ρ ((1)) = , ρ ((132)) = √2 2 , ρ ((12)) = √2 2 2 2 3 1 2 3 1 0 1 − − 2 2 2 2 √ 1 3 1 0 − − ρ ((23)) = , ρ ((13)) = √2 2 . 2 2 3 1 0 −1 − 2 2 This representation is certainly two-dimensional but is it irreducible? The character of a representation tells whether the representation is irreducible or not. To check that ρ2 is irreducible we will consider hχ2, χ2i. If this inner product is anything other than one we know that the representation is reducible over the complex numbers. In order to evaluate the bilinear form h, i, one must first recall the size of each conjugacy 19 class, |C(123)| = 2, |C(12)| = 3 and the identity is in its own class and the value of the character over each. We then have
1 X hχ2, χ2i = χ2(φ)χ2(φ) |S3| φ∈S3 1 = 1 · χ (1)χ (1) + 2 · χ ((123))χ ((123)) + 3 · χ ((12))χ ((12)) 6 2 2 2 2 2 2 1 = 1 · 2 · 2 + 2 · (−1) · (−1) + 3 · 0 · 0 6 1 = (4 + 2 + 0) = 1. 6
Hence χ2 is the character of an irreducible representation, ρ2 is irreducible. Notice that the trace of the representations for the three cycles are both nega- tive one. This is consistent with our findings using the fact about the sum of the columns weighted by dimensions. The trace of each of the transpositions is zero. The completed character table is then,
(1) (2) (3) C1 C(123) C(12)
χ0 1 1 1
χ1 1 1 -1
χ2 2 -1 0
Table 2.7: Character table for S3
Its good to check your work by summing the characters over a conjugacy class, weighted by their dimension, to find either the order of the group or zero. Recall, you will only get the order of the group when summing the characters over the trivial conjugacy class. Check also that the sum of the rows weighted by the size of the
20 conjugacy class is also zero or the order of the group. Recall again, you will only get the order of the group when you sum over the group using the trivial character.
Let us look at another non-abelian group the dihedral group of order four, D4 = {1, (1234), (13)(24), (1432), (13), (24), (12)(34), (14)(23)}.
Problem 2.2.2. Find the all irreducible representations and characters for D4. Con- struct the character table.
Begin in a similar manner as in the previous problem, find the conjugacy classes of D4. They are,
C1 = {1},C(1234) = {(1234), (1432)},C(13)(24) = {(13)(24)},
C(13) = {(13), (24)},C(12)(34) = {(12)(34), (14)(23)}.
Since there are five conjugacy classes there are also five irreducible representations.
The order of D4 is eight. We should be able to write eight as the sum of five squares according to the fact about the sum of the squares of the dimensions. If one represen- tation is three-dimensional the square is nine, too large, hence the representations are all two dimensional or less. If two representations are two-dimensional then the sum of their squares is eight, this leaves nothing for the other three representations. They all cannot be one dimensional, hence there must be one two-dimensional representation with four one-dimensional representations,
2 2 2 2 2 1 + 1 + 1 + 1 + 2 = 8 = |D4|.
Call the five representations ρ0, ρ1, . . . , ρ4 with their respective characters χ0, χ1,..., χ4 where ρ0 is the trivial representation and ρ4 is the two-dimensional representation.
We know that for S4 there are only two one-dimensional representations, the trivial and the sign representations. The sign representation from S4 can be restricted to the elements of D4, still preserving the homomorphic properties and thus is a represen- tation of D4 and since ρ1(13) = −1 we know ρ0 6= ρ1. At this point we know two of 21 the four one-dimensional representations, to find the remaining two one-dimensional representations takes some work. Geometrically it is a fact that a rotation is a com- position of two reflections and the composition of two rotations can be written as a single rotation, using this fact yields another one dimensional representation. The el- ements of D4 can be thought of as the symmetries of a square, C1,C(1234), and C(13)(24) are rotations while C(13) and C(12)(34) are reflections. For two reflections φ and ψ, the character of their product must be the character of a rotation, χ(φψ) = χ(rotation). If the characters of reflections are −1 then the characters of rotations must be 1, this re- lationship defines another one-dimensional representation ρ2. It is important to check that this gives a character different from ρ1, since ρ1((1234)) = −1 and ρ2((1234)) = 1 we have two unique characters. The last one-dimensional representation, ρ3, is the tensor product of the other two nontrivial one-dimensional representations, ρ1 and
ρ2, hence the product of characters.
The two-dimensional representation, ρ4 of D4 can be realized by placing the four corners of the square 1,2,3, and 4 on the points (1, 0), (0, 1), (−1, 0), and (0, −1) respectively. Again, choose an arbitrary 2 × 2 matrix (aij) to be the representation of a specific group element and then apply the coordinate vectors of each point and set the resulting vector equal to the respective coordinates of the vertex determined by the group action. This process gives a system to determine the arbitrary matrix.
22 ρ4((1234))(1) = (2) ρ4((1234))(2) = (3) a a 1 0 0 a 0 −1 11 12 12 = = a21 a22 0 1 1 a22 1 0 a 0 a −1 11 12 = . = . a21 1 a22 0 0 −1 Therefore ρ4(1234) gives the matrix representation . 1 0
Exercise 4. Find the remaining matrix representations for ρ4. Once you have found the other matrix representations, calculate the trace of each of the matrices to find the character χ4 over the five conjugacy classes. Find hχ4, χ4i to show that ρ4 is irreducible. The values should be what are shown in the partial character Table 2.8.
Recall, χ0 is the trivial character, χ1 is the sign character, χ2 is the reflec- tion/rotation character, and χ4 is the two-dimensional character. We still lack the last one-dimensional representation. We can use the partially completed character table to help us find the remaining character. Note: The remaining character is one-dimensional, the character is the same as the representation.
To find the value of the character over each of these conjugacy classes we will sum the value of the columns weighted by the dimension of the character. This value is the sum of the characters over a non-trivial conjugacy class which is zero. For φ in
C(1234) we have X deg(χi) · χi(φ) = χ0(φ) + χ1(φ) + χ2(φ) + χ3(φ) + χ4(φ)
χi irreducible
= 1 · (1) + 1 · (−1) + 1 · (1) + 1 · χ3(φ) + 2 · (0)
= 1 + χ3(φ).
Hence 1 + χ3(φ) = 0 or χ3(φ) = −1. The value of χ3 over C(1234) is −1. 23 (1) (2) (1) (2) (2) C1 C(1234) C(13)(24) C(13) C(12)(34)
χ0 1 1 1 1 1
χ1 1 -1 1 -1 1
χ2 1 1 1 -1 -1
χ3 1
χ4 2 0 -2 0 0
Table 2.8: Partial character table for D4
Exercise 5. Find the value of χ3 over the the remaining conjugacy classes of D4 using the above outlined method to complete the character table. Are your findings consistent with the fact that ρ3 = ρ1 ⊗ ρ2, in other words χ3 = χ1 · χ2?
The dihedral groups have combinations of the cycle reflection, σ, and cycle rota-
2 n n−1 tion, τ, such that Dn = {σ, τ|σ = 1, τ = 1, and στ = τ σ}. This structure and the order of the of an element is important when determining the various represen- tations for the dihedral group. When considering powers of τ without the reflection,
σ, the representation is the the same as the cyclic groups of order n.
Exercise 6. Find the dimension of all the irreducible representations of D5. What are the characters of the one-dimensional representations?
Problem 2.2.3. Find the character table for
A4 = {1, (123), (124), (132), (142), (143), (243), (234), (12)(34), (13)(24), (14)(23)}.
Geometrically this group is the reflections and rotations of a tetrahedron. First
find the conjugacy classes, they are
C1 = {1} C(123) = {(123), (243), (134), (142)}
C(132) = {(132), (234), (143), (124)} C(12)(34) = {(12)(34), (14)(23), (13)(24)}. 24 This means that there are four irreducible representations, ρ0, ρ1, ρ2, and ρ3. Let
ρ0 be the trivial. The square of their dimensions is equal to |A4| = 12. Hence
2 2 2 2 1 + d1 + d2 + d3 = 12 the only sum of three integer squares that is 11 is 1, 1, and
3. Let ρ1 and ρ2 be the two one-dimensional representations. Notice that inverses of the elements in C(123) are in C(132), each one of these cycles has order three. It will suffice for one-dimensional representations to map each of these elements in C(123) to a cube root of unity. Now for this potential representation to act homomorphically on the group we must check that the elements in C(12)(34) preserve the group relation. Now, (123)(234) = (12)(34) and for a cube root of unity ω and i = 1, 2 we have
2 ρi((12)(34)) = ρi((123) · (234)) = ρi(123) · ρi(234) = ωω = 1.
Exercise 7. Check that the remaining two-two-cycles, (13)(24) and (14)(23), are also homomorphically mapped to 1.
We now have all three one-dimensional representations for ρ0((123)) = 1, ρ1((123)) =
2 ω, and ρ2((123)) = ω . We can now find the character of the three-dimensional representation by summing over the irreducible characters weighted by their dimen- sion. Let ρ3 be the three-dimensional representation and its character χ3. Now for
φ ∈ C(123)
X 0 = deg(χi) · χi(φ) = 1 · χ0(φ) + 1 · χ1(φ) + 1 · χ2(φ) + 3 · χ3(φ)
χiirreducible 2 = 1 · 1 + 1 · ω + 1 · ω + 3 · χ3(φ) √ √ 1 3 1 3 = 1 + (− + i ) + (− − i ) + 3 · χ (φ) 2 2 2 2 3
= 3 · χ3(φ).
Hence we have 3 · χ3(φ) = 0 hence the character is zero over C(123). Similarly, for
ψ ∈ C(132) one will find χ3(ψ) = 0. Finally, for µ ∈ C(12)(34) we find χ3(µ) = −1.
The completed character table for A4 is shown below. The three-dimensional matrix 25 (1) (4) (4) (3) C1 C(123) C(132) C(12)(34)
χ0 1 1 1 1
2 χ1 1 ω ω 1
2 χ2 1 ω ω 1
χ3 3 0 0 -1
Table 2.9: Character table for A4
representation can be a challenge to work out. In Michael Artin’s book, Algebra [1], he states that the logical basis to realize this matrix representation has coordinate axes passing through the midpoint of three of the edges that share a vertex of the tetrahedron. This basis gives a representation that is composed entirely of unit values of the basis elements. Consider vertices 1, 2, 3 and 4 mapped to (1, 1, 1), (−1, 1, −1),
(1, −1, −1), and (−1, −1, 1) with this basis. Use the behavior of the group elements to determine the individual three by three matrices for an element. Below I have completed the work to find the matrix representation for the element ρ3((123)). Let
ρ3((123)) be the 3 × 3 matrix with entries (aij) then, 1 −1 a11 + a12 + a13 = −1 ρ3((123)) 1 = 1 =⇒ a + a + a = 1 21 22 23 1 −1 a31 + a31 + a31 = −1
−1 1 −a11 + a12 − a13 = 1 ρ3((123)) 1 = −1 =⇒ −a + a − a = −1 21 22 23 −1 −1 −a31 + a31 − a31 = −1
26 1 1 a11 − a12 − a13 = 1 ρ3((123)) −1 = 1 =⇒ a − a − a = 1 21 22 23 −1 1 a31 − a31 − a31 = 1. Solving this system of nine equations with nine unknowns we find the matrix rep- 0 0 −1 resentation for the element (123), ρ3((123)) = 1 0 0 . One can check that 0 −1 0 tr(ρ3((123))) = 0, which is consistent with the previously calculated χ3((123)).
Exercise 8. Find the other 11 matrix representations for ρ3 using the outlined method above. Calculate the trace of each matrix to check for consistency with the character table. Check irreducibility with hχ3, χ3i.
Problem 2.2.4. Find the character table for the Hamiltonian Quaternions, Q8 = {±1, ±i, ±j, ±k}
The conjugacy classes for the quaternions are,
C1 = {1}, C−1 = {−1}, Ci = {±i}, Cj = {±j}, Ck = {±k}.
Hence there are five irreducible representations, ρr for 0 ≤ r ≤ 4, with dimensions dr. 4 X 2 We know dr = 8. A little calculation and elimination concludes that there must r=0 be four one-dimensional representations and one two-dimensional representation. Let
ρ0 be the trivial representation and ρ4 be the single two-dimensional representation. There are three remaining one-dimensional representations. Each i, j, and k have
27 order two, the representations must reflect this property. Consider a non-trivial one- dimensional representation ρr for r = 1, 2, 3. We have that,
2 ρr(−1) = ρr(i )
= ρr(i)ρr(i) ←− Representations are homomorphisms
= ρr(jk)ρr(i)
= ρr(j)ρr(k)ρr(i)
= ρr(k)ρr(j)ρr(i) ←− One dimensional representations commute
= ρr(kj)ρr(i)
= ρr(−i)ρr(i)
2 = ρr(−i )
= ρr(1) = 1.
Hence the values of all one-dimensional representations and characters of −1 are 1.
Suppose that ρ1(i) = 1 then
ρ1(j) = ρ1(ki) = ρ1(k)ρ1(i) = ρ1(k).
Since each j and k are of order two the only non-trivial one-dimensional represen- tation here takes the value of −1. Hence ρ1(j) = −1 and ρ1(k) = −1. Similarly, suppose ρ2(j) = 1 then ρ2(i) = −1 and ρ2(k) = −1 yielding a new one-dimensional representation. Again, suppose ρ3(k) = 1 then ρ3(i) = −1 and ρ3(j) = −1. We will check that the purposed ρ1 is a representation, note we only need to check the positive values since the value of −1 is 1 for all one-dimensional representations of Q8.
ρ1(i) = ρ1(jk) = −1 · −1 = 1, ρ1(i) = ρ1(−kj) = 1 · −1 · −1 = 1
ρ1(j) = ρ1(ki) = −1 · 1 = −1, ρ1(j) = ρ1(−ik) = 1 · 1 · −1 = −1
ρ1(k) = ρ1(ij) = 1 · −1 = −1, ρ1(k) = ρ1(−ji) = 1 · −1 · 1 = −1.
28 Exercise 9. Verify that ρ2 and ρ3 are also representations of Q8.
Now we can complete the character table for the quaternions since we only lack the two-dimensional representation. Let the characters of ρr be χr for r = 0, 1, 2, 3, 4
. Then let q ∈ Ci and we know that the one-dimensional characters are the one- dimensional representations so,
X 0 = deg(χr) · χr(q) = 1 · χ0(q) + 1 · χ1(q) + 1 · χ2(q) + 1 · χ3(q) + 2 · χ4(q)
χr
= 1 · 1 + 1 · 1 + 1 · (−1) + 1 · (−1) + 2 · χ4(q)
= 2 · χ4(q).
Hence χ4(q) = 0 for q ∈ Ci. Similarly, χ4(C−1) = −2, χ4(Cj) = 0 and χ4(Ck) = 0. The character table is given in Table 2.10. One may wonder if character tables
(1) (1) (2) (2) (2) C1 C−1 Ci Cj Ck
χ0 1 1 1 1 1
χ1 1 1 1 -1 -1
χ2 1 1 -1 1 -1
χ3 1 1 -1 -1 1
χ4 2 -2 0 0 0
Table 2.10: Character table for Q8
capture all of the information about a specific group. With the tables for Q8, Table
2.10, and D4, Table 2.8, known one observes that these two tables are identical up to column and row changes. Since Q8 and D4 are non-isomorphic groups and they have the same character table we know that these tables do not quite capture everything 29 about the group. Therefore, two non-isomorphic groups can have the same character table.
Even though the character table is complete some may still wonder what that two-dimensional representation looks like. To find the representation first one must be aware that for any q ∈ H, the algebra of Hamilton’s quaternions, one can write,
q = a0 + a1i + a2j − a3k where a0, a1, a2, a3 ∈ R. Now, we can split this form up in a special way to realize a complex representation,
q = (a0 + a1i) + j(a2 + a3i).
Furthermore, we can now find each 2 × 2 matrix representation by multiplying this relation by each of the quaternions,
iq = i(a0 + a1i) + j(−i(a2 + a3i))
= (−a1 + a0i) + j(−a3 − ia2)
2 We can express this as a vector relation in C , i 0 a + a i −a + a i 0 1 1 0 iq = = 0 −i a2 + a3i −a2 − a3i
Similarly, we find 0 −1 a + a i −a − a i 0 1 2 3 jq = = 1 0 a2 + a3i a0 + a1i and 0 −i a + a i −i(a + a i) 0 1 2 3 kq = = . −i 0 a2 + a3i −i(a0 + a1i)
30 Therefore the eight matrix representations are, 1 0 −1 0 ρ4(1) = ρ4(−1) = 0 1 0 −1 i 0 −i 0 ρ4(i) = ρ4(−i) = 0 −i 0 i 0 −1 0 1 ρ4(j) = ρ4(−j) = 1 0 −1 0 0 −i 0 i ρ4(k) = ρ4(−k) = . −i 0 i 0
One can, and should, check that the trace of these matrices matches the respective value of χ4 in the character table.
Exercise 10. Check that the character for each of the above matrix representation matches the Table 2.10. Also, show that ρ4 is an irreducible representation! (Hint:
Use hχ4, χ4i.)
Now that we have some experience with representations and characters let us look at the proof of the properties that we have been using to find the character tables for these groups.
31 CHAPTER 3
PROOFS
In this section of the paper we will give rigorous definitions for the previously defined terms for the purpose proving the results used in Chapter 2. Also, these definitions are to provide the reader with a deeper understanding of the relationship between the defined terms.
Definition 1. A representation ρ of a group G on a vector space V over the field C is a group homomorphism from G to GL(V ), that is
ρ : G → GL(V )
such that ρ(g1g2) = ρ(g1)ρ(g2) for all g1, g2 ∈ G.
Recall that an irreducible representation is a representation that cannot be written as the direct sum of two or more representations. While this is true there is a more formal idea defined here that we avoided, G-invariance. If the representation is understood we will write gv for ρ(g)v.
Definition 2. Let ρ be a representation of a group G on a vector space V . A subspace
W of V is called G-invariant if gw ∈ W , for all w ∈ W and g ∈ G.
Definition 3. If a representation ρ of a group G on a nonzero vector space V has no proper G-invariant subspace, it is called an irreducible representation. If there is a proper G-invariant subspace, ρ is said to be reducible 32 Definition 4. When V is the direct sum of G-invariant subspaces V = W1 ⊕ W2, the representation ρ on V is said to be the direct sum of its restrictions ρi to Wi, and we write ρ = ρ1 ⊕ ρ2.
Theorem 3.0.5. Let ρ be a representation of a finite group G on a complex vector space V . If h, i is any positive definite hermitian form on V then,
1 X hv, wi = hgv, gwi ρ |G| g∈G is hermitian, positive definite and G-invariant form on V , in other words, one can always find a G-invariant form.
Proof. First to show that h, iρ is a hermitian product for v, w ∈ V . To do this we need to show linearity in the second component and conjugate linearity in the first component.
1 X hv, cwi = hgv, gcwi ρ |G| g∈G c X = hgv, gwi |G| g∈G
= chv, wiρ.
1 X hv, w + w i = hgv, g(w + w )i 1 2 ρ |G| 1 2 g∈G 1 X = hgv, gw + gw i |G| 1 2 g∈G 1 X = (hgv, gw i + hgv, gw i) |G| 1 2 g∈G 1 X 1 X = hgv, gw i + hgv, gw i |G| 1 |G| 2 g∈G g∈G
= hv, w1iρ + hv, w2iρ. 33 Therefore h, iρ is linear in the second component. Now to show conjugate linear in the first component. I will leave this for the reader to do as an exercise:
Exercise 11. Show that h, iρ is conjugate linear in the first component.
One can see at this point that the hermitian properties of h, iρ are directly inherited from the hermitian properties of h, i. The last property we need to check is Hermitian symmetry.
1 X hv, wi = hgv, gwi ρ |G| g∈G 1 X = hgw, gvi |G| g∈G 1 X = hgw, gvi |G| g∈G
= hw, viρ.
Let v ∈ V be nonzero. Then
1 X hv, vi = hgv, gvi > 0 ρ |G| g∈G since hv, vi > 0 for all v ∈ V . Hence we have h, iρ is positive definite.
Now we must show that h, iρ is a G-invariant form. To show this one must show that hg0v, g0wiρ = hv, wiρ for all v, w ∈ V and g0 ∈ G. By definition we have,
1 X hg v, g wi = hgg v, gg wi 0 0 ρ |G| 0 0 g∈G 1 X = hg0v, g0wi |G| g0∈G
= hv, wiρ.
Take special notice of the change in group element g to g0. As g ranges over G so does
0 0 gg0. We can choose g = gg0 and the element g ranges over G. Therefore if a vector 34 space V has a positive definite, hermitian form one can always find a G-invariant positive definite, hermitian form on V [1].
Theorem 3.0.6. Let ρ be a representation of G on a hermitian vector space V with a G-invariant form, h, iρ, and let W be a G-invariant subspace. The orthogonal complement W ⊥ is also G-invariant, and ρ is a direct sum of its representations on
W and W ⊥.
Proof. Theorem 3.0.5 and ρ : G → GL(V ) gives the G-invariant inner product h, iρ
⊥ on V . Let W be that G-invariant subspace of V . Consider W = {v ∈ V |hv, wiρ = 0 for all w ∈ W }. From linear algebra we know that V = W ⊕ W ⊥. We want to
⊥ ⊥ ⊥ ⊥ show here that W is also G-invariant. Let w ∈ W , we show hgw , wiρ = 0 for all w ∈ W .
⊥ ⊥ −1 hgw , wiρ = hgw , gg wiρ
⊥ −1 = hw , g wiρ
= 0 since g−1w ∈ W . Hence gw⊥ ∈ W ⊥ and W ⊥ is G-invariant. Therefore every unitary representation on a hermitian vector space with a G-invariant subspace is a direct sum of its representations on the subspace and its complement.
Theorem 3.0.7. Every representation ρ : G → GL(V ) on a vector space V is a direct sum of irreducible representations.
Proof. By Theorem 3.0.5 we can find a G-invariant form on V . We will proceed by induction on the dimension of the vector space V . Case I: dim(V ) = 1 then there are no proper subspaces of V and ρ is an irreducible representation. Case II: dim(V ) = 2 if there are no proper G-invariant subspaces then ρ is irreducible. If there is a proper G-invariant subspace let W1 be that G-invariant subspace where 35 dim(W1) = 1 and W2 is its orthogonal complement with dim(W2) = 1. By Theorem
3.0.6 ρ is therefore the direct sum of its restrictions to W1 and W2, call ρ’s restriction to W1, ρ1 and call ρ’s restriction to W2, ρ2. Then we have ρ = ρ1 ⊕ ρ2. Since W1 and W2 are both one dimensional then by case I we have that ρ1 and ρ2 are both irreducible representations.
Induction: Assume that if dim(V ) ≤ n then ρ is the direct sum of irreducible representations. Let dim(V ) = n + 1. If there is no proper G-invariant subspace then
ρ is of course irreducible. Let W be a proper G-invariant subspace then V = W ⊕W ⊥ where dim(W ) ≤ n and W ⊥ ≤ n. Then ρ restricted to W is a direct sum of irreducible representations and ρ restricted to W ⊥ is a direct sum of irreducible representations.
Therefore ρ is the direct sum of irreducible representations.
Definition 5. Two representations ρ : G → GL(V ) and ρ0 : G → GL(V 0) of a group
G are called isomorphic or equivalent, if there is an isomorphism of the vector spaces
T : V → V 0 which is compatible with the operation of G:
gT (v) = T (gv) or ρ0(g)(T (v)) = T (ρ(g)(v)) for all v in V and g in G.
Definition 6. The character χ of a representation ρ is the map
χ : G → C defined by χ(g) = trace(ρ(g))
Definition 7. The degree of a character χ is defined to be the dimension of the representation ρ, which is defined to be the dimension of the vector space V where,
ρ : G → GL(V )
Theorem 3.0.8. Let χ be the character of a representation ρ of a finite group G on a vector space V . 36 (a) χ(1) is the degree of the character.
(b) χ(g) = χ(hgh−1) for all g and h in G. This shows that the character is constant
over each conjugacy class.
(c) χ(g−1) = χ(g).
(d) If χ0 is the character of another representation ρ0, then the character of the direct
sum ρ ⊕ ρ0 is χ + χ0.
Proof. (a) 1 ∈ G is the identity χ(1) = trace(ρ(1)) = tr(I) = dim V , where I repre-
sents the identity matrix.
(b) Since ρ is a representation it is by definition a homomorphism and ρ(hgh−1) =
ρ(h)ρ(g)ρ(h−1) we have
tr(ρ(hgh−1)) = tr(ρ(h)ρ(g)ρ(h−1))
= tr(ρ(h−1)ρ(h)ρ(g))
= tr(ρ(g)).
This is because cyclically permuting products of matrices preserves the trace.
(c) If the eigenvalues of ρ(g) are λ1, λ2, . . . , λn where n = dim(ρ) then
det(ρ(g) − λI) = p(λ) ← characteristic polynomial
= (λ − λ1)(λ − λ2) ··· (λ − λn)
n n−1 n = λ − (λ1 + λ2 + ··· + λn)λ + ··· + (−1) λ1λ2 ··· λn.
From Linear Algebra we have that
det(ρ(g) − λI) = λn − (tr(ρ(g)))λn−1 + ··· + det(ρ(g)).
37 Therefore the character χρ(g) is the sum of the the eigenvalues of ρ(g). Now
−1 −1 −1 −1 −1 the eigenvalues of ρ(g ) = ρ(g) are therefore λ1 , λ2 , . . . , λn .
Since G is a finite group every element g of G has a finite order. If gr = 1, then
ρ(g) is a matrix of order r, so its eigenvalues λ1, λ2, . . . , λn are therefore roots of −1 χ −1 χ unity implying |λi| = 1 and therefore λi = λi for each i. Hence (g ) = (g) and
−1 −1 −1 −1 χ(g ) = λ1 + λ2 + ··· + λn = λ1 + λ2 + ··· + λn = χ(g).
(d) The trace of a block matrix is the sum of the traces of the blocks.
Definition 8. Let ρ : G → GL(V ) and ρ0 : G → GL(V 0) be two representations. We call a linear transformation T : V → V 0 G-invariant if it is compatible with the two operation of G on V and V 0, that is,
gT (v) = T (gv) or ρ0(g)(T (v)) = T (ρ(g)(v)) for all g in G and v in V .
Consider the commutative diagram below which displays the relationship between T ,
ρ and ρ0, T V V 0
ρ(g) ρ0(g)
V V 0 T
It is also important to note that an isomorphism of representations is a bijective
G-invariant transformation.
Theorem 3.0.9. The kernal and image of a G-invariant linear transformation T :
V → V 0 are G-invariant subspaces of V and V 0 respectively. 38 Proof. We know ker T and imT are subspaces of V and V 0 respectively. Let v ∈ ker T
T (gv) = gT (v) = g · 0 = 0 hence gv ∈ ker T . Also, if v0 ∈ imT , then v0 = T (v) for some v ∈ V and
gv0 = gT (v) = T (gv) hence gv0 ∈ imT . Therefore both ker T and imT are G-invariant subspaces.
Theorem 3.0.10. (Schur’s Lemma) Let ρ and ρ0 be two irreducible representations of G on vector spaces V and V 0. Also, let T : V → V 0 be a G-invariant linear transformation.
(a) Either T is an isomorphism, or else T = 0.
(b) If V = V 0 and ρ = ρ0, then T is a multiplication by a scalar.
Proof. (a) Since ρ is irreducible it has no proper G-invariant subspaces and since
ker T is a G-invariant subspace, ker T = V or else ker T = 0. Case I: If ker T =
V then T (v) = 0 for all v in V , T = 0. Case II: If ker T = 0 then T is injective
and maps V is isomorphically to its image. Then imT is nonzero. Since ρ0 is
also an irreducible representation and imT a G-invariant subspace it must be
that imT = V 0. Therefore T is an isomorphism.
(b) Suppose that V = V 0 with T a linear transformation on V . Since ρ is a complex
representation we may choose λ to be an eigenvalue of T . Then (T − λI) = T1
is another G-invariant transformation. Now the corresponding eigenvector vλ
0 satisfies T1(vλ) = 0 therefore ker T1 is nonzero. Since ρ = ρ is irreducible there
is no proper G-invariant subspace and ker T1 = V , which implies T1 = 0. We now have T − λI = 0 and now T = λI. Hence the transformation T is a
multiplication by a scalar.
39 Proposition 3.0.11. The sum of the character of a one-dimensional representation over a group G is either zero or |G|. χ χ X 0 if 6= 0 χ(g) = g∈G |G| if χ = χ0
Proof. We know that for a one dimensional character χ,
χ(g1g2) = χ(g1)χ(g2).
X If χ 6= χo then there exists h in G such that χ(h) 6= 1. Let S = χ(g) then, g∈G X χ(h)S = χ(h) χ(g) g∈G X = χ(h)χ(g) g∈G X = χ(hg) g∈G X = χ(g0) g0∈G = S
X where g0 = hg and as g ranges over G so does hg. Therefore the sum χ(g) must g∈G X X be zero for χ 6= χ0. If χ = χ0 then, χ0(g) = 1 = |G|. g∈G g∈G Most of the major theorems proved after this point use the result that irreducible characters are orthonormal proved in Theorem 3.0.13. The next Lemma helps us prove this fact.
Lemma 3.0.12. Let ρ and ρ0 be two nonisomorphic irreducible representations of G.
Then X ρ0(g−1)Aρ(g) = 0 g∈G for every matrix A that has compatible size with ρ and ρ0.
40 X 0 −1 Proof. Call this sum B so that B = ρ (g )Aρ(g) and B : Vρ → Vρ0 . I claim that g∈G B is a G-invariant linear transformation. Lets confirm this,
X Bρ(h) = ρ0(g−1)Aρ(g)ρ(h) g∈G X = ρ0(g−1)Aρ(gh) g∈G X = ρ0(h(g0)−1)Aρ(g0) g0∈G X = ρ0(h) ρ0((g0)−1)Aρ(g0) g0∈G = ρ0(h)B.
Hence B is a G-invariant linear transformation and by Theorem 3.0.10 B = 0.
Theorem 3.0.13. Let G be a group of order |G|, let ρ1, ρ2,... represent the distinct isomorphism classes of irreducible representations of G, and let χi be the character of ρi. The characters χi are orthonormal.
The theorem can be stated more plainly with hχi, χji = 0 if i 6= j, and hχi, χii = 1 for each i.
Proof. Let χ and χ0 be two non-isomorphic irreducible characters, corresponding to the representations ρ and ρ0 of G. Using Theorem 3.0.8 (c) we know that
χ(g−1) = χ(g) and we can rewrite the condition for orthogonality as,
1 X hχ0, χi = χ0(g−1)χ(g) = 0 |G| g∈G
Let Eαβ be the matrix whose (i, j) entry is given by i6=α 0 if j6=β (Eαβ)ij = i=α 1 if j=β
41 1 X By Lemma 3.0.12 we have that ρ0(g−1)E ρ(g) = 0. We can find the (i, j) |G| αβ g∈G entry to be
X 0 −1 X 0 −1 0 = (0)ij = ρ (g )Eαβρ(g) ij = ρ (g )iαρ(g)βj. g∈G g∈G Now,
1 X hχ0, χi = χ0(g)χ(g) |G| g∈G 1 X = χ0(g−1)χ(g) |G| g∈G ! ! 1 X X X = ρ0(g−1) ρ(g) |G| ii jj g∈G i j ! 1 X X X = ρ0(g−1) ρ(g) |G| ii jj i j g∈G 1 X X = (0) |G| i j = 0.
If χ0 = χ then we must consider the linear transformation B from Lemma 3.0.12.
Recall that in Schur’s Lemma, Theorem 3.0.10, if ρ = ρ0 the linear transformation
0 is a scalar matrix. So, on one hand B = aId where d = dim ρ = dim ρ , on the other
X 0 −1 hand B = ρ (g )Aρ(g). Now tr B = tr aId = ad. But we can also compute tr B g∈G by using,
X tr B = tr(ρ(g−1)Aρ(g)) g∈G X = tr(Aρ(g)ρ(g−1)) g∈G X = tr(Aρ(1)) g∈G X = tr(A) g∈G = |G| tr(A). 42 X −1 Now consider A = Eij so that B = ρ(g )Eijρ(g) = aId. Then ad = tr B = g∈G |G| tr E |G| tr E so we have a = ij . When i 6= j we have a = 0 but if i = j then ij d |G| a = . Now, consider the inner product of χ with itself, d ! ! 1 X X X hχ, χi = ρ(g−1) ρ(g) |G| ii jj g∈G i j ! 1 X X X = ρ(g−1) ρ(g) |G| ii jj i j g∈G ! 1 X X X = ρ(g−1)E ρ(g) |G| ij ij i j g∈G ! 1 X X = ρ(g−1)E ρ(g) (3.0.1) |G| ii ii i g∈G 1 X = (B ) |G| ii i 1 X = a |G| i 1 X |G| = |G| d i 1 X d = 1 = = 1. d d i In equation 3.0.1 the sum over j is performed and the only nonzero entries are when i = j. Therefore the characters irreducible characters are orthonormal.
Matrix representations are usually very difficult to construct. They often involve geometric interpretations and some amount of intuition when choosing a basis. One matrix representation that is easily found and can be useful is the regular represen- tation.
Definition 9. The regular representation of G denoted ρreg has basis {vg|g ∈ G} and action ρreg(h)vg = vhg. Since the basis is determined by the elements of G then dim(ρreg) = |G|. 43 Proposition 3.0.14. Let G be a group then character of the regular representation evaluated at a non-trivial conjugacy class is zero, evaluated at the trivial conjugacy class its value is |G|. |G| if g = 1 χ ρreg (g) = . 0 if g 6= 1
In Definition 9, the regular representation is defined by assigning each group element to a basis element of the vector space of dimension |G|. If the matrix repre- sentation has a nonzero value on the diagonal then for some g ∈ G we have gvg = vg or g = 1. The only element with a nonzero value on the diagonal of its regular repre- sentation is the identity. The character of the regular representation evaluated at a non-trivial conjugacy class is zero. Since the dimension of the regular representation is |G| the trace of ρreg(1) = |G|.
Definition 10. A complex-valued function φ : G → C which is constant on each conjugacy class of G is called a class function.
It is important to note that the character χ of a representation is a class function.
One might look at a class function in a couple of different ways:
φ : {C | C ⊂ G, where C is a conjugacy class} → C
φ : G → C, such that φ(hgh−1) = φ(g) for all g and h in G. Either way one views a class function it is a mapping from a group to the complex numbers that is constant on each conjugacy class. Theorem 3.0.15 shows just how important conjugacy classes are to representations and their corresponding characters.
Theorem 3.0.15. Let G be a group of order N, let ρ1, ρ2, ... represent the distinct
44 isomorphism classes of irreducible representations of G, and let χi be the correspond- ing character of ρi, then there are finitely many isomorphism classes of irreducible representations, the same number of conjugacy classes of the group G.
Proof. The first task is to realize the class functions are spanned by the irreducible characters. Let C be the space of all class functions and C[G] be the set of conjugacy classes of G with [g] the corresponding conjugacy class of g in G.[g1] = [g2] if there
−1 exists h in G such that g2 = hg1h . Now one can write the space of class functions as,
C = {φ : C[G] → C}.
A basis of C[G] is given by {e[g] | [g] ∈ C[G]} where 1 if [h] = [g] e[g](h) = 0 if [h] 6= [g]
X is the conjugacy class indicator. If φ ∈ C then φ = φ([g])e[g]. We can put an [g]∈C[G] inner product on C such that for φ1 and φ2 in C
1 X hφ , φ i = φ (g)φ (g). 1 2 |G| 1 2 g∈G
For this inner product the irreducible characters are orthonormal. Let χ1 and χ2 be irreducible characters, then 1 if χ1 = χ2 hχ1, χ2i = 0 if χ1 6= χ2
We want to show that C = Cχ1+···+Cχr. Let X = span(χ1,..., χr) so that X ⊂ C as a vector space. Since we have an inner product we can write C = X ⊕ X ⊥. We now want to show that X ⊥ = 0 and then we have the desired result. Let φ ∈ X ⊥, it is enough to show that φ = 0. Let ρ be any matrix representation of G on the vector
45 space of ρ which we will call Vρ. The character of ρ is in the span of X , χρ ∈ X .
⊥ Hence we have hφ, χρi = 0 since φ ∈ X . Form the linear transformation, T , on Vρ. 1 X T : V → V defined by T = φ(g)ρ(g). Now, ρ ρ |G| g∈G
1 X tr(T ) = φ(g) tr(ρ(g)) |G| g∈G 1 X = φ(g)χ (g) |G| ρ g∈G
= hφ, χρi
= 0.
We want to show that T and ρ commute. In other words T ρ(h) = ρ(h)T for all h ∈ G, but this is if and only if T = ρ(h−1)T ρ(h) for all h ∈ G. So now consider
1 X ρ(h−1)T ρ(h) = φ(g)ρ(h−1)ρ(g)ρ(h) |G| g∈G 1 X = φ(g)ρ(h−1gh). |G| g∈G
Let h−1gh = g0 then we have g = hg0h−1 and
1 X ρ(h−1)T ρ(h) = φ(g)ρ(h−1gh) |G| g∈G 1 X = φ(hg0h−1)ρ(g0) |G| g0∈G 1 X = φ(g0)ρ(g0) |G| g0∈G = T.
This result is true since φ is a class function and is therefore constant over conjugacy classes. Therefore we have tr(T ) = 0 and T ρ = ρT for all representations ρ.
Case I: Suppose ρ is irreducible. By Theorem 3.0.10, Schur’s lemma, if T ρ = ρT
46 then T = cIVρ . In plain terms, if a linear transformation commutes with an irreducible representation then the transformation is a scalar matrix. c 0 ··· 0 0 c ··· 0 T = . . . . . . .. . 0 0 ··· c But we know tr(T ) = 0. Therefore tr(T ) = c dim(ρ) implies c = 0 so T = 0, for ρ irreducible.
Case II: Any ρ is the direct sum of irreducible representations. We consider the case of the sum of two irreducible representations, but the same argument works for any number of irreducible summands. Hence ρ = ρ1 ⊕ ρ2 and,
ρ1(g) 0 ρ(g) = . 0 ρ2(g) Now again consider the linear transformation, T , for ρ,
1 X T = φ(g)ρ(g) |G| g∈G 1 X ρ1(g) = φ(g) 0 |G| g∈G 0 ρ2(g) 1 X φ(g)ρ (g) |G| 1 0 g∈G = 1 X φ(g)ρ (g) 0 |G| 2 g∈G T ρ1 0 = . T 0 ρ2
From case I we know that for irreducible ρi, Tρi = 0. Then we have, 0 Tρ 0 1 0 .. T = = . , T . 0 ρ2 0 0 47 since each Tρi = 0. Therefore Tρ = 0 for all representations ρ. Consider Tρreg , by case
II we have that Tρreg = 0. Now,
1 X 0 = T v = φ(g)ρ (g)v ρreg 1 |G| reg 1 g∈G 1 X = φ(g)v . |G| g g∈G
The only way for this to be zero is for φ(g) = 0. This result is so because the set
{vg|g ∈ G} is an orthonormal basis. Hence φ(g) = 0 for all g in G so φ(g) = 0 and
X ⊥ = 0. Therefore C = X ⊕ X ⊥ = X . The number of irreducible characters is equal to the number of conjugacy classes of G.
Theorem 3.0.16. For the regular representation ρreg = d1ρ1 ⊕ · · · ⊕ drρr, where di is the dimension of ρi and diρi is the direct sum of di ρi’s. Also, χreg = d1χ1 +···+drχr.
The statement of the theorem is that ρreg = ⊕ dim(ρ)ρ, and irreducible
χ X χ ρreg = (dim ρ) ρ. irreducible We can use the fact that the dimension of the regular representation is the order of the group G to discover an important relationship between the order of the group and the dimension of the irreducible representations.
χ X χ X 2 |G| = ρreg (1) = (dim ρ) ρ(1) = (dim(ρ)) irreducible irreducible One can see this is a powerful statement about the irreducible representations of a group. This fact is used many times in finding the irreducible representations in the examples of this paper. Now to begin the proof.
χ Proof. We have that the character of the regular representation ρreg is a class func- χ tion and that ρi for irreducibles ρi are an orthonormal basis for the space of class χ functions. Therefore we may write ρreg as a linear combination of the irreducible 48 χ X χ N characters, ρreg = ai ρi , for ai ∈ . By Theorem 3.0.13 the irreducible char- χ χ acters are orthogonal and we have h ρi , ρreg i = ai. Now by Theorem 3.0.7 we now have that ρreg = ⊕ aiρi, this is because the trace of a block matrix is equal irreducible ρi to the sum of the traces of the individual matrices. Using the inner product on the space of characters we can find the coefficient ai,
1 X a = hχ , χ i = χ (g)χ (g) i ρi ρreg |G| ρi ρreg g∈G 1 = χ (1)|G| |G| ρi χ = ρi
= dim(ρi).
Hence ρi occurs dim(ρi) times in ρreg and ρreg = ⊕ dim(ρi)ρi. irreducible ρi In Chapter 2 we used the fact that the sum of the irreducible characters over a non-trivial conjugacy class weighted by their dimension is zero to find unknown characters. At this point we have all of the tools necessary to arrive at this fact.
Theorem 3.0.17. Let ρ0, ρ2, . . . , ρr be the irreducible representations of a group G and χ0, χ1,..., χr their respective characters. Then for g ∈ G, r X |G| if g = 1 deg χi · χi(g) = i=1 0 if g 6= 1
r χ X χ Proof. By Theorem 3.0.16 we know that ρreg = ⊕ dim ρiρi and ρreg (g) = deg i· i=1 χ χ i(g). Consider the case where g = 1. Again in Theorem 3.0.16 we saw ρreg (1) = |G| hence we arrive at the desired result,
r X deg χi · χi(1) = |G|. i=1 Next consider the case where g 6= 1. By Proposition 3.0.14 the regular matrix representation has a nonzero value on the diagonal then the element must fix itself. 49 The only element with a nonzero value on the diagonal of its regular representation is χ the identity. Hence for g 6= 1 we have ρreg (g) = tr(ρreg(g)) = 0. Therefore for g 6= 1,
r X χ χ χ deg i · i(g) = ρreg (g) = 0. i=1
Lastly, we need to show that the sum of a non-trivial character over a group is equal to zero. This fact uses the orthogonality of the characters.
Theorem 3.0.18. Let ρ be an irreducible representation of a group G and χ its character. Then χ χ X |G| if = 0 χ(g) = . g∈G 0 if χ 6= χ0
Proof. By Theorem 3.0.13 we have that hχi, χji = 0 if i 6= j, and hχi, χii = 1 for two irreducible characters χi and χj of G. We may fix χi to be χ0 and χj to be χ. The result then becomes, χ χ 1 X 1 X 1 if = 0 hχ , χi = χ (g)χ(g) = χ(g) = . 0 |G| 0 |G| g∈G g∈G 0 if χ 6= χ0
Multiplying by |G| yields the desired result, χ χ X |G| if = 0 χ(g) = . g∈G 0 if χ 6= χ0
Theorem 3.0.19. A representation ρ is irreducible if and only if hχρ, χρi = 1.
Proof. Assume ρ is irreducible, by Theorem 3.0.13 hχρ, χρi = 1.
50 Assume hχρ, χρi = 1. By Theorem 3.0.7 ρ can be written as a direct sum X of irreducible representations ρi, so that ρ = aiρi, where ai is the number of i χ X χ occurrences of the irreducible representation ρi. Then ρ = ai ρi . Therefore, i
1 = hχρ, χρi 1 X X X = a χ (g) a χ (g) |G| i ρi j ρj g∈G i j X X 1 X = a a χ (g)χ (g) i j |G| ρi ρj i j g∈G X 2 = ai . i
Therefore it must be that i = 1 and a1 = 1 and ρ is irreducible.
51 BIBLIOGRAPHY
[1] Artin, M. (1991). Algebra. Englewood Cliffs, New Jersey: Prentice Hall.
[2] Curtis, C. W. (1992). Representation Theory of Finite Groups: from Frobenius to Brauer. In R. Wilson & J. Gray (Eds.), Mathematical Conversations (Vol. 14, pp. 128-141). New York: Springer-Verlag.
[3] Curtis, C. W. (1999). Pioneers of Representation Theory: Frobenius, Burnside, Schur, and Brauer. American Mathematical Society.
[4] Isaacs, I. M. (1976). Character Theory of Finite Groups. New York: Academic Press.
[5] Conway, J. H., Curtis, R. T., Norton, S. P., Parker, R. A., Wilson, R. A., & Thackray, J. G. (1985). Atlas of Finite Groups: Maximal Subgroups and Ordinary Characters for Simple Groups. In New York: Oxford University Press.
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