Representation Theory of Algebras · Introduction These notes are based on lectures given by Peter McNamara on the Representation Theory of Algebras at the University of Sydney in

Representation Theory of Algebras

Lectures by Peter McNamara

Notes by the Students

The University of Sydney, 2014

Lecture 1 (2014-04-01) 1

Lecture 2 (2014-04-02) 2

Lecture 3 (2014-04-08) 3

Lecture 4 (2014-04-09) 4

Lecture 5 (2014-04-15) 7

Lecture 6 (2014-04-16) 10

0.1 Density Theorem . . 10

0.2 Induction . . . . . . 11

0.2.1 Bimodules . . 11

0.2.2 Tensor Prod-ucts . . . . . 12

Lecture 7 (2014-04-29) 130.3 Tensor-Hom Adjunc-

tion . . . . . . . . . 130.4 Categories . . . . . . 140.5 Induction and Re-

striction . . . . . . . 15

Lecture 8 (2014-05-14) 160.6 Bar Resolution . . . 16

0.6.1 Application . 170.7 Baer’s Theorem . . . 18

Lecture 9 (2014-05-21) 200.8 Characters . . . . . . 200.9 Character Tables . . 220.10 Character Orthogo-

nality . . . . . . . . 23

Introduction

These notes are based on lectures given by Peter McNamara on the Representation Theory ofAlgebras at the University of Sydney in 2014. As part of the assessment for the course, eachstudent has contributed several lectures to these notes. We are responsible for all faults in thisdocument, mathematical or otherwise; any merits of the material here should be credited to thelecturer, not to us.

Please email any corrections or suggestions to [email protected].

mailto:[email protected]

Lecture 1 (2014-04-01)

Last edited23-07-2014

Representation Theory of Algebras Page 1Lecture 1

Lecture 2 (2014-04-02)



Lecture 3 (2014-04-08)



Lecture 4 (2014-04-09)

More on Finite Groups

In the following, we let k be a field, G a finite group and we write k[G] for the group algebra. Forthis lecture, we additionally stipulate that char(k) ∤ |G|. In this case, Maschke’s theorem applies.

Theorem 1 (Maschke’s theorem). Let G be a finite group, and k a field such that char(k) ∤ |G|.Suppose thatW is a k[G]-module (equivalently, a representation of G over k), and V is a submoduleof W . Then there exists a submodule U of W such that W = U ⊕ V .

Last lecture, we showed that for a unital algebra A, we have EndA(A) ∼= Aop. Here, Aop is theopposite algebra, equal to A as an abelian group, but equipped with the opposite multiplication(a, b) 7→ ba (as opposed to (a, b) 7→ ab in A).

Since all groups have an identity element, the group algebra k[G] is unital, hence Endk[G](k[G]) ∼=k[G]op.

Remark. The map g 7→ g−1 extends linearly to an isomorphism of algebras between k[G] andk[G]op.

Before progressing further, we prove a few results about the structure of k[G]-modules.

Theorem 2. Any k[G]-module that is finite dimensional (as a vector space over k) can be writtenas a direct sum of irreducible k[G]-modules.

Proof. Let M be a finite dimensional k[G]-module. We use induction on the dimension of M (as avector space over k).

In the base case, dim(M) = 1. As a a submodule of M is a vector subspace of M , the onlysubmodules of M are {0} and M . In particular, M is itself irreducible (hence it is trivially a sumof irreducible modules).

If M is not irreducible, it has a proper nontrivial submodule U . Maschke’s theorem guarantees theexistence of a submodule V of M , such that M = U ⊕ V . As U is a nontrivial submodule of M ,it has dimension at least 1, and at most dim(M) − 1; the same must be true of V . Applying theinduction hypothesis to U and V shows that each is a direct sum of irreducible modules. Therefore,M , being the direct sum of U and V must be a direct sum of irreducible modules. This completesthe induction.

Corollary 1. Under the hypotheses of Maschke’s theorem, k[G] (considered as a k[G]-module) canbe decomposed into a direct sum of irreducible k[G]-modules.

Proof. As G is assumed to be finite, k[G] is a finite dimensional k[G]-module (it has dimension|G|).

Theorem 3. Let A, B, and C be k[G]-modules. Then we have the following isomorphisms:

Hom(A,C)⊕Hom(B,C) ∼= Hom(A⊕B,C) (1)

Hom(A,B)⊕Hom(A,C) ∼= Hom(A,B ⊕ C) (2)

Proof. The obvious maps (f, g) 7→ ((a, b) 7→ f(a) + g(b)) and (f, g) 7→ (a 7→ f(a) + g(a)) turn outto be isomorphisms.



Remark. These immediately generalise to any direct sum with finitely many summands, but canfail for infinite direct sums.

We let the irreducible k[G] modules be Mi, for i ∈ I (I being some index set). By what we alreadyknow, k[G] decomposes into a direct sum of irreducible modules when considered as a k[G]-module.We write k[G] ∼=

⊕i∈I Mi

⊕ni , for some multiplicities ni (i ∈ I). Note that since the dimension ofk[G] is finite, all but finitely many ni must be zero. By equations (1) and (2), we have:

k[G]op ∼= Endk[G](k[G])

∼= Homk[G](⊕

i∈I

Mi⊕ni ,

⊕

j∈I

Mj⊕nj )

∼=⊕

i,j∈I

Homk[G](Mi⊕ni ,Mj

⊕nj )

We recall Schur’s Lemma.

Theorem 4 (Schur’s Lemma). Let M , N be simple A-modules (where A is an algebra over a fieldk), and ϕ : M → N be a homomorphism. Then either ϕ is the zero map, or it is an isomorphism.Furthermore, for any simple A-module M , EndA(M) is a division algebra. When k is algebraicallyclosed, this division algebra is isomorphic to k.

It immediately follows that for i 6= j, Homk[G](Mi⊕ni ,Mj

⊕nj ) ∼=⊕ni

a=1

⊕nj

b=1Homk[G](Mi,Mj) ∼={0}. This simplifies our earlier equation (the ‘‘off-diagonal’’ terms vanish). Hence, k[G]op ∼=

⊕i,∈I Endk[G](Mi

⊕ni).

To identify the remaining summands, let Di = Endk[G](Mi). Also, let M(r)i denote the rth copy of

Mi in M⊕ni

i . Then we have:

Endk[G](Mi⊕ni) ∼=

ni⊕

a,b=1

Homk[G](M(a)i ,M

(b)i )

Each summand is isomorphic to Di by definition; this determines the structure as a k[G]-module

(in particular, as an abelian group). To find the algebra structure, let f : M(a)i → M

(b)i act via

multiplication by d1 ∈ Di, and g : M(c)i → M

(d)i act via multiplication by d2 ∈ Di (by abuse of

notation, we extend these to all of M⊕ni

i , acting trivially on the other summands). Then we clearly

have the following composition rule: g ◦ f maps M(a)i to M

(d)i via multiplication by d2d1 when

b = c, and acts as zero otherwise. This means that these endomorphisms mapping M(r)i to M

(s)i via

multiplication by d ∈ Di compose in the same way as the matrices {δa,sδb,rd}ni

a,b=1 multiply. Hence,

Endk[G](M⊕ni

i ) ∼=Matni(Di). We conclude:

k[G]op ∼=⊕

i∈I

Matni(Di)

This in turn gives an expression for k[G]:

k[G] ∼=⊕

i∈I

Matni(Di

op)

In the case where k is algebraically closed (so by Schur’s lemma, Di∼= Dop

i∼= k), k[G] ∼=⊕

iMatni(k). Comparing the dimensions of each side of the equation shows that |G| =

∑i ni

2.



Remark. Additionally, it is true that dim(Mi) = ni, and that everyMi occurs as a direct summandof k[G]. These results are self-consistent because of the Krull-Schmidt theorem, which guaranteesthe uniqueness of direct sum decompositions of finite dimensional k[G]-modules. We will discussthis later.

Theorem 5. Let k be an algebraically closed field, and G a finite group with char(k) ∤ |G|. Thenthe number of irreducible representations of G is equal to the number of conjugacy classes of G.

Proof. Assume for now the fact that every irreducible k[G]-module occurs as a direct summand ofk[G]. We compute the dimension of the centre of k[G] (the set of elements that commute with allelements of k[G]); denote the centre of an algebra A by Z(A). On one hand, an easy calculationshows that Z(Matni

(k)) consists precisely of scalar matrices (and is hence isomorphic to k). SinceZ(A1 ⊕ A2) ∼= Z(A1) ⊕ Z(A2), it follows that Z(k[G]) ∼=

⊕i∈I Z(Matni

(k)) ∼= k|I|. Therefore thedimension of the centre is equal to the number of irreducible representations of k[G] (we assumedeach such representation occurs within k[G]).

Now, z =∑

g∈G f(g)g is an element in Z(k[G]) if and only if zh = hz for all h ∈ G (this is clearly

a necessary condition, and it is sufficient by linearity). This in turn is equivalent to z = hzh−1 forall h ∈ G. This condition holds precisely when f(g) = f(h−1gh) for all h ∈ G; that is, when f isconstant on conjugacy classes of G. It is clear that the set of elements

∑g∈C g (for each conjugacy

class C) define a basis of Z(k[G]). The dimension of the space is therefore equal to the number ofconjugacy classes of G.

Let us consider an example: the alternating group on four letters, A4. It is known that there is anormal subgroup of A4 given by V = {Id, (12)(34), (13)(24), (14)(23)}, and that A4/V ∼= C3 (thecyclic group with 3 elements). A short computation shows that A4 has four conjugacy classes andtherefore has four isomorphism classes of irreducible representations.

Now, C3 has 3 irreducible representations, each of which is one dimensional. In each case, a generatorof C3 acts by scalar multiplication by a third root of unity. Each of the irreducible representationsof C3 lifts to an irreducible representation of A4, where V acts trivially. This means that oneirreducible representation of A4 remains to be identified. It turns out that it is given by the actionof A4 on {(x1x2, x3, x4) ∈ k4 | x1+x2+x3+x4 = 0} by permutation of coordinates. This is a threedimensional representation, and we can verify that |A4| = 12 = 12 + 12 + 12 + 32, as we expect.

For the Symmetric group on four letters, S4, we have a similar situation; the conjugacy classesare determined by partitions of 4 (in correspondence with the possible cycle types). There arefive conjugacy classes. There are two one dimensional representations, where a permutation actsby scalar multiplication. For the trivial representation, the multiplication is by 1, and for the signrepresentation, it is by the sign of the permutation. It turns out that V (defined above as a subgroupof A4) is a normal subgroup of S4, with S4/V ∼= S3. Therefore the two dimensional representationof S3 lifts to S4 (with V acting trivially). We know that there are two remaining (isomorphismclasses of) irreducible representations; if they have dimensions d1 and d2, then they must satisfy12 + 12 + 22 + d1

2 + d22 = 24. The only solution to this equation is d1 = d2 = 3.

One of these is the representation on {(x1x2, x3, x4) ∈ k4 | x1 + x2 + x3 + x4 = 0} by permutation(which was irreducible as a representation of A4, so it is irreducible as a representation of S4). Thefinal representation is on the same space, but the action of an element of S4 is to permute thecoordinates, and multiply them by the sign of the permutation.



Lecture 5 (2014-04-15)

Definition 1 (Quotients of representations). Let N be a submodule of an A-module M . Thenthe quotient module is M/N as a quotient of a vector space or an abelian group with A-actiona(m+N) = a(m) +N , for all a ∈ A, m+N ∈M/N .

Remark. The A-action is well defined. ∀a, b ∈ A, ∀m+N ∈M/N

a((m+ n) +N) = a(m) + a(n) +N = a(m) +N = a(m+N)

(ab)(m+N) = (ab)(m) +N = a(b(m)) +N = a(b(m) +N) = a(b(m+N))

Definition 2. Let k be a field, A be a k algebra. A composition series of A-moduleM is a sequenceof submodules

M =Mn )Mn−1 ) · · · )M1 )M0 = 0

such that Mi/Mi−1 is simple ∀1 ≤ i ≤ n.The factors Mi/Mi−1 are called the composition factors.

Theorem 1 (Jordan-Holder Theorem). Let k be a field, A be a k-algebra andM be a representationof A. If

M =Mn )Mn−1 ) · · · )M1 )M0 = 0 and,

M =M ′m )M ′

m−1 ) · · · )M ′1 )M ′

0 = 0

are two composition series, then m=n and there exists a permutation σ ∈ Sn such that

Mi/Mi−1∼=M ′

σ(i)/M′σ(i)−1 ∀1 ≤ i ≤ n

Proof. We will prove the theorem using induction on n. Firstly, if n = 1 then M is simple so clearlythe composition series is unique.Now assume the theorem is true for less than n. Then we can assume Mn−1 6=M ′

m−1 otherwise weare done using the induction hypothesis.

Now, we have a quotient map

ψ :M ′m−1 →M/Mn−1 with kerψ =Mn−1 ∩M

′m−1

So we can induce an injective homomorphism

ψ :M ′m−1/(Mn−1 ∩M

′m−1) →M/Mn−1

M/Mn−1 is simple and im ψ is a non-zero submodule of it, so ψ is surjective and hence an isomor-phism. So we have

M ′m−1/(Mn−1 ∩M

′m−1)

∼=M/Mn−1 and Mn−1/(Mn−1 ∩M′m−1)

∼=M/M ′m−1 (3)

by the same argument.



M

Mn−1

Mn−2

M ′m−1

M ′m−2

Mn−1 ∩M′m−1

... ...

...A D

B C

The composition series A in the diagram has length n − 1, so by the induction hypothesis, B alsohas length n− 1. Then C also has length n− 1, and we have the length of D equals to n− 1 usingthe induction hypothesis again. This also shows that the composition factors of A and B, and Cand D are isomorphic so combined with (3), we have the isomorphism of the composition factorsof the original composition series.

Lemma 1. LetM be a simple A-module. Then there exist a surjective homomorphism of A-modules

ϕ : A→M

Proof. Pick m0 ∈M , m0 6= 0. Define ϕ by ϕ(a) = am0. Then ϕ is a homomorphism since

ϕ(ab) = (ab)(m0) = a(bm0) = aϕ(b)

Now m0 6= 0 so imϕ 6= 0 and M is simple, so ϕ is surjective.

Corollary 1. Let A be finite dimensional k-algebra. Then there exist a composition series

A =Mn )Mn−1 ) · · · )M1 )M0 = 0

with every simple A-module is isomorphic to at least one of the composition factor.

Proof. The composition series exists as A is finite dimensional, and by the lemma, every simpleA-module is isomorphic to M ′

n/M′n−1 for some composition series of A by taking M ′

n−1 = kerϕ.Then Jordan-Holder Theorem implies that the simple module must be isomorphic to one of thecomposition factors in the original composition series.

Example 1. Let A = k[G] where G is a finite group with char(k) ∤ |G|. Then A ∼= ⊕M⊕ni

i and bythe Corollary, every simple A-module appears in the direct sum.

Example 2. For q ∈ k, a Hecke algebra Hq has a presentation

Hq =< T1, T2 | (Ti − q)(Ti + 1) = 0, T1T2T1 = T2T1T2 >

We can also present Hq as

C[GL3(Fq)] = {f : GL3(Fq) → C}⋃

Hq ={f | f(b1gb2) = f(g), ∀g ∈ GL3(Fq) ∀b1, b2 ∈

( ∗ ∗ ∗· ∗ ∗· · ∗

)}



For simplicity’s sake, take q = 0. Then the relations become T 2i = −Ti and T1T2T1 = T2T1T2 and

we can write the action on the basis {1, T1, T2, T1T2, T2T1, T1T2T1} as

1

T1 T2

T2T1 T1T2

T1T2T1

T1 T2

T2

T1

T1

T2

T1

T2

T2

T1

T1, T2

Now it is clear that by taking the basis elements from the bottom of the diagram we can constructa composition series

H0 =M6 )M5 ) · · · )M1 )M0 = 0 with dimMi = i

Therefore every simple H0-module is 1 dimensional by Jordan-Holder Theorem.

So every simple H0-module can be represented as ϕ : Hq → Endk(k) = k.If we consider the first relation, we get (ϕ(Ti)− q)(ϕ(Ti)− 1) = 0 =⇒ ϕ(Ti) = −1, qCombined with the second relation ϕ(T1T2T1) = ϕ(T2T1T2) we getϕ(T1) = q and ϕ(T2) = qϕ(T1) = −1 and ϕ(T2) = −1ϕ(T1) = q and ϕ(T2) = −1 if −q2 = qϕ(T1) = −1 and ϕ(T2) = q if −q2 = qFor q = 0 all 4 cases satisfy the relations so there are 4 simple H0-modules.



0.1 Density Theorem

Lecture 6 (2014-04-16)

Definition 1 (Semisimple). A module is called semisimple if it is the direct sum of simple modules.

Lemma 1. Let M be a submodule of⊕n

i=1M⊕kii where each Mi is simple, then M ∼=

⊕ni=1M

⊕ℓii

for 0 ≤ ℓi ≤ ki. That is, semisimple modules have no ‘‘interesting’’ submodules.

Proof. We prove this only for finite dimensional modules, but it is true in general. The proof willprocedure by induction on

∑ki and dimM .

Let S be a simple submodule ofM then Hom(S,Mi) = 0 unlessMi∼= S. We have Hom(S,

⊕M⊕ki

i ) 6=0, so WLOG we can take S ∼=M1. We have the following diagram

S M⊕M⊕ki

i

M1

i

ϕπ

project onto first summand

As it is a homomorphism of simple modules, we know ϕ = 0 or ϕ is an isomorphism. In the lattercase, π is surjective and

Si

−→Mp

−−→ S

where p = ϕ−1 ◦ π, note that p ◦ i = idS .

We will show thatM ∼= i(S)⊕kerϕ. If v ∈ i(S)∩ker p, chose s such that v = i(s), we have p(v) = 0so (p ◦ i)(s) = 0, but p ◦ i = idS , so s = 0 hence v = 0 too. Now, if m ∈M , then we can write

m = (i ◦ p)(m)︸︷︷︸∈i(s)

+ [m− (i ◦ p)(m)]︸︷︷︸∈ker p

i(s) is simple, so we can use the inductive hypothesis on ker p.

Returning to the two cases for ϕ, we now assume ϕ = 0. This implies π = 0, so M is in the kernelof the projection, that is,

M ⊂M⊕(k1−1)1 ⊕

⊕

i≥2

M⊕kii

and so we can again use the inductive hypothesis.

The requirement for finite dimensionality can be removed with more effort; that assumption is usedfor the existence of S, and for the induction with i(S)⊕ ker p.

0.1 Density Theorem

Theorem 1. Let A be a unital k-algebra, and let M be a simple A-module. Let D = EndA(M)(which is a division algebra, and equal to k in nice cases). Take v1, . . . , vn ∈M , linearly independentover M , and w1, . . . , wn ∈M arbitrary, then there is a ∈ A such that avi = wi for 1 ≤ i ≤ n.

Theorem (Alternate statement). Preserving the notation above, let f : A→ Endk(M) take a ∈ Ato the endomorphism acting by a. If k = k and dimM <∞, then f is injective.



0.2 Induction

Proof. Define an A-module homomorphism ϕ : A→M⊕n,

a 7→

av1...avn

We wish to show that ϕ is surjective. Lemma 1 implies imϕ ∼=M⊕k for some k, giving the followingdiagram

A M⊕n

M⊕k = imϕ

ϕ

∈ Hom(M⊕k,M⊕n)

We have

HomA(M⊕k,M⊕n) =

k⊕ n⊕HomA(M,M)︸︷︷︸

D

so every x ∈ HomA(M⊕k,M⊕n) can be written

x

m1...mn

= X

m1...mn

where X is an (n × k)-matrix with entries in D. By way of contradiction, suppose k < n, thenthere exists c = (c1, . . . , cn) 6= 0 ∈ Dn such that cX = 0. If v ∈ imX then cv = 0. We have

im(ϕ) = im(X) and so ϕ(1) =

( v1...vn

)∈ imX, meaning

(c1, . . . , cn)

v1...vn

= 0

contradicting linear independence, hence k = n.

x is injective, so rankX = k. Hence x is an isomorphism, so ϕ is injective.

0.2 Induction

Our short-term goal is to show give a construction of ‘‘induction’’ for turning A-modules into B-modules if A → B is an inclusion of algebras (works for general homomorphisms too). For examplek[H] → k[G] if H < G are groups.

0.2.1 Bimodules

Definition 2 (Bimodule). Let A, B be algebras. An (A,B)-bimodule M is a left A-module and aright B-module.

An (A,B)-bimodule has action maps

A×M →M (a,m) 7→ am



0.2 Induction

B ×M →M (b,m) 7→ mb

such that (a1a2)m = a1(a2m),m(b1b2) = (mb1)b2 and the actions commute, that is, (am)b = a(mb),for all a, a1, a2 ∈ A, m ∈M , b, b1, b2 ∈ B.

Remark. A right B-module is the same as a left Bop-module.

Example 1. An algebra A is an (A,A)-bimodule, and also a (B,C)-bimodule for any subalgebrasB and C.

0.2.2 Tensor Products

Definition 3 (Tensor product). LetM be a right A-module, and N be a left A-module. The tensorproduct, denoted

M ⊗AN

is the Abelian group generated by the symbols m⊗n, m ∈M , n ∈ N , with relations

(m1 +m2)⊗n = m1⊗n+m2⊗n

m⊗(n1 + n2) = m⊗n1 +m⊗n2

(ma)⊗n = m⊗(an)

for all m,m1,m2 ∈M , n, n1, n2 ∈ N and a ∈ A.

The tensor product is often written just M ⊗N , if the algebra A is clear.

Example 2. Take A = k a field, meaning modules are vector spaces. If M has basis {bi}i∈I , andN has basis {cj}j∈J , then M ⊗N has basis {bi⊗ cj}i∈I,j∈J .

Example 3. Z/2Z⊗Z Z/3Z = 0 since 3 = 1 ∈ Z/2Z, so

m⊗n = (m3)⊗n = m⊗(3n) = m⊗ 0 = 0

Remark. IfM is a (B,A)-bimodule andN an (A,C)-bimodule, thenM ⊗AN is a (B,C)-bimodule,with action b(m⊗n)c = (bm)⊗(nc).

We now have enough machinery to define the induced representation of an module.

Definition 4 (Induced representation). Let A → B be an inclusion of algebras, so B is a (B,A)-bimodule, and let M be an A-module. The induced representation is defined as

IndBA(M) = B⊗AM



0.3 Tensor-Hom Adjunction

Lecture 7 (2014-04-29)

0.3 Tensor-Hom Adjunction

Let A and B be unital algebras, and let P be a (B,A)-bimodule. Then we can map A-modules intoB-modules and B-modules into A-modules via the maps Φ and Ψ respectively, defined by

Φ :M 7→ P ⊗A M

Ψ : N 7→ HomB(P,N).

Remark. We need to justify the fact that Ψ maps to A-modules, i.e. that HomB(P,N) is anA-module. For a B-module N and λ ∈ HomB(P,N) define the action (aλ)(p) = λ(pa) for alla ∈ A, p ∈ P . It can be easily checked that this action is well defined. For example, since P is aright A-module we have: ((a1a2)λ)(p) = λ(p(a1a2)) = (a2λ)(pa1) = (a1(a2λ))(p).

Theorem 1. Let M and N be A and B-modules respectively. Then we have

HomB(P ⊗A M,N) ∼= HomA(M,HomB(P,N)).

Proof. We will establish mutually inverse correspondences between the two structures. Given aB-module homomorphism f : P ⊗A M → N , define a corresponding A-module homomorphismg :M → HomB(P,N) by

[g(m)](p) := f(p⊗m).

For any m ∈ M , [g(m)] is B-module homomorphism since f is, so it suffices to check that g isan A-module homomorphism. Additivity follows easily, and to see ag(m) = g(am) for all a ∈ A,m ∈M , note that

[ag(m)](p) = [g(m)](pa)

= f(pa⊗m)

= f(p⊗ am)

= [g(am)](p).

Conversely, given any g : M → HomB(P,N) it suffices to define f : P ⊗A M → N on simpletensors by

f(p⊗m) := [g(m)](p)

and extend by linearity. We need to check that f is indeed a B-module homomorphism. Thisfollows from

bf(p⊗m) = b[g(m)](p)

= [g(m)](bp)

= f(bp⊗m)

= f(b(p⊗m)).

The correspondences are mutually inverse by construction as required.



0.4 Categories

0.4 Categories

Definition 1. (Category). A category C consists of a class of objects, denoted ob(C), and a classof morphisms between the objects (often denoted hom(C)), such that the following axioms hold:-(Identity) For every object X ∈ ob(C) there exists a morphism idX : X → X.-(Composition) If X,Y, Z ∈ ob(C) with morphisms f : X → Y , g : Y → Z, then there exists acomposition g ◦ f : X → Z. Composition is associative, and satisfies f ◦ idX = f , idY ◦ f = f .

Example. Categories are ubiquitous in mathematics. Some commonly encountered categories(written here as class-morphism pairs) include (sets, functions), (groups, homomorphisms), (vectorspaces, linear maps) and (A-modules, A-module homomorphisms).

Definition 2. (Functor). Let C and D be categories. A functor F from C to D is a mapping whichsatisfies the following:-For all objects X of C, F (X) is an object of D.-For all morphisms f : X → Y in C, F (f) : F (X) → F (Y ) is a morphism in D which is compatiblewith identity and associative.

Definition 3. (Adjoint Functors). Let C and D be categories. The functors F : C → D andG : D → C are called adjoint if there is a family of isomorphisms such that

HomD(FX, Y ) ∼= HomC(X,GY )

for all X ∈ Ob(C) and Y ∈ Ob(D). In this case, F is called the left adjoint and G is called theright adjoint. Given a morphism f : X → X ′ in C the following diagram commutes:

HomD(FX, Y ) HomC(X,GY )

HomD(FX′, Y ) HomC(X

′, GY )

≃

≃

Similarly, given a morphism g : Y → Y ′ in D the following diagram commutes:

HomD(FX, Y ) HomC(X,GY )

HomD(FX, Y′) HomC(X,GY

′)

≃

≃

The vertical arrows in these diagrams are those induced by composition with f and g.

Example. Theorem 1 proved that P ⊗A − and HomB(−, N) are adjoint functors.



0.5 Induction and Restriction

0.5 Induction and Restriction

Let B be a unital algebra with A a subalgebra, and let M and N be A and B-modules respectively.Then B is a (B,A)-module by left/right multiplication, so that

IndBAM = B ⊗A M.

Furthermore, the right adjoint of Ind is

ResBAN = HomB(B,N) ∼= N,

where the (A-module) isomorphism above follows from the fact that any λ ∈ HomB(B,N) iscompletely determined by its value on 1 ∈ B.

Remark. ResBA also has a right adjoint; it is HomA(B,−), often called coinduction. We have

HomA(ResBAM,N) ∼= HomB(M,HomA(B,N)).

For A and B finite group algebras we will see later that IndBA− = HomA(B,−), although this doesnot hold in a general setting.

Example 1. Let H ≤ G be finite groups, and let B = k[G], A = k[H]. We wish to understand thestructure of k[G]⊗k[H] M . Let us choose representatives {gi}1≤i≤n for cosets of H in G, and write

G =n⊔

i=1

giH.

Since the union is disjoint it follows that any f ∈ k[G] can be written uniquely as

f =n∑

i=1

gihi, hi ∈ k[H]. (4)

Now pick a basis {mj} ofM . We claim that {gi⊗mj} is now a basis for the induced representationk[G]⊗k[H] M . This follows easily from (1).We now have a great deal of information about the structure of the induced representation. Indeed,our work has shown

dim(k[G]⊗k[H] M) = [G : H] dimM.

Our final task is to find a representation for the action of g ∈ G on IndGH in terms of matrices forthe H-action on M . Since the cosets of H partition G we have that for any g ∈ G and any cosetrepresentative gi, there is a unique j such that g−1

j ggi ∈ H. Therefore

g(gi ⊗mk) = (ggi)⊗mk

= gj(g−1j ggi)⊗mk

= gj ⊗ (g−1j ggi)mk

= some linear combination of gj ⊗ml where j is fixed and l varies.

Therefore, the action of g corresponds to a block permutation matrix, with submatrices given bythe behaviour of some element of H on M .



0.6 Bar Resolution

Lecture 8 (2014-05-14)

0.6 Bar Resolution

Definition 1. Let A be a k-algebra and M be a A-module. Define

Pn = A⊗kA⊗

k· · · ⊗

kA

︸︷︷︸n+1 times

⊗kM

then Pn is a free A-module where A acts by multiplication on first tensor factor. Then the barresolution of M is

· · · −→ P2∂2−→ P1

∂1−→ P0∂0−→M → 0

with

∂n : Pn →Pn−1

∂n(a0⊗ a1⊗ a2 . . .⊗ an⊗m) =n∑

i=0

(−1)ia0⊗ . . .⊗ aiai+1⊗ . . .⊗ an⊗m

+ (−1)n+1a0⊗ a1⊗ . . .⊗ an−1⊗ anm

Proposition 1. Bar resolution is a projective resolution

Proof. Firstly, we know Pn is a free A-module, so Pn is a projective module. To show the abovesequence is exact, we will show it is exact at Pn.

Let a0⊗ . . .⊗ an+1⊗m ∈ Pn+1. Then

∂n+1∂n(a0⊗ . . .⊗ an+1⊗m) = 0

Since expanding the left hand side involves a huge sum, we are not going to explicitly work it out.However the idea is, every term on the left hand side is obtained by merging a tensor factor twiceand multiplied by 1 or −1 depending on how they are merged. Since the factors can merged in twodifferent order, there are exactly two ways to obtain the same term if we disregard ±1 at the front.But merging in different order shifts the index by 1, so we have +1 in one of them and −1 in theother, so every terms cancels and we get 0 in the end. So we have ker ∂n ⊇ im ∂n+1

Now to show ker ∂n ⊆ im ∂n+1, define

hn : Pn → Pn+1, hn(a0⊗ . . .⊗ an⊗m) = 1⊗ a0⊗ . . .⊗ an⊗m

Then for all a0⊗ . . .⊗ . . .⊗ an⊗m ∈ Pn

(hn−1∂n + ∂n+1hn)(a0⊗ . . .⊗ an⊗m)

=hn−1∂n(a0⊗ . . .⊗ an⊗m) + ∂n+1hn(a0⊗ . . .⊗ an⊗m)

=hn−1

(n−1∑

i=0

(−1)ia0⊗ . . .⊗ aiai+1⊗ . . .⊗ an⊗m+ (−1)na0⊗ . . .⊗ an−1⊗ anm

)

+ ∂n+1(1⊗ a0⊗ . . .⊗ an⊗m)



0.6 Bar Resolution

=n−1∑

i=0

(−1)i1⊗ a0⊗ . . .⊗ aiai+1⊗ . . .⊗ an⊗m+ (−1)n1⊗ a0⊗ . . .⊗ an−1⊗ anm

+ a0⊗ . . .⊗ . . .⊗ an⊗m

+n−1∑

i=0

(−1)i1⊗ a0⊗ . . .⊗ aiai+1⊗ . . .⊗ an⊗m+ (−1)n+11⊗ a0⊗ . . .⊗ an−1⊗ anm

=a0⊗ . . .⊗ . . .⊗ an⊗m

Hence hn−1∂n(x) + ∂n+1hn(x) = x for all x ∈ Pn.Now if x ∈ ker ∂n, then hn−1∂n(x) = 0 so ∂n+1h(n) = x. So we get x ∈ im ∂n+1 and henceker ∂n ⊆ im ∂n+1.

Therefore ker ∂n = im ∂n+1, so the sequence is exact and hence the sequence is a projective resolu-tion.

0.6.1 Application

Now using the bar resolution, we will compute Ext1(M,N). After applying HomA( · , N) to thebar resolution, we get

0 HomA(A⊗kM,N) HomA(A⊗k A⊗kM,N) . . .

Homk(M,N) Homk(A⊗kM,N)

d0

∼=φ1

d1

∼=φ2

Where the isomorphism φ1 maps f : A⊗kM → N to

g :M → N, g(m) = f(1⊗m)

and φ−11 maps g :M → N to

f : A⊗kM → N, f(a⊗m) = ag(m)

And similarily, the isomorphism φ2 maps f : A⊗k A⊗kM → N to

g : A⊗kM → N, g(a⊗m) = f(1⊗ a⊗m)

and φ−12 maps g : A⊗kM → N to

f : A⊗kA⊗

kM → N, f(a⊗ b⊗m) = ag(b⊗m)

Now we will first compute ker d1 in order to compute Ext1(M,N) = ker d1im d0

. So suppose Y ∈ ker d1.Then Y : A⊗kM → N , so for all a ∈ A, there exist a linear map Y (a) : M → N given byY (a)m = Y (a⊗m)



0.7 Baer’s Theorem

A⊗k A⊗k A⊗kM 1⊗ a⊗ b⊗m

1⊗ a⊗m A⊗k A⊗kM a⊗ b⊗m− 1⊗ ab⊗m+ 1⊗ a⊗ bm

Y (a)m N aY (b)m− Y (ab)m+ Y (a)bm

∂2

Y

And d1(Y ) = Y ◦ ∂2, so Y ∈ ker d1 if and only if Y ◦ ∂2 = 0. Now if A(a) is how a acts in N andB(b) is how b acts in M then Y ◦ ∂2 = 0 is equivalent to, for all m ∈M ,

[A(a)Y (b)− Y (ab) + Y (a)B(b)] (m) = 0

Hence Y ∈ Z1(M,N). We know Z1(M,N) ⊆ ker d1 from the previous lecture, so Z1(M,N) =ker d1.

Now we need to show B1(M,N) = im d0. So suppose Y ∈ im d0 ⊆ Homk(A⊗kM,N). That is ifand only if there exist Z ∈ Homk(M ⊗N) such that for all a⊗m ∈ A⊗kM ,

Y (a⊗m) = aZ(m)− Z(am)

If we consider Y (a⊗m) = Y (a)m, and then the above equation is equivalent to

Y (a)m = A(a)Z(m)− Z(B(a)m) = A(a)Zm− ZB(a)m

Which is the definition of element in B1(M,N). Hence B1(M,N) = im d0.

Therefore Ext1(M,N) = ker d1im d0

and hence Ext1(M,N) classifies short exact sequence

0 → N → X →M → 0

In particular, 0 ∈ Ext1(M,N) corresponds to a short exact sequence

0 → N → N ⊕M →M → 0


Theorem 1 (Zorn’s lemma). Let (E , <) be a poset. If every chain has an upper bound then E hasa maximal element.

Theorem 2 (Baer’s Theorem). Let E be an A-module. Then E is injective if and only if for allJ ⊆ A an A-submodule and f : J → E, there exists f ′ : A → E such that f ′ ◦ i = f , wherei : J → A is an embedding. This can be summarised to the following commutative diagram.

∀J E

A

f

∃f ′




Proof. If E is injective, then such extension exists by the definition.

On the other hand, suppose f ′ exists for all J and f . LetN be an A-module andM be a A-submoduleof N . Define (E , <) to be a poset

E := {(M ′, f ′) |M ⊆M ′ ⊆ N, f ′ :M ′ → E extends f}

and the partial order < to be such that

(M ′, f ′) < (M ′′, f ′′) if

1)M ′ ⊆M ′′, and

2) f ′′|M ′ = f ′

Now we will show that this poset satisfies the assumption of Zorn’s Lemma. Suppose (Mi, fi)i∈I bea chain in (E , <). Then let

M =⋂

i∈I

Mi, f : M → E

such that f(m) = fi(m) if m ∈Mi. Now if m ∈Mi and m ∈Mj , then either Mi < Mj or Mj < Mi,so f is well defined. Clearly (M, f) ∈ E , so every chain in (E , <) has an upper bound. Hence byZorn’s Lemma there exist a maximal element

(M, f) ∈ E

Now suppose by contradiction that N 6= M . Then there exists n ∈ N \ M .

Define J := {a ∈ A | an ∈ M} be an A-submodule of A. Then by the assumption, there existh : A→ E such that the diagram below commutes.

J M E

A

a 7−→ an f

h

Now if extend f to

g : 〈M, n〉 −→ E

m+ an 7−→ f(m) + h(a)

But this implies (〈M, n〉, g) ∈ E so by the maximality of M , M = 〈M, n〉 a contradiction. HenceN = M , so E is an injective module.



0.8 Characters

Lecture 9 (2014-05-21)

0.8 Characters

Definition 1. Let G be a finite group, V a finite dimensional vector space over C. If π : G→ GL(V )is a representation then the character of π is the map χ : G → C defined by χV (g) = Tr(π(g)),where Tr is the trace.

Remark 1. We collect some simple but important properties of characters here:

χV (g−1) = χV (g)

χV (h−1gh) = χV (g)

χV⊕W (g) = χV (g) +χW (g).

Here the second property follows from the fact that for n×nmatricesA andB we have Tr(AB) =Tr(BA).

Lemma 1. Let V ∗ be the dual space of V . Then χV ∗(g) = χV (g−1) = χV (g).

Proof. Let v1, v2, . . . , vn be a basis of V, with λ1, λ2, . . . , λn a dual basis of V ∗, so that λi(vj) = δij .For g ∈ G define aij(g) to be the (i, j)-entry of the matrix representation of the action of g withrespect to the dual basis λ1, . . . , λn. Then we have gλj =

∑i aij(g)λi. Hence

aij(g) = (gλj)(vi)

= λj(g−1vi)

= λj(∑

k

bki(g−1)vk)

= bji(g−1),

where g−1 acts on V by the matrix [bij(g)]ni,j with respect to the basis v1, . . . , vn. Therefore we have

χV ∗(g) =∑

i

aii(g) =∑

i

bii(g−1) = χV (g

−1) = χV (g).

Lemma 2. Let V and W be two representations of G. Then χV⊗W (g) = χV (g)χW (g).

Proof. Select bases v1, v2, . . . , vn of V and w1, w2, . . . , wm of W . Suppose that g ∈ G acts by thematrices [aij(g)]

ni,j on V and [bij(g)]

mi,j on W with respect to the aforementioned bases. Now G acts

on V ⊗W by g(v ⊗ w) = gv ⊗ gw, and the set {vi ⊗ wj} is a basis for V ⊗W . Hence

g(vi ⊗ wj) = gvi ⊗ gwj

=(∑

k

akivk

)⊗(∑

l

bljwl

)

=∑

k,l

akiblj(vk ⊗ wl).

ThereforeχV⊗W (g) =

∑

i,j

aiibjj =(∑

i

aii

)(∑

j

bjj

)= χV (g)χW (g).



0.8 Characters

Remark 2. Combining the two lemmas gives the following result:

χV ∗⊗W (g) = χV (g−1)χW (g) = χV (g)χW (g).

Definition 2. Hom C(V,W ) is a representation of G with action defined by

(gf)(v) := gf(g−1v),

for all g ∈ G, f ∈ Hom C(V,W ) and v ∈ V .

Proposition 1. Suppose that V and W are finite dimensional vector spaces over C. Then asC[G]-modules

V ∗ ⊗C W = Hom C(V,W ).

In particularχV ∗⊗W (g) = χHom C(V,W )(g).

Proof. Let v1, . . . , vn be a basis of V with corresponding dual basis λ1, . . . , λn of V ∗, and letw1, . . . , wm be a basis of W . Extend the map

Φ : V ∗ ⊗C W → Hom C(V,W )

defined by[Φ(λ⊗ w)](v) = λ(v)w

linearly. We claim this is the required C[G]-module isomorphism. Indeed, the map is injective: if∑i,j aijλi ⊗ wj ∈ kerΦ then for all v ∈ V we would have

∑i,j aijλi(v)wj = 0. By setting v = vi

we deduce aij = 0 for all i, j, showing that Φ has trivial kernel. Since Φ is injective and linear ona finite dimensional space it is an isomorphism. Finally Φ is a C[G]-module homomorphism:

[gΦ(λ⊗ w)](v) = g[Φ(λ⊗ w)](g−1v)

= g(λ(g−1v)w)

= (gλ)(v)gw

= [Φ(gλ⊗ gw)](v)

= [Φ(g(λ⊗ w))](v).

Lastly we can work out the characters for induced representations using Example 1.

Lemma 3. Let G be a finite group with subgroup H, and let V be a finite dimensional C[H]-module.If {g1, g2, . . . , gn} is a complete set of left coset representatives of H in G then

χIndGH(V )(g) =∑

i | g−1

i ggi∈H

χV (g−1i ggi).

Proof. From Example 1 we have

g(gi ⊗mk) = some linear combination of gj ⊗ml where j is fixed and l varies.

Diagonal entries in the matrix representation of g are zero, except possibly those where j = i. Sinceg acts as a block permutation matrix with submatrices given by the action of g−1

j ggi onM we have

χIndGH(V )(g) =∑

i | g−1

i ggi∈H

χV (g−1i ggi).



0.9 Character Tables

0.9 Character Tables

Definition 3. Let G be a finite group with t irreducible representations. The character table of G isthe t×t matrix with rows indexed by the irreducible representations with characters χ1,χ2, . . . ,χt,and columns indexed by the conjugacy classes C1, C2, . . . , Ct and with (i, j)th-entry χi(gj) wheregj ∈ Cj .

Remark 3. From our earlier remark characters of conjugate elements are equal, so the entiresare well defined (that is, they do not depend on the conjugacy class representative). Futher, it isknown from a theorem in Lecture 4 that for fields of characterstic zero the number of irreduciblerepresentations of G is equal to the number of conjugacy classes of G. Therefore the character tableis indeed square, with dimension equal to that of the dimension of Z(C[G]).

Example 1. Consider the character table of S3, the symmetric group on three letters. We have:

S3 (1) (12) (123)triv 1 1 1sgn 1 -1 1std 2 0 -1

Example 2. As a further example consider the character table of C3 = 〈u〉, the cyclic groupof order three with generator u. Here ζ is a third root of unity, and Xi is the (one dimensional)representation where u acts as multiplication by ζi for i = 1, 2.

C3 1 u u2

triv 1 1 1X1 1 ζ ζ2

X2 1 ζ2 ζ

Example 3. Lastly, consider the character table of S4. We can quickly calculate the charactersfor the standard representation using the fact that Cn = triv⊕ std, since this implies

χstd(g) = χCn(g)−χtriv(g) = #fixed points− 1.

From our earlier work the characters of sgn ⊗ std can be found by multiplying the correspondingcharacters of sgn and std together.

S4 (1) (12) (123) (1234) (12)(34)

triv 1 1 1 1 1sgn 1 -1 1 -1 1std 3 1 0 -1 -1sgn ⊗ std 3 -1 0 1 -1X5 2 0 -1 0 2

Remark 4. Since the sum of the squares of the dimensions of the irreducible characters is theorder of the group, this forces χX5

(1) = 2. The row for X5 can be filled out using the fact thatcolumns of the character table are orthogonal.



0.10 Character Orthogonality

0.10 Character Orthogonality

We now exhibit a characterisation of the elements of Hom C(V,W ) which are morphisms of G-modules, i.e. those f ∈ Hom C(V,W ) such that f(gv) = gf(v), for all v ∈ V , g ∈ G. To begin wedefine

e :=1

|G|

∑

g∈G

g.

It is a straightforward exercise to check that e2 = e. We have the following lemma.

Lemma 4. Let V and W be finite dimensional representations of the finite group G. Then

HomG(V,W ) = eHom C(V,W ).

Moreover,Hom C(V,W ) = eHom C(V,W )⊕ (1− e)Hom C(V,W ).

Proof. Let f ∈ HomG(V,W ). Then gf = f for all g ∈ G since

(gf)(v) = gf(g−1v) = f(g(g−1v)) = f(v),

for all g ∈ G and v ∈ V . It follows that f = ef ∈ eHom C(V,W ). Conversely if f ∈ eHom C(V,W )then f = ef ′ for some f ′ ∈ Hom C(V,W ) whence

gf = g(ef ′) = (ge)f ′ = ef ′ = f,

for all g ∈ G. This implies gf(v) = f(gv) for all g ∈ G and v ∈ V so that f ∈ HomG(V,W )as required. In order to establish the latter equality, first note that for any f ∈ Hom C(V,W ) wetrivially have

f = ef + (1− e)f ∈ eHom C(V,W ) + (1− e)Hom C(V,W ).

Lastly if f ∈ Hom C(V,W ) satisfies ef = (1− e)f , acting on both sides of the equation by e showsthat ef = 0, since e2 = e. Hence f = 0 and the sum is direct as needed.

In view of the lemma we can now prove the main result of this section.

Proposition 2.

dim(HomG(V,W )) =1

|G|

∑

g∈G

χV (g)χW (g).

In particular, if V,W are irreducible then

∑

g∈G

χV (g)χW (g) =

{|G| if V ∼=W0 if V 6∼=W.

Proof. Let P be the matrix representation of e. From the lemma after picking bases for the directsummands we see dim(eHom C(V,W )) = tr(P ). Therefore

dim(HomG(V,W )) = dim(eHom C(V,W ))

= tr(P )

=1

|G|

∑

g∈G

χHom C(V,W )(g)

=1

|G|

∑

g∈G

χV (g)χW (g).

The remaining statement follows from Schur’s Lemma.



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Representation Theory of Algebras · Introduction These notes are based on lectures given by Peter McNamara on the Representation Theory of Algebras at the University of Sydney in