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REPRESENTATIONS OF THE SYMMETRIC GROUP

DAPHNE KAO

Abstract.I will introduce the topic of representation theory of nite groups by investigating representations ofS3andS4using character theory. Then I will generalize these examples by describing all irreducible representations of any symmetric group onnletters. Finally, I will brie y discuss how to dis- cover irreducible representations of any group using Schur Functors, which are constructed using the irreducible representations ofSn. This paper assumes familiarity with group theory,FG-modules, linear algebra, and category the- ory.

Contents

1. Introduction 1

2. Basic Denitions and Complete Reducibility 2

3. Characters and Hands-on Examples Involving Them 4

4. Representations ofSn9

5. Schur Functors 13

Acknowledgments 14

References 14

1.Introduction

Arepresentationof a groupGis a homomorphismfromGinto the general linear group of some vector spaceV. When such a map exists, if we writegv rather than(g)(v) to denote the image of a vectorvinVunder the automorphism (g), then we can think ofVas aCG-module, since for any vectorsvandw, any group elementsgandh, and any scalarc, the following properties hold: (1)(g)(v) =gvis inV. (2)(1) =I2GL(V) by the homomorphism property, so(1)(v) =I(v) =v, which means 1v=v. (3) ((g)(h))(v) =(g)((h)(v)), so (gh)v=g(hv). (4)(g)(cv) =c(g)(v) by linearity, sogcv=c(gv). (5)(g)(v+w) =(g)(v) +(g)(w) by linearity, sog(v+w) =gv+gw. When we know the particularCG-module structure ofV, we often callVitself a representation ofG. A subspaceWofVis asubrepresentationif it is invariant under the action ofG. A representationVofGisirreducibleif it has no proper nontrivial subrepresentations. Otherwise, it isreducible. As we will soon see, every representation of a nite group over a nite-dimensional complex vector space can

be expressed as a direct sum of irreducible representations (over nite-dimensionalDate: September 1, 2010.

1

2 DAPHNE KAO

complex vector spaces) of the group. Two main goals of representation theory are to describe all irreducible representations of a given groupGand to nd a sure-re method of decomposing any arbitrary representation ofGinto a direct sum of irreducible representations. We will accomplish both of these goals for the symmetric group onnlettersSn.

2.Basic Definitions and Complete Reducibility

We begin by describing some common representations and a few ways of con- structing new representations from known ones. Assume that all groups mentioned in this paper (except in Section 5) are nite and that all vector spaces mentioned are nite-dimensional overC. Denition 2.1.Thetrivial representationof a nite groupGisCequipped with the trivial action ofG:gx=xfor everyxinCand for everyginG. Note that every nite group has the trivial representation, and sinceChas no proper nontrivial subspaces, it is irreducible, as is any one-dimensional representation. Denition 2.2.LetXbe any niteG-set. LetWbe the vector space generated by the basisfexjx2Xg. Dene the action ofGonWby g(a1ex1+a2ex2+:::+amexm) =a1egx1+a2egx2+:::+amegxm: Such aWis called apermutation representationofG. Notice that the subspace spanned by the vectorex1+ex2++exmis invariant under the action ofG because each element ofGsimply \shues" the addends but does not change the sum. Thus, every permutation representation has a nontrivial subrepresentation and is therefore reducible. Denition 2.3.In Denition 2.2, replace the arbitraryG-setXbyGitself under the action of left multiplication. In this case,Wis called theregular representation of G. Denition 2.4.For a symmetric groupSn, thealternating representationisC equipped with the action v=( v;ifis an even permutation v;ifis an odd permutation or equivalently,() = sgn()Ifor everyinSn. Note that anySnwheren2 has the alternating representation, and since this representation is one-dimensional, it is irreducible. Denition 2.5.For anyn, letfe1;e2;:::engbe the standard basis forCn. Dene the action ofSnonCnto be (a1e1+a2e2++anen) =a1e(1)+a2e(2)++ane(n): This is a permutation representation ofSn. Again, notice that the one-dimensional subspace ofCspanned bye1+e2++enis invariant under the action ofSn. Therefore,he1+e2++eniis a subrepresentation ofCn. Its orthogonal comple- mentV=f(x1;x2;:::;xn)jx1+x2++xn= 0gis also invariant and therefore a subrepresentation.Vis called thestandard representationofSn.

REPRESENTATIONS OF THE SYMMETRIC GROUP 3

Remark2.6.It can be shown that given any representationVof a groupG, its dual space, its symmetric powers, and its alternating powers are also representations of G. Furthermore, given two representationsVandWofG, their tensor product and direct sum are also representations, as is the vector space Hom(V,W). In Section

5, we will see how we can construct new representations (of any group) from old

using special functors. The following results show that the irreducible representations of a given group Gare the \building blocks" for all of its other representations. Lemma 2.7.Any representationVof a nite groupGcan be given aG-invariant inner product, meaning that for anyhinGand for anyv1;v2inV,hhv1;hv2i= hv1;v2i. Proof.Leth;ibe any positive-denite Hermitian inner product on V. Dene a new Hermitian inner product on V in the following way: hv1;v2i=1jGjX g2Ghgv1;gv2ifor anyv1;v22V: Now see that for anyhinGand for anyv1;v2inV, we have: hhv1;hv2i=1jGjX g2Ghg(hv1);g(hv2)i 1jGjX g2Gh(gh)v1;(gh)v2i 1jGjX g2Ghgv1;gv2ibecause for everyg2G,ghis inGalso =hv1;v2i: Lemma 2.8.LetVbe a representation of a nite groupG, and letWbe a sub- representation ofV. ThenW?, the orthogonal complement toWinsideVunder theG-invariant inner product described above, is also a subrepresentation ofV. Proof.Fix anxinW?=fx2Vj hw;xi= 0 for everyw2Wg:Because our inner product isG-invariant,hw;gxi=hw;xi= 0 for anyginGand for anywin W. Therefore, for anyxinW?,gxis inW?also. This shows thatW?is invariant under the action ofG. Exercise 2.9.Using Lemma 2.8, show that any representation of a nite groupG is the direct sum of irreducible representations. Denition 2.10.LetVandWbe representations of a nite groupG. A map betweenVandWas representations is a vector space map:V!Wsuch that g(v) =(gv) for everyvinVand for everyginG. Proposition 2.11.The kernel ofis a subrepresentation ofV, and the image of is a subrepresentation ofW. Proof.Fix aginG. Sinceis a map between representations, for anyxinVwe have(gx) =g((x)) =g(0) =(g)(0) = 0 by linearity of the transformation (g). Thus, keris invariant under the action ofG.

4 DAPHNE KAO

Again, x aginG. Letzbe any vector in the image of. Thengz=g(x) for somexinV. And sinceis a map between representations, we haveg(x) = (gx), which is in the image since for everyxinV,gxis also inV. Therefore, imis also invariant under the action ofG. Lemma 2.12.(Schur's Lemma)LetXandYbe two distinct, irreducible rep- resentations ofG, and let:X!Ybe a map of representations between them.

Thenis either an isomorphism or the zero map.

Proof.By Proposition 2.11, kerand imare subrepresentations ofXandY, respectively. SinceXis irreducible, it can have no proper nontrivial subrepresenta- tions. Therefore, keris eitherf0gor all ofX. If ker=f0g, thenis one-to-one. SinceYis also irreducible, again, either im=f0gor im=Y. If the image is all ofY, thenis onto, making it an isomorphism. If the image isf0g, thenis the zero map. If, on the other hand, ker=X, thensends every vector inXto 0, making it the zero map. Theorem 2.13.Any representationVof a nite groupGcan be written uniquely as a direct sum of the form

V=Va11Va22 Vakk;

where theVi's are distinct irreducible representations ofVand the multiplicityai of eachViis unique.

Proof.SupposeVcan also be expressed as

V=Wb11Wb22 Wbmm:

Let:V!Vbe the identity map. Then by Schur's Lemma, restricted to each irreducible componentVaii,is an isomorphism betweenVaiiand the component W bj jfor whichViis isomorphic toWj.

3.Characters and Hands-on Examples Involving Them

This section introduces characters and how they can be used to nd irreducible representations and decompositions of arbitrary representations. Denition 3.1.LetVbe a representation of a nite groupG, and letbe the associated group homomorphism. ThecharacterVofVis the function fromG intoCgiven by

V(g) = Tr((g))

Remark3.2.Since conjugate matrices have the same trace, for a xed elementg inG, Tr((g)) = Tr((h)(g)(h1)) = Tr((hgh1)) for anyhinG. This follows from the linearity of. Therefore, the character of any representationVofGis, in fact, aclass function, which is a function that is constant on conjugacy classes. It turns out that computing the characters of permutation representations is very easy if we use the next theorem: Theorem 3.3.(The Fixed Point Formula)LetGbe a nite group, and let Xbe a niteG-set. LetVbe the associated permutation representation ofGas described in Denition 2.2. Then for every elementginG,V(g)is the number of elements inXleft xed by the action ofg.

REPRESENTATIONS OF THE SYMMETRIC GROUP 5

Proof.The matrixMassociated with the action ofgis a permutation matrix. For example, suppose thatXhas four elementsx1;x2;x3, andx4and that(g) permutes the basis vectors ofVby sendingex1toex3,ex2to itself,ex3toex1, and e x4to itself. ThenMis the 44 matrix2 6

640 0 1 0

0 1 0 0

1 0 0 0

0 0 0 13

7 75:
In general, ifgexi=exjfor somexiinX, thenMwill have a 1 in thei-th column andj-th row, and zeroes in all other entries of that column. In particular, if gx i=xi, thengexi=egxi=exi, meaning thatMhas a 1 in thei-th row andi-th column. Therefore, the trace ofMis the number of 1's along the diagonal, which is exactly the number of points left xed byg. Proposition 3.4.For any representationsVandWof a groupG,VW=

V+WandV

W=VW. Proof.This is left as an exercise, but it simply involves counting the number of eigenvalues of any transformation(g). The set of class functions on a nite groupGis a vector space. We can dene the following inner product on this space by: h;i=1jGjX g2G(g)(g) Theorem 3.5 and Corollaries 3.6 through 3.9 below allow us to decompose any representation of a nite group by applying this inner product to characters. Proofs can be found in [3] on pages 16, 17, and 22 and in [4] on pages 137 through 144. Theorem 3.5.The set of character functions of the irreducible representations of Gis orthonormal with respect to this inner product. Corollary 3.6.Any representation ofGis determined by its character. Corollary 3.7.A representationVofGis irreducible if and only ifhV;Vi= 1. Corollary 3.8.LetWbe any representation ofG. As we know,Wcan be written uniquely as a direct sum of irreducible representations in the form

W=Va11Va22 Vakk:

In this decomposition,

ai=hW;Viifor everyi, hW;Wi=Pk i=1a2i, Pk i=1(dim(Vi))2=jGj, and for anyginGthat is not the identity,Pk i=1(dimVi)Vi(g) = 0. Corollary 3.9.The number of irreducible representations of a nite groupGis equal to the number of conjugacy classes inG. Example 3.10.Let's see what these results can tell us about the groupS3. We know right away from Corollary 3.9 thatS3has exactly three irreducible representa- tions. This is because in a symmetric group, each equivalence class of permutations

6 DAPHNE KAO

of a certain cycle type constitutes a conjugacy class, andS3has three cycle-types; namely, the identity, transpositions, and 3-cycles. The trivial and alternating rep- resentations (let's denote them byUandU0, respectively) ofS3are irreducible because they are one-dimensional. Since in the trivial representation(g) is the

11 identity matrix [1] for everyg, we know thatU(g) = Tr[1] = 1 for everyg. In

the alternating representation, we haveU0() = Tr[1] = 1,U0(12) = Tr[1] =1, andU0(123) = Tr[1] = 1. There now remains only one more irreducible represen- tation for us to nd. It cannot be any sort of permutation representation, because those are all reducible. Thus, it is worth checking the standard representation V=f(x1;x2;x3)jx1+x2+x3= 0g. By denition,C3=he1+e2+e3i V. Note thathe1+e2+e3iunder the permutation action ofS3is isomorphic toC under the trivial action ofS3, so it is the trivial representationU. SinceS3acts onC3by permuting the three standard basis vectors, by the Fixed-Point Formula we haveC3() = 3,C3(12) = 1, andC3(123) = 0. Also, becauseC3=UV, by Proposition 3.4 we haveC3=U+V. Therefore,C3has values 31 = 2,

11 = 0, and 01 =1 on the conjugacy classes of, (12) and (123), respectively.

Notice that

hC3;C3i=1jS3j(221 + 023 + (1)22) =16

6 = 1:

Therefore, by Corollary 3.7,Vis irreducible. We found it! We can summarize the work we did into the following simple array called thecharacter tableforS3:

Conjugacy Class Sizes1 3 2

Representatives(12) (123)Trivial RepresentationU1 1 1

Alternating RepresentationU01 -1 1

Standard RepresentationV2 0 -1

Remark3.11.Notice that for any representationWof a nite groupG,W(1) = Tr((1)) = Tr(I). This is simply the number of columns ofI, which is the dimension ofW. Thus, the rst column of the character table above shows the dimension of each irreducible representation. We can use this and other handy information from the table to decompose any other representation ofS3. Example 3.12.SinceS3acts on the vertices of the equilateral triangle by per- muting its three diagonals, there is a permutation representation associated with the resulting action on the set of vertices. If we label our triangle as shown, A

AAAAAAAA

A CB Q

QQQQQQQ3 1

2 then we see that the transposition (12) acts on the triangle by interchanging diag- onals 1 and 2. This is a re ection about diagonal 3, and it interchanges vertices

REPRESENTATIONS OF THE SYMMETRIC GROUP 7

A and B but leaves vertex C xed. Similarly, the other transpositions inS3each leave exactly one vertex xed. The element (123) rotates the triangle by 120 de- grees clockwise and therefore leaves no vertices xed. Finally, the identity element leaves all three vertices xed. By the Fixed Point Formula, then, the character of this representation has values 3, 1, and 0 on the conjugacy classes consisting of the identity, transpositions, and 3-cycles, respectively. Therefore, we have hW;Wi=1jS3j(321 + 123 + 022) =16 (9 + 3 + 0) =16

12 = 2:

By Corollary 3.8,hW;Wiis the sum of the squares of the multiplicities of the irreducible representations in the decomposition ofW. Since 2 = 12+ 12, and this is the only way to write 2 as the sum of squares of nonnegative integers, it must be thatWis the direct sum of exactly two distinct irreducible representations of S

3. To nd out which two, we simply take the product ofWwith the character

of each irreducible representation as follows: hW;Ui=16 (311 + 113 + 012) = 1; hW;U0i=16 (311 + 1(1)3 + 012) = 0; hW;Vi=16 (321 + 103 + 0(1)2) = 1 By Corollary 3.8, the trivial representationUappears in the decomposition ofW once, the standard representationVappears once, and the alternating representa- tionU0does not appear at all. Therefore, we have the decompositionW=UV. Example 3.13.As mentioned before, any tensor power of a known representa- tion of a group is again a representation. Therefore, then-th tensor power of the standard representationVofS3is another representation ofS3. To nd its decom- position, rst note that Proposition 3.4 impliesV n= (V)n. Therefore, we have V n() = 2n;V n((12)) = 0n= 0, andV n((123)) = (1)n. Take the product ofV nwith the character of each irreducible: hV n;Ui=16 (2n11+013+(1)n12) =16 (2n+2(1)n) =13 (2n1+(1)n); hV n;U0i=16 (2n11+0(1)3+(1)n12) =16 (2n+2(1)n) =13 (2n1+(1)n); hV n;Vi=16 (2n21+003+(1)n(1)2) =16 (2n+1+2(1)n+1) =13 (2n+(1)n+1):

This gives us the decomposition

V n=U13 (2n1+(1)n)U013 (2n1+(1)n)V13 (2n+(1)n+1): Example 3.14.Now let's compute the character table ofS4. Since there are ve cycle-types (i.e., conjugacy classes) inS4, we know right away thatS4has exactly ve irreducible representations. The trivial and alternating representationsUand U

0are irreducible, and we know their characters. The next easy candidate to check

is the standard representationV. It is left as an exercise to show thatVand the representationV0=V Uare both irreducible. (You will need Proposition 3.4 and Corollary 3.7.) It remains now to nd the character values of the nal irreducible representation, which we will callW. Leta;b;c;d, andedenote the values ofWon

8 DAPHNE KAO

the conjugacy classes of;(12);(123);(1234) and (12)(34), respectively. Corollary

3.8 and the entries in the rst column that we know so far give us

5 X i=1(dim(Vi))2=jS4j=)12+ 12+ 32+ 32+a2= 24 =)a2= 4 =)a= 2: Now we can use the last part of Corollary 3.8 to computeb: 5 X i=1(dimVi)Vi(g) = 0 =)11+(1)1+13+(1)3+b2 = 0 =)2b= 0 =)b= 0: We can computec;dandein the same way, giving us the complete character table:1 6 8 6 3 S

4(12) (123) (1234) (12)(34)U1 1 1 1 1

U

01 -1 1 -1 1

V3 1 0 -1 -1

V

03 -1 0 1 -1

W2 0 -1 0 2

Although for now the only thing we know aboutWis its character, in Section 4 we will learn the tools for explicitly describing the vector spaceWitself. Example 3.15.S4acts on the faces, vertices and edges of the cube by permuting the four long diagonals inside the cube and thus has a permutation representation associated with each action. We will decompose each representation. Label the cube as shown:P

PPPPPPPPPP

PPPPPPPPPa

bZ

ZZZZZZZZZZZA

A AA AA z 1 23
4 Faces:Call this permutation representationX. The identity permutation does not move the cube at all, leaving the 6 faces xed. The transposition (12) rotates the cube 180 degrees about the axis connecting the midpoints of edgesaandb. This motion moves every face to a new location and therefore xes no faces. The 3-cycle (123) rotates the cube 120 degrees about long diagonal number 4 and also xes no faces. The 4-cycle (1234) rotates the cube 90 degrees about the axisz, leaving the top and bottom faces xed but moving the other four around. Finally, (12)(34) rotates the cube 180 degress aboutz, again leaving only the top and bottom faces xed. Therefore, by the Fixed Point Formula,Xhas the values 6, 0, 0, 2, 2 on the conjugacy classes of, (12), (123), (1234), and (12)(34), respectively. This gives us hX;Xi=1jS4j(621+026+028+226+223) =124 (36+24+12) =124

72 = 3:

REPRESENTATIONS OF THE SYMMETRIC GROUP 9

Since the only way to express 3 as the sum of squares is 3 = 1

2+12+12, we know

thatXis the direct sum of exactly three irreducible representations. See that hX;Ui=1jS4j(611+016+018+216+213) =124 (6+12+6) =124

24 = 1;

hX;U0i=1jS4j(611+0(1)6+018+2(1)6+213) =124 (612+6) =124

0 = 0;

hX;Vi=1jS4j(631+016+008+2(1)6+2(1)3) =124 (18126) =124

0 = 0;

hX;V0i=1jS4j(631+0(1)6+008+216+2(1)3) =124 (18+126) =124

24 = 1:

At this point, we know thatXmust contain a copy ofW. Therefore, we have the decompositionX=UV0W. Vertices:Call this permutation representationY. The identity permutation xes all 8 vertices. Transpositions x no vertices. Each 3-cycle xes 2 vertices. Each 4-cycle xes no vertices. Finally, elements in the conjugacy class of (12)(34) do not x any vertices, either. Therefore,Yhas the values 8, 0, 2, 0, 0 on the respective conjugacy classes, which gives us hY;Yi=1jS4j(821+026+228+026+023) =124 (64+32) =124

96 = 4:

Since the possible ways of writing 4 as the sum of squares of nonnegative integers are 4 = 2

2and 4 = 12+ 12+ 12+ 12,Yis either the direct sum of four irreducible

representations or the direct sum of two copies of one irreducible representation.

To nd out which case it is, we simply compute:

hY;Ui=1jS4j(811+016+218+016+013) =124 (8+16) =124

24 = 1;

hY;U0i=1jS4j(811+0(1)6+218+0(1)6+013) =124 (8+16) =124

24 = 1;

hY;Vi=1jS4j(831+016+208+0(1)6+0(1)3) =124

24 = 1;

hY;V0i=1jS4j(831+0(1)6+208+016+0(1)3) =124

24 = 1:

Again, we don't need to computehY;Wi. We already know the decomposition:

Y=UU0VV0.

Edges:Call this permutation representationZ. By following the same process as described above, we can nd thatZhas the values 12, 2, 0, 0, 0 on the respective conjugacy classes, and thereforeZdecomposes asZ=UV2V0W.

4.Representations ofSn

Recall thatSnhas exactly as many irreducible representations as it does con- jugacy classes, and each conjugacy class is a cycle-type equivalence class. Now, there is a one-to-one correspondence between the set of cycle-types and the ways to writenas the sum of positive integers. For example,S4has the following ve cycle-types:

10 DAPHNE KAO

Permutation of the FormCan Also be Written AsCorresponds to the Sum (1)(2)(3)(4) 1 + 1 + 1 + 1 (12) (12)(3)(4) 2 + 1+ 1 (123) (123)(4) 3 + 1 (12)(34) (12)(34) 2 + 2 (1234) (1234) 4 A fundamental tool for studying the representations ofSnis theYoung Diagram, which is an array of boxes: for a given partitionn=1+2++k(which we denote by= (1;2;:::;k)) whereii+1for every 1ik, we draw a row of1boxes. Beneath it, we draw a row of2boxes, and so on so that the last row containskboxes and each row is as long as or shorter than the one above it.

Thus, the Young Diagrams corresponding to the partitions of 4 are:Now that we see how each conjugacy class ofSncorresponds to a Young Diagram,

let's look at an algorithm that actually generates all of the irreducible representa- tions ofSn. (For a proof of why this method works, see Section 7.2 of [2] and Section 4.2 of [3].) First, let's pick a particular partition= (1;2;:::;k) of nand label the boxes of the corresponding Young Diagram in order, from left to right and from the top down, with the integers 1 throughn. For example, if we were working inS6and had chosen to nd the irreducible representation associated with the partition= (3;2;1), then we would number our boxes like this: 1= 3 2= 2

3= 11 2 3

4 5 6 Dene the following two sets, which are, in fact, subgroups ofSn: P =f2Snjpreserves the set of numbers in each rowgand Q =f2Snjpreserves the set of numbers in each columng: For our particular numbered diagram above, we have: P Q

Before we proceed, we need:

Denition 4.1.LetFbe any eld, and letGbe a group. Thegroup algebraof GoverFis the vector space overFgenerated by the basisfexjx2Ggwith

REPRESENTATIONS OF THE SYMMETRIC GROUP 11

multiplication dened by X x2Gc xx! 0 X y2Gd yy1 A =X x;y2Gcquotesdbs_dbs15.pdfusesText_21

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