Tensor products of vector spaces

8/11/2019 Tensor products of vector spaces

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On the application of tensor products of vector spaces to the

representation theory of finite groups

R. A. Hayden

[email protected]

1 The tensor product of vector spaces

1.1 Multilinear maps, definition and basic properties

Let V and Wbe finite-dimensional vector spaces over some field K. If the function : V

W

satisfies

(v1+ v2) = (v1) + (v2)

for all v1, v2 V and , K, then we term a linear map.If we fix bases of V and W, elementary linear algebra lets us represent every such linear mapuniquely as an n m matrix (where n = dimK(W) and m = dimK(V)). Furthermore, all suchmatrices uniquely determine a linear map. That is, linear maps from V to W over K are inone-to-one correspondence with n m matrices over K.We now proceed to generalise the notion of a linear map to allow for functions of several variables,which are linear in each variable separately. Motivated by the special case of linear maps, naturalquestions will be how all such multilinear maps can be determined and how one might expressthem. The method employed is to convert multilinear maps into linear maps on a different vectorspace, the tensor productof the constituent vector spaces.

Definition 1.1 (Multilinear map) SupposeV1, . . . , V kandWare vector spaces over some fieldK. A function f : V1 . . . Vk W is called multilinear if it is linear in each of its variablesseparately, that is if fori = 1, . . . , k,

f(v1, . . . , vi1, vi+ v

i, vi+1, . . . , vk) = f(v1, . . . , vi1, vi, vi+1, . . . , vk)

+ f(v1, . . . , vi1, v

i, vi+1, . . . , vk)

for alla, b K, vj Vj wherej = 1, . . . , k andv

i Vi.The first step is to construct the (rather large) K-vector space, which we will call F(V1, . . . , V k),whose basis consists formally of the elements ofV1 . . . Vk. The key important property of thisvector space is given by the following lemma.

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g[(v1, . . . , vi1, vi+ v

i, vi+1, . . . , vk) (v1, . . . , vi1, vi, vi+1, . . . , vk) (v1, . . . , vi1, vi, vi+1, . . . , vk)] = 0

That is, if and only if g vanishes on R.

We are now ready to define the vector space discussed in the introduction.

Definition 1.4 (Tensor product of vector spaces) Suppose thatV1, . . . , V k are vector spacesover K. Then the quotient vector space F(V1, . . . , V k)/R and the map : V1 . . . Vk F(V1, . . . , V k)/R defined by (v1, . . . , vk) : = (v1, . . . , vk) +R is called the tensor product ofV1, . . . , V k. It is denoted byV1 . . . Vk and the image(v1, . . . , vk)byv1 . . . vk.

We have the following immediate lemma.

Lemma 1.5 The map in Definition 1.4 is a multilinear map.

Proof. We require to show that for i = 1, . . . , k,

(v1 . . . vi1 (vi+ v i) vi+1 . . . vk) (v1 . . . vi1 vi vi+1 . . . vk) (v1 . . . vi1 vi vi+1 . . . vk) = 0

for all a, b K, vj Vj wherej = 1, . . . , k andv i Vi.This is equivalent to the following statement in F(V1, . . . , V k)

(v1, . . . , vi1, vi+v

i, vi+1, . . . , vk)

(v1, . . . , vi1, vi, vi+1, . . . , vk)

(v1, . . . , vi1, vi, vi+1, . . . , vk) R

which is immediately true by definition ofR.

Before proceeding, we require a lemma regarding quotient spaces.

Lemma 1.6 Let V, U andWbe vector spaces overK and let f : V U be a linear map withW ker(f). Then there is a unique linear map f : V/W U for which f = f q, whereq: V V /W is the natural quotient map, i.e. q(v) := v+ W.

Proof. Define fbyf(v+W) := f(v), which is well-defined since ifv+W=v +W, thenvv W,hence

f(v) = f((v v) + v) = f(v v) + f(v) = f(v)

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Definition 1.11 (Symmetric and exterior products of a vector space) LetVbe a vectorspace over some fieldK. Letv1, . . . , vn be a basis ofVand for some permutation of {1, . . . , k},define the linear transformation T :

ki=1 V

ki=1 V by

T(vi1 . . . vik) := v1(i1) . . . v1(ik)

for all 1 i1, . . . , ik n and extending linearly. Note that if w1, . . . , wk V, T acts onw1 . . . wk in the same manner, i.e.

T(w1 . . .wk) := w1(1) . . . w1(k)

Define thek-fold symmetric product ofVto be the subspace

S

ki=1

V

:=

x

ki=1

V|T(x) = x for all permutations

and thek-fold exterior product ofVto be the subspace

A

ki=1

V

:=

x

ki=1

V|T(x) = sgn()x for all permutations

From here on we only consider the special case of the 2-fold symmetric and exterior products. Intheir case, we may write

S(V V) = {x V V|T(x) =x}A(V V) = {x V V|T(x) = x}

forT :V V V V given by

T(vi vj) := vj vi

for a basis v1, . . . , vn ofVand extended linearly. The next theorem provides the aforementioneddecomposition of the 2-fold tensor product.

Theorem 1.12 LetVbe a vector space over some fieldK. Then

V V =S(V V) A(V V)

Proof. Ifx S(V V) A(V V), then x = T(x) = x so x = 0.Furthemore, for all x V V, we have

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x=1

2(x + T(x)) +

1

2(x T(x))

SinceT2 is simply the identity, 12(x + T(x)) S(V V) and 12(x T(x)) A(V V).

We now determine bases for S(V

V) andA(V

V).

Lemma 1.13 Letv1, . . . , vn be a basis ofV . Then

1. The vectorsvi vj +vj vi (1 i j n) form a basis ofS(V V). The dimension ofS(V V)isn(n + 1)/2.

2. The vectorsvi vj vj vi (1 i < j n) form a basis ofA(V V). The dimension ofA(V V) isn(n 1)/2.

Proof. It is immediate that the vectors vi vj +vj vi are linearly independent elements ofS(V V). Similarly, the vectors vi vj vj vi are linearly independent elements ofA(V V).From this we may deduce that

dim(S(V V)) n(n + 1)/2 and dim(A(V V)) n(n 1)/2

However Theorem 1.12 informs us that

dim(S(V V)) + dim(A(V V)) = dim(V V) = n2

Therefore we must have

dim(S(V V)) = n(n + 1)/2 and dim(A(V V)) = n(n 1)/2

as required.

2 The tensor product of group representations

2.1 Definition and basic properties

Consider a finite groupG. Our aim now is given linear representations of this group, say (1, V1)and (2, V2), to construct a linear representation on the tensor productV1 V2. We will see thatthe character of this representation is simply that obtained by multiplying the characters of theconstituent representations, giving a very useful way to find new group characters from given groupcharacters. We motivate this first by considering a simple special case, for which one does not needto be concerned explicitly with tensor products.

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Theorem 2.1 Alinear characterof a group is a character of dimension one. Suppose that is acharacter ofG and is a linear character ofG. Then the product of characters defined

(g) := (g)(g)

for allg

Gis a character ofG. Furthermore, if is irreducible, so is.

Proof. Choose some representation (, V) with character . Define a map : G Aut(V) by

(g) := (g)(g)

for all g G. From the fact that and are homomorphisms, we have immediately that is.Its linearity is also immediate. Hence (, V) is a representation ofG. Since(g) is just a scalar,it is clear that this new representation has character as defined above.

Assume that (, V) is reducible. LetW V be a-invariant subspace ofV. By linearity, it isalso a -invariant subspace ofV, so we have that the irreducibility of implies that of.

It would be very useful if we could multiply arbitrary characters and know that we were going toget another character as a result. Such an extension is not obvious and requires the definition ofthe tensor product representation.

Definition 2.2 (Tensor product representation) LetG be a finite group and let (, V) and(, W) be representations of G. Let v1, . . . , vm be a basis of V and w1, . . . , wn of W. ThenTheorem 1.9 states that the elementsvi wj V Wgive a basis of the tensor product. Definea multiplication onV Wby an elementg Gas follows

g.(vi wj) := (g)vi (g)wj

for alli, j and by linear extension to the whole ofV W. That is,

g.

i, j

ij(vi wj) :=

i, j

ij((g)vi (g)wj)

forij C. This multiplication byG is clearly a group action and linear by definition. Hence itinduces a linear representation(, V W)on the tensor product vector space.Lemma 2.3 Let G be a finite group and let (, V) and (, W) be representations of G. Let(, VW)be the tensor product representation. Then for allv V,w W andg G, we have

(g)(v w) =(g)v (g)w

Proof. Let v1, . . . , vm be a basis of V and w1, . . . , wn of W. Write v =m

i=1 ivi and w =nj=1 jwj . Then

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(g)(v w) = (g)

i, j

ij(vi wj) by Lemma 1.5

=

i, jij((g)vi (g)wj)

=

i

i(g)vi

j

j(g)wj

=(g)v (g)w

The following theorem gives the anticipated result regarding the character of this new representa-tion.

Theorem 2.4 Let (, V) and(, W) be representations of some finite group G with characters and respectively. Let (, V W) be the tensor product representation. Then the characterof the tensor product representation is the product of the two characters and , that is

(g) := (g)(g)

for allg G.

Proof. Fixg G. Choose bases e1, . . . , emofV andf1, . . . , f n ofWsuch that(g)ei= iei and(g)fj =jfj fori, j C (i.e. such that the matrices of(g) and(g) are diagonal). Then

(g) =

m

i=1 i and(g) =n

j=1 jNote that

(g)(ei fj) =(g)ei (g)fj =ij(ei fj)

by Lemma 2.3. Also by Theorem 1.9, the vectorsei fj are a basis ofVW. Therefore, we maywrite the character of the tensor product representation (, V W) as

i, jij =

i

i

jj

=(g)(g)

as required.

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Example 2.5 By way of example, we now present the character table ofS4, which is easily cal-culable without resorting to techniques such as those introduced here. However, we have includedtwo more (reducible) characters of tensor product representations ofS4, calculated by multiplyingcharacters.

gi (1) (12) (123) (12)(34) (1234)

|C(gi)| 24 4 3 8 41 1 1 1 1 12 1 -1 1 1 -13 2 0 -1 2 0 4 3 1 0 -1 -15 3 -1 0 -1 1

34 6 0 0 -2 0 44 9 1 0 1 1

And the generated characters decompose as

34 = 4+ 544 = 1+ 3+ 4+ 5

2.2 Decomposing squares of characters

So given two characters for a finite group G, we may obtain another by multiplying them together.However, this is not of much use in finding more irreducible characters ofG unless we are able todecompose the resulting character into its irreducible components, in which case it might be thatwe have obtained some new ones.

To this end, recall Theorem 1.12, which said that ifVis a vector space,

V V =S(V V) + A(V V)

It in fact turns out that if we have a group representation defined on V , both of these subspacesare invariant with respect to the tensor product representation as shown in the next theorem.

Theorem 2.6 Let G be a finite group and let (, V) be a linear representation ofG. Considerthe 2-fold tensor power representation (2, V V). Then the subspacesS(V V) andA(V V)are2-invariant.

Proof. Let ij C and g G, then

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2(g)

T

i, j

ij(vi vj) =

i, j

ij((g)vj (g)vi) by Lemma 2.3

=

i, jijT((g)vi (g)vj)

=T

2(g)

i, j

ij(vi vj)

So T is a 2-endomorphism on V V. It thus follows that ifx S(V V) and y A(V V),then for g G, we have

T(2(g)x) =2(g)T(x) = 2(g)xand

T(2(g)y) = 2(g)T(y) = 2(g)y

so2(g)x S(V V) and2(g)y A(V V) as required.

If we let be the character of the representation (, V) ofG, the character of the 2-fold tensorpower representation is 2 by Theorem 2.4. Combining Theorems 2.6 and 1.12, we may deducethat2 =S+AwhereSis the character of the sub-representation on the

2-invariant subspaceS(VV) andAis the character of the sub-representation on the 2-invariant subspaceA(VV).The next theorem gives the values of these characters explicitly in terms of.

Theorem 2.7 Let G be a finite group and let(, V)be a linear representation ofGwith character. Let Sbe the character of the sub-representation of(

2, V V) on the2-invariant subspaceS(VV)andAbe the character of the sub-representation on the2-invariant subspaceA(VV).Then forg G, we have

S(g) =1

2(2(g) + (g2))and

A(g) =1

2(2(g) (g2))

Proof. Choose a basise1, . . . , en ofVsuch that(g)ei= iei for i C (i.e. such that the matrixof(g) is diagonal). Then it follows that

(g)(ei ej ej ei) = ij(ei ej ej ei)

and by Lemma 1.13, we have that

A(g) =i


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gi (1) (12) (12)(34) (12)(345) (123) (1234) (12345)|C(gi)| 1 10 15 20 20 30 24

1 1 1 1 1 1 1 1234567

An easy irreducible representation is the so-called signrepresentation ofS5 on theC-vector spaceC1. Let e1 be a basis of this vector space and let S5 act on it by

.e1 = sgn()e1

and extend this to a linear action on C1. Let the character of this representation be a and notethata() = sgn() by considering the matrices of this action with respect to the given basis. Sothe charactera takes values

gi (1) (12) (12)(34) (12)(345) (123) (1234) (12345)|C(gi)| 1 10 15 20 20 30 24

a 1 -1 1 -1 1 -1 1

and is thus 1-dimensional and irreducible. Hence we have another row of the character table.

gi (1) (12) (12)(34) (12)(345) (123) (1234) (12345)|C(gi)| 1 10 15 20 20 30 24

1 1 1 1 1 1 1 1a 1 -1 1 -1 1 -1 1

34567

SinceS5 is a permutation group, an obvious next step is to consider the natural permutation actionofS5 on a basis of the C-vector space C5. Let e1, . . . , e5 be such a basis and let S5 act onindividual basis elements by

.ei= e(i)

and extend this to a linear action on C5. Let the character of this representation be p and notethatp() = |fix()| for S5 by considering the matrices of this action with respect to the givenbasis. So the character p takes values

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gi (1) (12) (12)(34) (12)(345) (123) (1234) (12345)|C(gi)| 1 10 15 20 20 30 24

1 1 1 1 1 1 1 1a 1 -1 1 -1 1 -1 1p 4 2 0 -1 1 0 -1

ap 4 -2 0 1 1 0 -1567

We now wish to apply Theorem 2.7 to p to obtain two more charactersS and A. To do thiswe need to establish what squaring the elements of the conjugacy classes ofS5 does to them sothat we can calculate p(g

2) easily. Since gxg1 =y gx2g1 = (gxg1)2 =y2 forg, x, y S5,squares of elements in the conjugacy class with representative (12) go to the conjugacy class of (1),with representative (12)(34) go to the conjugacy class of (1), with representative (12)(345) go tothe conjugacy class of (123), with representative (123) stay in the conjugacy class of (123), withrepresentative (1234) go to the conjugacy class of (12)(34) and with representative (12345) stay inthe conjugacy class of (12345). So we can compute

gi (1) (12) (12)(34) (12)(345) (123) (1234) (12345)|C(gi)| 1 10 15 20 20 30 24

S12(16 + 4)

12(4 + 4)

12(0 + 4)

12(1 + 1)

12(1 + 1)

12(0 + 0)

12(1 + (1))

A12(16 4) 12(4 4) 12(0 4) 12(1 1) 12(1 1) 12(0 0) 12 (1 + 1)

which is

gi (1) (12) (12)(34) (12)(345) (123) (1234) (12345)|C(gi)| 1 10 15 20 20 30 24

S 10 4 2 1 1 0 0A 6 0 -2 0 0 0 1

Calculate

A|A = 1120

(1.6.6 + 15.(2).(2) + 24.1.1)= 1

soA is irreducible and we have another row of the character table.

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gi (1) (12) (12)(34) (12)(345) (123) (1234) (12345)|C(gi)| 1 10 15 20 20 30 24

1 1 1 1 1 1 1 1a 1 -1 1 -1 1 -1 1p 4 2 0 -1 1 0 -1

ap 4 -2 0 1 1 0 -1A 6 0 -2 0 0 0 1

67

Note that

S|S = 1120

(1.10.10 + 10.4.4 + 15.2.2 + 20.1.1 + 20.1.1)

= 3

soS is not irreducible.

Calculate

S|1 = 1120

(1.10.1 + 10.4.1 + 15.2.1 + 20.1.1 + 20.1.1)

= 1

S|a = 1120

(1.10.1 + 10.4.(1) + 15.2.1 + 20.1.(1) + 20.1.1)= 0

S|p = 1120

(1.10.4 + 10.4.2 + 20.1.(1) + 20.1.1)= 1

SoS decomposes as S= 1+ p+ d for some new characterd taking values

gi (1) (12) (12)(34) (12)(345) (123) (1234) (12345)|C(gi)| 1 10 15 20 20 30 24

d 5 1 1 1 -1 -1 0

Calculate

d|d = 1120

(1.5.5 + 10.1.1 + 15.1.1 + 20.1.1 + 20.(1).(1) + 30.(1).(1))= 1

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sod is irreducible and we have another row of the character table.

gi (1) (12) (12)(34) (12)(345) (123) (1234) (12345)|C(gi)| 1 10 15 20 20 30 24

1 1 1 1 1 1 1 1a 1 -1 1 -1 1 -1 1

p

4 2 0 -1 1 0 -1ap 4 -2 0 1 1 0 -1A 6 0 -2 0 0 0 1d 5 1 1 1 -1 -1 07

We now compute the tensor product character ofa andd to get the new character ad

gi (1) (12) (12)(34) (12)(345) (123) (1234) (12345)|C(gi)| 1 10 15 20 20 30 24

ad 5 -1 1 -1 -1 1 0

Calculate

ad|ad = 1120

(1.5.5 + 10.(1).(1) + 15.1.1 + 20.(1).(1) + 20.(1).(1) + 30.1.1)= 1

so ad is irreducible and we have another row of the character table. Therefore the completecharacter table ofS5 is

gi (1) (12) (12)(34) (12)(345) (123) (1234) (12345)

|C(gi)| 1 10 15 20 20 30 241 1 1 1 1 1 1 1a 1 -1 1 -1 1 -1 1p 4 2 0 -1 1 0 -1

ap 4 -2 0 1 1 0 -1A 6 0 -2 0 0 0 1d 5 1 1 1 -1 -1 0

ad 5 -1 1 -1 -1 1 0

2.2.2 The character table ofA5

We now use the techniques developed in the previous section to compute the character table of the

alternating group of degree 5 (A5). This is the group of the 60 even permutations of five objects,a subgroup of index two in S5, the full symmetric group of degree 5.

The first stage is of course to determine the conjugacy classes ofA5. Recall that the conjugacyclasses of a symmetric group are exactly the sets of permutations of identical cycle structure. So a

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good starting point is to note that in S5, representatives of the four conjugacy classes of the evenpermutations are

(1) (12)(34) (123) (12345)

comprising

1 15 20 24

elements respectively. Unfortunately, if two elements x, y S5 are conjugate in S5, it does notnecessarily follow that they are also conjugate inA5. However, let

t1xt= y (1)

where t S5 and suppose that x commutes with some odd permutation, say sS5 (s / A5), sothat

s1xs= x (2)

Then in addition to (1), we have that

(st)1x(st) =y

and eithert or st is even. Hence x and y remain conjugate if (2) holds.

So since (12)(34) commutes with the odd permutation (12), the conjugacy class of (12)(34) is thesame inA5as inS5. Likewise, (123) commutes with (45) so that this class also remains unchanged.However, this is not the case for the element v = (12345). Recall that ifC(v) is the conjugacy

class ofv andZ(v) is the centraliser ofv inA5, then

|C(v)| = [A5 : Z(v)] (3)

Similarly, ifC(v) is the conjugacy class ofv andZ(v) is the centraliser ofv inS5, then

|C(v)| = [S5 : Z(v)]

Since|S5| = 120 and|C(v)| = 24, it thus follows that|Z(v)| = 5. The five distinct elements1, v, v2, v3, v4 trivially commute with v , so we determine that they constitute Z(v) exactly. Asall these elements are even, we deduce that Z(v) = Z(v) and thus by (3), we determine that

|C(v)| = 12. So it is clear that the class ofv in S5 splits in A5. Indeed, the elements v and v2conjugate in S5 are not conjugate in A5. This is the case since we have

t1vt = v2 (4)

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with t = (2354), which is an odd permutation. If there were an even permutationw such that

w1vw = v2

then tw1, which is odd, would commute with v, which is not possible since we have seen thatZ(v) consists entirely of even permutations. Iterating (4), we find that

t2vt2 =t1v2t= u4

Thusv and v4 are conjugate inA5 and so arev2 andv3. Note by the same argument used to show

that|C(v)| = 12, we have that|C(v)| = 12 and thus representatives for the five conjugacy classesofA5 are

(1) (12)(34) (123) (12345) (13524)

comprising

1 15 20 12 12

elements respectively. Thus there are five irreducible characters ofA5.

Having determined representatives of the conjugacy classes ofA5, we presentA5s character table,yet to be completed apart from the trivial character.

gi (1) (12)(34) (123) (12345) (13524)|C(gi)| 1 15 20 12 12

1 1 1 1 1 123

45

Since A5 is a permutation group, the obvious first step is to consider the natural permutationaction ofA5 on a basis of the C-vector space C5. Let e1, . . . , e5 be such a basis and let A5act on individual basis elements by

.ei= e(i)

and extend this to a linear action on C5. Let the character of this representation be p and notethat p() =|fix()| for A5 by considering the matrices of this action with respect to thegiven basis. So the character p takes values

gi (1) (12)(34) (123) (12345) (13524)|C(gi)| 1 15 20 12 12

p 5 1 2 0 0

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Note thatXA|X1 = XA|Xp = XA|XS = 0, so we have n4 = n5 = 1 and n1 = n2 = n3 = 0.SoA= 4+ 5.

Now consider degrees,

4(e) + 5(e) = A(e) = 6

but we know that

60 = |A5| = 12 + 42 + 52 + 24(e) + 25(e)

Which is

4(e) + 5(e) = 6

24(e) + 25(e) = 18

implying that4(e) = 5(e) = 3. Our character table now looks like

gi (1) (12)(34) (123) (12345) (13524)|C(gi)| 1 15 20 12 12

1 1 1 1 1 1p 4 0 1 -1 -1S 5 1 -1 0 04 3 2 3 4 55 3 2 3 4 5

Where the i and i are still to be determined. By the column orthogonality relations betweencolumn 1 and i, we get

2+ 2 = 2, 3+3= 0, 4+ 4 = 5+ 5 = 1

Since every element ofA5 is conjugate to its inverse, all the numbers in the character table mustbe real. The orthogonality relation for columni with itself (2 i 5) gives

4 = 2 + 22+ 22

3 = 3 + 23+ 23

5 = 2 + 24+ 24 = 2 +

25+

25

Thus we deduce

2 = 2 = 1 and 3 = 3 = 0

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References

G. James and M. Liebeck. Representations and characters of groups. Cambridge University Press,2001.

W. Ledermann. Introduction to group characters. Cambridge University Press, 1987.

J. P. Serre. Linear representations of finite groups. Springer Verlag, 1996.

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Tensor products of vector spaces