August 25

References

1. (CR1) Curtis and Reiner: Methods of Representation Theory, I and II

2. (CR2) Curtis and Reiner: Representation Theory of Finite Groups and Associative Algebras

3. (DB) D. J. Benson, Representations and Cohomology I, pp. 1-19

4. (DF) Dummitt and Foote: Abstract Algebra

The modern approach to representation theory involves group-rings over certain rings, not necessarily fields. Let G be a finite group an R a commutative ring with 1.

Definition 1 An R-algebra A is a ring such that there is a homomorphism ψ : R → Z(A) with ψ(1) = 1.

Observe that if R = K is a field, that is K ⊂ Z(A) then A is a vector space over K and is also a ring. That is, A is an R-module as

ra = ψ(r)a, r ∈ R, a ∈ A

as well as a ring.

For arbitrary rings, the structure of A can be quite complicated.

Example 1 The Group-ring RG is a free R-module with basis {g : g ∈ G} (of finite rank since G is finite). That is,

RG = {∑_{x∈G} a_{x}x : a_{x} ∈ R}

with multiplication inherited from the group operation.

If R = K is a field, then KG is a group algebra.

Definition 2 An R-subalgebra of A is an R-submodule plus a subring containing 1. A map ψ : A → B is an R-algebra homomorphism if it is a ring homomorphism and a R-module homomorphism (ψ(ra) = rψ(a)).

Definition 3 A representation of A is a A-algebra homomorphism ρ : A → M_{n}(R), (n × n matrices over R) Similarly, a group homomorphsim ρ : G → GL(n, R), (n × n invertible matrices over R) is a representation of G over R.

If A = RG, a representation ρ of G gives rise to a representation of A and conversely.

Definition 4 Two representations ρ and ρ′ of A (or G) are equivalent if there exists T ∈ GL(n, R) such that ρ(a) = T ρ′(a)T^{−1} for all a ∈ A, and similarly for G.

Example 2 Representations equivalent over ℚ but not over ℤ (check this) [CR, p.205] 1. S_{3} = ⟨a, b : a^{2} = b^{2} = 1, a^{−1}ba = b^{−1}⟩ can be represented as

ρ_{1} : a ↦ ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) b ↦ ( \begin{matrix}0 & −1 \\ 0 & −1 \end{matrix} )

ρ_{2} : a ↦ ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) b ↦ ( \begin{matrix}0 & 1 \\ −1 & −1 \end{matrix} )

2. D_{8} = ⟨a, b : a^{2} = b^{4} = (ab)^{2} = 1⟩ can be represented as

ρ_{1} : a ↦ ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) b ↦ ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )

ρ_{2} : a ↦ ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) b ↦ ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )

Definition 5 Given ρ : A → M_{n}(R), we get an A-module M which is a free R-module of rank n as a · m = ρ(a)m. We define dim ρ = rank M = n.

Conversely, given an A-module M (free over R, finite rank), if we pick an R-basis of M, we get a representation of ρ. Furthermore, ρ, ρ′ are equivalent iff M, M′ are isomorphic. (see DF, p. 812)

Definition 6 An A-module M is irreducible or simple if the only submodules of M are M and {0}. M is indecomposable if we cannot write M as the direct sum of proper, non-zero submodules, i.e., M = M_{1} ⊕ M_{2}

Now if R = K is a field, M is irreducible (ρ is irreducible) means that we cannot find T with

T ρ(a)T^{−1} = ( \begin{matrix}∗ & ∗ \\ 0 & ∗ \end{matrix} )

for all a. Similarly, ρ is indecomposable means we can’t have

T ρ(a)T^{−1} = ( \begin{matrix}∗ & 0 \\ 0 & ∗ \end{matrix} ) .

Example 3 G = Z_{2} × Z_{2} = ⟨a, b : a^{2} = b^{2} = 1, ab = ba⟩. Consider representations over a field K of characteristic 2

KG = K[x, y]/ ⟨(x − 1)^{2}, (y − 1)^{2}⟩

and note that in this case (x − 1)^{2} = x^{2} − 1 and (y − 1)^{2} = y^{2} − 1. If J is a n × n matrix in indecomposable rational canonical form

ρ : x → ( \begin{matrix}I & I \\ 0 & I \end{matrix} )

ρ : y → ( \begin{matrix}I & J \\ 0 & I \end{matrix} )

is a 2n-dimensional indecomposible representation. So there are an infinite number of indecomposable representations over K. This is proved in Heller and Reiner, Illinois J. Math 5 (1961); we will a special case later.

Theorem 1 If p||G|, K a field of characteristic p, then G has a finite number of indecomposable representations over K (up to equivalence) iff G has a cyclic Sylow-p subgroup. (D. G. Higman, Duke Math. J. 21 (1954))

Then, given the hopelessness of enumerating such representations current research looks at KG = ⨁_{i} M_{i}, M_{i} indecomposable.

August 27

Aug 27

The R-algebra A is the basic object of study where R is commutative with 1, and the basic example we should have in mind is the group-ring A = RG, G a finite group. Normally, we want A to be free over R with finite rank. When we look at representations, this is the same as looking at A-modules or G-modules M. We study representations because matrices are more concrete than abstract groups; hence, we hope to obtain info about groups via representations.

Example 4 G = ⟨a : a^{p} = 1⟩. By Higman’s theorem, there is only a finite number of representations. Take K to be of characteristic p. Since KG ≅ K[x]/(x − 1)^{p}, the indecomposable representations of KG are given by

ρ(a) = ( 1 1  ⋯ 0 0 ; 0 1 1  0 0 ; ⋮ ; 0 0  1 1 ; 0 0  1 ) (r × r)

where 1 ≤ r ≤ p. We get precisely p indecomposable representations. A KG-module has the form

M = M_{0} ⊃ M_{1} ⊃ ⋯ ⊃ M_{r} = {0}

where M_{i} are KG-submodules with dim M_{i}/M_{i+1} = 1 so that M_{i}/M_{i+1} is a trivial KGmodule, i.e. gives the representation ρ(a) = 1. This follows from the Jordan form.

Definition 7 The trivial RG-module is R with gr = r for all r ∈ R, g ∈ G.

Proposition 1 Let |G| = p^{n} and char(K) = p. Then the only irreducible KG-module is the trivial one.

Proof. Let V be an irreducible KG-module. Let H ≤ G be the kernel of the representation, that is,

H = {x ∈ G : ρ(x) = I}

If H ≠ 1, consider G/H. We have |G/H| < |G| and apply induction as follows. If W is a G/H module, W can be regarded as a KG-module by g · x = (gH) · x. This is well defined because the representation is trivial on H. W is irreducible as a G/H-module implies W is irreducible as a G-module also, and the result follows by induction.

Now suppose H = 1. Choose x ∈ Z(G) with x ≠ 1 (recall that the center of a p-group is always ≥ 1). Consider (x − 1)V. We show this is a submodule of V. If v ∈ V, g ∈ G, g(x − 1)v = (x − 1)gv because x ∈ Z(G) so that g(x − 1)v ∈ (x − 1)V. Also, (x − 1)V is not 0 since x ≠ 1. Therefore, since V is irreducible, we have that V = (x − 1)V. But similarly we have V = (x − 1)^{2}V = (x − 1)^{3}V = ⋯ = (x^{p^{n}} − 1)V = 0, a contradiction. Note that 0 is not irreducible by definition.

Definition 8 Let A be any ring with 1 and let M be an A-module. Then a series of submodules

M = M_{0} ⊃ M_{1} ⊃ ⋯ ⊃ M_{n} = 0

is called a composition series if each M_{i}/M_{i+1} is irreducible.

Not every module has a composition series. For example, ℤ does not; for example, the series

ℤ ⊃ (p) ⊃ (p^{2}) ⊃ ⋯ ⊃ (p^{n}) ⊃ ⋯

does not terminate.

Proposition 2 (Butterfly Lemma) If M^{∗} ≤ M, N^{∗} ≤ N are submodules of some module, then M^{∗} + (M ∩ N^{∗}) ≤ M^{∗} + (M ∩ N), N^{∗} + (M^{∗} ∩ N) ≤ N^{∗} + (M ∩ N), and

(M^{∗} + (M ∩ N) )/(M^{∗} + (M ∩ N^{∗}) ) ≅ ( (M ∩ N) + N^{∗})/( (M^{∗} ∩ N) + N^{∗})

[figure: M; N; M^{∗} + (M ∩ N); (M ∩ N) + N^{∗}; =; M ∩ N; =; M^{∗} + (M ∩ N^{∗}); =; (M^{∗} ∩ N) + N^{∗}; M^{∗}; (M^{∗} ∩ N) + (M ∩ N^{∗}); N^{∗}; M^{∗} ∩ N; M^{∩}N^{∗}]

Proposition 3 (Schreider’s Theorem) If

M = M_{1} ≥ M_{2} ≥ ⋯ M_{r} = {0}
N = N_{1} ≥ N_{2} ≥ ⋯ N_{s} = {0}

are two series for an A-module M, and if M has a composition series, we can refine the two series to two composition series such that the factors in the first series are isomorphic in some order to the factors in the second series.

[figure: M_{i}; N_{j}; M_{i,j}; N_{j,i}; =; M_{i} ∩ N_{j}; =; M_{i,j+1}; N_{j,i+1}; M_{i+1}; N_{j+1}]

where M_{i,j} = M_{i+1} + (M_{i} ∩ N_{j}) and N_{j,i} = N_{j} + (M_{i} ∩ N_{j}).

Corollary 1 (Jordan Hölder Theorem) Any two composition series of M have the same composition factors up to isomorphism.

Proposition 4 M has a composition series iff M satisfies the Ascending Chain Condition and the Descending Chain Condition.

August 29

Proposition 5 (1) If A is any ring, M has a composition series iff (2) M has ACC and DCC for submodules iff (3) every series of submodules of M can be refined to a composition series. [DF, p. 438, 637-638, 717]

(1) ⟺ (3) follows from the Butterfly lemma. For (1) ⟺ (2), see e.g.[CR2, §13].

Example 5 ℤ as a ℤ-module has ACC but not DCC and the ℤ-module

ℤ_{p∞} = ⟨a_{1}, a_{2} … a_{n} …: a_{1}^{p} = 1, a_{2}^{p} = a_{1} … a_{i+1}^{p} = a_{i} …⟩

is an abelian p-group such that

⟨a_{1}⟩ < ⟨a_{2}⟩ < ⋯ < ℤ_{p∞}.

This group has DCC but not ACC as a ℤ-module.

We see from this example that to get a finite series, we need to be able to go both up and down from a submodule and end in a finite number of steps.

Now given an R-algebra A, our goal is to find and classify the irreducible A-modules. However, this is obviously too ambitious!

Definition 9 A itself as a left A-module, sometimes written _{A}A, gives rise to the regular representation.

In this situation, the submodules of _{A}A are the left ideals.

Let M be a simple A-module (recall that this doesn’t include the zero module) and let m ∈ M, m ≠ 0. Then Am = {am : a ∈ A} is a non-zero submodule, so Am = M. Also φ : A → M defined a ↦ am is a surjective A-module homomorphism, that is, φ(am) = aφ(m) for a ∈ A, m ∈ M. The kernel of φ is a maximal left ideal L (since M is simple) such that A/L ≅ M. From this we explore the idea that simple modules correspond to maximal left ideals. This will lead to the Jacobson radical.

Next, we will want to extend concepts from commutative rings to non-commutative rings. Recall from [DF p.650] that if R is a commutative ring with 1, we have the nilradical

N = ⋂_{P prime} P = {x ∈ R : x^{n} = 0, some n ≥ 1}

and the radical

rad R = ⋂_{L maximal} L.

Then N ⊂ rad R. Recall also that x ∈ rad R iff 1 − rx is a unit for all r ∈ R.

Now let A be any ring with 1, possibly non-commutative.

Definition 10 Define the Jacobson radical J(A), sometimes written rad A,

J (A) = ⋂_{L a maximal ; left ideal} L

Definition 11 If M is an A-module, ann M = {a ∈ A : aM = 0}. If m ∈ M, ann(m) = {a ∈ A : am = 0}.

Note that ann M is a two-sided ideal and ann(m) is a left ideal.

Proposition 6 J(A) is the annihilator of all the simple A modules, i.e.,

J(A) = ⋂_{M simple} ann M

Note that if M is an A-module, we have a ring homomorphism A → End_{R}(M) (= Hom_{R}(M, M)) whose kernel is two-sided ideal. (This expresses the fact that we have a representation.) So the proposition shows that J(A) is a two-sided ideal and not simply a left ideal.

Proof. Suppose M is simple and as before, M = Am for some m ≠ 0. We have a homomorphism φ : A → Am defined a ↦ am. Then φ is a surjective A-module homomorphism so that A/L ≅ M where L = ker φ is a maximal left ideal. That is

L = {a ∈ A : am = 0} = ann(m).

Then

⋂_{M simple} ann M = ⋂_{M simple} ⋂_{0≠m∈M} ann(m) ⊃ ⋂_{L maximal} L = J(A).

Conversely, if aM = 0, then am = 0 so a ∈ ann(m) = L, where L is a maximal left ideal, for all m ∈ M. Also every maximal left ideal arises in this way. Then

⋂_{M simple} ann(M) ⊂ ⋂_{L a maximal ; left ideal} L = J(A)

Example 6 Let G be a cyclic group of order p and let A = KG for K with char K = p. M is simple implies that M is the trivial module. Then J(A) is the ideal of A spanned by g − 1, g ∈ G. Then dim J(A) = p − 1 = dim A − 1.

Exercise 1 Look at the example in DF on page 828 of G = S_{3} in which a simple module of dimension 2 is constructed.

September 3

Note that we generally use the symbol R for a commutative ring with 1 and the symbol A for any ring with 1.

Additional reference: [H] T. Hungerford, Algebra. Look at the section on Jacobson radicals in chapter 9. Note also that when Jacobson introduced the radical, he was considering any ring, not necessarily with 1.

Recall that

J(A) = ⋂_{M a maximal left ideal} M = ⋂_{M simple} ann(M)

If we only want to consider simple modules, we may as well study A/J(A). Since these descriptions involve maximal ideals, which we don’t want to compute, and simple modules, which are what we want to determine in the first place, we develop an intrinsic description of J(A). (CR, §5 or H, Ch. 9)

Theorem 2

J(A) = {x ∈ A : 1 − axb is a unit for all a, b ∈ A}
= {x ∈ A : 1 − ax has a left inverse for all a ∈ A} .

Proof. We prove first that J(A) = {x ∈ A : 1 − ax has a left inverse for all a ∈ A}. Let x ∉ J. Then x ∉ L, some maximal left ideal L. then ⟨L, x⟩ = A (the ideal generated by L and x). Therefore, 1 = y + ax for some y ∈ L, a ∈ A. Therefore 1 − ax = y ∈ L. Then 1 − ax does not have a left inverse else y ∈ L has a left inverse and then 1 ∈ L.

Alternately, assume 1 − ax has a left inverse for all a ∈ A. If xW ≠ 0 for some W simple, then xw ≠ 0 for some w ∈ W. Then W = Axw. Then w = axw for some a so that (1 − ax)w = 0. Then, since 1 − ax has a left inverse,w = 0, a contradiction. So xW = 0 for all simple W , and x ∈ J. For the converse, it suffices to show that 1−x has a left inverse for x ∈ J, i.e. A(1 − x) = A. If not, A(1 − x) ⊂ L for some maximal left ideal L. Then 1 − x ∈ L, but since x ∈ L since x ∈ J, we have that 1 ∈ L, a contradiction.

Next, to prove the first equality, note that from the above proof, 1 − axb is a unit implies x ∈ J. To show the converse, it suffices to show that if x ∈ J, then 1 − x is a unit.

Again, by the first part, take some t with t(1 − x) = 1. Then t − tx = 1 so that 1 − t = −tx ∈ J. So 1 − (1 − t) = t has a left inverse u, that is ut = 1. Then u = ut(1−x) = 1−x so that tu = 1. Thus, t(1−x) = 1, (1−x)t = 1, i.e. t is a two-sided inverse of 1 − x.

Remark 1 Write t = 1−y. Then (1−y)(1−x) = 1. Expanding, we have 1−y−x+xy = 1 so that xy−x−y = 0 and note that we’ve eliminated the reference to 1. Now for any ring, not necessarily with 1, define x to be left-quasi-regular if there exists y with x+y+xy = 0. Then J can be defined in this case using this concept, see e.g. (H, p.426).

Recall now that if A has a 1, every proper left ideal is contained in a maximal left ideal (DF, p.255, H, p. 128, CR2). Therefore we know that there exist maximal ideals in our case and hence there exist simple A-modules. It is therefore non-vacuous to say that J “kills” all simple modules. If however a ∉ A and there are no simple A-modules, we say that J(A) = A is a radical ring.

Now our goal is to study A/J(A). Recall the example where G is cyclic of order p^{n} and K is a field is a field of characteristic p. If A = KG, dim A = p^{n} and dim J(A) = p^{n} − 1.

Example 7 Let A = M_{n}(K) for K a field. Our aim is to write this as a direct sum of simple modules, i.e. as a sum of minimal left ideals. Write

E_{i,j} = ( 0 ⋯ 0 ⋯ 0 ; ⋮ ; 0 1 ; ⋮ ; 0 )

where the 1 is in position i, j. Then E_{i,j}E_{k,l} = δ_{j,k}E_{i,l} and

AA = ⨁_{i=1}^{n} L_{i}

where

L_{i} = \{∑_{j} a_{j}E_{j,i} : a_{j} ∈ K\} = ( \begin{matrix}& ∗ \\ & ∗ \\ 0 & ∗ & 0 \\ & ∗ \\ & ∗ \end{matrix} )

L_{i} is an irreducible module as follows. Let x ∈ L_{i} with x ≠ 0. Let x = ∑_{j} a_{j}E_{j,i}, and suppose a_{k} ≠ 0. Then E_{j,k}x ∈ L_{i} and has a_{k,i} in the jth row and 0 in the other rows. Thus we get all the E_{j,i} in L_{i}, showing that any non-zero element in L_{i} generates all of L_{i}. Thus L_{i} is irreducible. (Remark: We used the fact that K is a field here.)

L_{i} ≅ L_{j} for each i, j: This follows from the matrix representation given by L_{i} as x → x for x ∈ A. Also, note that J(A) = 0 since we have maximal left ideals L′_{j} = ⨁_{i≠j}^{n} L_{i}, whose intersection is 0.

September 5

Example 8 Let A = M_{n}(D) forD a division ring. The decomposition of A into the direct sum of simple modules can be done in the same way as for M_{n}(K) with K a field, as in the previous example. Again A = L_{1} ⊕L_{2} ⊕⋯⊕L_{r} for L_{j}, the left ideals and simple A-modules constructed before. Also as before, J(A) = 0 since x ∈ J implies xA = 0 since it kills all the simple modules, but A has 1, so x = 0. Also note that if we set e_{i} = E_{i,i}, then e_{1}^{2} = e_{1} and L_{i} = Ae_{i}. This says that each L_{i} is generated by e_{i} an idempotent.

Digression. Let T : V → V be a linear transformation. T^{2} = T means that T is a projection. With a suitable basis, T has matrix

( \begin{matrix}1 \\ & 1 \\ & & . & . & . \\ & & & & & 0 \\ & & & & & & 1 \end{matrix} )

and in fact V = ker T ⊕ im T. Now if e ∈ A with e^{2} = e, then A = Ae ⊕ A(1 − e) since if x ∈ A, x = xe + x(1 − e) so that if x ∈ Ae, then xe = x and if x ∈ A(1 − e), then xe = 0.

Definition 12 A is called (left) artinian if _{A}A has the DCC on submodules.

Then A is artinian if and only if every non-empty set of left ideals has a minimal element.

Proposition 7 Assume A is artinian. Then every nilpotent left ideal is contained in J and J itself is nilpotent. (CR1)

Proof. We have a chain

J ⊃ J^{2} ⊃ ⋯ ⊃ J^{i} ⊃ ⋯

which terminates as

J^{m} = J^{m+1} = ⋯ = J^{2m} = ⋯

If J^{m} ≠ 0, then J^{2m} = J^{m}J^{m} ≠ 0. Choose a minimal left ideal I such that J^{m}I ≠ 0. Then for some a ∈ I, we have J^{m}a ≠ 0 and J^{m}a ⊂ I, but J^{m}(J^{m}a) = J^{m}a ≠ 0 so that J^{m}a = I since I is the minimal ideal for which J^{m}I ≠ 0. Now since a ∈ I we can write a = xa for some x ∈ J^{m}. Then (1 − x)a = 0 implies a = 0 since 1 − x has a left inverse. But this is a contradiction, so J^{m} = 0, that is J is nilpotent.

Now let N be a nil left ideal, that is, every element is nilpotent. Then x ∈ N implies x^{k} = 0 for some k ≥ 1 so that 1 − x is invertible. This means that x ∈ J so that N ⊂ J.

Example 9 If A = KG, then J(A) is nilpotent. If G is some p-group, then J is the span of {x − 1 : x ∈ G} and (x − 1)^{p^{n}} = 0

Proposition 8 Assume A is artinian. If I is a non-nilpotent left ideal, then I contains an idempotent (CR2, 24.2)

In particular if J = 0, then A would be non-nilpotent and we would be able to do the decomposition as in Example 8. Proof. Let I_{1} be a minimal left ideal among the non-nilpotent left ideals contained in I. Then I_{1}^{2} = I_{1} by minimality. Let L_{1} be minimal in {L ⊂ I_{1} : I_{1}L ≠ 0}. Then since I_{1}x ≠ 0 for some x ∈ L_{1} and I_{1}x ⊂ L_{1}, we have by minimality that I_{1}x = L_{1}. Then x = ax for some a ∈ I_{1}. But then x = ax = a^{2}x = ⋯ so that (a^{2} − a)x = 0. Let N = {u ∈ I_{1} : ux = 0} ⊂ I_{1} and N ≠ I_{1} for if N = I_{1} then

0 = Nx = I_{1}x = L_{1} ≠ 0,

a contradiction. Then a^{2} − a ∈ N and N is nilpotent since N ⊊ I_{1} and I_{1} was chosen to minimal among non-nilpotent ideals. Let n_{1} = a^{2} − a. If n_{1} = 0, we’re done as a would be an idempotent. If not, let a_{1} = a + n_{1} − 2an_{1} ∈ I_{1} and repeat the above process producing n_{2} = a_{1}^{2} −a_{1}. We will get a sequence which will eventually give an idempotent.

September 8

Sketch of proof of the previous proposition (for details see the last lecture):

We have found I_{1} ≤ I, non nilpotent, and N ⊂ I_{1} with N nilpotent. We have an asequence in I and an n-sequence in N. As before, take n_{1} = a^{2} − a. If this is zero, we’re done. If not, recursively define a_{i} = a_{i−1} + n_{i} (1 − 2a_{i−1}) and n_{i+1} = a_{i}^{2} − a_{i}. In fact n_{2} = 4n_{1}^{3} − 3n_{1}^{2} so n_{1}^{2}|n_{2} in A. Inductively, n_{i+1} has n_{1}^{2^{i}} as a factor. N is nilpotent implies that n_{1}^{2^{k}} = 0 for some k. Hence n_{k+1} = 0 and a_{k}^{2} = a_{k}, that is, a_{k} is an idempotent.

The definition of a_{i} may have been motivated by calculus as follows (folklore). We’re trying to solve f(x) = x^{2} − x. Thus

a_{1} = a + n_{1}(1 − 2a)
= a − (a^{2} − a)(2a − 1)
= a − f(a)f′(a)

Recall that M_{n}(D) = ⨁_{j=1}^{n} L_{j} is the direct sum of simple left ideals L_{j} with L_{i} ≅ L_{j}. Also, note that simple left ideals are the same as irreducible left ideals and the same as minimal left ideals.

Definition 13 Let A be any ring with 1. We say that A is left-semi-simple or completely reducible if _{A}A is the direct sum of simple left ideals. Similarly, an A-module M is leftsemi-simple or completely reducible if M is the direct sum of simple submodules.

Proposition 9 (H, Ch. 9, Th 3.6 or CR2, §15.3) The following are equivalent:

1. M = ⨁_{i∈I} M_{i} with M_{i} simple

2. M = ∑_{j∈J} M_{j} with M_{j} simple

3. For any M′ a submodule of M, we have M = M′ ⊕ M′′ for some submodule M′′ of M.

Observe that if N ≤ M and L ≤ M with L simple, then N ∩ L is either L or 0. Hence N + L is either N or N ⊕ L. Proof. The proof uses Zorn’s Lemma over and over. (1) ⇒ (2) is obvious. For (2) ⇒ (1), among all partial sums ∑_{i∈I} M_{i}, I ⊂ J, which are direct, choose a maximal L and then show that L = M by observing that if some j ∉ I, then L⊕M_{j} is direct by the observation above, a contradiction.

For (2) ⇒ (3), given M′, choose a maximal M′′ = ∑_{i∈I} M_{i} with M′′ ∩ M′ = 0 by Zorn’s lemma. Then show that M = M′ ⊕ M′′. If not, pick m ∈ M with m ∉ M′ ⊕ M′′. Then m is in a sum of a finite number of simple submodules, and at least one of them, say M_{k}, is not contained in M′ ⊕ M′′. This would contradict maximality of M′′.

For (3) ⇒ (2), first show that if N ≤ M with N ≠ 0, then N has a simple submodule. Pick some 0 ≠ n ∈ N and choose some N_{0} maximal such that n ∉ N_{0}. By the assumption, we have N = N_{0} ⊕N_{1} for some N_{1}. Then N_{1} is simple, for if not, N_{1} = N_{2} ⊕N_{3}, but then n ∉ N_{0} + N_{2} or n ∉ N_{0} + N_{3} which contradicts maximality. In particular, M contains simple modules. Among all ∑_{i∈T} M_{i} with M_{i} simple, find a maximal L. If L ≠ M, then M = L ⊕ L′ and L′ contains a simple submodule, which contradicts the maximality of L

Finally, we have the main theorem on Artinian rings.

Theorem 3 If A is an Artinian ring with 1, then J(A) = 0 if and only iff _{A}A is semisimple.

Proof. If J = 0, let L_{1} be a minimal left ideal. Then L_{1} is not nilpotent, so L_{1} = Ae for some e with e^{2} = e so that A = Ae ⊕ A(1 − e) = L_{1} ⊕ L′_{1}. If L′_{1} is non minimal, let L_{2} ≤ L′_{1} be minimal and A = L_{1} ⊕ L_{2} ⊕ L′_{2}, etc. Then L′_{1} ⊃ L′_{2} ⊃ ⋯ terminates so that A = ⊕_{j=0}^{n}L_{j}.

Conversely, let _{A}A be semi-simple and write A = J ⊕ J′ by criterion (3). We then have that 1 = x + x′ with x ∈ J and x′ ∈ J′. Then x = x^{2} + xx′ which means that x^{2} − x ∈ J ∩ J′ = 0. But x − x^{2} = x(1 − x) = 0 means that x = 0 since 1 − x is a unit. This means that x′ = 1 so that J′ = A and J = 0.

A counterexample for this in non-artinian rings is J(ℤ) = 0 since the maximal ideals are {(p) : p prime}, but ℤ is not semi-simple (in our sense). For a structure theorem on arbitrary rings A with J(A) = 0 see H, Chapter 9, 3.2.

September 10

We proved last time that if A is artinian, then J (A) = 0 if and only if _{A}A is semi-simple. From this, it follows that if J (A) ≠ 0, then A/J (A) is semi-simple since J (A/J (A)) = 0.

Proposition 10 Let A be any ring with 1. Then submodules and quotient modules of completely reducible modules are completely reducible.

Proof. (Schur’s Lemma) First note that if L and M are simple A-modules and φ : L → M is a A-module homomorphism, then φ(L) is either 0 or M and ker φ is either 0 or L, so φ is ether 0 or an isomorphism. In particular, End_{A}(M) is a division ring.

If M is completely reducible and N ≤ M, then M = ∑_{i∈I} M_{i} for M_{i} simple so that M/N = ∑_{i∈I} M_{i}/N_{i}. If φ : M → M/N is the canonical map, then φ(M) is the sum of those M_{i} such that φ(M_{i}) ≠ 0 and such that φ(M_{i}) is simple.

Now if M is completely reducible and N ≤ M, then M = N ⊕ N′ for some N′ and so N ≅ M/N′.

Proposition 11 Assume _{A}A is semisimple. Then

1. _{A}A is the direct sum of a finite number of left A-ideals.

2. Every finitely generated A-module is completely reducible.

3. Every A-module is completely reducible.

Proof. If _{A}A is semisimple, then _{A}A = ∑_{i∈I} L_{i} with L_{i} minimal left ideals. But since the ring has 1, we have 1 ∈ L_{i_{1}} + L_{i_{2}} + ⋯ + L_{i_{n}} and similarly for a ∈ A so that A = L_{i_{1}} + L_{i_{2}} + ⋯ + L_{i_{n}}.

Suppose M is finitely generated by {m_{1}, m_{2}, ⋯ m_{r}}. Then we have a surjective Amodule homomorphism

φ : \underbrace{A ⊕ A ⊕ ⋯ ⊕ A}_{r} → M given by φ (a_{1}, a_{2}, ⋯ a_{r}) = ∑_{i=1}^{r} a_{i}m_{i}.

Since A ⊕ A ⊕ ⋯ ⊕ A is semisimple, M is semisimple.

Finally, suppose M is an A-module. Then

M = AM = (∑_{i∈I} L_{i}) M = ∑_{i∈I} ∑_{m∈M} L_{i}m.

But by the above proposition, each L_{i}m is either 0 or simple, so M is completely reducible

The following theorem provides the main example of an artinian, semisimple ring.

Theorem 4 (Maschke’s Theorem) If G is a finite group and K is a field of characteristic either 0 or p with p |̸ |G|, then KG is semisimple (DF, p. 815).

Note that KG is artinian as it is a finite dimensional algebra and a vector space over K Remark: If M is an A-module and N ≤ M, then N is a direct summand of M if and only if there exists a A-module homomorphism π : M → N with π|N = i, the identity map. Confirm that if we have such a π, then M = N ⊕ ker π. Proof. Let M be a KG-submodule and N ≤ M. Since M, N are vector spaces over K, find subspace W of M such that M = N ⊕ W where ⊕ denotes direct sum as vector spaces. Then we have a K-homomorphism π : M → N with π|_{N} = i. Let π′ : M → M be given by

π′(u) = \frac{1}{|G|} ∑_{g∈G} gπ (g^{−1}u) .

Show that π′ is a KG-module homomorphism, π′|_{N} = i, and Im (π′) = N. For example, if u ∈ N,

π′(u) = \frac{1}{|G|} ∑_{g∈G} g (g^{−1}u) = u

as g^{−1}u ∈ N, and if x ∈ G, u ∈ M,

π′(xu) = \frac{1}{|G|} ∑_{g∈G} gπ (g^{−1}xu) = \frac{1}{|G|} ∑_{g∈G} x (x^{−1}g) π (g^{−1}xu) = xπ′(u)

so that π′ is a G-module homomorphism, hence a KG-module homomorphism.

Exercise 1 (DF p.828) Let G = S_{3} and let V = ⟨e_{1}, e_{2}, e_{3}⟩ be the natural module defined ge_{i} = e_{g(i)}. Let V_{1} = ⟨e_{1} + e_{2} + e_{3}⟩ ≤ V. Then V = V_{1} ⊕ W as K-spaces where the characteristic of K is 0 and W = ⟨e_{2}, e_{2}⟩. However, W is not a KG-module. Go through Maschke’s Theorem to start from W to find a KG-module W_{1} with V = V_{1} ⊕ W_{1}

Preview: If A is an artinan ring and semisimple, we know that A = ∑ L_{i}. What is the structure of A as a ring?

September 12

We’ve shown that if A is artinian with 1 and semisimple, then _{A}A is completely reducible. Consider the ring A = M_{n}(D) where D is a division ring. We know that A = ⨁_{j=1}^{n} L_{j} where L_{j} = AE_{j,j} and L_{i} ≅ L_{j} for all i, j. Now we’re interested in recovering D knowing the above decomposition.

Recall that Hom_{A} (M, N) is an abelian group when A is a ring, but if A is an R-algebra where R is a commutative ring, then Hom_{A} (M, N) becomes an R-module.

Consider

Hom_{A} (L_{i}, L_{j}) = Hom_{A} (AE_{i,i}, AE_{j,j}) .

By Schur’s lemma, every element of this vector space over D is either 0 or an isomorphism.

Lemma 1 If A is any ring, M is an A-module, and e ∈ A with e^{2} = e, then Hom_{A} (Ae, M) ≅ eM.

Proof. Let φ : eM → Hom_{A} (Ae, M) be given by em ↦ α : e ↦ em for m ∈ M. Now let ψ : Hom_{A} (Ae, M) → eM be given by α ↦ α (e). Note that if α (e) = m, then α (e · e) = em so m = em ∈ eM. Then, φ and ψ are inverses.

Corollary 2 End_{A} (Ae)^{op} ≅ eAe where for rings B, B^{op} is the abelian group B with multiplication reversed.

Proof. Check that the isomorphism given in the theorem reverses the multiplication in the two rings.

Then Hom_{A} (AE_{i,i}, AE_{j,j}) ≅ E_{i,i}AE_{j,j}. So if x ∈ A, x = ∑_{k,l} a_{k,l}E_{k,l} so that E_{i,i}xE_{j,j} = a_{i,j}E_{i,j} (check this). Hence Hom_{A} (AE_{i,i}, AE_{j,j}) ≅ D. In particular, End (AE_{i,i}) ≅ D^{op}. Then we have that if A = M_{n} (D) and L is a minimal left ideal, then End_{A} (L) ≅ D^{op}.

Suppose now that V_{1}, V_{2}, W_{1}, W_{2} are A-modules for A any ring. Consider

Hom_{A} (V_{1} ⊕ V_{2}, W_{2} ⊕ W_{2}) ≅ Hom_{A} (V_{1}, W_{1}) ⊕ Hom_{A} (V_{1}, W_{2}) ⊕
Hom_{A} (V_{2}, W_{1}) ⊕ Hom_{A} (V_{2}, W_{2})

θ → ( \begin{matrix}θ_{1,1} & θ_{2,1} \\ θ_{1,2} & θ_{2,2} \end{matrix} )  (1)

where θ_{i,j} = P_{j}θL_{i}, the L_{i} being injections of V_{i} into V_{1} ⊕ V_{2} and the P_{j} being projections of W_{1} ⊕ W_{2} on W_{j}.

Formally, if θ ∈ Hom_{A} (V_{1} ⊕ V_{2}, W_{1} ⊕ W_{2}), then θ corresponds to the matrix as in (1)

θ : ( \begin{matrix}v_{1} \\ v_{2} \end{matrix} ) → ( \begin{matrix}θ_{1,1} & θ_{2,1} \\ θ_{1,2} & θ_{2,2} \end{matrix} ) ( \begin{matrix}v_{1} \\ v_{2} \end{matrix} )

for v_{1} ∈ V_{1} and v_{2} ∈ V_{2}. This gives an additive group isomorphism End_{A} (V_{1} ⊕ V_{2}) ≅Group of matrices as in (1). Finally, if V_{1} = V_{2} = V, then

End_{A} (V ⊕ V) ≅ M_{2} (End_{A} (V))

Check that if θ → (θ_{j,i}) and Φ → (φ_{j,i}), then θΦ → (θ_{j,i}) (φ_{j,i}).

The same proof shows that End_{A} (nV) ≅ M_{n} (End_{A} (V)).

If V is simple, then End_{A} (V) = D a division ring and then End_{A} (nV) ≅ M_{n} (D).

Lemma 2 End_{A} (_{A}A) ≅ A^{op} Proof. Let φ ∈ End_{A} (_{A}A). Then φ : A → A and if 1 → x, so that a → ax, φ is just right multiplication by x. Conversely, if x ∈ A, φ : a → ax is an A-module homomorphism _{A}A →_{A} A. Now if φ ⟷ x, i.e., φ(1) = x, and ψ ⟷ y, i.e., ψ(1) = y, then (φψ) (1) = φ (y) = yx

Finally, let A be artinian and semisimple. Then_{A}A ≅ ⨁_{j=1}^{r} n_{j}S_{j} for S_{j} simple A-modules with S_{i} not isomorphic to S_{j} if i ≠ j. Then

End_{A} (_{A}A) ≅ ⨁_{j=1}^{r} End_{A} (n_{j}S_{j})

≅ ⨁_{j=1}^{r} M_{n_{j}} (D′_{j})

Note that (M_{n} (D))^{op} ≅ M_{n} (D^{op}) (take transposes of matrices).

Finally, A ≅ ⨁_{j=1}^{r} M_{n} (D_{j}) and this is the Wedderburn decomposition with D_{j} ≅ End_{A} (S_{j})^{op}.

September 15

Last time we proved the Wedderburn Theory, that is, if A is artinian with 1 and semi-simple, then A ≅ ⨁_{j=1}^{r} M_{n_{j}} (D_{j}). Here, if _{A}A ≅ ⨁ n_{j}S_{j} for S_{j} simple, then D_{j} ≅

End_{A} (S_{j})^{op}

Corollary 3 If A is commutative, then A is the direct sum of fields.

This follows since if M_{n_{j}} (D_{j}) is a field, then n_{j} will have to be 1 and D_{j} has to be commutative. Remark: If A is a finite dimensional D-algebra with D a division ring and

A = L_{0} ⊋ L_{1} ⊋ ⋯ ⊋ ⋯

is a chain of left ideals, the chain has to terminate as dim_{D} L_{i} > dim_{D} L_{i+1}. Compare this with ℤ ⊃ (p) ⊃ (p^{2}) ⋯ where the rank does not decrease.

Definition 14 A ring A is simple if it has no two-sided ideals except {0} and A.

We know that M_{n} (D) is simple so that the Wedderburn theory says that if A is a semisimple artinian ring with 1, then A is a direct sum of simple rings.

Proposition 12 (CR1 §3.24) A simple artinian ring is semi-simple.

Proof. Let L be a minimal left ideal of A. Take B = ∑_{a∈A} La. Then B is a two-sided ideal since if x ∈ A, then (La) x is either 0 or of the same form. Then B = A so that A is semi-simple.

Now we want to study the internal structure of a semi-simple artinian ring A. Write A = A_{1} ⊕ A_{2} ⊕ ⋯ ⊕ A_{r} for A_{j} subrings of A with A_{j} ≅ M_{n_{j}} (D_{j}). In fact, A_{j} is the sum of all the minimal left ideals which are isomorphic to S_{j}. First, observe that A_{j} is a two-sided ideal of A as follows.

If x, y ∈ A, then

x = x_{1} + x_{2} + ⋯ + x_{r}  (2)

and

y = y_{1} + y_{2} + ⋯ + y_{r}  (3)

for x_{j}, y_{j} ∈ A_{j}. Then xy = x_{1}y_{1}+⋯ x_{r}y_{r} so xy_{j} = x_{j}y_{j} ∈ A_{j} as A_{j} is a subring. Similarly, y_{j}x = y_{j}x_{j} ∈ A_{j}.

Now let

1 = e_{1} + e_{2} + ⋯ e_{r}  (4)

for e_{j} ∈ A_{j}. From (2) and (4), we have

x = x1 = ∑ x_{j}e_{j} = ∑ x_{j}

so that x_{j}e_{j} = x_{j}. Similarly, e_{j}x_{j} = x_{j} which says that e_{j} is the identity element for A_{j}. Also, e_{j}^{2} = e_{j} and e_{j} ∈ Z (A) as e_{j}x = xe_{j} for all x ∈ A. Note also that e_{i}e_{i} = e_{i} and e_{i}e_{j} = 0 for i ≠ j as e_{i}e_{j} ∈ A_{i} ∩ A_{j}. Thus we say the e_{i} are orthogonal idempotents. They are also central, that is, contained in the center of A. Furthermore, A_{j} = Ae_{j}.

Theorem 5 Let A be a semi-simple, artinian ring. Then A is the direct sum A = ⨁ A_{j} of two-sided ideals which are simple rings. Each A_{j} itself is the direct sum of minimal left ideals, all isomorphic to a simple A-module S_{j}. Furthermore, every simple A-module is isomorphic to some S_{j}.

Example 10 Let n = 3 and G = ⟨a : a^{3} = 1⟩. Take A = ℚG. Then dim_{ℚ} A = 3. We can take e_{1} = \frac{1}{3} (1 + a + a^{2}). Then e_{1} is a central idempotent (as similarly defined elements are for all group-rings) Then Ae_{1} ≅ M_{1} (ℚ). Next, we need to fill in the blank in A ≅ M_{1} (ℚ) ⊕ with some algebra of dimension 2.

September 17

In the last decomposition, 1 = e_{1} + e_{2} + ⋯ e_{r} for e_{j} ∈ A_{j}. The e_{j} are central idempotents, i.e., xe_{j} = e_{j}x for all x ∈ A and x_{j} is the identity element of A_{j} = Ae_{j}.

Definition 15 Idempotents e and f are orthogonal if ef = fe = 0. An idempotent e is primitive if e can’t be written as the sum e_{1} + e_{2} of orthogonal idempotents.

Then the e_{j} above are primitive central and orthogonal idempotents (i.e. can’t be written as a sum of orthogonal central idempotents), for otherwise if e_{j} = e′_{j} + e′′_{j}, then Ae_{j} = Ae′_{j} ⊕ Ae′′_{j} contradicts the simplicity of A_{j} = Ae_{j}.

Remark: The A_{j} and the e_{j} are unique with respect to a decomposition of A into the direct sum of simple rings. This is because the A_{j} are defined as the sum of all the minimal left ideals isomorphic to one fixed simple module S_{j}. Note that a simple artinian ring has only one simple module up to isomorphism (this follows from the proof given earlier that a simple artinian ring is semisimple). However, _{A}A = ⨁ L_{j} is not unique. For example, in M_{n} (D) there are many ways of writing 1 as a sum of orthogonal idempotents.

Example:Returning to G = S_{3} = ⟨(12), (23)⟩ V the standard module ⟨v_{1}, v_{2}, v_{3}⟩, and A = KG for K a field, we have two cases to consider.

Case 1. char K ≠ 3. Then V = V_{1} ⊕ V_{2} where V_{1} = ⟨v_{1} + v_{2} + v_{3}⟩ and V_{2} = ⟨v_{1} − v_{2}, v_{2} − v_{3}⟩ as before. Case 2. char K = 3. If

(123) ↦ ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} ) ∼ ( \begin{matrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix} ) ,

then the eigenvalues are all 1. We need to find a basis for V to exhibit a composition series. We begin with v_{1} + v_{2} + v_{3} as the first basis element and extend to a full basis of V. Consider the basis {v_{1} + v_{2} + v_{3}, v_{1} − v_{2}, v_{1}} and check that with respect to this basis,

(12) ↦ ( \begin{matrix}1 & 0 & 0 \\ 0 & −1 & 1 \\ 0 & 0 & 1 \end{matrix} )

(23) ↦ ( \begin{matrix}1 & −1 & 0 \\ 0 & −1 & 0 \\ 0 & 0 & 1 \end{matrix} )

(123) ↦ ( \begin{matrix}1 & −1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix} )

Then

V_{1} = ⟨v_{1} + v_{2} + v_{3}⟩ ⊂ V_{2} = ⟨v_{1} + v_{2} + v_{3}, v_{1} − v_{2}⟩ ⊂ V

This chain is called a flag and is fixed by G. Also, it is a composition series with simple factors giving the trivial and the sign representations as follows. On V_{1}, G has the trivial representation. This corresponds to the 1,1 entry of the matrices. On V_{2}/V_{1}, G has the sign representation, i.e., g ↦ sgn g, This corresponds to the 2,2 entry of the matrices. On V/V_{2}, G again has the trivial representation. In the previous example, we initially took a representation of G in char 0, and then regarded the matricies ρ(g) as matricies over K with char K = 3. This method is known as the method of going mod p.

From now on, A will be a group ring. We will consider K a field of charactersitic 0 containing a suitable ring R which has a prime ideal 𝔭, such that R/𝔭 = k is a field of charactersitc p. Now if we have a representaion ρ of G over K, we write the matrices ρ(g) over R and then reduce the entries mod 𝔭 to get a modular representation \overline{ρ} of G over k of charactersitc p. For example, going mod 3 on the two dimensional representation of S_{3} gave rise to two one-dimensional representations over 𝔽_{3}. In that case, K = ℚ, R = ℤ, and k = 𝔽_{3}.

The course bifurcates from this point into two branches, one branch involving further development in characterstic 0, and the other further exploring going mod p. The latter path leads to current research in representation theory.

September 19

We continue with group algebras of characteristic 0. Let G be a finite group and A=CG.

Remark 2 If K is an algebraically closed field and D ⊃ K is a division algebra with

[D : K] < ∞, then D = K 16

Proof. Let a ∈ D. Then a is algebraic over K, so let f(x) be the minimal polynomial of A over K. If f(x) = h(x)k(x) with deg h, deg k < deg f. Then

0 = f(a) = h(a)k(a)

so that one of h(a) or k(a) is 0 since D is a division algebra. This contradicts minimality of f. Hence, f(x) is irreducible. This together with the fact that K is algebraically closed imply that deg f = 1 so that f(x) = x − a. Thus, a ∈ K.

We therefore have

ℂG ≅ ⨁_{j} M_{n_{j}} (ℂ) .

Note that each factor of this sum corresponds to one simple algebra A_{j} in the decomposition A = ⨁ A_{j}. We know that M_{n_{j}} is the sum of n_{j} irreducible left ideals, each of which has dimension n_{j}. (Recall that these irreducible left ideals are columns of the matrix algebras and hence have dimension n_{j}.) Therefore, each A_{j} corresponds to one irreducible A- module of dimension n_{j} repeated n_{j} times (in the sense that it occurs n_{j} times in a decomposition of A as left A-module) . So we get ∑_{j} n_{j}^{2} = |G| where the n_{j} are the dimensions of the irreducible representations of G. For example, with G = S_{3}, we have 6 = 1 + 1 + 2^{2}.

Next, we want to describe r in terms of the group. We have

A = ℂG = A_{1} ⊕ A_{2} ⊕ ⋯ ⊕ A_{r}.

Consider the center Z of A. We then have

Z = Z_{1} ⊕ Z_{2} ⊕ ⋯ ⊕ Z_{r}

with Z_{j} ≅ Z (A_{j}). However, since each A_{j} is a matrix algebra over ℂ, we have r = dim_{ℂ} Z. We construct a basis of Z over ℂ. If C_{1}, C_{2} ⋯ C_{s} are the conjugacy classes of G, we claim that the set

{k_{j} = ∑_{g∈C_{|}} g : j = 1, 2, …, s}

forms a basis of Z. The k_{j} are clearly independent. To see that they generate Z, let x ∈ ∑_{g} a_{g}g ∈ Z and let h ∈ G. Then hxh^{−1} = ∑_{g} a_{g} (hxh^{−1}) = ∑_{g} a_{g}g = ∑ a_{hgh^{−1}}hgh^{−1} so that a_{g} = a_{hgh^{−1}} for all g, h ∈ G. That is, if x ∈ Z, then conjugates must have the same coefficients (see DF, p.827). This says that x is a linear combination of the k_{j}.

Therefore r = s and the number of irreducible representations of G is equal to the number of conjugacy classes of G. Note also that this holds only when the field is ℂ. This was not the case in the homework problem about ℤ_{n} over ℚ, for example.

We now develop the theory of characters. Let ρ : G → GL (n, ℂ) be a group homomorphism and ρ : A = ℂG → M_{n} (ℂ) be an algebra homomorphism. Define the function χ : G → ℂ by χ(g) = Tr (g) for g ∈ G, and similarly χ(x) = Tr (x) for x ∈ A. Now since Tr (X) = Tr (P XP^{−1}) for matrices X and P, we have that equivalent representations have the same character. Also χ (hgh^{−1}) = χ (g) for g, h ∈ G so that χ is constant on conjugacy classes. We call such functions class functions.

Recall that Tr (x + y) = Tr x + Tr y and Tr (xy) = Tr (yx).

Remark 3 What if the field has characteristic p? For example, S_{3} has a representation of dimension 2 given by V_{2} = ⟨v_{1} − v_{2}, v_{2} − v_{3}⟩. Consider V_{2} as a module over a field K of characteristic 2, then we get

1 ↦ ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) .

But then χ(1) = 0, but we would want χ(1) to be the dimension of the representation. There is a way to get around this and define characters in characteristic p due to Brauer which we will see later.

September 22

Hint for homework: Produce e_{1} and e_{2}, the trivial and sign idempotents. Take e_{3} = 1 − e_{1} − e_{2}, write Ae_{3} = M_{2} (ℚ), and figure out explicitly what the isomorphism is.

Next we study characters. Let A = RG for R a commutative ring with 1. Let M and N be A-modules which are finitely generated and free as R-modules. Construct the following new A-modules from M and N.

1. M^{∗} = Hom_{R} (M, R), the dual of M, that is, R linear maps of M into R. If M has an R- basis {v_{1}, v_{2} ⋯ v_{n}}, then M^{∗} has the dual basis {f_{1}, f_{2} ⋯ f_{n}} where f_{i} (v_{j}) = δ_{i,j}. Then M^{∗} can be made into a left A-module by

(g · f) (m) = f (g^{−1}m) .

Note also that we can make M^{∗} into a right A-module by

(f · g) (m) = f (gm) .

(The latter right module is more natural, but we convert it into a left module by using the “antipode” g → g^{−1}.) If χ is the character of the representation of G on M and ψ is the character of the representation of G on M^{∗}, we see that χ (g) = ψ (g^{−1}). Hence, if R = ℂ, then ψ = \overline{χ} since the eigenvalues of ρ (g) where ρ is the representation are roots of unity and for roots of unity, inverses are complex conjugates.

2. M ⊗_{R} N is an A-module by

g (m ⊗ n) = gm ⊗ gn.

Note that M ⊗_{R} N is naturally a (G × G)-module by

(g, g′) (m ⊗ n) = gm ⊗ g′n.

Then we have the diagonal embedding G → G×G given by G ↦ (g, g). Also, there is a nice isomorphism M^{∗} ⊗_{R} N ≅ Hom_{R} (M, N) given by

f ⊗ n ↦ [m ↦ f(m) · n ∈ R].

Using this we make Hom_{R} (M, N) into an A-module by (g · f) (m) = g · f((g^{−1}m)).

September 24

Continuing from last time, Hom, Dual, and the tensor product are three functors deriving from an A-module. If A = RG and M is a finitely generated free R-module, we make M^{∗} = Hom_{R} (M, R) and M ⊗_{R} N into A-modules as in the last lecture. Note that

M^{∗} ⊗_{R} N ≅ Hom_{R} (M, N) .  (5)

Hom_{R} (M, N) is an A-module by

(gf) (m) = g · f (g^{−1}m)

and this definition is compatible with (5), that is, the map given to induce (5) is Gequivariant i.e., φ : M^{∗} ⊗_{R} N → Hom_{R} (M, N) is such that gφ = φg for g ∈ G. Check this. Also M ⊕ N is an A-module by

g (m, n) = (gm, gn) .

Finally, recall that if A and B are R-algebras, then so is A ⊗_{R} B as in DF, p. 355. Then if M is an RG-module and N is a RH module, then M ⊗_{R} N is an RG ⊗ RH-module by

(g ⊗ h) (m ⊗ n) = gm ⊗ hn

In fact, this comes from the following. Let M, N, M′ and N′ be R-modules . If φ : M → N and φ′ : M′ → N′ are R-module homomorphisms then we get an R- homomorphism φ ⊗ φ′ : M ⊗_{R} M′ → N ⊗_{R} N′ by m ⊗ m′ ↦ φ (m) ⊗ φ (m′). Now if M and N are RG-modules and M′ and N′ are RH-modules, φ : M → N is an RG-homomorphism, and φ′ : M′ → N′ is an RH-homomorphism, then φ ⊗ φ′ : M ⊗ M′ → N ⊗ N′ is an RG ⊗ RH-homomorphism. We note that RG ⊗ RH ≅ R (G × H). Now if we have ρ : RG → End_{R} (M) and ρ′ : RG → End_{R} (N), then we get ρ⊗ρ′ : RG → End_{R} (M ⊗_{R} N) using the diagonal embedding G → G × G given by g → (g, g).

Digression: (CR2, p.69) Let S = (α_{i,j}) for 1 ≤ i, j ≤ n and T = (β_{i,j}) for 1 ≤ i, j ≤ m be matrices of linear transformations ρ, ρ′ of a vector space V and a vector space W respectively with respect to bases (v_{i} : 1 ≤ i ≤ n} and {w_{j} : 1 ≤ j ≤ m} of V and W . Then {v_{i} ⊗ w_{j} : 1 ≤ i ≤ n, 1 ≤ j ≤ m} is a basis for V ⊗ W, and the matrix S ⊗ T for ρ ⊗ ρ′ with respect to a suitable ordering of the above basis looks like

( β_{j,k}S )

Then Tr (S ⊗ T) = Tr S × Tr T.

Let M be an A-module, corresponding to a representation ρ of A = ℂG with character χ. Then ρ : A → End_{ℂ} (M) and χ (g) = Tr (ρ (g)) = Tr (g, M) = χ_{M} and similarly for N. We then have the following correspondence:

Representation

Character

M^{∗}  \overline{χ} i.e. χ_{M^{∗}}(g) = \overline{χ(g)} = χ (g^{−1})
M ⊗ N  χ · ψ i.e. χ_{M⊗N} (g) = χ (g) ψ (g)
M ⊕ N  χ + ψ 19

Now if n_{i} = dim χ_{i} for 1 ≤ i ≤ r with χ_{i} the irreducible characters, then we know ∑_{i} n_{i}^{2} = |G| and r is the number of conjugacy classes of G. Pick a representative g_{i} ∈ C_{i} where {C_{i} : 1 ≤ i ≤ r} are the conjugacy classes of G with C_{1} = 1

We record all the irreducible characters in a table, called the character table of G:

Character C_{1}  C_{2}  ⋯ C_{j}  ⋯ C_{r}
χ_{1}  1  1  1  1
⋮
χ_{i}  χ_{i} (1) χ_{i} (g_{2})  χ_{i} (g_{j})  χ_{i} (g_{r})
⋮

For example, the character table for S_{3} is given by

1 (123) (12)
χ_{1} 1 1  1
χ_{2} 1 1 −1
χ_{3} 2 −1 0

Now look at KS_{3} for K a field of characteristic 3. Only the columns for 1 and (12) are relevant since the eigenvalues of (123) are all 1. We get two modular irreducible characters, the trivial and the sign.

We have so far considered left A-modules, and right A-modules are similar.

For A and B algebras, we can also consider bimodules. Thus M is an (A, B)-bimodule if A acts on the left, B on the right, and a (m · b) = (a · m) b for a ∈ A,m ∈ M, and b ∈ B. (DF p. 347). For example, if A = RG, then A is a bimodule whose submodules are 2-sided ideals. Consider A = M_{n} (C). What is its structure as an A-bimodule, that is, what is the representation of A ⊗ A^{op} on A?

September 26

Exercise 2 Let A = M_{n} (K) for K a field. Consider the linear transformation

(x, y) ↦ [φ_{x,y} : s ↦ xsy]

for x, y ∈ A. Then this is a map A ⊗ A → End_{K} (A). Check that Trφ_{x,y} = TrxTry by considering the natural basis {E_{i,j} : 1 ≤ i, j ≤ n} of A and computing each xE_{i,j}y.

Let A = ℂG. We have the left regular map

a ↦ [ρ_{a} : x ↦ ax]

and the right regular map

a ↦ [ψ_{a} : x ↦ xa].

The right regular map is an anti-representation because ψ_{a}ψ_{b} = ψ_{a} (xb) = xba whereas ψ_{ab} (x) = xab. We therefore consider A as an A ⊗ A-module or an A-bimodule by

g ⊗ h ↦ [x ↦ gxh^{−1}].  (6)

Remark 4 The trace of the regular representation R afforded by ρ_{a} is

R (g) = \{ \begin{matrix}|G| & g = 1 \\ 0 & g ≠ 1 \end{matrix}

We compute the trace of (6) in two ways.

1. The trace of g ⊗ h is the number of x ∈ G such that x = gxh^{−1}, that is, x^{−1}gx = h. In particular, if g is not conjugate to h, then the trace is 0. Fix x ∈ G with x^{−1}gx = h. Then for any y ∈ G, we have

y^{−1}gy = h ⇔ xy^{−1} ∈ C_{G} (g) ⇔ yx^{−1} ∈ C_{G} (g) ⇔ y ∈ C_{G} (g) · x

so that

Tr (g ⊗ h) = \{ \begin{matrix}|C_{G} (g)| & g, h conjugate \\ 0 & otherwise \end{matrix}

2. We have A = ⨁_{j=1}^{r} A_{j} for A_{j} simple two sided ideals. Then the trace of g ⊗ h is

∑_{j=1}^{r} χ_{j} (g) χ_{j} (h^{−1})

where χ_{j} is the irreducible character corresponding to A_{j} by Exercise 2. We therefore have

∑_{j=1}^{r} χ_{j} (g) χ_{j} (h^{−1}) = \{ \begin{matrix}|C_{G} (g)| & g, h conjugate \\ 0 & otherwise \end{matrix}

This is known as the second orthogonality relation. It gives a relation between the columns of the character table. Example 11 The character table for S_{3}.

1 (123) (12)
χ_{1} 1  1 1
χ_{2} 1 −1 1
χ_{3} 2 −1 0

We can rewrite the second orthogonality relation as

∑_{j=1}^{r} χ_{j} (g) \overline{χ_{j} (g)} = |C_{G} (g)|  (7)

since

\overline{χ_{j} (g)} = \overline{∑_{k} ε_{k}} = ∑_{k} \overline{ε_{k}} = ∑_{k} ε_{k}^{−1} = χ_{j} (g^{−1})

where ε_{k} are the eigenvalues of ρ_{j} (g).

Take {g_{j} ∈ C_{j} : 1 ≤ j ≤ r} to be representatives of the conjugacy classes of G and write (7) in matrix form as follows.

( χ_{1} (g_{1}) χ_{2} (g_{1}) ⋯ χ_{r} (g_{1}) ; χ_{1} (g_{2}) χ_{2} (g_{2}) ; ⋮ ; χ_{1} (g_{r})  χ_{r} (g_{r}) ) ( \overline{χ_{1} (g_{1})} \overline{χ_{2} (g_{1})} ⋯ \overline{χ_{r} (g_{1})} ; \overline{χ_{1} (g_{2})} \overline{χ_{2} (g_{2})} ; ⋮ ; \overline{χ_{1} (g_{r})}  \overline{χ_{r} (g_{r})} )

= ( \begin{matrix}|C_{G} (g_{1})| & & & & & 0 \\ & |C_{G} (g_{2})| \\ & & . & . & . \\ 0 & & & & & |C_{G} (g_{r})| \end{matrix} )

But since AB = D for D diagonal implies that BA = D, we have

( \frac{\overline{χ_{1}(g_{1})}}{m_{1}}  \frac{\overline{χ_{2}(g_{1})}}{m_{2}} ⋯ \frac{\overline{χ_{r}(g_{1})}}{m_{r}} ; \frac{\overline{χ_{1}(g_{2})}}{m_{1}}  \frac{\overline{χ_{2}(g_{2})}}{m_{2}} ; ⋮ ; \frac{\overline{χ_{1}(g_{r})}}{m_{1}}  \frac{\overline{χ_{r}(g_{r})}}{m_{r}} ) ( χ_{1} (g_{1}) χ_{2} (g_{1}) ⋯ χ_{r} (g_{1}) ; χ_{1} (g_{2}) χ_{2} (g_{2}) ; ⋮ ; χ_{1} (g_{r})  χ_{r} (g_{r}) ) = I_{r}

where m_{j} = |C_{G} (g_{j})|. Then

∑_{j=1}^{r} \frac{\overline{χ_{k} (g_{j})}χ_{l} (g_{j})}{m_{j}} = \{ \begin{matrix}1 & k = l \\ 0 & otherwise \end{matrix}

so that

\frac{1}{|G|} ∑_{j=1}^{r} |C_{j}| \overline{χ_{k} (g_{i})}χ_{l} (g_{j}) = \frac{1}{|G|} ∑_{g∈G} \overline{χ_{k} (g)}χ_{l} (g) = δ_{k,l}

and this is known as the first orthogonality relation. This is a relation between the rows of the character table.

Now if N ◁ G and ψ is a representation of G/N, we get a representation of G by ψ̃: g ↦ ψ (gN). Then ψ̃ is a pullback as in

G ⟶ G/N ⟶ GL (n, k)

and ψ is irreducible if and only if ψ̃ is irreducible. We can see this more easily via the orthogonality relations.

September 29

Definition 16 C (G) is the vector space over ℂ of class functions on G.

There is a natural inner product on C (G) given by

⟨φ_{1}, φ_{2}⟩ = \frac{1}{|G|} ∑_{g∈G} φ_{1} (g) \overline{φ_{2} (g)}

Then the set of irreducible characters {χ_{1}, χ_{2} ⋯ χ_{r}} form an orthonormal basis of C (G). Now if χ is any character, we can write χ = ∑_{j} n_{j}χ_{j} for n_{j} ≥ 0, n_{j} ∈ ℤ and ⟨χ, χ_{j}⟩ = n_{j}. Note also that any character χ is irreducible iff ⟨χ, χ⟩ = 1. In terms of modules, if χ corresponds with M and χ_{j} corresponds with S_{j}, then M ≅ ⨁_{j} n_{j}S_{j} where n_{j} = dim Hom_{A} (S_{j}, M) = dim Hom_{A} (M, S_{j}) by Schur’s lemma since Hom_{A} (S_{i}, S_{j}) = δ_{i,j}.

Proposition 13 If χ, ψ are irreducible characters of G and H respectively, then χ × ψ defined (χ × ψ) (x, y) = χ (x) ψ (y) is an irreducible character of G × H and all the irreducible characters of G × H are obtained in this way.

Proof. Show that

\frac{1}{|G × H|} ∑_{(x,y)} (χ × ψ) (x, y) \overline{(χ × ψ) (x, y)} = 1

and also note that the number of conjugacy classes of G × H is the number of conjugacy classes of G times the number of conjucacy classes of H.

If χ is a virtual character, that is, χ = ∑_{j} n_{j}χ_{j} with n_{j} ∈ ℤ, then ±χ is an irreducible character iff ⟨χ, χ⟩ = ±1. This will not be true, however, if n_{j} ∉ ℤ. For example, if χ = \frac{1}{2}χ_{1} + \frac{1}{2}χ_{2} + \frac{1}{2}χ_{3} + \frac{1}{2}χ_{4}, then ⟨χ, χ⟩ = 1.

Let ρ be a representation of G with character χ and let g ∈ G. Then the eigenvalues of ρ (g) are {ε_{j} : 1 ≤ j ≤ n} where ε_{j} is an mth root of unity if g has order m. Note that χ (g) = ∑_{j} ε_{j} and that

|χ (g)| = |∑_{j} ε_{j}| ≤ ∑_{j} |ε_{j}| = n

with equality iff each ϵ_{j} is real. Hence, χ (g) = n = χ (1) iff ρ (g) = I. Note that if we consider the subgroup ⟨g⟩ ⊂ G and look at an irreducible representation ρ on ⟨g⟩, we have that ρ (g) are scalar matrices. From the character table, we see that if χ (g) = n, then g ∈ ker ρ. Hence, we can identify normal subgroups of G from the character table.

October 1

Comments on homework 3.

If M is a completely reducible A-module and e is a central idempotent in A, then M = eM ⊕ (1 − e) M. Note that since e is central, we have that eM and (1 − e) M are submodules, and e^{2} = e. Therefore e acts as a projection of M on eM. eM is the sum of all the simple submodules of M isomorphic to the simple A-module corresponding to e. For example, A = ⨁ Ae_{j} is a special case with M =_{A} A.

In the homework problem on finding f_{3}, take H = ⟨(12)⟩ and take e to be \frac{1}{2} (1 + (12)). This is an idempotent corresponding to the trivial representation of the subgroup H. Now take M to be Ae. Then Ae = Aee_{3} ⊕Ae (1 − e_{3}) and apply the above comments. In other words, we find the part of the module Ae which gives the representation of dimension 2, which is Aee_{3} and is irreducible. Thus we can take f_{3} to be ee_{3}.

We consider characters of Abelian groups. Let G be Abelian and let ρ be an irreducible representation of G over ℂ as usual. By Schur’s lemma, if g ∈ G, ρ (g) = λ_{g}I. But since block diagonal matrices can’t be irreducible, we must have that dim ρ = 1, that is, ρ (g) = λ_{g} so that ρ : G → ℂ^{×} is a homomorphism. (We can also show that each irreducible representation of an Abelian group is of dimension 1 by Wedderburn theory; see e.g. [DF, p.827].)

For general G, if g ∈ G and ρ is any representation of G, consider ⟨g⟩. Since ρ is completely reducible, we have ρ (g) is diagonalizable since ⟨g⟩ is Abelian. Last time we saw that χ (g) = χ (1) implies each ε_{j} is 1. Then ρ (g) = I so that g ∈ ker ρ. The converse is also true, that is, if g ∈ ker ρ, then χ (g) = χ (1).

We therefore have that normal subgroups can be detected from the character table. In fact, N ◁ G iff

N = ⋂_{j} { ker χ_{j} : χ_{j} irreducible with N ⊂ ker χ_{j}}  (8)

Proof. First note that the intersection of the kernels of all the χ_{j} is precisely {1} by the Wedderburn decomposition A = ⨁_{j} Ae_{j}. Then if N ◁ G, consider G/N and note that the irreducible representations of G/N are obtained from the irreducible representations of G which contain N in their kernel. Conversely, every irreducible representation of G/N gives rise to a representation of G containing N in its kernel. This proves (8).

Next, we want to construct the central idempotents of A = ℂG. Consider Z = Z (A). (Incidentally, note that x ∈ Z if x is a linear combination of class sums; this can be used in homework 3.) Then Z has dimension r and a basis {C_{j} : 1 ≤ j ≤ r} where C_{j} = ∑_{g∈C_{j}} g. Let h_{j} = |C_{j}|. Also, note that Z = ⨁_{j=1}^{r} Ze_{j} with Ze_{j} ≅ ℂ. So {e_{j} : 1 ≤ j ≤ r} is another basis of Z. Let C_{i} = ∑_{j} a_{i,j}e_{j} and e_{i} = ∑_{j} b_{i,j}C_{j}. We want b_{i,j} ∈ ℂ.

Observe that Ze_{j} is the center of Ae_{j} and χ_{j} is the character of the representation corresponding to Ae_{j} by our notational convention.

Note that χ_{i} (e_{j}) = 0 when i ≠ j and χ_{i} (e_{j}) = n_{i} = χ_{i} (1). Now C_{i} are represented (in the representation corresponding to χ_{j}), by scalar matrices, say

C_{i} ↦ (\begin{matrix}ω_{j} (C_{i}) \\ & ω_{j} (C_{i}) \\ & & . & . & . \\ & & & & & ω_{j} (C_{i}) \end{matrix}) .

Taking traces, h_{i}χ_{j} (g_{i}) = n_{j}ω_{j} (C_{i}). Here g_{i} ∈ C_{i}. So we let

ω_{j} (C_{i}) = \frac{h_{i}χ_{j} (g_{i})}{n_{j}}.

The notation ω_{j} (C_{i}) reflects the fact that C_{i} ↦ ω_{j} (C_{i}) is a homomorphism of Z into ℂ. The ω_{i} are the characters (1-dimensional representations) of Z into ℂ.

Then for g ∈ C_{i}, χ_{j} (C_{i}) = h_{i}χ_{j} (g_{i}) = a_{i,j}n_{j} so that

a_{i,j} = \frac{h_{i}χ_{j} (g_{i})}{n_{j}} = ω_{j} (C_{i}) .

Substituting, we have

C_{i} = ∑_{j} \frac{h_{i}χ_{j} (g_{i}) e_{j}}{n_{j}} ,

\frac{1}{|G|} ∑_{i=1}^{r} \overline{χ_{k} (g_{i})}C_{i} = \frac{1}{|G|} ∑_{j} ∑_{i} \frac{h_{i}}{n_{j}} χ_{j} (g_{i}) \overline{χ_{j} (g_{i})}e_{j} = ∑_{j} \frac{1}{n_{j}} e_{j}δ_{j,k} = \frac{e_{k}}{n_{k}}

so that

e_{k} = \frac{n_{k}}{|G|} ∑_{i=1}^{r} \overline{χ_{k} (g_{i})}C_{i}.

See DF, p.836 for a different way of getting the e_{k}. This method is important because we will be using the ω_{j} later. This proof can be found in [CR2, p.236].

October 3

We explain the idea behind the calculation of the central idempotents. Let C_{i} be the class sum, g_{i} ∈ C_{i}, n_{k} = χ_{k} (1), and h_{i} the number of elements of C_{i}.

We have two bases of Z(A) given by

C_{i} = ∑_{j} a_{i,j}e_{j}  (9)

e_{i} = ∑_{j} b_{i,j}C_{j}.  (10)

The a_{i,j} are easier to find. We have χ_{j} (C_{i}) = ∑_{j} a_{i,j}χ_{j} (e_{j}) = a_{i,j}n_{j} since χ_{j} (e_{i}) = δ_{i,j}n_{j}, but on the other hand, χ_{j} (C_{i}) = h_{i}χ_{j} (g_{i}). So h_{i}χ_{j} (g_{i}) = a_{i,j}n_{j} or a_{i,j} = h_{i}χ_{j} (g_{i}) /h_{j}. Invert the equation (9) to get (10) by multiplying (9) by \overline{χ_{k} (g_{i})} and summing over i to get the idempotents e_{j}.

Let

ω_{j} (C_{i}) = \frac{h_{i}χ_{j} (g_{i})}{n_{j}}  (11)

Then ω_{j} : Z → ℂ given by (11) is a character of Z obtained by restricting χ_{j} to Z, that is, in the representation of A corresponding to Ae_{j}, C_{i} → ω_{j} (C_{i}) I

We digress to consider algebraic integers [see, e.g. DF p. 853]. If K ⊃ ℚ is an extension field, an element a ∈ K is algebraic over ℚ if a is a root of a monic polynomial p (x) ∈ ℚ [x]. If p (x) ∈ ℤ [x], then a is an algebraic integer.

We have [e.g. DF, p 853] that

1. sums and products of algebraic integers are algebraic integers,

2. if a ∈ ℂ, a is an algebraic integer over ℚ iff ℤ [a] is a finitely generated ℤ=module,

3. if a is an algebraic integer and is in ℚ, then a ∈ ℤ.

Returning to characters, we have that χ_{j} (g_{i}) is an algebraic integer, since it is a sum of roots of unity. We want to show that ω_{j} (C_{i}) is an algebraic integer

Now

ω_{j} (C_{i}) = \frac{h_{i}χ_{j} (g_{i})}{n_{j}}.

Consider the basis {C_{i}} of Z = Z (A) and write C_{k}C_{j} = ∑_{l} c_{k,j,l}C_{l} for some c_{k,j,l}. In fact, c_{j,k,l} is the number of pairs g, g′ with gg′ = h a fixed element in C_{l} where g ∈ C_{k} and g′ ∈ C_{j}. In particular, c_{j,k,l} is an integer. Since ω_{i} is a character of Z, ω_{i} (C_{k}) ω_{i} (C_{j}) = ∑_{l} c_{k,j,l}ω_{i} (C_{l}). Hence, ℤ [1, ω_{i} (C_{k}) : 1 ≤ i, k ≤ r] is finitely generated. Since ℤ is a PID, ℤ [ω_{j} (C_{i})] is finitely generated so that ω_{j} (C_{i}) is an algebraic integer.

Theorem 6 If χ_{i} (1) = n_{i}, then n_{i}| |G| for each i.

Proof.

∑_{j=1}^{r} ω_{i} (C_{j}) \overline{χ_{i} (g_{j})} = ∑_{j=1}^{r} \frac{h_{j}χ_{i} (g_{j}) \overline{χ_{i} (g_{j})}}{n_{i}} = \frac{1}{n_{i}} ∑_{j=1}^{r} χ_{i} (g_{j}) \overline{χ_{i} (g_{j})} = \frac{|G|}{n_{i}}.

Now we use the fact that \frac{|G|}{n_{i}} is an algebraic integer and is rational, hence an integer.

Remark 5 n_{i}| |G| is not true in fields of characteristic p where p| |G|

Our last topic in this section is that of induced representations. Many of our computations have involved lifting characters from normal subgroups. When we don’t have normal subgroups, we want to be able to do something similar for arbitrary subgroups. If H ≤ G and ρ is a representation of G, then we get a restriction of ρ to H, written ρ_{H}, ρ|_{H}, or Res_{H}^{G} (ρ). Can we go the other way?

Now if ρ is a representation of H, we want to construct Ind_{H}^{G} (ρ) a representation of G and correspondingly for characters, that is, if ψ is a character of H, we want to construct a character Ind_{H}^{G} (ψ). and we would like

⟨Ind_{H}^{G} (ψ) , ζ⟩ = ⟨Res_{H}^{G} (ζ) , ψ⟩ ,

where ζ is an arbitrary character of G. That is, Ind and Res should be adjoint operators.

October 6

Today we develop the theory of induced representations, which is discussed in CR1 §10, p.227 (see also DF, p.858). As before R is a commutative ring with 1. Let H ≤ G and let A = RG and B = RH.

In fact, let A be a ring with 1 and B a subring of A containing 1. Restriction gives a way to produce B-modules from A-module as follows. If M is an A-module, we get a B-module M|_{B} or Res_{B}^{A} (M) obtained by restricting the scalars.

The reverse operation is called induction, which allows us to produce A-modules from Bmodules. If L is a B-module, define an A-module, written Ind_{B}^{A} (L), by Ind_{B}^{A} (L) = A⊗_{B}L. It is then regarded as a A module by

a (x ⊗ l) = ax ⊗ l

for a, x ∈ A and l ∈ L.

Now let A = RG. In this situation, this construction is known as Frobenius induction. If M is an A-module, then Res_{H}^{G} (M) is a B-module as before, and if L is a B-module, then Ind_{H}^{G} (L), also written L^{G}, is an A module by

g (x ⊗ l) = gx ⊗ l

for g, x ∈ G and l ∈ L.

We have alternate descriptions of Ind_{H}^{G} (L) in the context of modules, matrix representations, and characters.

1. (Modules) Let {x_{1}, x_{2} ⋯ x_{n}} be left coset representatives for H in G. Then G = ⋃_{i} x_{i}H, a disjoint union. Then {x_{i} ⊗ l : 1 ≤ i ≤ n, l ∈ L} generate L^{G} since g ∈ G implies g = x_{i}h for some x_{i} and some h, so that

g ⊗ l = x_{i}h ⊗ l = x_{i} ⊗ hl.

We could express this as

L^{G} = ⨁_{i=1}^{n} x_{i} ⊗ L

and under this identification, 1 ⊗ L ≅ L so that L can be regarded as an RHsubmodule of L^{G}.

2. (Matrices) Let L correspond to the matrix representation L of H, that is, h ↦ L (h). Now since g (x_{i} ⊗ l) = gx_{i} ⊗ l for g, x_{i} ∈ G we have gx_{i} = x_{j}h for some coset representative x_{j} and some h ∈ H, and x_{j} is unique given g and x_{i}. Then

gx_{i} ⊗ l = x_{j}h ⊗ l = x_{j} ⊗ \underbrace{hl}_{∈L} .

Note that h = x_{j}^{−1}gx_{i}

Now if L^{G} corresponds to the matrix representation L^{G},

L^{G} (g) = ( \begin{matrix}... & & 0 \\ & ... & 0 \\ 0 & 0 & L (x_{j}^{−1}gx_{i}) \end{matrix} )

This can be seen as follows. Given x_{i}, g, we have x_{j}^{−1}gx_{i} ∈ H for precisely one j and we get L (x^{−1}gx_{i}) in the (j, i) block of L^{G}.

Example 12 When G = S_{3} and H = ⟨(123)⟩, let L be the trivial representation of H. then G = H ∪ (12) H with representatives x_{1} = 1 and x_{2} = (12). We want to compute L ((13)). Write g = (13). Then

gx_{1} = (13) 1 = (13) = (12) (132) = x_{2} · h

for h = (132) and

gx_{2} = (13) (12) = (123) = x_{1}h′

for h′ = (123). Then L^{G} ((12)) = ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ). Similarly, L^{G} ((1)) = (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}) and

L^{G} ((123)) = (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}).

Example 13 Let H ≤ G. Then Ind_{H}^{G} (1) is the permutation representation of G on the cosets of H. Conversely, if π is a transitive permeation representation of G on a set S, let H be the stabilizer of a fixed element x ∈ S. Then π is equivalent to Ind_{H}^{G} (1).

3. (Characters) Let A = KG and B = KH. Let λ be the character of the representation L corresponding to L. We want to find λ^{G}, the character of L^{G}. It is sufficient to look at blocks of the form L (x_{i}^{−1}gx_{j}). Thus,

λ^{G} (g) = ∑_{i=1}^{n} λ (x_{i}^{−1}gx_{k})

with the convention that λ (x_{i}^{−1}gx_{i}) = 0 if x_{i}^{−1}gx_{i} ∉ H. Alternately, set

λ^{·} (x_{i}^{−1}gx_{i}) = \{ \begin{matrix}λ (x_{i}^{−1}gx_{i}) & x_{i}^{−1}gx_{i} ∈ H \\ 0 & otherwise \end{matrix}

Then λ^{G} (g) = ∑_{i=1}^{n} λ^{·} (x_{i}^{−1}gx_{i}).

Alternately, take

λ^{G} (g) = \frac{1}{|H|} ∑_{x∈G} λ^{·} (x^{−1}gx)

and this is choice free.

Note that λ^{G} (1) = [G : H]

Remark 6 (a) Ind is additive, that is, Ind_{H}^{G} (L_{1} ⊕ L_{2}) ≅ Ind_{H}^{G} (L_{1}) ⊕ Ind_{H}^{G} (L_{2}).

(b) Ind is transitive, that is, If K ≤ H ≤ G, then

Ind_{K}^{G} (L) = Ind_{H}^{G}Ind_{K}^{H} (L) .

Next time we consider Frobenius reciprocity, that is

Hom_{RG} (L^{G}, M) ≅ Hom_{RH} (L, M|_{H})

for L an RH-module, and M and RG-module.

October 8

Theorem 7 (Adjoint Associativity, CR1 §2.19) If A and B are two rings, L′ a left Amodule, M′ is a (B, A)-bimodule and N′ is a left B module. Then

Hom_{A} (L′, Hom_{B} (M′, N′)) ≅ Hom_{B} (M′ ⊗_{A} L′, N′)

as abelian groups.

Theorem 8 (Frobenius Reciprocity CR1 p. 232) Let H ≤ G and let L be a RH-module and M an RG-module. Then

Hom_{RH} (L, M_{H}) ≅ Hom_{RG} (L^{G}, M)

Proof. Set A = RH, B = RG L′ = L, M′ = RG, and N′ = M as in (7). Then

Hom_{RG} (L, Hom_{RG} (RG, M)) ≅ Hom_{RG} (RG ⊗_{RH} L, M)

Then the right hand side is as required. For the left hand side, we have Hom_{RG} (RG, M) is an RH-module by

(bf) (a) = f (ab)

for b ∈ RH, a ∈ RG, and f ∈ Hom_{RG} (RG,). Then τ : Hom_{RG} (RG, M) → M which maps f to f(1) is an RH-isomorphism as follows.

τ (bf) = (bf) (1) = f(b) = bf(1) = bτ (f)

since f is an RG-module homomorphism. Then the left hand side is as required.

Corollary 4 If R = ℂ or some other field of charactersic 0, ψ is a character of H and ζ is a character of G, then

⟨Ind_{H}^{G} (ψ) , ζ⟩_{G} = ⟨ψ, Res_{H}^{G} (ζ)⟩_{H} .

This says that Ind and Res are adjoint functors.

Returning to characters,

λ^{G} (g) = \frac{1}{|H|} ∑_{x∈G} λ^{·} (x^{−1}gx)

where λ is a character of H and g ∈ G. Now we want to simplify this to a more manageable form.

First we digress to consider double cosets (DF p. 119, problem 10) If H, K ≤ G, then G = ⋃_{x} HxK for disjoint double cosets HxK. We count

|HxK| = \frac{|H|}{[K : K ∩ x^{−1}Hx]}
= \frac{|K|}{[H : H ∩ xKx^{−1}]}
= \frac{|K| |H|}{[H ∩ xKx^{−1}]}

We fix the following notation. Write C for the conjugacy class of g in G and

C ∩ H = ⋃_{i=1}^{s} C′_{i}

where C′_{i} are the conjugacy classes of H contained in C. Let h_{i} ∈ C′_{i}. Then

λ^{G} (g) = |C_{G} (g)| ∑_{i=1}^{s} \frac{1}{|C_{H} (h_{i})|}λ (h_{i})

where λ^{G} (g) = 0 if C ∩ H = ∅. Proof. Let

X_{i} = {x ∈ G : x^{−1}gx conjugate to h_{i} inH}

Then

λ^{G} (g) = \frac{1}{|H|} ∑_{\begin{matrix}x∈G \\ x^{−1}gx∈H \end{matrix}} λ (x^{−1}gx) = \frac{1}{|H|} ∑_{i=1}^{s} |X_{i}| λ (h_{i}) .  (12)

For each i, fix t_{i} ∈ X_{i} such that t_{i}^{−1}gt_{i} = h_{i}.

Then

x ∈ X_{i} ⇔ x^{−1}gx = h^{−1}h_{i}h(h ∈ H)
⇔ hx^{−1}gxh^{−1} = h_{i} = t_{i}^{−1}gt_{i}
⇔ (t_{i}hx_{i}^{−1}) g (xh^{−1}t_{i}^{−1}) = g
⇔ xh^{−1}t_{i}^{−1} ∈ C_{G} (g)
⇔ x ∈ C_{G} (g) t_{i}H

so

|X_{i}| = |C_{g} (g) t_{i}h| = \frac{|C_{G} (g)| |H|}{|H ∩ t_{i}^{−1}C_{G} (g) t_{i}|}

Substituting this into (20) and noting that |H ∩ t_{i}^{−1}C_{G} (g) t_{i}| = |C_{H} (h_{i})| we get the result.

Example 14 Let H = A_{4} and G = S_{4}. Then

Ind_{H}^{G} (χ_{2}) = χ_{3},
Ind_{H}^{G} (χ_{3}) = χ_{3},
Ind_{H}^{G} (χ_{4}) = χ_{4} + χ_{5},
Ind_{H}^{G} (χ_{1}) = χ_{1} + χ_{2}.

Now let K = S_{3} (acting on {1,2,3}), regarded as a subgroup of G. Then

Ind_{K}^{G} (χ_{1}) = χ_{1} + χ_{4}.

Thus we obtain all the characters of G.

October 10

We discuss the example of G = S_{4}, H = A_{4}, further. By Frobenius reciprocity, χ_{3}|_{H} = χ_{2} + χ_{3}, χ_{1}|_{H} = χ_{2} + χ_{3}, and χ_{4}|_{H} = χ_{5}|_{H} = χ_{4}.

As a demonstration of induction by the simplified formula, we compute χ_{2}^{G} ((123)). Note that the class of (123) splits in A_{4} into two classes, those of (123) and (124). Then χ_{2}^{G} ((123)) = 3 (\frac{1}{3}ζ + \frac{1}{3}ζ^{2}) = −1.

We discuss the characters of GL(3, 2). We first make some general comments about G = GL_{n} (𝔽_{q}). We have |G| = q^{\frac{n(n−1)}{2}} (q − 1) (q^{2} − 1) ⋯ (q^{n} − 1). G acts on a vector space V over 𝔽_{q} of dimension n. A flag in V is a sequence of subspaces

0 ⊂ V_{1} ⊂ V_{2} ⊂ ⋯ ⊂ V_{m} = V. 30

Then G permutes the flags transitively. A flag is complete if n = m and dim V_{j} = j. For example,

0 ≤ ⟨v_{1}⟩ ≤ ⟨v_{1}, v_{2}⟩ ≤ ⋯ ≤ ⟨v_{1}, v_{2} ⋯ v_{n}⟩ = V

is a complete flag where {v_{1}, v_{2} ⋯ v_{n}} is a basis of V. Then the stabilizer B (for Borel) of a complete flag is the set of upper triangular matrices. The stabilizer of an arbitrary flag is called a parabolic subgroup.

When n = 3, take {v_{1}, v_{2}, v_{3}} to be a basis. Then B is the stabilizer of

0 ≤ ⟨v_{1}⟩ ≤ ⟨v_{1}, v_{2}⟩ ≤ v.

and |B| = q^{3} (q − 1)^{3}. G acts transitively on the complete flags, equivalently on the cosets of B.

The permutation representation of G on the flags is Ind_{B}^{G} (1) is a representation of degree \frac{|G|}{|B|} = (q + 1) (q^{2} + q + 1).

When q = 2 as in the homework, |B| = 8. Then Ind_{B}^{G} (1) (g) should be the number of flags fixed by g ∈ G. This can be used to compute the values of the induced character. (See the example below.) The trivial character will be a constituent of this by Frobenius reciprocity.

For P the stabilizer of 0 ≤ ⟨v_{1}⟩ ≤ V, Ind_{P}^{G} (1) should be the sum of the trivial character and one other.

We determine that P is a semidirect product (GL_{2} × GL_{1}) U, U ◁ P and use this to compute Ind_{P}^{G} (ϵ).

As an example, if g = ( \begin{matrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ), then consider g as an element of GL(3, \overline{𝔽}_{q}) where \overline{𝔽}_{q} is the algebraic closure of 𝔽_{q}. The set of complete flags in a corresponding vector space over \overline{𝔽}_{q}, fixed by g is an algebraic variety which is the intersection of two projective lines. The number of such flags over 𝔽_{q} is 2q + 1. To compute this last number, work with the basis given above and write down the possible flags fixed by g.

Remark 7 If G = GL_{n} (𝔽_{q}) ⊂ \overline{G} = GL_{n} (\overline{𝔽}_{q}) then B ⊂ \overline{B}. The variety over \overline{𝔽}_{q} of flags fixed by g ∈ G is an important one.

We now connect ordinary (characteristic 0) and modular (characteristic p for p| |G|) representations of G. The representations over ℂ can be written (by taking matrices over ℂ, which are finite in number) in a finite extension of ℚ, that is, an algebraic number field K.

Let O ⊂ K be the ring of algebraic integers in K, 𝔭 a prime ideal in O containing p. Then O/𝔭 is a field of characteristic p. The ring O , however, is not a PID, but we can fix this by taking the localization of O at 𝔭, which is the set O_{𝔭} = \{\frac{a}{b} : a, b ∈ O, b ∉ p\}. Then O_{𝔭} is a DVR.

For the properties of a DVR, see DF p.721 or CR1 p. 81. Start with a discrete valuation on K, i.e. a map ν : K^{∗} → ℤ with, among other properties, ν (x + y) ≥ min (ν (x) , ν (y)). Then R = {x ∈ K^{∗} : ν (x) ≥ 0} ∪ {0} is a DVR.

The valuation induces a metric on K and we can form the completion K̂ of K and R̂ of R with respect to this metric. [See DF p. 725, or CR1, p.83.]

Our goal is to consider KG, RG, and kG where K is a field of characteristic 0, R is a DVR in K, and 𝔭 is a prime ideal in R such that k = R/𝔭 is a field of characteristic p.

October 13

Final Homework Hint. The funny number \frac{−1+i\sqrt{7}}{2} that appears in the character table is a sum of three seventh roots of unity. Final Remarks on ℂG.

1. The number of linear characters of G is [G : G′], the index of the commutator subgroup of G (DF p.827).

2. To compute characters of abelian groups, we write the group as a direct product of cyclic groups. We know how to compute characters for cyclic groups and for direct products.

3. An application of the theory of representations over ℂG is the classical theorem of Burnside that a group of order p^{a}q^{b} is solvable for p and q primes (see DF p. 852). This theorem later lead to the famous theorem of Feit and Thompson from 1963 that every group of odd order is solvable. This year is the 40th anniversary of that famous theorem.

4. Hossein is going to Feit’s retirement conference, which also celebrates that anniversary.

Let p be a fixed prime. We call (K, R, k) a p-modular system when K is a field of characteristic 0, usually an algebraic number field, that is, a finite extension of ℚ, in which we can write the complex irreducible representations of G and its subgroups. R ⊂ K is a DVR, sometimes taken to be complete, 𝔭 is a prime ideal of R containing p, and k = R/𝔭, a field of characteristic p.

We can take 𝔭 = Rπ to be a principal ideal. If K is complete and a ∈ K, then

a = π^{−k} (a_{0} + a_{1}π + ⋯)

and a ∈ R if and only if a = a_{0} + a_{1}π + · · ·. The a_{i} ∈ S, a set of representatives of R over 𝔭.

Consider KG, RG, and kG. We have RG ⊂ KG, KG = K ⊗_{R} RG (known an extension of scalars) and kG = k ⊗_{R} RG (known as reduction mod p). Also, KG and kG are finite-dimensional algebras over K and k respectively and hence are artinian. We don’t know much about the structure of RG, but KG and kG/J are semisimple. Write

kG/J = ⨁_{i}(kG/J)f_{i}

for f_{i} primitive, orthogonal idempotents. Surprisingly, we can lift this decomposition to kG and provided that R is complete, we can lift the decomposition of kG to RG.

Let V be a KG-module with basis {v_{1}, v_{2} ⋯ v_{n}}. Let M be the RG-module defined by

M = ∑_{i=1}^{n} RGv_{i} = ∑_{g∈G} ∑_{i=1}^{n} Rgv_{i}.

Then M is a finitely generated, torsion-free R-module and R a PID. Hence, M is a free R-module. Pick a basis {u_{1}, u_{2}, …, u_{m}} of M (over R). Then

V = K ⊗_{R} M 32

so that m = n and {u_{1}, u_{2}, …, u_{m}} is a also a basis of V over K. This all means that we can find a basis of V which generates, over R, an RG-module M, that is, starting with a representation ρ of G such that ρ(g) has entries in K, we find an equivalent representation

ρ̂ of G which has entries in R. Then reducing mod 𝔭, we have that \overline{ρ̂ (g)} are matrices over k. Note that M is not unique since we could have selected a different basis for V initially. We call M an RG-lattice in V.

Furthermore, we can take

\overline{M} = k ⊗_{R} M

giving us modules V over K, M over R, and \overline{M} over k.

Preview of Idempotent Lifting Let A be an R-algebra for R a commutative ring with 1. If N is an ideal of A the N-adic topology on A consists of the system of neighborhoods of a ∈ A given by a + N^{k} for k = 0, 1, 2 · · ·.

Theorem 9 Let N be an ideal of A such that A is complete in the N-adic topology. Let f be an idempotent of A/N. Then there is an idempotent e ∈ A such that \overline{e} = f where \overline{·} denotes reduction modN. Furthermore, e is primitive if and only if \overline{e} is primitive.

October 15

Further Hint for Homework, if you want to use MAPLE

We have P ≤ G with |P | = 24. If we take a suitable ε, the induced character will be irreducible. Select coset representatives as follows. Let S be the 7-Sylow subgroup generated by A = ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} ). Now for G/P (not a group, a set of cosets) take representatives

{A^{j} : 0 ≤ j ≤ 6}. We need to determine if x_{j}^{−1}gx_{i} ∈ H and then put in ε (x_{j}^{−1}gx_{i}) in the formula for the induced character. As g runs over the representatives of conjugacy classes of G, find out whether A^{−j}gA^{i} ∈ P using Maple.

Let A be a ring, N a two sided ideal in A such that A is complete in the N-adic topology. This is explained as follows. That is, if a ∈ A, the neighborhoods of A are {a + N^{k} : k = 0, 1, 2 ⋯ }. A Cauchy sequence {a_{n} : n ≥ 1} is a sequence where given N^{k}, we have a_{m} − a_{n} ∈ N^{k} for m, n > n_{0} for some n_{0}. A is complete if every Cauchy sequence converges to some point in A, that is, there exists l ∈ A such that given N^{k}, a_{n} − l ∈ N^{k} for n > n_{0}.

For example, if A is Artinian and N ⊂ J, then A is complete since N^{k} = 0 for sufficiently large k.

Theorem 10 Let A be a ring and let N be a two-sided ideal such that A is complete in the N-adic topology. Write \overline{A} = A/N.

1. If f is an idempotent of \overline{A}, then there exists an idempotent e of A with \overline{e} = f.

2. Furthermore, if N ⊂ J(A), e is primitive if and only if f is primitive.

For example, if A were Artinian, then A/J would be semisimple so that the idempotents of A/J we found previously would lift to idempotents of A.

Proof. In ℤ [x], write

1 = x + (1 − x) = (x + (1 − x))^{2n} = ∑_{j=0}^{2n} \binom{2n}{j}x^{2n−j} (1 − x)^{j} .

Let

f_{n} (x) = ∑_{j=0}^{n} \binom{2n}{j}x^{2n−j} (1 − x)^{j} ∈ ℤ [x] .

We make the following preliminary computations.

1. f_{n} ≡ 0 (modx^{n}), since every term in f_{n} is divisible by x^{n}.

2. f_{n} ≡ 1 (mod (x − 1)^{n}) since

1 = f_{n} (x) + ∑_{j=n+1}^{2n} \binom{2n}{j}x^{2n−j} (1 − x)^{j} .

3. f_{n} (x)^{2} ≡ f_{n} (x) (modx^{n} (1 − x)^{n}) from above.

4. f_{n} (x) ≡ f_{n−1} (x) (modx^{n−1} (x − 1)^{n−1}) since

f_{n} − f_{n−1} = \binom{2n}{n}x^{n−1} (x − 1)^{n−1} .

5. Finally, f_{1} (x) ≡ x (modx − x^{2}) since

f_{1} = x^{2} + 2x (1 − x) = 2x − x^{2}

so that f_{1} − x = x − x^{2}.

Returning to \overline{A}, suppose f^{2} = f in \overline{A}. Pick a ∈ A with \overline{a} = f. Then a (a − 1) = a^{2} − a ∈ N. Since a and a − 1 commute, this means that for any j,

a^{j−1} (a − 1)^{j−1} ∈ N^{j−1}

so that

f_{j} (a) ≡ f_{j−1} (a) (modN^{j−1})

and

f_{j+k} (a) ≡ f_{k} (a) (modN^{k})

for any k so that {f_{n} (a) : n ≥ 1} is a Cauchy sequence in A. This sequence has a limit e ∈ A. By part 3 above, we have that e^{2} = e , since what part 3 means is that the sequence f_{n} (a)^{2} − f_{n} (a) has the limit 0. Also we have e − a ∈ N and so \overline{e} = f by the last two calculations.

October 17

Final, Final Homework Hint. P ≅ S_{4}. You will use the character table of GL(3, 2) to verify the Alperin-Mckay and Isaacs-Navarro conjectures for this group.

Last time we had A complete in the N-adic topology and f an idempotent of \overline{A} = A/N. We picked a ∈ A with \overline{a} = f. Then a^{2} − a ∈ N. We defined a sequence {f_{n} (a) ∈ A : n ≥ 1} and showed that

f_{1} (a) − a ∈ N
⋮
f_{n} (a) − f_{n−1} (a) ∈ N^{n−1} 34

and that f_{n} (a)^{2} − f_{n} (a) ∈ N^{n}. We let lim_{n→∞} f_{n} (a) = e so that e^{2} = e and \overline{e} = \overline{a} = f.

We now consider the second part of the claim. Assuming N ⊂ J (A), we want to show that f is primitive if and only if e is primitive.

Proof. If e = e_{1} + e_{2} for e_{1}^{2} = e_{2}, e_{2}^{2} = e_{2}, e_{1}e_{2} = e_{2}e_{1} = 0, then \overline{e} = \overline{e_{1}} + \overline{e_{2}}. If \overline{e} is primitive, then one \overline{e_{1}} = 0 say, that is, e_{1} ∈ N ⊂ J. But then 1 − e_{1} has a left inverse which is not possible as (1 − e_{1}) e_{1} = 0. Conversely, if \overline{e} = e′_{1} + e′_{2} for e_{1}^{′2} = e_{1} etc, pick a ∈ A with \overline{a} = e′_{1}. Let b = eae. Then

be = eb = b and \overline{b} = \overline{e}e′_{1}\overline{e} = e′_{1}  (13)

Let lim_{n→∞} f_{n} (b) = e_{1} ∈ A. Then \overline{e_{1}} = e′_{1} (= \overline{b}) and ee_{1} = e_{1}e = e_{1} by 13. Let e − e_{1} = e_{2} so that e = e_{1} + e_{2} and e_{1}e_{2} = 0 and e_{2}^{2} = e_{2}. Then e is not primitive.

Apply the lifting of idempotents to the group algebras RG and kG where (K, R, k) is a p-modular system with R/P ≅ k. We apply the theorem in the following cases. The notation below is insane and is not intended to match that of Theorem (3).

1. \overline{A} = kG and N = J (\overline{A}). The completeness follows from the fact that \overline{A} is artinian, so that J is nilpotent.

In this situation, \overline{A}/J (\overline{A}) is semisimple so that

\overline{A}/J (\overline{A}) ≅ ⨁_{i} \overline{A}/J (\overline{A}) ˜ \overline{e_{i}}

where the ˜ \overline{e_{i}} are primitive, orthogonal idempotents.

Then

\overline{A} = ⨁_{i} \overline{A}\overline{e_{i}}

where the \overline{e_{i}} are primitive, orthogonal idempotents. The \overline{A}\overline{e_{i}} are generally not irreducible and have composition factors. We have

J (\overline{A}) \overline{e_{i}} ⊂ \overline{A}\overline{e_{i}}

with \overline{A}\overline{e_{i}}/J (\overline{A}) e_{i} completely reducible.

2. We go from kG to RG = A. Note that A/P A ≅ \overline{A}.

Recall Nakayama’s Lemma (DF p. 717, CR1 5.7)

Lemma 3 If A is a ring, M is a finitely generated A-module, and L ≤ M, then L + J (A) = M implies L = M.

We use this to show the following lemma.

Lemma 4 P A ⊂ J (A)

Proof. Since M is a simple A-module, M = Am for some 0 ≠ m ∈ M. Then (P A) M ≤ M so that (P A) M is M or 0. If (P A) M = M, then by Nakayama’s lemma, M = 0 since P = J (R) and M is finitely generated as R-module. Hence (P A) M = 0, P A kills every simple module, that is, P A ⊂ J (A)

Finally, A/P A = \overline{A} is artinian and

J (A/P A) = J (A) /P A

so that J (A)^{t} ⊂ P A for some t. A is a finitely generated R-module and R is complete in the P -adic topology imply that A is complete in the P A-adic topology. This implies that A is also complete in the J-adic topology.

We have, by lifting from A/P A to A,

A

J (A)

P A

\overline{A} = A/P A = ⨁ \overline{A}\overline{e_{i}} leads to A = ⨁ Ae_{i} for A = RG where the e_{i} are a set of primitive orthogonal idempotents. Thus we have lifted idempotents from characteristic p to characteristic 0. We can also lift from A/J(A) to A, and this will be done next time.

October 20

Further discussion on Homework 5. Let G = GL_{3} (𝔽_{2}). Then B, the subgroup of upper triangular matrices, has order 8 and is the stabilizer of a “standard flag”

⟨v_{1}⟩ ≤ ⟨v_{1}, v_{2}⟩ ≤ V

where {v_{1}, v_{2}, v_{3}} is a basis of the vector space V over 𝔽_{2} on which G acts. G acts transitively on the set of flags, and thus the induced representation Ind_{B}^{G} (1) is just the permutation representation of G on the set of flags. Given an element g ∈ G, we can compute the number of flags fixed by g, and this gives the value of this induced character at g. For example if g is unipotent (eigenvalues 1) with Jordan blocks of sizes 2 and 1, we compute the number of flags fixed by g to be 5 (1 + 2q if we replace 𝔽_{2} by 𝔽_{q}).

From last time, we have the rings A = RG which is neither semisimple nor artinian and \overline{A} = A/P A = kG, which is not semisimple but artinian. We have

\overline{A} = ⨁_{i} \overline{A}\overline{e_{i}}

where the \overline{e_{i}} are lifted from idempotents ˜ \overline{e_{i}} in \overline{A}/J\overline{A} and

A = ⨁_{i} Ae_{i}

where the e_{i} are lifted from the idempotents \overline{e_{i}}.

In fact we can also look at A/JA which is semisimple so that

A/JA = ⨁_{i} (A/JA) ẽ_{i}.

We want to consider the structure of Ae_{i} and \overline{A}\overline{e_{i}} when \overline{e_{i}} and e_{i} are primitive.

Proposition 14 For A any ring with e a primitive idempotent, Ae is an indecomposable left A-module.

Proof. Let Ae = L_{1} ⊕ L_{2} for left ideals L_{1} and L_{2} and let e = e_{1} + e_{2} for e_{1} ∈ L_{1}, e_{2} ∈ L_{2}. For any x ∈ Ae, xe = x for x = ye, xe = ye^{2} = ye = x so e_{1}e = e_{1} so e_{1} = e_{1} (e_{1} + e_{2}) = e_{1}^{2} + e_{1}e_{2} where e_{1}^{2} ∈ L_{1} and e_{1}e_{2} ∈ L_{2}. Hence e_{1} = e_{1}^{2} and e_{1}e_{2} = 0. Similarly, e_{2}^{2} = e_{2} and e_{2}e_{1} = 0. Therefore e_{1}, e_{2} are orthogonal idempotents. Thus e is not primitive. Hence Ae is indecomposable.

Conversely, if Ae is indecomposable, suppose e = e_{1} + e_{2} with e_{1}^{2} = e_{1}, e_{2}^{2} = e_{2} and e_{1}e_{2} = e_{2}e_{1} = 0. Then x ∈ Ae implies x = xe = xe_{1} + xe_{2} ∈ Ae_{1} + Ae_{2}. Also, Ae_{1} ∩ Ae_{2} = 0 since xe_{1} = ye_{2} implies xe_{1}^{2} = ye_{2}e_{1} = 0 so that xe_{1} = 0

Hence irreducible in the semisimple case is replaced by indecomposable in the general case. In the semisimple case, if M is an irreducible A-module, then End_{A} (M) is a division ring. In the general case, we will see that M is indecomposable implies that End_{A} (M) is a local ring.

Recall that a local ring is a ring which has a unique maximal left ideal and that a ring A is local if and only if J (A) is maximal. Also, M is indecomposable if and only if End_{A} (M) has no idempotents except 0 and 1 since and idempotent in End_{A} is a projection on a direct summand.

October 22

Let A be any ring and e an idempotent in A. We showed last time that e is primitive if and only if Ae is indecomposable. We also know that if Ae is irreducible, then End_{A} (Ae) ≅ eAe is a division ring by Schur’s lemma. We want to prove the corresponding claim that Ae is indecomposable implies End_{A} (Ae) is a local ring. If B is a ring, recall that the following are equivalent.

1. B is a local ring

2. B has a unique maximal left ideal

3. J (B) is maximal

4. the non-units form a left ideal

5. B/J (B) is a division ring.

The proof of this claim is similar to the proof of the commutative case given in DF, p.684. For the general case see CR§5.21.

Example 15 Let A = kG for k a field of characteristic p and G a p-group. Then J (A) is spanned by {1 − x : x ∈ G} since (1 − x)^{p^{n}} = 0 where p^{n} = |G|. Also, dim_{k} A/J (A) = 1 so J (A) is maximal and A is a local ring.

Now let A be any ring. We begin with the following important lemma.

Lemma 5 (Fitting’s Lemma) Let M a an A-module with ACC and DCC. Then M has a composition series. Let f ∈ End_{A} (M). Then M = Im (f^{n}) ⊕ ker (f^{n}) for sufficiently large n. (DB, Lemma 1.4.4; DF p. 646 in some form)

Proof. Since M has ACC and DCC, we can find n with Im (f^{n}) = Im (f^{n+k}) and ker (f^{n}) = ker (f^{n+k}) for k > 0. Let x ∈ M. Then f^{n} (x) = f^{2n} (x) and

x = \underbrace{f^{n} (x)}_{∈Im(f^{n})} + \underbrace{(x − f^{n} (x))}_{∈ker f^{n}(x)} .

If x ∈ Im (f^{n}) ∩ ker (f^{n}), then x = f^{n} (y) so that 0 = f^{n} (x) = f^{2n} (y) = f^{n} (y) = x.

Proposition 15 Let M be an indecomposable A-module with ACC and DCC. Then End_{A} (M) is a local ring.

Proof. Let E = End_{A} (M) and let I be a maximal ideal in E. Let a ∉ I. We show that a is a unit in E. We have E = Ea + I as I is maximal so we have 1 = λa + μ for 0 ≠ λ ∈ E and μ ∈ I. Then 1 − μ = λa and μ is not a unit since μ ∈ I. Hence Im (μ) ≠ M. By Fitting’s Lemma, M = Im (μ^{n}) ⊕ ker (μ^{n}). Now since M is indecomposable, we have Im (μ^{n}) = 0 and ker (μ^{n}) = M so that μ^{n} = 0. Then 1−μ has the inverse 1+μ+⋯+μ^{n−1} so that λa is invertible and a is a unit.

Remark 8 Fitting’s Lemma is sometimes stated as the following. If M is indecomposable and μ ∈ End_{A} (M), then μ is an isomorphism or is nilpotent. Note the analogy with Schur’s Lemma.

Theorem 11 (Krull-Schmidt Theorem, CR1 p.128) Let A be any ring and M a finitely generated A-module which is a direct sum M = ⨁_{i=1}^{r} M_{i} for M_{i} indecomposable and End_{A} (M_{i}) local. Then if also M = ⨁_{j=1}^{s} N_{j} with N_{j} indecomposable, then r = s and M_{i} ≅ N_{j} after some reordering.

Proof. We prove this by induction on r. When r = 1 the statement is clear. Assume r ≥ 2. We then have

M = M_{1} ⊕ M_{2} ⊕ ⋯ ⊕ M_{r}
M = N_{1} ⊕ N_{2} ⊕ ⋯ ⊕ N_{s}.

Let μ_{i} : M → M_{i} and ν : M → N_{i} be projections. Then ∑_{j=1}^{s} ν_{j} = 1. On M_{1}, we have

∑_{j=1}^{s} μ_{1}ν_{j} = 1_{M}.

Now since End_{A} (M_{1}) is a local ring so that a sum of non-units is a non-unit, we have that some μ_{1}ν_{j} is a unit, say μ_{1}ν_{1} after some reordering.

Let φ = μ_{1}ν_{1} : M_{1} → M_{1}. Then

ν_{1} : M_{1} → N_{1}
φ^{−1}μ_{1} : N_{1} → M_{1}  (14)

are inverses of one another, that is,

(φ^{−1}μ_{1}) ν_{1} = 1_{M_{1}},

that is, N_{1} is a direct summand of M_{1}. Since M_{1} is indecomposable, we have N_{1} = ν_{1} (M_{1}).

Consider M′ = N_{1} ⊕ M_{2} ⊕ ⋯ ⊕ M_{r} ≤ M. If n_{1} + m_{2} + ⋯ + m_{r} = 0, then μ_{1} (n_{1}) = 0 so that n_{1} = 0 since μ_{1} is injective by (14). Then m_{2} = ⋯ = m_{r} = 0, that is, the sum is direct.

Next we show that M′ = M and that we have an automorphism ρ : M → M which takes M_{1} → N_{1} and hence induces an isomorphism M/M_{1} → M/N_{1}. We then apply induction.

October 24

The philosophy of modular representations is to obtain global information from local information. Having fixed a prime p, obtaining local or p-local information means gathering information from subgroups of G such as number of characters or character values, which is related to p-subgroups such as normalizers of p-subgroups.

We complete the proof of the Krull-Schmidt Theorem: Proof. Given

M = M_{1} ⊕ M_{2} ⊕ ⋯ ⊕ M_{r}
M = N_{1} ⊕ N_{2} ⊕ ⋯ ⊕ N_{r}

we defined projections

μ_{i} : M → M_{i}
ν_{i} : M → N_{i}
μ_{1} : M → M_{i}
φ^{−1}μ_{1} : N_{1} → M_{1}

with the last two being isomorphisms, and we defined

M′ = N_{1} ⊕ M_{2} ⊕ ⋯ ⊕ M_{r} ≤ M

Now ∑_{i} μ_{i} = 1_{M}. If x ∈ N_{1},

x = μ_{1} (x) + μ_{2} (x) + ⋯ + μ_{r} (x)
μ_{1} (x) = \underbrace{x}_{∈N_{1}} − \underbrace{μ_{2} (x) − ⋯ − μ_{r} (x)}_{∈M_{2}⊕···⊕M_{r}}

so that μ_{1} (x) ∈ M′, that is, μ_{1} (N_{1}) ⊂ M′, that is, M_{1} ⊂ M′ so M = M′

Then ν_{1}μ_{1} + μ_{2} ⋯ μ_{r} is an action on M which takes M_{1} → N_{1} and M_{i} → M_{i} for i > 1 and this induces an isomorphism

M_{2} ⊕ ⋯ ⊕ M_{r} ≅ M/M_{1} → M/N_{1} ≅ N_{2} ⊕ ⋯ ⊕ N_{s}.

Applying induction, we are done.

Hence we have uniqueness of decomposition into indecomposable submodules of a module M if M has ACC and DCC.

This can be applied to the A-modules A = KG and A = kG, which are finite dimensional algebras. It remains to show that Krull-Schmidt Theorem for RG. For this, we need to show that if A = RG and M is a finitely generated A-module, then End_{A} (M) is a local ring.

The following discussion is in CR1 p. 105 and p. 112.

Lemma 6 If f : B → B′ is a surjective map of rings, then f (J (B)) ⊂ J (B′) and we get a surjective map \overline{f} : B/JB → B′/JB′.

This can be verified by considering maximal left ideals.

Proposition 16 If R is a commutative, local ring, P ⊂ R A a finitely generated Ralgebra which is finitely generated as an R-module, then A/JA is artinian semisimple.

Proof. We proved in Lemma 4 that P A ⊂ JA. Therefore, A/JA ≅ \frac{A/P A}{JA/P A} by the isomorphism theorems. We then have a map

f : A/P A → A/JA

which induces a surjective homomorphism as in the Lemma 6

\frac{A/P A}{J (A/P A)} → A/JA

since J (A/JA) = 0. We also have an onto homomorphism A → A/P A inducing

A/JA → \frac{A/P A}{A (A/P A)}.

Therefore, A/P A is a finite dimensional algebra over R/P = k, so by comparing dimensions, we have JA = J (J/P A).

We now want to apply this to our situation KG, RG, kG, V a KG-module, M an RG-module, and V = K ⊗_{R} M. Let A = RG and E = End_{A} (M) ⊂ End_{R} (M). Then since End_{R} (M) is finitely generated, we have End_{A} (M) is finitely generated as an Rmodule. Now let M be an indecomposable R-complete A-module. It remains to show the following.

Proposition 17 E is a local ring.

Proof. Consider E/J (E). If this is not a division ring, suppose it has a proper non-zero left ideal L. E/JE is artinian semisimple, so there exists an idempotent 1 ≠ ε ∈ L. Since R is complete, we can lift ε to an idempotent e ≠ 1 of E. But since M is indecomposable, this is not possible.

Hence, the Krull-Schmidt Theorem holds for A-modules when A is KG, RG, and kG. We make the following comparison

General case, kG, and RG  The semisimple case and KG
Indecomposable module M  Irreducible module M
End_{A} (M) is local  End_{A} (M) is a division ring

Krull-Schmidt Theorem

Jordan-Hölder Theorem

A = ⨁ Ae_{i} for indecompasble left ideals A = ⨁ Af_{i} for minimal left ideals

Finally, review Proposition 30 on p. 369 of DF. on projective modules.

October 31

Homework Comments. There is a typographic error in the character table for S_{5} given in DF. The character of degree 6 has the value 1 at the class of elements of order 5. Also, recall that χ (x^{−1}) = \overline{χ (x)}. From this it follows that if x and x^{−1} are conjugate in a finite group G, then χ (x) = χ (x^{−1}) = \overline{χ (x)} so that χ (x) is real. Also note that

N_{A_{5}} (⟨(12345)⟩) = ⟨(12345) , (25) (34)⟩ ≅ D_{10}.

Remark 9 The connection between A_{5} and the simple group of order 168 is that A_{5} ≅ P SL (2, 5) and the simple group of order 168 is isomorphic to P SL (2, 7), that is, both are P SL (2, q) for some q.

Let A = RG and \overline{A} = kG where (K, R, k) is a p-modular system with R/𝔭 ≅ k. We have the decomposition

A = Ae_{1} ⊕ Ae_{2} ⊕ ⋯ ⊕ Ae_{r}
\overline{A} = \overline{A}\overline{e}_{1} ⊕ \overline{A}\overline{e}_{2} ⊕ ⋯ ⊕ \overline{A}\overline{e}_{s}.

The Ae_{i} are indecomposable left ideals of A and the \overline{A}\overline{e}_{i} are are indecomposable left ideals of \overline{A}. Hence, they are projective A-modules and \overline{A}-modules respectively, being direct summands of free modules. They are called principal indecomposable modules or PIMs for A and \overline{A} respectively.

One of the main problems of modular representation theory is the following. Given G and p, describe the PIMs. This is a difficult problem. To do this, we typically take K to be some finite extension of ℚ in which we can write all the (complex) irreducible characters of G, O the ring of integers in K, R = O_{𝔭}, the localization at a prime ideal 𝔭 which contains p, and k = O_{p}/𝔭, a field of characteristic p.

We want to consider now the central idempotents of A and \overline{A}. Consider A, Z (A) and Z (\overline{A}). We can apply the theory of idempotent lifting to Z (A) and Z (\overline{A}), since in the natural homomorphism A → \overline{A}, we have Z(A) → Z (\overline{A}). This follows from the observation that we have a basis for Z (A) given by class sums.

Suppose in \overline{A} we have

1 = \overline{E}_{1} + \overline{E}_{2} + ⋯ + \overline{E}_{k}  (15)

for \overline{E}_{i} central primitive idempotents, and in Z (A) we have

1 = E_{1} + E_{2} + ⋯ + E_{k}

for E_{i} central primitive idempotents. Then

A = AE_{1} ⊕ AE_{2} ⊕ ⋯ ⊕ AE_{k}
\overline{A} = \overline{A}\overline{E}_{1} ⊕ \overline{A}\overline{E}_{2} ⊕ ⋯ ⊕ \overline{A}\overline{E}_{k}

where the AE_{i} are indecomposable two-sided ideals of A and similarly for \overline{A}.

Remark 10 Just as for the \overline{e}_{i}, we could look at \overline{A}/J (\overline{A}), which is semisimple and lift its central idempotents. Then we get (15).

The AE_{i} are called blocks or p-blocks of A. Similarly, \overline{A}\overline{E}_{i} are p-blocks of \overline{A}. Each AE_{i} is a sum of indecomposable left ideals of A. Hence, we have a (possibly easier) problem of finding the p-blocks of A. This is done for S_{n} and the solution is attractive.

We revisit the example G = S_{3}. We computed the central idempotents in ℚG to be

f_{1} = \frac{1}{6}( (1) + (12) + ⋯ + (123) )
f_{2} = \frac{1}{6}( (1) − (12) + ⋯ + (123) )
f_{3} = \frac{2}{3} (1) − \frac{1}{3}( (123) + (132) )

Let p = 2. Let E_{1} = f_{1} + f_{2}, E_{2} = f_{3}. Then E_{1} and E_{2} are central primitive idempotents in A and \overline{E_{1}} and \overline{E_{2}} are central primitive idempotents in \overline{A}. Thus \overline{A}\overline{E}_{1} and \overline{A}\overline{E}_{2} are the 2-blocks of \overline{A}.

November 3

We consider Proposition 30 from DF, which corresponds with CR1 p.29. Let R be a ring and P an R-module. The following are equivalent.

1. P is projective

2. P is a direct summand of a free R-module

3. Given an exact sequence M →^{φ} N → 0 and an R-module homomorphism f : P → N, we have an R-module homomorphism F : P → M such that the diagram below commutes.

\xymatrix{ & P \ar[dl]_{F} \ar@{^{(}->}[d]^{f} \\ M \ar[r]^{φ} & N \ar[r] & 0}

42 4. Any short exact sequence of the following form splits. That is, given

\xymatrix{0 \ar[r] & L \ar[r] & M \ar[r]^{ψ} & P \ar[r] & 0}

we have an R-module homomorphism μ : P → M such that ψ ∘ μ = i.

\xymatrix{0 \ar[r] & L \ar[r] & M \ar[r]^{ψ} & P \ar[l]^{μ} \ar[r] & 0}

5. Given a short exact sequence

\xymatrix{0 \ar[r] & L \ar[r] & M \ar[r] & P \ar[r] & 0}

we have that

\xymatrix{0 \ar[r] & Hom (P, L) \ar[r] & Hom (P, M) \ar[r] & Hom (P, N) \ar[r] & 0}

is exact.

Returning to (K, R, k) with A = RG and \overline{A} = kG, we recall that the PIM’s of A are Ae_{i} for e_{i} primitive, and the PIM’s of \overline{A} are \overline{A}\overline{e_{i}} for \overline{e_{i}} primitive. A large open problem is to describe the PIM’s of \overline{A}.

For this section, let A be a finite dimensional algebra over k and assume A-modules have composition series. Fitting’s lemma therefore holds in this case.

Proposition 18 If two indecomposable, projective A-modules P_{1} and P_{2} have isomorphic quotients, then P_{1} ≅ P_{2}. [CR]

Proof. Lifting the map P_{1} ⟶^{π_{1}} V_{1} ⟶^{α} V_{2}, we get P_{1} ⟶^{ξ} P_{2} with π_{2}ξ = απ_{1} by part 3 above as in the following diagram.

\xymatrix{ & P_{1} \ar[dl]_{ξ} \ar@{^{(}->}[d]^{απ_{1}} \\ P_{2} \ar[r]^{π_{2}} & V_{2} \ar[r] & 0}

Similarly, we get μ : P_{2} → P_{1} with π_{1}μ = α^{−1}π_{2} as in the diagram below.

\xymatrix{P_{1} \ar[d]_{π_{1}} \ar[r]^{ξ} & P_{2} \ar[l]^{μ} \ar[d]^{π_{2}} \\ V_{1} \ar[d] \ar[r]^{α} & V_{2} \ar[d] \ar[l]^{α^{−1}} \\ 0 & 0}

Then

π_{1}μξ = α^{−1}π_{2}ξ = α^{−1}απ_{1} = π_{1}

and similarly π_{2}ξμ = π_{2}.

Now μξ ∈ End_{A} (P_{1}) which is local since P_{1} is indecomposable. Let β = μξ. Then β is either nilpotent or an automorphsim by Fitting’s Lemma. We have

π_{1}β = π_{1} ⟹ π_{1}β^{2} = π_{1}β = π_{1} ⟹ … ⟹ π_{1}β^{k} = π_{1}

for k ≥ 1. This means that β^{k} ≠ 0 for all k ≥ 1 since otherwise, we would have π_{1} = 0. Thus, β is an automorphsim. Similarly, ξμ is an automorphism of P_{2} so that P_{1} ≅ P_{2}.

Proposition 19 If Q is an indecomposable, projective A-module, then any quotient is also indecomposable

\xymatrix{Q \ar[d]^{π} \ar[r]^{η} & Q \ar[d]^{π} \\ U \ar[d] \ar[r]^{ε} & U \ar[d] \\ 0 & 0}

If U is not indecomposable, there exists ε ∈ End_{A} (U) with ε^{2} = ε and ε ≠ 0, 1. Since Q is projective, there exists η ∈ End_{A} (Q) with πη = επ. Then

πη^{2} = επη = ε^{2}π = επ = πη

and similarly πη^{k} = πη for k ≥ 1. If η^{k} = 0, then πη = 0. But then η = 0 since π is surjective. Hence, η is an automorphsim of Q so that

(πη) (Q) = π (Q) = U
επ (Q) = ε (U) ⊊ U,

a contradiction. Therefore, U is indecomposable.

Definition 17 If M is any A-module, define

Rad M = ⋂_{M maximal} M.

Corollary 5 If Q is indecomposable and projective, then Q has a unique maximal submodule Rad Q.

Proof. If M_{1} and M_{2} are maximal submodules of Q, then

Q/M_{1} ∩ M_{2} ≅ Q/M_{1} ⊕ Q/M_{2},

a contradiction since Q/M_{1} ∩ M_{2} is indecomposable.

Returning to \overline{A} = kG which is artinian and semisimple, we have

\overline{A} = ⨁_{i} \overline{A}\overline{e_{i}}

is the sum of indecomposable projective modules, and

˜\overline{A} = \overline{A}/J (\overline{A}) = ⨁_{i} ˜ \overline{A}˜\overline{e_{i}}

is the sum of irreducible modules. In fact,

˜\overline{A}˜\overline{e_{i}} ≅ \overline{A}\overline{e_{i}}/J\overline{e_{i}}.

Then J\overline{e_{i}} is the unique maximal submodule of \overline{A}\overline{e_{i}} and

\overline{A}\overline{e_{i}}/J\overline{e_{i}}

is a simple \overline{A}-module. Also,

\overline{A}\overline{e_{i}} ≅ \overline{A}\overline{e_{j}}

if and only if

\overline{A}\overline{e_{i}}/J\overline{e_{i}} ≅ \overline{A}\overline{e_{j}}/J\overline{e_{j}}.

44 Remark 11 Dualizing, we have a similar theory for injective modules. In fact, the \overline{A}\overline{e_{i}} are also injective, and have a unique minimal submodule.

November 5

Here are some comments on the homework. To construct the normalizer of the Sylow 7-subgroup in GL (3, 2), recall the construction for groups of order pq with q|p − 1. Let S_{1} be a Sylow p-subgroup and S_{2} a Sylow q-subgroup. Then we have G = S_{1}S_{2} with S_{1} ◁ G and

G = ⟨a, b : a^{p} = b^{q} = 1, b^{−1}ab = a^{r} for r with r^{q} ≡ 1 (mod p)⟩ .

with S_{1} = ⟨a⟩ and S_{2} = ⟨b⟩.

For groups of order 21, we have a^{7} = b^{3} = 1 and b^{−1}ab = a^{2} and the conjugacy classes and their sizes are as follows.

element 1 a a^{3} b ab ; size 1 3 3 7 7

Also note that a Sylow 2-subgroup of GL(3, 2) is the group of all upper triangular matrices with diagonal entries 1, and that S_{3} is naturally embedded in GL(3, 2) as the group of permutation matrices. Note that the normalizer of a Sylow 3-subgroup is isomorphic to S_{3}.

Returning to \overline{A} = kG, we have

\overline{A} = ⨁_{i} \overline{A}\overline{e_{i}}

and each \overline{A}\overline{e_{i}} has a unique maximal submodule J (\overline{e_{i}}) with

\overline{A}\overline{e_{i}}/J (\overline{e_{i}}) ≅ \overline{A}\overline{e_{j}}/J (\overline{e_{j}}) iff \overline{A}\overline{e_{i}} ≅ \overline{A}\overline{e_{j}}.

This says that there is a bijection

{PIMs (up to isomorphism)} ⟷ {Irreducible modules (up to isomorphism)}

A similar result holds for A = RG as in (CR1 §6.6, p. 123). If P_{1} and P_{2} are finitely generated A-modules and we have N ⊂ J (A), V_{1} = P_{1}/NP_{1} and V_{2} = P_{2}/NP_{2}, then V_{1} ≅ V_{2} if and only if P_{1} ≅ P_{2}. The proof consists of diagram arguments and an application of Nakayama’s Lemma.

I’ll do this later

There exists a map ξ : P_{1} → P_{2} with π_{2}ξ = απ_{1}. Let T = P_{2}/ζ (P_{1}). We need to show that NT = T, that is, that P_{2} ≡ NP_{2} (mod ζ (P_{1})). To do this, let x ∈ P_{2}. Then α^{−1}π_{2} ∈ V_{1} so there exists y ∈ P_{1} with

π_{1} (y) = α^{−1}π_{2} (x) so that απ_{1} (y) = π_{2} (x)  (16)

Then π_{2} (x − ξ (y)) = π_{2} (x) − π_{2}ξ (y) = 0 from (16) Then T = NT and T is finitely generated, so by Nakayama’s lemma, we have T = 0. Then P_{2} = ξ (P_{1}) implies ξ is surjective. Consider

0 → ker ξ → P_{1} → P_{2} → 0 45

which splits since P_{2} is projective, so ker ξ is finitely generated. Now if x ∈ ker ξ, then π_{2}ξ (x) = 0 so that π_{1} (x) = 0 and x ∈ NP_{1} so that ker ξ = N ker ξ and ker ξ = 0 again. Hence P_{1} ≅ P_{2}.

Thus we have, also for A, a bijection

{PIMs (up to isomorphism)} ⟷ {Irreducible modules (up to isomorphism)} .

In addition we have Ae_{i} ≅ Ae_{j} as A-modules if and only if \overline{A}\overline{e_{i}} ≅ \overline{A}\overline{e_{j}} as \overline{A}-modules. This follows by taking N = J (A) or N = P A.

Write Â = KG, a semisimple algebra. Then we study Â for the ordinary irreducible representation of G, A = RG (not semisimple) for the principal indecomposable representations of G over R, and \overline{A} = kG (also not semisimple) for the principal indecomposable and modular irreducible representations of G over k.

As in the case of Â we want to study characters of modular representations, i.e. representations of \overline{A} = kG. The usual definition of “trace” leads to difficulties. For example, if we have a representation of dimension divisible by p, then the trace of the identity matrix would be 0. The concept of Brauer characters (discussed in CR1 §17) handles this difficulty.

Definition 18 Let (K, R, k) be a p-modular system and G a group with |G| = p^{a}m and (p, m) = 1. Then g ∈ G is p-regular if the order of g is prime to p and hence divides m.

Choose K and k large enough to contain the mth roots of unity and choose a bijection

{Group of m^{th} roots of unity in K^{×}} ⟷ {m^{th} roots of 1 in k^{×}}

Write ω ⟷ \overline{ω}. Given a p-regular g, suppose that in a representation ρ of G over k, the eigenvalues of ρ (g) are \{\overline{ω}^{i_{1}}, \overline{ω}^{i_{2}}, … \overline{ω}^{i_{n}}\}. Define a Brauer character λ of ρ to be given by

λ (g) = ω^{i_{1}} + ω^{i_{2}} + ⋯ + ω^{i_{n}} ∈ K.

Thus λ is a function from G to K.

November 7

Example 16 (See CR2, p.648.) For G = S_{4} and p = 2, consider

G_{p′} = {1 and the elements of order 3} ≤ G

We have

1 (123)
χ_{1} 1 1 ⟵ remains irreducible
χ_{2} 1 1
χ_{3} 2 −1 ⟵ irreducible as follows
χ_{4} 3 0 ⟵ χ_{3} + χ_{1}
χ_{5} 3 0

If the two-dimensional representation was reducible mod 2, then all matrices would be

of the form (\begin{matrix}1 & ♣ \\ 0 & 1 \end{matrix}) hence would have trace 0. This would contradict the values in the

character table. So it must be irreducible. So the irreducible Brauer characters are χ_{1} and χ_{3}. Note that the degrees of the characters need not divide the order of the group. A difficult problem is to find the Brauer characters, or at least their degrees.

Next we consider blocks. Let Â = KG, A = RG, and \overline{A} = kG. Let \{V_{j} : 1 ≤ j ≤ s\} be irreducible Â-modules up to isomorphism. Let M_{i} ≤ V_{i} so that K ⊗_{R} M_{i} ≅ V_{i} for 1 ≤ i ≤ s. Write \overline{M_{i}} for the reduction of M_{i} mod P for P a prime ideal of R. Then \overline{M_{i}} ≅ k ⊗_{R} M_{i} and

Z \binom{}{A} = ⨁_{i=1}^{s} ÂF_{i}

for F_{i} central primitive idempotents which can be computed from the character by the formula

F_{i} = \frac{χ_{1} (1)}{|G|} ∑_{g∈G} χ_{i} (g) g^{−1}.

In Z (A), write 1 = E_{1} + E_{2} + ⋯ + E_{t} where each E_{i} is a sum of some of the F_{j}. The E_{i} are central primitive idempotents in Z (A) but not primitive in Z \binom{}{A}. Then

A = AE_{1} ⊕ AE_{2} ⊕ ⋯ ⊕ AE_{t}

is the sum of indecomposable two-sided ideals. The AE_{i} are called the blocks or p-blocks of G.

Also,

\overline{A} = \overline{A}\overline{E_{1}} ⊕ \overline{A}\overline{E_{2}} ⊕ ⋯ ⊕ \overline{A}\overline{E_{t}}

and the \overline{A}\overline{E_{i}} are also called p-blocks of G. Also

Â = ÂE_{1} ⊕ ÂE_{2} ⊕ ⋯ ⊕ ÂE_{t}

. Each ÂE_{i} is the sum of matrix algebras, hence corresponds to a set of ordinary characters. Thus, {χ_{1}, χ_{2}, …, χ_{s}} are divided into disjoint subsets, also called p-blocks of G.

Example 17 For G = S_{3} and p = 2, the 2-blocks are {χ_{1}, χ_{2}} and {χ_{3}}.

Recall that

ω_{i} (C_{j}) = \frac{χ_{i} (g_{j})}{χ_{i} (1)}h_{j}

for 1 ≤ i ≤ s where the χ_{i} are the characters of Z \binom{}{A}, the C_{j} are the class sums of C_{j}, h_{j} = |C_{j}|, and g_{j} ∈ C_{j} is a representative of its conjugacy class. Take K to be sufficiently large.

Now since ω_{i} (C_{j}) is an algebraic integer, we can assume ω_{i} (C_{j}) ∈ R and hence we can reduce mod P. Let

\overline{ω_{i}} : Z (\overline{A}) → k be given by C_{j} ↦ \overline{ω_{i} (C_{j})}.

Then the \overline{ω_{i}} are all the characters of Z (\overline{A}) as follows.

ker (\overline{ω_{i}}) = ⨁_{l≠i} Z (\overline{A}) \overline{E_{l}} ⊕ J (Z (\overline{A})) \overline{E_{i}}.

Z (\overline{A}) = Z (\overline{A}) \overline{E_{1}} ⊕ Z (\overline{A}) \overline{E_{2}} ⊕ ⋯ ⊕ Z (\overline{A}) \overline{E_{t}}

.

(To be continued next time.)

November 10

Recall that Â = KG, A = RG and \overline{A} = kG with K and k sufficiently large so that the irreducible representations of G over ℂ can be realized over K.

To study p-blocks, consider

Z (A) = Z (A) E_{1} ⊕ … Z (A) E_{t}

and in fact

A = AE_{1} ⊕ … AE_{t}

where each AE_{i} is a p-block. The E_{i} are primitive in RG, but not in KG. Write

E_{i} = F_{i,1} + F_{i,2} + ⋯ + F_{i,α_{i}}

where F_{i,1} = F_{i}, and the F_{i,j} are primitive central idempotents in KG. Then

\overline{A} = \overline{A}\overline{E}_{1} ⊕ ⋯ ⊕ \overline{A}\overline{E}_{t}.

The \overline{A}\overline{E}_{t} are also p-blocks. Then

Z (\overline{A}) = Z (\overline{A}) \overline{E}_{1} ⊕ ⋯ ⊕ Z (\overline{A}) \overline{E}_{t}.  (17)

Each F_{i} and hence χ_{i} corresponds to a character ω_{i} of Z \binom{}{A} and

ω_{i} (C_{j}) = \frac{χ_{i} (g_{j}) h_{j}}{χ_{i} (1)} ∈ R  (18)

where C_{j} are the class sums of C_{j}, g_{j} ∈ C_{j}, and h_{j} = |C_{j}|. Then ω_{i} (F_{i}) = 1 and ω_{i} (F_{j}) = 0 for i ≠ j. Hence ω_{i} (E_{i}) = 1 and ω_{i,j} (E_{i}) = 1 for 1 ≤ j ≤ α_{j} where ω_{i,j} is the character corresponding to F_{i,j} and ω_{i} = ω_{i,1}.

Define \overline{ω} : Z (\overline{A}) → k by \overline{ω} (C_{j}) = \overline{ω_{i} (C_{j})}. Then we get t linear characters of Z (\overline{A}). We can compute ker \overline{ω}_{i} as follows. We have from (17) that

\overline{ω}_{i,l} (\overline{E_{j}}) = 0

if i ≠ j. Also, we have ker \overline{ω}_{i} is a maximal ideal of Z (\overline{A}). Hence,

ker \overline{ω}_{i} = ⨁_{l≠i} Z (\overline{A}) \overline{E}_{l} ⊕ \underbrace{J (Z (\overline{A})) \overline{E}_{i}}_{unique maximal ideal of Z(\overline{A})\overline{E}_{i}} .

Since (17) is the unique decomposition of Z (\overline{A}) into indecomposable ideals, any maximal ideal is of this form. Therefore, \overline{ω_{i}} are the only irreducible characters of Z (\overline{A}). Hence χ_{i} and χ_{j} are in the same p-block if and only if \overline{ω_{i}} = \overline{ω_{j}} if and only if ω_{i} ≡ ω_{j} (mod P) as functions. Thus, blocks are identified by characters of the center of KG mod 𝔭. (Note that here we are going back to the old notation of characters χ_{i}, i = 1, 2, … s of G.)

Definition 19 B_{p} (G) is the block containing the trivial character, called the principal block.

Example 18 (CR2, p. 649) Work out the ω_{i} (C_{j}) from the character table. For S_{4}, we have the following table.

\begin{matrix}(1) & (12) & (123) & (1234) & (12)(34) \\ 1 & 1 & 1 & 1 & 1 \\ 1 & −1 & 1 & −1 & 1 \\ 2 & 0 & −1 & 0 & 2 \\ 3 & 1 & 0 & −1 & −1 \\ 3 & −1 & 0 & 1 & −1 \end{matrix}

We compute the character table for Z (\overline{A}) using formula 18. For example, ω_{4} (C_{2}) = \frac{1}{3}6 = 2 and we have the following table.

\begin{matrix}1 & 6 & 8 & 6 & 3 \\ 1 & −6 & 8 & −6 & 3 \\ 1 & 0 & −4 & 0 & 3 \\ 1 & 2 & 0 & −2 & −1 \\ 1 & −2 & 0 & 2 & −1 \end{matrix}

Mod 2, this table becomes

1 0 0 0 1
1 0 0 0 1
1 0 0 0 1
1 0 0 0 1
1 0 0 0 1

so we have only one block. Mod 3, we have

1 0 2 0 0
1 0 2 0 0
1 0 2 0 0
1 2 0 1 2
1 1 0 2 2

so that the first three form one block and the last two each form one block.

Look next at the restrictions of the ordinary characters to the 3-regular classes. We have four Brauer characters, two arising from the first three characters of S_{4} and one each for the last two characters. (Recall that by a Brauer character we mean the character of a p-modular irreducible representation, regarded as a complex function by Brauer’s procedure.) Note that the first two characters of S_{4} (trivial and sign) give two Brauer characters, whereas the third character of S_{4} is the sum of the first two.

Example 19 For G = A_{5}, B_{5} (G) = {1, χ_{4}, χ_{3}, χ′_{3}}. Note that the degree of the character is the subscript here. Note also that χ_{3} and χ′_{3} are not rational. Write ζ = e^{2πi/5}. Then χ_{3} ((12345)) = ζ + ζ^{4}, χ′_{3} ((12345)) = ζ^{2} + ζ^{3}. To show that these two characters are in the same 5-block, check that

(1 − ζ) (1 − ζ^{2}) (1 − ζ^{3}) (1 − ζ^{4}) = 5

which follows from ∑_{j=0}^{4} ζ^{j} = 0. For p = 5, take P = (1 − ζ), K = ℚ (ζ) and R the ring of integers ℤ [\frac{1+ζ}{2} ] (see e.g. DF, p.230). Then ζ ≡ ζ^{2} (mod P), ζ^{3} ≡ ζ^{4} (mod P). From this we can show that ω_{3} ≡ ω_{3′} (mod P). (This example will be continued next time.)

Continuing with ω_{i} (C_{j}) for A_{5}, we have

\begin{matrix}ω_{1} & 1 & 20 & 15 & 12 & 12 \\ ω_{4} & 1 & 5 & 0 & −3 & −3 \\ ω_{5} & 1 & −4 & 3 & 0 & 0 \\ ω_{3} & 1 & 0 & −5 & 2 (1 + \sqrt{5}) & 2 (1 − \sqrt{5}) \\ ω′_{3} & 1 & 0 & −5 & 2 (1 − \sqrt{5}) & 2 (1 + \sqrt{5}) \end{matrix}

For this example, we took R = ℤ [ζ], P = (1 − ζ). Note that in this case, for our purpose of calculating the blocks, we didn’t need to take the localization of R at P or the completion.

Note: We have ζ + ζ^{4} + 1 = \frac{1+\sqrt{5}}{2} and ζ^{2} + ζ^{3} + 1 = \frac{1−\sqrt{5}}{2}. This is used in computing the blocks, using the calculations of last time.

The Isaacs-Navarro conjecture looks at the degrees of characters of G and N_{G} (P). We refine this in the following ways.

1. Take B_{p} (G)

2. There should be a bijection of characters, not just equality of number of characters. That is, χ is attached to ±γ in some way. The handout calls this map I_{p}, an isotypy.

3. We want to know the significance of the sign in the map in 2. There is a sign involved in the Isaacs-Navarro conjecture, where you look at congruence classes k or −k mod p. There is a sign involved in isotypies. Is there a connection?

Note that in the table above ω_{5} is in a block by itself. This isn’t an accident as explained in the following proposition.

Proposition 20 Let A = RG and Â = KG. Let F be a primitive central idempotent of Â such that F corresponds to χ, an irreducible ordinary character of G and p^{a}|χ (1) where |G| = p^{a}m for (m, p) = 1. Then F ∈ A and AF is a block of A, that is, χ is the only irreducible character in its block.

Proof. Take

F = \frac{χ (1)}{|G|} ∑_{g∈G} χ (g^{−1}) g = \frac{p^{a}l}{p^{a}m} ∑_{g∈G} χ (g^{−1}) g ∈ RG

Then AF ⊂ ÂF so that AF ≅ M_{n} (R) and \overline{A}\overline{F} ≅ M_{n} (k).

Corollary 6 χ is irreducible mod p.

Returning now to A_{5} with p = 5, we can use this information to compute the Brauer characters. Restricting the ordinary characters to G_{5′}, we have the following.

1 (123) (12) (34)
φ_{1} 1 1  1
4 1  0
φ_{3} 5 −1  1
φ_{2} 3 0  −1
(3 0  −1) / 50

The first and third are irreducible. We discard the last character, and it remains to consider which is the other Brauer character, since there can be only three. It turns out that the fourth character is irreducible as we will discover next time. Then the Brauer characters are the ones labeled φ_{1}, φ_{2} and φ_{3} in the table above. Normally it won’t be so easy to compute the Brauer characters. Then we have

χ_{1}|_{G_{5′}} = φ_{1}
χ_{3}|_{G_{5′}}, χ_{3′}|_{G_{5′}} = φ_{2}
χ_{4}|_{G_{5′}} = φ_{1} + φ_{2}
χ_{5}|_{G_{5′}} = φ_{3}

Writing this information in matrix form, we have

φ_{1} φ_{2} φ_{3}
χ_{1} 1 0 0
χ_{2} 0 1 0
χ_{3} 0 1 0
χ_{4} 1 1 0
χ_{5} 0 0 1

This matrix is called the decomposition matrix, to be discussed below.

Next,we want to consider the Cartan invariants and the decomposition numbers. Let K be sufficiently large and A = RG. Write

A = Ae_{1} ⊕ Ae_{2} ⊕ ⋯ = ⨁_{i=1}^{s} Ae_{1} ⊕ ⨁_{i=s+1} Ae_{2}.

where the Ae_{i} in the first expression are the representatives of the isomorphism classes of the Ae_{j}. Write Q_{i} = Ae_{i}. Therefore, the Q_{i} are the projective indecomposable Amodules. Also, write P_{i} = \overline{A}\overline{e_{i}} for the projective, indecomposable \overline{A} modules. Write Q̂_{i} = Âe_{i} and S_{i} = P_{i}/J (\overline{A}) P_{i} for 1 ≤ i ≤ s. The S_{i} are simple \overline{A}-modules. Write V_{i} for the irreducible Â modules for 1 ≤ i ≤ r. Let M_{i} be an A-module with M_{i} ≤ V_{i} such that V_{i} = K ⊗_{R} M_{i}. Then d_{i,j}, the i, j entry in the decomposition table above, are the multiplicity of S_{j} as a composition factor of \overline{M_{i}}. Our shorthand for this will be d_{i,j} = (\overline{M} : S_{j}).

November 14

Homework Addition: Add Part 6 to Homework 8: Do for p = 7 what you were asked to do in Part 5 for p = 3.

Remark 12 The Brauer characters are known for SL (2, p) for p prime (see CR1, 17.17) This is used for A_{5}, p = 5, as mentioned last time, and for the 168-group, which is isomorphic to P SL(2, 7), for p = 7.

Remark 13 If ζ = e^{2πi/5}, then ζ + ζ^{−1} = 2 cos (\frac{2π}{5}) ∈ ℝ and

1 + ζ + ζ^{−1} = \frac{1 + \sqrt{5}}{2} 51

and

1 + ζ^{2} + ζ^{−2} = \frac{1 − \sqrt{5}}{2}

Use this information to confirm the character table of A_{5}.

We fix the following notation, introduced last time. A reference for this section is DB, p.19.

Â = KG  A = RG  \overline{A} = kG
{V_{i} simple : i = 1, 2, …, s}  M_{i} ≤ V_{i}  {S_{i} simple; i = 1, …, r}
V_{i} ⟷ χ_{i}  RG-lattice in V_{i}
V_{i} = K ⊗_{R} M_{i}  Q_{i} projective, indecomposable lift of P_{i} S_{i} ⟷ P_{i} projective, indecomposable
Q̂_{i} = K ⊗_{R} Q_{i}  d_{i,j} = (\overline{M}_{i} : S_{j})  c_{i,j} = (P_{j} : S_{i})

For any KG-module L, write (L : V_{i}) for the multiplicity of V_{i} as a composition factor of L. Similarly, for any kG-module L, write (L : S_{i}) for the multiplicity of S_{i} as a composition factor of L.

Definition 20 d_{i,j} = (\overline{M}_{i} : S_{j}) are the decomposition numbers and c_{i,j} = (P_{j} : S_{i}) are the Cartan invariants.

Another major problem in modular representation theory is to determine the decomposition numbers and Cartan invariants, given G and p.

D = (d_{i,j}) is a s × r matrix. It connects the characteristic 0 and the characteristic p representations, and C = (c_{i,j}) is an r × r matrix which connects projective and simple modules in characteristic p. The matrix D can be expressed as a transition matrix between ordinary and Brauer characters as

χ_{i} = ∑_{j=1}^{r} d_{i,j}φ_{j}

restricted to G_{p′}, the set of p regular elements.

We are now curious as to the connection between C and D. Note K, R, and k are chosen sufficiently large as before.

Proposition 21

Hom\overline{A} (P_{i}, S_{j}) = \begin{cases}k if i = j \\ 0 if i ≠ j \end{cases}

Proof. We showed that S_{i} is the unique top composition factor of P_{i}. Since any homomorphism of P_{i} → S_{i} can be regarded as the canonical homomorphism composed with an element of Hom\overline{A} (S_{i}, S_{i}), we get the proposition.

Remark 14 Also we have that c_{i,j} = dim_{k} Hom\overline{A} (P_{i}, P_{j}) This follows from a more general proposition below.

Proposition 22

dim_{k} Hom\overline{A} (P_{i}, M) = (M : S_{i})

for any \overline{A}-module M.

Proof. Suppose S_{j} is a top composition factor of M. Then we have an exact sequence

0 → M′ → M → S_{j} → 0

now since P_{i} is projective, we have

0 → Hom\overline{A} (P_{i}, M′) → Hom\overline{A} (P_{i}, M) → Hom\overline{A} (P_{i}, S_{j}) → 0

is exact. If i ≠ j, we have Hom\overline{A} (P_{i}, S_{j}) = 0 which implies that Hom\overline{A} (P_{i}, M′) ≅ Hom\overline{A} (P_{i}, M). Apply induction on the composition length of M. Then

dim_{k} Hom\overline{A} (P_{i}, M) = dim_{k} Hom\overline{A} (P_{i}, M′) = (M′ : S_{j}) = (M : S_{j})

Next if j = i,

dim_{k} Hom\overline{A} (P_{i}, M) = dim_{k} Hom\overline{A} (P_{i}, M′) + 1

and (M : S_{i}) = (M′, S_{i}) + 1 and we have the result.

We can now compute the c_{i,j}. We have

d_{i,j} = (\overline{M}_{i}, S_{j}) = dim_{k} Hom\overline{A} (P_{j}, \overline{M}_{i}) = rank_{R}Hom_{A} (Q_{j}, M_{i})

since Q_{j} is projective. That is, a homomorphism of P_{j} into \overline{M}_{i} can be lifted to a homomorphism of Q_{j} into M_{i}. Note also that rank_{R}Hom_{A} (Q_{j}, M_{i}) = (Q̂_{j} : V_{i}).

\xymatrix{Q_{j} \ar@{-->}[dr] \ar[r] & P_{j} \ar[dr] \ar[r] & 0 \\  & M_{i} \ar[r] & _{\overline{M}_{i}} \ar[r] & 0}

Now consider

d_{k,i}d_{k,j} = (\overline{M}_{k} : S_{i}) (Q_{j} : M_{k})

where (Q_{j} : M_{k}) is interpreted as rank_{R}Hom_{A} (Q_{j}, M_{k}) ) = (\overline{M}_{k} : S_{i}) (Q̂_{j} : V_{i})). Then

∑_{k} d_{k,i}d_{k,j} = ∑_{k} (Q̂_{j} : V_{k}) (\overline{M}_{k}, S_{i}). This counts the number of composition factors

isomorphic to S_{i} where Q̂_{j} is taken mod P which is (P_{j} : S_{i}) = c_{i,j}.

Hence we have the following.

Theorem 12 C =^{t} DD.

November 17

Recall that we have the following situation.

ÂKG  A = RG  \overline{A} = kG
V_{i} irreducible  M_{i} ≤ V_{i}  S_{i} ireducible
(χ_{i})  (φ_{i})
Q̂_{i}  Q_{i} projective, P_{i} projective,

indecomosable

indecomposable

Theorem 13 C = D^{T} D. 53

Proof. Write d_{ij} = (\overline{M}_{i} : S_{j}). (Then note χ_{i}|_{G′_{p}} = ∑_{j} d_{i,j}φ_{j}.) The last proposition showed that

d_{i,j} = dim_{k} Hom\overline{A} (P_{j}, \overline{M_{i}}) = rank_{R}Hom_{A} (Q_{j}, M_{i}) = dim HomÂ (Q̂_{j}, V_{i}) = (Q̂_{j} : V_{i})

Then ∑_{k} d_{k,i}d_{k,j} = ∑_{k} (Q̂_{j} : V_{k}) (\overline{M}_{k}, S_{i}). This counts the number of composition fac-

tors isomorphic to S_{i} where Q̂_{j} is taken mod P, which is (P_{j} : S_{i}) = c_{i,j}

Returning to A_{5} with p = 5, we computed

D = (\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{matrix})

The third is the same as the second mod 5, the fourth is a linear combination of the first two, and the last character is in a block by itself. Then

C =^{t} DD = (\begin{matrix}2 & 1 & 0 \\ 1 & 3 & 0 \\ 0 & 0 & 1 \end{matrix})

The block (\begin{matrix}2 & 1 \\ 1 & 3 \end{matrix}) shows that P_{1} corresponds with the trivial module S_{1} and the only

possible composition series is

S_{1}

S_{2}

S_{1}

P_{2} has one of the following composition series.

\xymatrix{S_{2} \ar[d] & S_{2} \ar[d] \\ S_{1} \ar[d] & S_{2} \ar[d] \\ S_{2} \ar[d] & S_{1} \ar[d] \\ S_{2} & S_{2}}

We have not yet shown that the irreducible ordinary character mod 3 is irreducible mod 5. This would be the same as showing that there is a 5-modular irreducible representation of degree 3. Recall for the following discussion that A_{5} ≅ P SL (2, 5).

We can show (see CR1, 17.17, hand-out) that G = SL (2, p) has p irreducible modular

representations over a field k of characteristic p. Take g = (\begin{matrix}α & β \\ γ & δ \end{matrix}) ∈ G. Then G can be

regarded as acting on a vector space of dimension 2, and then on k [x, y] by

g · x = αx + βy g · y = γx + δy.

Take M_{d} to be the subspace of k [x, y] spanned by x^{d}, x^{d} − 1y, …, y^{d}. We can show that M_{d} is irreducible. For example, when d = 2, we have M_{2} = ⟨x^{2}, xy, y^{2}⟩. Then

(\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}) : \begin{cases}x → x \\ y → x + y \end{cases}

(\begin{matrix}1 & 0 \\ 1 & 1 \end{matrix}) : \begin{cases}x → x + y \\ y → y \end{cases}

Now let V = ⟨v_{1}, v_{2}, v_{3}⟩ be a vector space of dimension k over k. In kG, take

(\begin{matrix}0 & 1 \\ 0 & 0 \end{matrix}) =A = (\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}) − (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix})
(\begin{matrix}0 & 0 \\ 1 & 0 \end{matrix}) =B = (\begin{matrix}1 & 0 \\ 1 & 1 \end{matrix}) − (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}) .

Then

A : \begin{cases}v_{1} → 0 \\ v_{2} → v_{1} \\ v_{3} → 2v_{2} + v_{1} \end{cases} B : \begin{cases}v_{1} → 2v_{2} + v_{3} \\ v_{2} → v_{3} \\ v_{3} → 0 \end{cases}

and

A → (\begin{matrix}0 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{matrix}) B → (\begin{matrix}0 & 0 & 0 \\ 2 & 0 & 0 \\ 1 & 1 & 0 \end{matrix})

on V. We then have

A^{2} : \begin{cases}v_{1} → 0 \\ v_{2} → 0 \\ v_{3} → 2v_{1} \end{cases} and B^{2} : \begin{cases}v_{1} → 2v_{3} \\ v_{2} → 0 \\ v_{3} → 0 \end{cases} .

It is easy to check that if 0 ≠ U ≤ V, and αv_{1} + βv_{2} + γv_{3} ∈ U with at least one of α, β, γ not 0, by applying A, A^{2}, B, and B^{2} to this element we get all of V. Thus V is irreducible.

Returning to GL (3, 2) = G, we look at the characters from the point of view of the general theory for GL (n, q). We had characters of degrees 1, 3, 3, 6, 7, and 8. We

obtained them by taking B = \{(\begin{matrix}∗ & ∗ & ∗ \\ 0 & ∗ & ∗ \\ 0 & 0 & ∗ \end{matrix})\} of size 8 and P = \{(\begin{matrix}∗ & ∗ & ∗ \\ ∗ & ∗ & ∗ \\ 0 & 0 & ∗ \end{matrix})\} of size

24. We found Ind_{P}^{G} (1) = χ_{1} + χ_{2} by counting lines fixed by P. χ_{2} is the character of degree 6. We found Ind_{B}^{G} (1) = 2χ_{1} + χ_{2} + χ_{4} by counting flags fixed by B. χ_{3} is the character of degree 8. Finally, Ind_{P}^{G} (ε) = χ_{3} is the character of degree 7. The remaining characters χ_{5} and χ_{6} were obtained from S, the Sylow 7-subgroup.

Complete the series of homeworks on GL (3, 2) by writing an intoduction to GL (3, 2) which begins “Let G = GL (3, 2). We compute… ”. This should be a one-page document.

Recall that in computing the characters of GL (3, 2), the characters of degree 1, 6, 7, and 8 came from Ind_{B}^{G} (1), Ind_{P}^{G} (1), and Ind_{P}^{G} (ϵ). The two characters of degree 3 remain and are explained in terms of Harish-Chandra Theory or Theory of Cusp Forms.

First let G be any finite group and let H ≤ G. Write A = KG for K sufficiently large. We want to decompose Ind_{H}^{G} (1) into a sum of irreducible representations of G. Recall that A is artinian and semisimple so that we have such a decomposition.

Recall from September 12 that

Hom_{A} (V_{1} ⊕ V_{2}, W_{1} ⊕ W_{2})

is isomorphic the group of matrices (\begin{matrix}θ_{1,1} & θ_{2,1} \\ θ_{1,2} & θ_{2,2} \end{matrix}) where θ_{i,j} ∈ Hom (V_{i}, W_{j}). The θ_{i,j}

involve injections and projections. This is relevant because decomposing V = Ind_{H}^{G}(1) involves projections in End_{A} (V).

More generally, if

V = V_{1} ⊕ V_{2} ⊕ ⋯ ⊕ V_{m}
W = W_{1} ⊕ W_{2} ⊕ ⋯ ⊕ W_{n},

then Hom_{A} (V, W) will be isomorphic to a group of matrices (θ_{i,j}) with θ_{i,j} ∈ Hom_{A} (V_{i}, W_{j}) and if V = W, the matrices are (θ_{i,j}) with θ_{i,j} ∈ Hom_{A} (V_{i}, V_{j}).

Returning to A = KG with A semisimple, suppose

V ≅ ⨁_{i=1}^{s} n_{i}S_{i}

for S_{i} simple modules. Then in this case, we have Hom_{A} (S_{i}, S_{j}) = δ_{i,j}K. ( Since K is sufficiently large, a matrix commuting with all the matrices representing elements of G in the representation corresponding to S_{i} is a scalar matrix. This is one way of saying that End_{A}(V) ≅ K. Hence

E := End_{A} (V) ≅ ⨁_{i} M_{n_{i}} (K)

This connects the representations of A and E given a fixed V. Let L_{i} be the simple M_{n_{i}} (K)- module. There is only one up to isomorphism. Hence, we have a bijection

{Irreducible A-modules in V } ⟷ {Irreducible E-modules}

given by S_{i} ⟷ L_{i} Also n_{i} = (V : S_{i}) = dim_{K} L_{i}

Example 20 Returning to G = GL (3, 2), we computed V = Ind_{B}^{G} (1) has character χ_{1} + 2χ_{2} + χ_{4} so E has irreducible representations of degree 1, 2, and 1. We observe that S_{3} also has irreducible representations of degree 1, 2, and 1 and in fact, E ≅ KS_{3} by the theory we will develop later.

Let H ≤ G, A = KG, and V = Ind_{H}^{G} (1). We want to consider E = End_{A} (V). In B = KH, the trivial representation is realized by Be where

e = \frac{1}{|H|} ∑_{h∈H} h.  (19)

Proposition 23 Let e be any idempotent in B. Then Ae is the module corresponding to the representation of A induced from the representation of B given by Be.

Proof. The module for the induced representation is

A ⊗_{B} Be ≅ Ae.

Now take e as in (19) and consider Ae. Recall that

Hom_{A} (Ae, M) ≅ M ≅ eM

and

End_{A} (Ae) ≅ (eAe)^{op} .

Hence, we want to study H := eAe, the ”opposite” of the endomorphism algebra of Ae which is called the Iwalori-Hecke Algebra. (A reference for this is CR1, 11D.)

Now eAe is a subalgebra of A where e = \frac{1}{|H|} ∑_{h∈H} h. Then a basis for H is given by

{exe : x runs over a set of representatives for H\G/H}

because H is spanned by {exe : x ∈ G} and exe = eye ⟺ x = hyh′ for h, h′ ∈ H.

Definition 21 indx = \frac{|H|}{|H∩^{x}H|}

Note that

|HxH| = \frac{|H| |H|}{|H ∩^{x} H|}.

Let {x_{j} : j = 1, 2, … } be a set of representatives of H\G/H.

Definition 22 a_{j} = (indx_{j}) ex_{j}e.

Then H has basis {a_{j}} with multiplication given by a_{i}a_{j} = ∑_{k} μ_{i,j,k}a_{k}. Next time we will consider how to describe the μ_{i,j,k}.

November 21

Recall that H ≤ G and H = eAe. A in this case is KG, thought this works sometimes in more general cases. Recall that 𝔼⋉_{A} (Ae) ≅ (eAe)^{op} and that e = \frac{1}{|H|} ∑_{h∈H} h. Then H is spanned by {exe : x a representative of H\G/H}. We also write

ind x = \frac{|H|}{|H ∩^{x} H|}.

Theorem 14 A basis for H is given by

a_{j} = \frac{1}{|H|} ∑_{x∈D_{j}} x

where D_{j} = Hx_{j}H for 1 ≤ j are the double cosets of H in G. Then a_{i}a_{j} = ∑_{k} μ_{i,j,k}a_{k} where

μ_{i,j,k} = |H|^{−1} |D_{i} ∩ x_{k}D_{j}^{−1}| .

Proof. We know that H is spanned by the ex_{j}e for 1 ≤ j and these are also independent as ex_{j}e and ex_{k}e for j ≠ k contain distinct elements of G. For simplicity, write x = x_{j}. Then

exe = \frac{1}{|H|^{2}} ∑_{h_{1},h_{2}∈H} h1xh_{2} \frac{1}{|H|^{2}} ∑_{h_{1}xh_{2}∈D} |H ∩^{x} H| h_{1}xh_{2}

where the sum is taken over distinct elements in D and this works because the number of elements in D is

\frac{|H|^{2}}{|H ∩^{x} H|}

In particular, exe ≠ 0 since K in this case has characteristic 0. This may not work so will for other fields. Let

a_{j} = (ind x_{j}) exe == \frac{|H|}{|H ∩^{x} H|}exe = \frac{|H|}{|H ∩^{x} H|} \frac{1}{|H|^{2}} |H ∩^{x} H| ∑_{y∈D} y = \frac{1}{|H|} ∑_{y∈D} y.

Then

a_{i}a_{j} = \frac{1}{|H|^{2}} (∑_{y∈D} y) (∑_{z∈D} z)

and count the number of times x_{k} appears in this product, that is, the number of pairs y ∈ D_{i}, z ∈ D_{j} with yz = x_{k}. That is, z = y^{−1}xk and y = x_{k}z^{−1} ∈ D_{k} ∩ x_{k}D_{j}^{−1} where D^{−1} = {d^{−1} : d ∈ D}. Hence, the number of such pairs is |D_{i} ∩ x_{k}D_{j}^{−1}| and the coefficient of x_{k} in a_{i}a_{j} is

\frac{1}{|H|^{2}} |D_{i} ∩ x_{k}D_{j}^{−1}| .

Hence,

\frac{1}{|H|}μ_{i,j,k} = \frac{1}{|H|^{2}} |D_{i} ∩ x_{k}D_{j}^{−1}|

so that μ_{i,j,k} = \frac{1}{|H|} |D_{i} ∩ x_{k}D_{j}^{−1}| as required. Consider G = GL (n, q) with H = B, the subgroup of upper triangular matrices. We need the double coset decomposition B\G/B. This can be accomplished by the famous Bruhat decomposition.

Theorem 15 G is the disjoint union

G = ⋂_{w} BwB

for w ∈ S_{n}.

Proof. Let X_{i,j} (α) be the elementary matrix

(\begin{matrix}1 \\ & 1 & α \\ . & . & . \\ 0 & & & 1 \end{matrix})

where the α is in position i, j. We want to show that if g ∈ G, g = b_{1}wb_{2} for b_{1}, b_{2} upper triangular and w a unique permutation matrix. By multiplying g on the left by matrices of the form X_{i,j} (α), we get every entry in the first column except the k_{1} position to be zero for some k_{1}. After all, we can never have all the elements zero or g wouldn’t be invertible. Repeat for the other columns giving a matrix for which the ith column is zero for every row except k_{i}th row. Since g is invertible, (k_{1}, k_{2}, …, d_{n}) is a permutation of (1, 2, …, n). We now have b_{1}^{−1}g for some b_{1} ∈ B since all the X_{i,j} (α) are upper triangular. Now left-multiplying by some w^{−1} ∈ S_{n} makes w^{−1}b_{1}^{−1}g =: b_{2} upper triangular. Then g = b_{1}wb_{2}.

This gives us the double coset decomposition since b_{1}wb_{2} ∈ BwB. Next, we want to see that this decomposition is unique. Proof. (Uniqueness) If we had

b_{1}wb_{2} = b′_{1}w′b′_{2},

then we would have

wb_{2}b_{1}^{′−1} = b_{1}^{−1}b′_{1}w′

so that

wb − 2b_{2}^{′−1} = b_{1}^{−1}b′_{1} ∈ B

If wb_{1}w^{′−1} is upper triangular, then w = w′. For example, if we had

\underbrace{(\begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix})}_{w_{1}} (\begin{matrix}b_{1} & ∗ & ∗ \\ 0 & b_{2} & ∗ \\ 0 & 0 & b_{3} \end{matrix}) = (\begin{matrix}0 & b_{2} & ∗ \\ 0 & 0 & b_{3} \\ b_{1} & ∗ & ∗ \end{matrix})

but the only w ∈ S_{3} which takes this matrix to an upper triangular matrix is w_{1}^{−1}.

November 24

If w_{1} = (132), then

w_{1} · b = (\begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}) (\begin{matrix}b_{1} & ♣ & ♣ \\ 0 & b_{2} & ♣ \\ 0 & 0 & b_{3} \end{matrix}) = (\begin{matrix}0 & b_{2} & ♣ \\ 0 & 0 & b_{3} \\ b_{1} & ♣ & ♣ \end{matrix})

Now if w_{1}bw_{2} ∈ B, then w_{2} = w_{1}^{−}. Note that this does not mean that w_{1}bw_{1}^{−1} is in B. It could be in B for some b. This shows that if G = GL (n, q) and B is the set of upper triangular matricies, then

G = ⋃_{w∈W} BwB

is a disjoint union with W = S_{n}. Then consider H = (eAe)^{op}, A = KG, and e =

\frac{1}{|B|} ∑_{g∈B} g. 59

Proposition 24 H has basis {a_{w} : w ∈ W } where a_{w} = ∑ 1|B|sum_{g∈BwB}g with multiplication constants as described earlier.

Lemma 7 (CR, Exercise 11.19) Let G be any finite group and H ≤ G and define H as before. Also, define ind (x) = \frac{|H|}{|H∩^{x}H|} as before for x ∈ H. Then the linear map

ind : H → K given by
a_{j} → ind (x_{j})

is a character of H. Recall that

a_{j} = \frac{1}{|H||} ∑_{x∈D_{j}} g and D_{j} = Hx_{j}H

Proof. Consider the trivial character ψ of KG restricted to H. We have

ψ (a_{j}) = \frac{1}{|H|} |D_{j}|
= \frac{1}{|H|} |Hx_{j}H|
= \frac{1}{|H|} \frac{|H|^{2}}{|H ∩^{x_{j}} H|}
= ind (x_{j})

Remark 15 Recall that the dimensions of the irreducible representations of H give the multiplicities of the multiplicities of the irreducible constituants of Ind_{H}^{G} (1).

Application. Let n = 2, G = GL (2, q), H = B, and W = S_{2} = {1, 2}. Then H has basis \{a_{1}, a_{w}\}. Recall that a_{i}a_{j} = ∑_{k} μ_{i,j,k}a_{k} and μ_{i,j,k} = \frac{1}{|H|} |D_{i} ∩ x_{k}D_{j}^{−1}|. Then a_{1} · a_{1} = a_{1} and a_{1} · a_{w} = a_{2} · a_{1} = a_{w} which is easy to check. It is more difficult to compute a_{w} · a_{w}. Let a_{w} · a_{w} = μa_{1} + λa_{w} for somr μ and λ. Then

μ = \frac{1}{|B|} ((BwB) ∩ (BwB)^{−1}) = \frac{1}{|B|} |BwB|

since w = w^{−1} But since G = B ∪ BwB, we have |G| = |B| + |BxB| giving |BwB| = q (q − 1) (q^{2} − q) = q^{2} (q − 1)^{2} so that

μ = \frac{q^{2} (q − 1)^{2}}{q (q − 1)^{2}} = q.

Then a_{2}^{2} = qa_{1} + λa_{w}. Apply the charaacter from the lemma to this formula giving

(ind a_{w})^{2} = q (ind a_{1}) + λ (ind a_{w})  (20)

so that ind (a_{1}) = 1 and ind (a_{w}) = q since B ∩^{w} B = T, the subgroup of diagonal matricies giving ind (a_{w}) = \frac{q(q−1)^{2}}{(q−1)^{2}} = q. From (20), we have

q^{2} = q + λq
λq = q (q − 1)
λ = q − 1

Thus, for n = 2 we have H = ⟨a_{1}, a_{w}⟩ with multipliciation given by a_{1}^{2} = a_{1}, a_{1}a_{w} = a_{w}a_{1} = a_{2} and a_{w}^{2} = qa_{1} + (q − 1) a_{w}. Note that if q = 1, then H is isomorphic to the group algebra of S_{2}.

Remark 16 In general, for G = GL (n, q), H is isomorphic to the group algebra of S_{n}, which explains why for GL (3, 2), the constituants of of Ind_{B}^{G} (1) occur with mulitpilicities 1, 2, and 1, the degrees of the characters of S_{3}.

The Harish-Chandra theory or Theory of Cusp Forms similarly develops H for characters other than the trivial character. Let G = GL (n, q) and B the set of upper- triangular matricies.

Definition 23 A Borel subgroup of G is any conjugate of B

Then the Borel subgroups are solvable subgroups and B = T U for T the diagonal matricies and U the unipotent matricies. Then |U| = q^{\frac{n(n−1)}{2}} is a p-group as q = p^{m}, so U is solvable. Also, T is abelian and U ◁ B so that B/U ≅ T.

Definition 24 A parabolic subgroup of G is one which contains a Borel subgroup.

November 26

Let G = GL (n, q) and T the subgroup of diagonal matrices. Then B = T U for U the subgroup of unipotent matrices in B. (Note that the set of unipotent matrices in G is not a subgroup.) A Borel subgroup is any conjugate of B and a parabolic subgroup is a subgroup containing a Borel subgroup. A parabolic subgroup is conjugate to a subgroup of the form

{( ♣  ♣ ; ♣ ; ... ; 0  ♣ )}

For n = 3, for example, there are, up to conjugacy, four parabolic subgroups B, P_{1}, P_{2}, and G

Insert picture here

and P_{1} is not conjugate to P_{2}.

We have a Levi decomposition of P given by P = LV, a semidirect product, with V the unipotent radical of P, that is, the maximal normal unipotent subgroup of P. Then L is a direct product GL (n_{1}, q) × GL (n_{2}, q) × ⋯ × GL (n_{r}, q). If P is as in the example above, we have

L = {( ♣  0 ; ♣ ; ... ; 0  ♣ )}

and

V = {( I  0 ; I ; ... ; 0  I )}

In fact, V is a p-group.

Remark 17 The characters of U are not known.

To conclude this section, we want to state the main result of the Harish-Chandra theory. We consider representations of G over a field K of characteristic 0.

If ψ is a character of L, we get a character ψ̃ of P since L ≅ P/V. Recall that this is accomplished as follows. If g ∈ P, then g = lv for some l ∈ L, v ∈ V and we take ψ̃(g) = ψ (l). Then we consider Ind_{P}^{G} \binom{}{ψ}. Note that Ind_{B}^{G} (1) is a special case.

Definition 25 An irreducible character χ of G is said to be cuspidal or in the discrete series if

⟨Ind_{P}^{G} \binom{}{ψ} , χ⟩ = 0

for any P < G with P ≠ G and any ψ.

Now cuspidal characters are defined for L where P = LV and

L = GL (n_{1}, q) × GL (n_{2}, q) × ⋯ × GL (n_{r}, q) .

Then a cuspidal character ψ of L is (ψ_{1}, ψ_{2}, …, ψ_{r}) for ψ_{i} cuspidal in the ith block of L

Definition 26 Let P_{1} and P_{2} be parabolic subgroups. Then P_{1} and P_{2} are associated if P_{1} = L_{1}V_{1} and P_{2} = L_{2}V_{2} for L_{1} conjugate to L_{2} in G.

Theorem 16 (CR1, Vol II, §70B, 70.15A and B but stated there in more generality) Let χ be an irreducible character of G. Then there is a unique pair (P, ψ) such that ψ is cuspidal for L where P = LV, and such that ⟨Ind_{P}^{G} \binom{}{ψ} , χ⟩ ≠ 0. That is, if (P_{1}, ψ_{1}) and (P_{2}, ψ_{2}) are two such pairs, then ^{x}L_{1} = L_{2} and ^{x}ψ_{1} = ψ_{2} for some x ∈ G.

Recall that the last statement means xL_{1}x^{−1} = L_{2} and ψ_{2} (xl_{1}x^{−1}) = ψ_{1} (l_{1}) for l_{1} ∈ L_{1}.

Hence, the characters of G are divided into disjoint sets called Harish-Chandra families, each family corresponding to a ”cuspidal pair” (L, ψ).

Example 21 Let G = GL (3, 2). One family is {χ_{1}, χ_{2}, χ_{4}} of degrees 1, 6, and 8, from the pair (T, 1) where T is the diagonal subgroup. The second family is {χ_{3}} from the pair (L, ε) where L ≅ GL (2, 2) × GL (1, 2) ≅ S_{3} and ε is the sign character on the GL (2, 2) block and is cuspidal for L. The remaining characters χ_{5} and χ_{6} of degree 3 are cuspidal for G.

The remaining characters in the homework discussed above were constructed an an ad-hoc manner. Indeed, there are two remaining problems.

1. Decompose Ind_{P}^{G} \binom{}{ψ}.

2. Construct cuspidals for G (and hence for L).

The following is from the file Hints518 on ftp://dirichlet.math.uic.edu

Hints for Homework Brauer characters for p = 3:

You can first use the following theorem. If p^{a} is the exact power of p dividing the order of the group, and if χ is an irreducible character of the group whose degree is divisible by p^{a} then χ is irreducible mod P, and is in a block by itself. (This was proved on November 12 )

Note that two characters in different blocks will not have any Brauer characters in common. This follows from the decomposition of A or \overline{A} as a sum of indecomposable two-sided ideals.

Using this one gets four Brauer characters for p = 3. You have to find only one more. Look at the character of degree 7. Can this decompose mod P? Or is it irreducible? That is what you have to answer.

Brauer characters for p = 7:

Here you have to use the fact that G ≅ P SL(2, 7). We described (and you have a hand-out) describing the representations of SL(2, 7) over a field of characteristic 7. The Brauer characters you get from these have degrees 1 through 7, of which the ones of degrees 1,3,5,7 will be Brauer characters for P SL(2, 7). You can take elements of orders 1, 2, 3, 4 (which are the ones you want) in SL(2, 7), and in fact since you only want the traces of these elements in the representations, you can take them to be diagonal in some extension field. For example, an element of order 4 in SL(2, 7) can be taken to be a diagonal matrix D with i, −i on the diagonal (here i is a fourth root of unity in an extension of F_{7}). Suppose you consider the representation of degree 3, which is in a space spanned by X^{2}, XY, Y^{2} (see the hand-out; this was also discussed in class, on November 17). How does D act on this space? It will have eigenvalues i^{2}, i(−i), (−i)^{2}, with trace −1 + 1 + (−1) = −1, which is exactly what you expect. (Why? Look at the reduction mod P of the character(s) of degree 3 of G to the 7-regular elements; you get the values 3, −1, 0, 1. Note that D modulo the center of SL(2, 7) is of order 2, and so you should expect −1 for the value of the character of degree 3 there, and this is what you got above.) For the element of order 4 in G you must start with a diagonal matrix with eighth roots of unity along the diagonal in SL(2, 7), and so on. So the solution is just computational, once you know the theory for SL(2, p).

December 1

Here is a summary of the techniques we used so far to produce characters.

1. Frobenius induction. Given a character ψ of H ≤ G (here G is any finite group), we can find a character Ind_{H}^{G}ψ of G. This also works for some groups other than finite groups.

2. Harish-Chandra induction. This works for groups G of Lie type such as GL (n, q). If G = GL (n, q) we take a parabolic subgroup P ≤ G and write P = LV where L is a product of smaller GL’s. We take a character ψ of L and inflate (or pull-back) ψ producing a character ψ̃ of P We then get the induced character Ind_{H}^{G} \binom{}{ψ} of G. This also works for some groups other than finite groups.

3. Cuspidal characters of G do not arise as constituents of Ind_{H}^{G} \binom{}{ψ} for any ψ From this procedure, the characters of G fall into Harish-Chandra families, one for each P up to associativity (which means the L’s are conjugate; see the definition given earlier) and a cuspidal character of L, or equivalently for each ”cuspidal pair” (L, ψ) up to Gconjugacy.

For GL (2, q), the conjugacy classes are described in terms of their ”types”. Suppose that in \overline{𝔽_{q}} we have selected a primitive (q^{2} − 1)^{st} root of unity ε, that is ε^{q^{2}−1} = 1 and ε is a generator of 𝔽_{q^{2}}^{×}. Let ρ = ϵ^{q+1}. Then ρ generates 𝔽_{q}^{×}. Then we can pick representatives for the conjugacy classes as follows.

The identity: (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix})

Classes arising from the decomposition B = T U: (\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}) from U and (\begin{matrix}ρ^{a} & 0 \\ 0 & ρ^{b} \end{matrix}) from

T with a ≠ b.

Classes containing elements of the form (\begin{matrix}ε^{a} & 0 \\ 0 & ε^{aq} \end{matrix}) for a ≢ 0 (mod q + 1), over 𝔽_{q^{2}}.

Over 𝔽_{q} such elements would be given in rational form.

Other classes are omitted such as (\begin{matrix}x & 0 \\ 0 & x \end{matrix}) and (\begin{matrix}x & 1 \\ 0 & x \end{matrix}).

We pick a complex primitive (q^{2} − 1)^{st} root of unity ε̃ and let ρ̃ = ε̃^{q+1}. We have a correspondence ρ^{a} ⟷ ρ̃^{a} and ε^{a} ⟷ ε̃^{a}.

Most importantly, we do the Harish-Chandra induction from B = T U writing

T = \{(\begin{matrix}ρ^{a} & 0 \\ 0 & ρ^{b} \end{matrix})\}

as a direct product of two cyclic groups of order q − 1. We get characters of G in Harish-Chandra families corresponding to pairs (T, ψ), as we see below.

We also have some cuspidal characters of G which don’t arise as constituents Ind_{B}^{G} \binom{}{ψ}. For example,

\{(\begin{matrix}ε^{a} & 0 \\ 0 & ε^{aq} \end{matrix})) ≤ GL (2, q^{2})

is a subgroup of order q^{2} − 1, which is conjugate in GL (2, q^{2}) to a subgroup of GL (2, q) in rational form. We expect some characters ”supported” by this subgroup.

We have the following table.

(\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}) (\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix})  (\begin{matrix}ρ^{a} & 0 \\ 0 & ρ^{b} \end{matrix})  (\begin{matrix}ε^{a} & 0 \\ 0 & ε^{aq} \end{matrix})
1  1  1  1  1  and other linerar characters
Ind_{B}^{G} \binom{}{ψ} , m ≠ n q + 1  1  ρ̃^{am}ρ̃^{bn} + ρ̃^{an}ρ̃^{bm}  0
S  q  0  1  −1  S is the Steinberg character of degree q

Here 1 and S arise as constituents of Ind_{B}^{G} (1).

To construct cuspidal characters, we use an important concept called the Brauer Lift, described in CR1, p. 436. For any G a finite group with (K, R, k) as before, we lift characters from kG to KG. Note that this is the reverse of what we did before. Given x ∈ G, we can write

x = su = us 64

where u has order a power of p and s has order prime to p. Then we have the following deep theorem.

Theorem 17 Let λ be the Brauer character of a kG module L. Then define a function

λ̂ : G → K by λ̂ (x) = λ (s) .

then λ̂ is a virtual character of G, that is, a ℤ-linear combination of irreducible characters.

Then we take λ to be the Brauer character of the ”natural representation” representing each element of G by itself as a matrix over k. From this we have the virtual character

2 2 ρ̃^{a} + ρ̃^{b} ε̃^{a} + ε̃^{aq}

If we subtract from this the character of degree q + 1 constructed above with m = n = 1, we have the character

q − 1 −1 0 − (ε̃^{a} + ε̃^{aq})

which is cuspidal. By using other Brauer characters we can construct all the missing cuspidal characters.

For the character tables of GL(2, q) and GL(3, q) see a paper of Steinberg [Canadian Journal of Math, 1951]. The Brauer lift is not used there, but was used in an important paper of J.A.Green (Trans. Amer. Math. Soc. 88 (1955)).

December 3

Consider G = GL (3, q) and T ≤ G the diagonal matricies. Then |T | = (q − 1)^{3}. For

Harish-Chandra induction, we use B = T U and P = LV where L = \{(\begin{matrix}♣ & ♣ & 0 \\ ♣ & ♣ & 0 \\ 0 & 0 & ♣ \end{matrix})\}

and P = \{(\begin{matrix}♣ & ♣ & ♣ \\ ♣ & ♣ & ♣ \\ 0 & 0 & ♣ \end{matrix})\}. This was accomplished in Homework 5, for q = 2. Note that

|B| = q^{3} (q − 1)^{3} Let T := \{(\begin{matrix}ρ^{a} & 0 & 0 \\ 0 & ρ^{b} & 0 \\ 0 & 0 & ρ^{c} \end{matrix})\} where ρ is a generator of 𝔽_{q}^{×} and let

ψ = ψ_{l,m,n} : \{(\begin{matrix}ρ^{a} & 0 & 0 \\ 0 & ρ^{b} & 0 \\ 0 & 0 & ρ^{c} \end{matrix})\} ↦ ρ̃^{la}ρ̃^{mb}ρ̃^{nc}

Here ρ̃ is as in the case of GL(2, q).

We must construct Ind_{B}^{G} \binom{}{ψ}. When l, m, n are distinct mod q − 1, we get an irreducible character of degree (q + 1) (q^{2} + q + 1). When l ≡ m ≢ n, we get characters of degree q^{2} + q + 1 and q (q^{2} + q + 1). When l ≡ m ≡ n, we get characters of degree 1, q^{2} + q with multiplicity 2, and q^{3}.

For example, when q = 2 as in our group, we have |B| = 8 and [G : B] = 21. We get charactes of degree 1, 6, and 8.

Now construct Ind_{P}^{G} \binom{}{φ} for φ a cuspidal character of L of degree q − 1. We have |P | = q^{3} (q − 1) (q^{2} − 1) (q − 1) and [G : P] = q^{2} + q + 1 and we get an irreducible character of degree (q − 1) (q^{2} + q + 1) = q^{3} − 1. For example, when q = 2 as in our group, this produces a character of degree 7.

Finally, G has cuspidal characters of degree (q − 1) (q^{2} − 1). For example, when q = 2 as in our group, these are characters of degree 3.

Remark 18 To construct cuspidal characters for groups other than GL (n, q), we need Deligne-Lusztig theory, which we will discuss next time.

Here is a summary of what we’ve done this semester, but viewing the material covered in a different order:

Let G be a finite group and F a field. We want to study the representation of G over F, which is the same as studying the representations of the group algebra F G, that is, the F G-modules, which are finite-dimensional vector spaces over F. Hence, it is natrual to consider A-modules with A an Artinian ring where we have the minimum condition on left ideals of A. Then

1. every non-nilpotent ideal contains an idempotent,

2. J is nilpotent

3. A/J is semi-simple or completely reducible.

We then develop the Wedderburn theory for semisimple Artinian rings. Now let A = F G and M an A-module. It is natural to ask whether M = M_{1} ⊕ M_{2}, and if so, whether we can decompose further. Since A is finite dimensional, we must get M = ⨁_{i} M_{i} for M_{i} indecomposable. Using the Krull-Schmidt theorem, we have that this decomposition is unique.

Next, we decompose _{A}A = ⨁_{i} L_{i} for L_{i} indecomposable left ideals. This produces e_{i} orthogonal primitive idempotents by taking 1 = ∑_{i} e_{i} so that A = ⨁_{i} Ae_{i}. We notice that Ae_{i} are projective A-modules. This implies that Ae_{i} has a unique maximal submodule Je_{i} and hence a unique, irreducible quotient Ae_{i}/Je_{i}. Furthermore, Ae_{i} ≅ Ae_{j} if and only if Ae_{i}/Je_{i} ≅ Ae_{j}/Je_{j}.

At this point, it is natural to consider the characteristic of F. If char F = 0, then A is semisimple by Maschke’s theorem. We develop a character theory over ℂ or over a sufficiently large field K of charactersic 0. We introduce the p-modular system (K, R, k) as a way to pass between charactersic 0 and charactersic p. The idea is that representations should give us information about the group G, hence the representations over fields of charactersic p should give us information about, and reflect, the p- structure of G. The homework has been intended to illustrate the ”global information” from ”local information” philosophy.

Further hints for the homework

This is a hint for the Brauer characters for p = 3 for GL(3, 2). The main question is whether the character of degree 7, restricted to the 3′-elements, is irreducible as a Brauer character. Consider the value of the character at the 7- elements, which is 0. See whether this is possible if the Brauer character decomposes, since the eigenvalues of these elements are 7-th roots of unity.

Further hints for the homework (added December 10)

With the hints I gave you earlier for the Brauer characters of the group of order 168 for p=3, you can get to the stage where the character of degree 7 is either irreducible (the correct answer) or the sum of the trivial character and a Brauer character of degree 6. I was too hasty in thinking that it is obvious to rule out the possibility that the character of degree 7 is reducible. Here is one way of doing this.

Recall that we constructed the character of degree 7 as an induced character Ind_{P}^{G}(ϵ) where P is a parabolic subgroup (isomorphic to S_{4}) and ϵ is the sign character of P. We can take an RG-module for this induced module and go mod P (here p=3) as usual, and get a kG-module L. Then L is still an induced module from P (the point is that the sign character of P is irreducible mod 3). If L contains the trivial module as a composition factor then Hom_{kG}(L, S) is non-zero, where S is a module for kG for the trivial representation, or Hom_{kG}(S, L) is non-zero. But both these are zero by Frobenius reciprocity (see the October 8 lecture, where it is proved that Hom_{RH}(L, M_{H}) = Hom_{RG}(L^{G}, M). In this formula put L to be a module for ϵ and M to be S). One other point is that from this formula you get Hom_{kG}(L, S) = 0. To show that Hom_{kG}(S, L) = 0 you use dual modules; L is dual to itself.

December 5

More on the nasty handout. We have |SL (2, p)| = p (p − 1) (p + 1). Also, a Sylow

p-subgroup is U = ⟨(\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix})⟩ . The elements of order p are unipotent. If x ∈ SL(2, p)

then x = su = us for s a p′-element and u a unipotent element. However, if u ∈ U, the only s such that su = us is s = 1. So every element is either a p′-element or of order p. Suppose s is a p′-element. If the order of s divides p − 1 then s is conjugate to an element of T, the subgroup of diagonal matrices. On the other hand if the order of s divides p + 1 then s can be put in rational form. In each case, two p′-elements are conjugate if and only if their characteristic polynomials are the same.

Brief Introduction to Deligne-Lusztig Theory [Carter, Finite Groups of Lie Type]

The philosophy is the following. Let G = GL (n, q). We have Ind_{B}^{G} \binom{}{ψ} irreducible in many cases. For example, when n = 3, we can take ψ = ψ_{l,m,n} for l, m, n distinct mod q − 1. Here B = T U, ψ is a character of T, and its lift to B is the character ψ̃ of B. But if S is a cyclic Sylow subgroup of order q^{n} − 1, then

S = ⟨A⟩ , for A = (0 0  ♣ ; 1 0  ♣ ; 0 1  ♣ ; ⋮ ; 0 0 1 ♣)

in rational form. Here S is not a subgroup of a bigger subgroup of G. Then Ind_{S}^{G} (θ) for θ a character of S will be hard to decompose. For example, in GL (3, 2), (S| = 7.

Now look at G ≤ G̃ = GL (n, \overline{𝔽_{q}}). In G̃, we have B̃ = T̃Ũ and T ≤ T̃ and B ≤ B̃. (Here B̃, T̃, Ũ are defined analogously to B, T, U in G.) Then g^{−1}Sg ≤ T̃ for some g ∈ G̃ so that S ≤ gT̃g^{−1} = T̃_{1} ≤ gB̃g^{−1} = B̃_{1}. Set gŨg^{−1} = Ũ_{1}.

Definition 27 Define the Frobenius map to be the homomorphism

F : G̃ → G̃ given by (a_{i,j}) → (a_{i,j}^{q})

Then

G = G̃^{F} = {F -fixed points of G̃}

and similarly, B = B̃^{F} and T = T̃^{F}, but B̃_{1} need not be F -stable.

We now return to Ind_{B}^{G} \binom{}{ψ}. Since B = T U, we have Ind_{U}^{G} (1) = Ind_{B}^{G} (Ind_{U}^{B} (1)) =

Ind_{B}^{G} (∑_{ψ} ψ̃) = ∑_{ψ} Ind_{B}^{G} \binom{}{ψ}. A module for Ind_{U}^{G} (1) is

KG ⊗_{B} (⨁ V_{ψ})

where V_{ψ} is a module for B for the 1-dimensional representation ψ̃. This is a bi-module with G acting on the left and T acting on the right. We have G ≤ G̃ and S ≤ B̃_{1} = T̃_{1}Ũ_{1}. Define an algebraic variety

X = {g ∈ G̃ : g^{−1}F (g) ∈ Ũ_{1}}

(This is analogous to the coset space G/U where F (gU) = F (g) U for g ∈ G and also F (gU) = gU. Hence, g^{−1}F (g) ∈ U.)

We have G acting on the left and S on the right of X. If h ∈ G, g ∈ X, we have to check that hg ∈ X. We have

(hg)^{−1} F (hg) = g^{−1} \underbrace{h^{−1}F (h)}_{1} F (g) ∈ X

Next, we have to check that if t ∈ S and g ∈ X, then gt ∈ X which follows from: Ũ_{1} is normal in B̃_{1}. We have

(gt)^{−1} F (gt) = \underbrace{t^{−1}g^{−1}F (g) t}_{∈Ũ_{1}} ∈ Ũ_{1}

since t normalizes Ũ_{1}.

We need an action of G on some vector space over a field of characteristic 0 associated with X. This is constructed using a deep theory of Grothendieck and others of l-adic cohomology. Take l prime with l not dividing q. Take \overline{Q_{l}} the algebraic closure of the field of l-adic numbers. We then have, corresponding to X, a vector space H_{c}^{i} (X, \overline{Q_{l}}) (for each i ≥ 0) over \overline{Q_{l}} (i.e. l-adic cohomology groups with compact support). Then G acts on the left and S acts on the right of the cohomology groups, by functoriality.

The final step is to take θ a character of S which produces R_{S}^{G} (θ), a virtual character of G, called a Deligne-Lusztig character, given by

R_{S}^{G} (θ) = ∑_{i≥0} (−1)^{i} H_{c}^{i} (X, \overline{Q_{l}})_{θ}

where H_{c}^{i} (X, \overline{Q_{l}})_{θ} is the part of H_{c}^{i} (X, \overline{Q_{l}}) on which S acts by the character θ. If g ∈ G, we get the character value

R_{S}^{G} (θ) (g) = ∑_{i≥0} (−1)^{i} Trace(g, H_{c}^{i} (X, \overline{Q_{l}})_{θ})

.

Finally, the construction given above generalizes to finite groups of Lie type (such as symplectic groups, orthogonal groups over finite fields). For details see the book by Carter referred to above, p.205.

Thus we have arrived at the end of the course, which showed you some of the beautiful aspects of the Representation Theory of Finite Groups. If you wish to go further, Bon Voyage and Bon Courage!