Book Rep
Book Rep
Book Rep
Preface 5
Chapter 1. Introduction to representation theory of finite groups. 9
1. Definitions and examples 9
2. Ways to produce new representations 11
3. Invariant subspaces and irreducibility 12
4. Characters 15
5. Examples. 22
6. Invariant forms 26
7. Representations over R 28
8. Relationship between representations over R and over C 29
Chapter 2. Modules with applications to finite groups 31
1. Modules over associative rings 31
2. Finitely generated modules and Noetherian rings. 34
3. The center of the group algebra k (G) 36
4. Generalities on induced modules 38
5. Induced representations for groups. 41
6. Double cosets and restriction to a subgroup 43
7. Mackey’s criterion 44
8. Hecke algebras, a first glimpse 45
9. Some examples 46
10. Some generalities about field extension 47
Chapter 3. Representations of compact groups 49
1. Compact groups 49
2. Orthogonality relations and Peter–Weyl Theorem 57
3. Examples 60
Chapter 4. Some results about unitary representations 67
1. Unitary representations of Rn and Fourier transform 67
2. Heisenberg groups and the Stone-von Neumann theorem 71
3. Representations of SL2 (R) 79
Chapter 5. On algebraic methods 83
1. Introduction 83
3
4 CONTENTS
5
6 PREFACE
If a topological group is not compact, for example, the group of real numbers
with operation of addition, the representation theory of such a group involves more
complicated analysis (Fourier transform instead of Fourier series). The representa-
tion theory of real non-compact groups was initiated by Harish-Chandra and by the
Russian school leaded by Gelfand. Here emphasis is on the classification of unitary
representations due to applications from physics. It is also worth mentioning that
this theory is closely related to harmonic analysis, and many special functions (such
as Legendre polynomials) naturally appear in the context of representation theory.
In the theory of finite groups one can drop the assumption that the characteristic
of the ground field is zero. This leads immediately to the loss of complete reducibility.
This representation theory was initiated by Brauer and it is more algebraic. If one
turns to algebras, a representation of an algebra is, by definition, the same as a
module over this algebra. Let k be a field. Let A be a k-algebra which is finite
dimensional as a vector space. It is a well-known fact that A-modules are not,
in general, completely reducible: for instance, if A = k[X]/X 2 and M = A, the
module M contains kX as a submodule which has no A-stable complement. An
indecomposable A-module is a module which has no non-trivial decomposition as a
direct sum. It is also interesting to attempt a classification of A-modules. It is a
very difficult task in general. Nevertheless, the irreduducible A-modules are in finite
number. The radical R of A is defined as the ideal of A which annihilates each of
those irreducible modules, it is a nilpotent ideal. Assume k is algebraically closed,
the quotient ring A/R is a product of matrix algebras over k, A/R = Πi Endk (Si )
where Si runs along the irreducible A-modules.
If G is a finite group, the algebra k(G) of k-valued functions on G, the composition
law being the convolution, is a finite dimensional k-algebra, with a zero radical as long
as the characteristic of the field k does not divide the cardinal of G. The irreducible
modules of k(G) are exactly the finite dimensional representations of the group G,
the action of G extends linearly to k(G). This shows that all k(G)-modules are
completely reducible (Maschke’s theorem).
In order to study finite dimensional k-algebras representations more generally,
it is useful to introduce quivers. Let A be a finite dimensional k-algebra, denote
S1 , . . . , Sn its irreducible representations, and draw the following graph, called the
quiver associated to A: the vertices are labelled by the Si s and we put l arrows
between Si and Sj , pointing at Sj , if Ext1 (Si , Sj ) is of dimension l (the explicit
definition of Ext1 requires some homological algebra which is difficult to summarize
in such a short introduction).
More generally, a quiver is an oriented graph with any number of vertices. Let Q
be a quiver, a representation of Q is a set of vector spaces indexed by the vertices of
Q together with linear maps associated to the arrows of Q. Those objects were first
systematically used by Gabriel in the early 70’s, and studied by a lot of people ever
since. The aim is to characterize the finitely represented algebras, or in other terms
the algebras with a finite number of indecomposable modules (up to isomorphism).
PREFACE 7
Today the representation theory has many flavors. In addition to the above
mentioned, one should add representations over non-archimedian local fields with its
applications to number theory, representations of infinite-dimensional Lie algebras
with applications to number theory and physics and representations of quantum
groups. However, in all these theories certain main ideas appear again and again
very often in disguise. Due to technical details it may be difficult for a neophyte to
recognize them. The goal of this book is to present some of these ideas in their most
elementary incarnation.
We will assume that the reader is familiar with usual linear algebra (including
the theory of Jordan forms and tensor products of vector spaces) and basic theory of
groups and rings.
CHAPTER 1
Let a group G act on a set X on the right. Let F(X) be the set of k-valued
functions on X. Then there is a representation of G in F(X) defined by
ρg ϕ(x) := ϕ(x · g)
x · g := g −1 · x,
and consider the representation ρ of G in F(X) associated with this action. Check
that ρ and σ are equivalent representations.
Remark 1.5. As one can see from the previous exercise, there is a canonical way
to go between right and left action and between corresponding representations.
2. WAYS TO PRODUCE NEW REPRESENTATIONS 11
for every v ∈ V, ϕ ∈ V ∗ .
Let V be a finite-dimensional representation of G with a fixed basis. Let Ag for
g ∈ G be the matrix of ρg in this basis. Then the matrix of ρ∗g in the dual basis of
V ∗ is equal to (Atg )−1 .
Exercise 2.2. Show that if G is finite, then its regular representation is self-dual
(isomorphic to its dual).
12 1. INTRODUCTION TO REPRESENTATION THEORY OF FINITE GROUPS.
Exercise 2.3. Show that if V and W are finite dimensional, then the represen-
tation τ of G on Homk (V, W ) is isomorphic to ρ∗ ⊗ τ .
Intertwining operators. A linear operator T : V → W is called an intertwining
operator if T ◦ ρg = σg ◦ T for any g ∈ G. The set of all intertwining operators will
be denoted by HomG (V, W ). It is clearly a vector space. Moreover, if ρ = σ,
then EndG (V ) := HomG (V, V ) has a natural structure of associative k-algebra with
multiplication given by composition.
Exercise 2.4. Consider the regular representation of G in k(G). Prove that the
algebra of intertwiners EndG (k(G)) is isomorphic to k(G). (Hint: ϕ ∈ EndG (k(G))
is completely determined by ϕ(1).)
The next statement follows from Maschke’s theorem and Theorem 3.9.
Proposition 3.12. Let G be a finite group and k be a field such that char k does
not divide |G|. Then every representation of G is completely reducible.
4. Characters
4.1. Definition and main properties. For a linear operator T in a finite-
dimensional vector space V we denote by tr T the trace of T .
For any finite-dimensional representation ρ : G → GL (V ) the function χρ : G → k
defined by
χρ (g) = tr ρg .
is called the character of the representation ρ.
Exercise 4.1. Check the following properites of characters.
(1) χρ (1) = dim ρ;
(2) if ρ ∼
= σ, then χρ = χσ ;
(3) χρ⊕σ = χρ + χσ ;
(4) χρ⊗σ = χρ χσ ;
(5) χρ∗ (g) = χρ (g −1 );
16 1. INTRODUCTION TO REPRESENTATION THEORY OF FINITE GROUPS.
X
χalt (g) = λi λj .
i<j
Hence
X
χsym (g) − χalt (g) = λ2i = tr ρg2 = χρ g 2 .
i
The following theorem is usually called the orthogonality relations for characters.
18 1. INTRODUCTION TO REPRESENTATION THEORY OF FINITE GROUPS.
Proof. Indeed,
r
X r
X
dim R = |G| = χR (1) = ni χρi (1) = n2i .
i=1 i=1
Example 4.15. Let G act on a finite set X and
( )
X
k (X) = bx x | bx ∈ k .
x∈X
Define ρ : G → GL (k (X)) by
!
X X
ρg bx x = bx g · x.
x∈X x∈X
where
Gx = {g ∈ G | g · x = x} .
|G|
Let X = X1 ∪ · · · ∪ Xm be the disjoint union of orbits. Then |Gx | = |Xi |
for each
x ∈ Xi and therefore
m
1 X X |G|
(1, χρ ) = = m.
|G| i=1 x∈X |Xi |
i
By construction πij ∈ EndG (Vj ). Corollary 3.7 (c) implies that πij = λ Id. By
Theorem 4.8
tr πij = ni (χi , χj ) = ni δij .
Now we write
r
X
πi = πij .
j=1
Hence
πi |Wj = δij Id .
The statement follows.
22 1. INTRODUCTION TO REPRESENTATION THEORY OF FINITE GROUPS.
5. Examples.
In the examples below we assume that the ground field is C.
Example 5.1. Let G = S3 . There are three conjugacy classes in G, each class
is denoted by some element in this class: 1,(12),(123). Therefore there are three
irreducible representations, denote their characters by χ1 , χ2 and χ3 . It is not difficult
to see that S3 has the following table of characters
1 (12) (123)
χ1 1 1 1
χ2 1 −1 1
χ3 2 0 −1
The characters of one-dimensional representations are given in the first and the
second row (those are the trivial representation and the sign representation, see Ex-
ercise 4.23), the last character χ3 can be obtained by using the identity
(1.5) χperm = χ1 + χ3 ,
where χperm stands for the character of the permutation representation, see Exercise
4.2.
Example 5.2. Let G = S4 . In this case we have the following character table (in
the first row we write the number of elements in each conjugacy class).
1 6 8 3 6
1 (12) (123) (12) (34) (1234)
χ1 1 1 1 1 1
χ2 1 −1 1 1 −1
χ3 3 1 0 −1 −1
χ4 3 −1 0 −1 1
χ5 2 0 −1 2 0
The first two rows are the characters of the one-dimensional representations. The
third one can again be obtained from (1.5). When we take the tensor product ρ4 :=
ρ2 ⊗ρ3 we get a new 3-dimensional irreducible representation, see Exercise 4.24 whose
character χ4 is equal to the product χ2 χ3 . The last character can be obtained through
Theorem 4.8. An alternative way to describe ρ5 is to consider S4 /K4 , where
K4 = {1, (12) (34) , (13) (24) , (14) (23)}
5. EXAMPLES. 23
is the Klein subgroup. Observe that S4 /K4 ∼ = S3 , and therefore the two-dimensional
representation σ of S3 can be lifted to a representation of S4 by
ρ5 = σ ◦ p,
where p : S4 → S3 is the natural projection.
Example 5.3. Now let G = A5 . There are 5 irreducible representations of G over
C. Here is the character table
1 20 15 12 12
1 (123) (12) (34) (12345) (12354)
χ1 1 1 1 1 1
χ2 4 1 0 −1 −1
χ3 5 −1 1 0√ 0√
1+ 5 1− 5
χ4 3 0 −1 2√ 2√
1− 5 1+ 5
χ5 3 0 −1 2 2
To obtain χ2 we use the permutation representation and (1.5) once more. In order
to construct new irreducible representations we consider the characters χsym and χalt
of the second symmetric and the second exterior powers of ρ2 respectively. Using
(1.1) we compute
We obtain n4 = n5 = 3.
Next, we use Theorem 4.8 to compute some other values of χ4 and χ5 . The
equations
(χ4 , χ1 + χ2 ) = 0, (χ4 , χ3 ) = 0
imply
χ4 ((123)) = 0, χ4 ((12) (34)) = −1.
The same argument applied to χ5 gives
χ5 ((123)) = 0, χ5 ((12) (34)) = −1.
Finally let us denote
x = χ4 ((12345)) , y = χ4 ((12354))
and write down the equation arising from (χ4 , χ4 ) = 1:
1
9 + 15 + 12x2 + 12y 2 = 1,
60
or more simply
(1.6) x2 + y 2 = 3.
On the other hand, (χ4 , χ1 ) = 0, which gives
3 − 15 + 12 (x + y) = 0,
or simply
(1.7) x + y = 1.
The system (1.6), (1.7) has two solutions
√ √ √ √
1+ 5 1− 5 1− 5 1+ 5
x1 = , y1 = , x2 = , y2 = .
2 2 2 2
They give the characters χ4 and χ5 .
Now that we have the character of A5 we would like to explain a geometric
construction related to it. First, we observe that the previous constructions work
over the grouond field R of real numbers. In particular, the representations ρ4 and ρ5
are defined over R. Indeed, they are subrepresentation of the second exterior power
of ρ2 and by Lemma 4.25 the corresponding projectors are defined over R. Therefore
we have an action of A5 in R3 . Our next step is to show that this action preserves
the scalar product. In a more general context we have the following result.
Lemma 5.4. Let V be a finite-dimensional vector space over R and ρ be a repre-
sentation of some finite group G in V . There exists a positive definite scalar product
B : V × V → R such that B(ρg u, ρg v) = B(u, v) for any u, v ∈ V and g ∈ G.
Remark. Such a scalar product is called invariant.
5. EXAMPLES. 25
6. Invariant forms
We assume here that char k = 0. Recall that a bilinear form on a vector space V
is a map B : V × V → k satisfying
(1) B (cv, dw) = cdB (v, w);
(2) B (v1 + v2 , w) = B (v1 , w) + B (v2 , w);
(3) B (v, w1 + w2 ) = B (v, w1 ) + B (v, w2 ).
One can also think about a bilinear form as a vector in V ∗ ⊗ V ∗ or as the homo-
morphism B : V → V ∗ given by the formula Bv (w) = B (v, w). A bilinear form is
symmetric if B (v, w) = B (w, v) and skew-symmetric if B (v, w) = −B (w, v). Every
bilinear form can be uniquely written as a sum B = B + + B − where B + is symmetric
and B − skew-symmetric form,
B (v, w) ± B (w, v)
B ± (v, w) = .
2
Such a decomposition corresponds to the decomposition
(1.8) V ∗ ⊗ V ∗ = S 2 V ∗ ⊕ Λ2 V ∗ .
A bilinear form is non-degenerate if B : V → V ∗ is an isomorphism, in other words
if B (v, V ) = 0 implies v = 0.
Let ρ : G → GL (V ) be a representation. We say that a bilinear form B on V is
G-invariant if
B (ρg v, ρg w) = B (v, w)
for any v, w ∈ V , g ∈ G. If there is no possible confusion we use the word invariant
instead of G-invariant.
Exercise 6.1. Check the following
(1) If W ⊂ V is an invariant subspace, then W ⊥ = {v ∈ V | B (v, W ) = 0} is
invariant. In particular, Ker B is invariant.
(2) B : V → V ∗ is invariant if and only if B ∈ HomG (V, V ∗ ).
(3) If B is invariant, then B + and B − are invariant.
Lemma 6.2. Let ρ : G → GL(V ) be an irreducible representation of G, then any
invariant bilinear form on V is non-degenerate. If k is algebraically closed, then such
a bilinear form is unique up to scalar multiplication.
Remark. Lemma 6.2 holds for a field of arbitrary characteristic.
Proof. Follows from Exercise 6.1 (2) and Schur’s lemma.
Lemma 6.3. Let ρ : G → GL(V ) be an irreducible representation of G. Then it
admits an invariant form if and only if χρ (g) = χρ (g −1 ) for any g ∈ G.
Proof. Since every invariant bilinear form establishes an isomorphism between
ρ and ρ∗ , the statement follows from Corollary 4.10.
6. INVARIANT FORMS 27
Lemma 6.4. (a) If k is algebraically closed, then every non-zero invariant bilinear
form on an irreducible representation ρ is either symmetric or skew-symmetric.
(b) Define
1 X
mρ = χρ g 2 .
|G| g∈G
Then mρ = 1, 0 or −1.
(c) If mρ = 0, then ρ does not admit an invariant form. If mρ = 1 (resp.
mρ = −1), then ρ admits a symmetric (resp. skew-symmetric) invariant form.
Proof. First, (a) is a consequence of Lemma 6.2 and Exercise 6.1.
Let us prove (b) and (c). Recall that ρ ⊗ ρ = ρalt ⊕ ρsym . Using 1.1 we obtain
1 X χ2ρ (g) + χρ (g 2 )
(χsym , χtriv ) = ,
|G| g∈G 2
1 X χ2ρ (g) − χρ (g 2 )
(χalt , χtriv ) = .
|G| g∈G 2
Note that
1 X 2
χ (g) = (χρ , χρ∗ ) .
|G| g∈G ρ
Therefore
(χρ , χρ∗ ) + mρ
(χsym , χtriv ) = ,
2
(χρ , χρ∗ ) − mρ
(χalt , χtriv ) = .
2
We have the folowing thrichotomy
• ρ does not have an invariant form, if and only if ρ is not isomorphic to ρ∗ .
In this case (χρ , χρ∗ ) = 0 and (χsym , χtriv ) = (χsym , χtriv ) = 0. Therefore
mρ = 0.
• ρ has a symmetric invariant form if and only if (χρ , χρ∗ ) = 1 and (χsym , χtriv ) =
1. This implies mρ = 1.
• ρ admits a skew-symmetric invariant if and only if (χρ , χρ∗ ) = 1 and (χalt , χtriv ) =
1. This implies mρ = 1.
Let k = C. An irreducible representation of a finite group G is called real if
mρ = 1, complex if mρ = 0 and quaternionic if mρ = −1.
Remark. Since χρ (s−1 ) = χ̄ρ (s), then χρ takes only real values for real and
quaternionic representations. If ρ is complex there is at least one g ∈ G such that
/ R. This terminology will become clear in Section 8.
χρ (g) ∈
28 1. INTRODUCTION TO REPRESENTATION THEORY OF FINITE GROUPS.
7. Representations over R
Let us recall that by Lemma 5.4 every representation of a finite group over R
admits an invariant scalar product. Assume the representation ρ : G → GL(V )
is irreducible. Denote by B(·, ·) an invariant scalar product and let Q(·, ·) denote
another invariant symmetric form on V . These two forms can be silmultaneously
diagonalized. Therefore there exists λ ∈ R, such that Ker (Q − λB) 6= 0. Since
Ker (Q − λB) 6= 0 is G-invariant and ρ is irreducible, this implies Q = λB. There we
have
Lemma 7.1. Let ρ : G → GL(V ) be an irreducible representation of G over R.
There is exactly one invariant symmetric form on V up to scalar multiplication.
Theorem 7.2. Let R ⊂ K be a division ring, which is finite-dimensional over R.
Then K is isomorphic to R, C or H.
Proof. If K is a field, then K ∼= R or C, because C = R̄ and [C : R] = 2.
Assume that K is not commutative. Then it contains a subfield isomorphic to C
obtained by taking x ∈ K\R and conidering R. Therefore without loss of generality
we may assume R ⊂ C ⊂ K.
Consider the involutive C-linear automoprhism of K defined by the formula
f (x) = ixi−1 .
Look at the eigenspace decomposition of K with respect to f
K = K1 ⊕ K−1 ,
where
K±1 = {x ∈ K | f (x) = ±x} .
One can easily check the following inclusions
K1 K1 ⊂ K1 , K−1 K−1 ⊂ K1 , K1 K−1 ⊂ K−1 , K−1 K1 ⊂ K−1 .
The eigenspace K1 coincides with the centralizer of C in K. Therefore K1 = C.
Choose a non-zero y ∈ K − . The left multiplication by y defines an isomorphism
of R vector spaces K1 and K−1 . Hence dimR K1 = dimR K−1 = 2 and dim K = 4.
For any z = a + bi ∈ K1 and any w ∈ K−1 , we have
wz̄ = wa − wbi = aw + biw = zw.
8. RELATIONSHIP BETWEEN REPRESENTATIONS OVER R AND OVER C 29
Proof. The statement (1) follows from Lemma 7.1. For (2) use Exercise 8.3.
Since χρR , χρR = 2 by Lemma 7.3, then χρ 6= χ̄ρ , and therefore ρ is complex.
Finally let us prove (3). Let j ∈ EndG V R = H, then j (bv) = b̄v for all b ∈ C.
Let H be a positive-definite invariant Hermitian form on V . Then
Q (v, w) = H (jw, jv)
is another invariant positive-definite Hermitian form. By Lemma 8.1 Q = λH and λ
should be positive because Q is also positive definite. Since j 2 = −1, one has λ2 = 1
and therefore λ = 1. Thus,
H (v, w) = H (jw, jv) .
Set
B (v, w) = H (jv, w) .
Then B is a bilinear invariant form, and
B (w, v) = H (jw, v) = H jv, j 2 w = −H (jv, w) = −B (v, w) ,
hence B is skew-symmetric.
Corollary 8.4. Let σ be an irreducible representation of G over R. There are
three possibilities for σ
(1) χσ = χρ for some real representation ρ of G over C;
(2) χσ = χρ + χ̄ρ for some complex representation ρ of G over C;
(3) χσ = 2χρ for some quaternionic representation ρ of G over C.
Theorem 8.5. Let G be a finite group, r denote the number of conjugacy classes
and s denote the number of classes which are stable under inversion. Then r+s
2
is the
number of irreducible representations of G over R.
Proof. Recall that C(G) is the space of complex valued class functions on G.
Consider the involution θ : C(G) → C(G) given by
θϕ(g) = ϕ(g −1 ).
An easy calculation shows dim C(G)θ = r+s 2
.
Denote by χ1 , . . . , χr the irreducible characters of G over C. Recall that χ1 , . . . , χr
form a basis of C(G). Observe that for any character χρ
θ(χρ ) = χρ∗ .
Therefore θ permutes irreducible characters χ1 , . . . , χr . Corollary 8.4 implies that
the number of irreducible representations of G over R equals the number of self-
dual irreducible representations over C plus half the number of those which are not
self-dual. Therefore this number is exactly the dimension of C(G)θ .
CHAPTER 2
(1) Every increasing chain of left ideals in R is finite, in other words for any
sequence I1 ⊂ I2 ⊂ . . . of left ideals, there exists n0 such that for all n > n0 ,
I n = I n0 .
(2) Every left ideal is a finitely generated R-module.
Proof. (1) ⇒ (2). Assume that some left ideal I is not finitely generated. Then
there exists an infinite sequence of xn ∈ I such that
xn+1 ∈
/ Rx1 + · · · + Rxn .
But then In = Rx1 + · · · + Rxn form an infinite increasing chain of ideals which does
not stabilize.
(2) ⇒ (1). Let I1 ⊂ I2 ⊂ . . . be an increasing chain of ideals. Consider
[
I := In .
n
Proof. We have to check (2.1). Since ρj (ei ) belongs to EndG (Vj ), Schur’s
Lemma implies ρj (ei ) = λ Id for some λ. Now we use orthogonality relations, Theo-
rem 4.8
ni X ni
tr ρj (ei ) = χi g −1 χj (g) = (χi , χj ) = δij ni .
|G| |G|
Therefore we have nj λ = δij ni which implies λ = δij .
Exercise 3.2. Define ωi : Z (G) → k by the formula
X 1 X
ωi as s = as χi (s) .
ni
and ω : Z(G) → k r by
ω = (ω1 , . . . , ωr ) .
Check that ω is an isomorphism of rings. Hint: check that ωi (ej ) = δij using again
the orthogonality relations.
For any conjugacy class C in G let
X
ηC := g.
g∈C
Clearly, the set ηC for C running the set of conjugacy classes is a basis in Z(G).
Lemma 3.3. For any conjugacy class C ⊂ G we have
X r
χi (g)
ηC = |C| ei ,
i=1
ni
where g is any element of C.
Proof. If we extend by linearity χ1 , ...χr to linear functionals on k(G), then
(2.1) implies χj (ei ) = ni δi,j . Thus, χ1 , . . . , χr form a basis in the dual space Z(G)∗ .
Therefore it suffices to check that
Xr
χi (g)
χj (ηc ) = |C| χj (ei ) = |C|χj (g).
i=1
n i
Lemma 3.4. If g, h ∈ G lie in the same conjugacy class C, we have
X r
|G|
χi (g)χi (h−1 ) = .
i=1
|C|
If g and h are not conjugate we have
Xr
χi (g)χi (h−1 ) = 0.
i=1
38 2. MODULES WITH APPLICATIONS TO FINITE GROUPS
Proof. The statement follows from Lemma 3.1 and Lemma 3.3. Indeed, if g is
in the congugacy class C, we have
r
X r
χi (g) |C| X X
ηc = |C| ei = χi (g)χi (h−1 )h.
i=1
ni |G| i=1 h∈G
generators are all elements of the Cartesian product A × M and the relations:
(2.3) (a1 + a2 ) × m − a1 × m − a2 × m, a1 , a2 ∈ A, m ∈ M,
(2.4) a × (m1 + m2 ) − a × m1 − a × m2 , a ∈ A, m1 , m2 ∈ M,
(2.5) ab × m − a × bm, a ∈ A, b ∈ B, m ∈ M.
This group has a structure of A-module, A acting on it by left multiplication. For
every a ∈ A and m ∈ M we denote by a ⊗ m the corresponding element in A ⊗B M .
Definition 4.1. The A-module A ⊗B M is called the induced module.
Exercise 4.2. (a) Show that A ⊗B B is isomoprhic to A.
(b) Show that if M1 and M2 are two B-modules, then there exists a canonical
isomorphism of A-modules
A ⊗B (M1 ⊕ M2 ) ≃ A ⊗B M1 ⊕ A ⊗B M2 .
(c) Check that for any n ∈ Z one has
Q ⊗Z (Z/nZ) = 0.
Theorem 4.3. (Frobenius reciprocity.) For every B-module M and for every
A-module N , there is an isomorphism of abelian groups
HomB (M, N ) ∼
= HomA (A ⊗B M, N ) .
Proof. Let M be a B-module and N be an A-module. Consider j : M → A⊗B M
defined by
j (m) := 1 ⊗ m,
which is a homomorphism of B-modules.
Lemma 4.4. For every ϕ ∈ HomB (M, N ) there exists a unique ψ ∈ HomA (A ⊗B M, N )
such that ψ ◦ j = ϕ. In other words, the following diagram is commutative
j
M II / A ⊗B M
II
II ψ
ϕ III
I$
N
Proof. We define ψ by the formula
ψ(a ⊗ m) := aϕ(m),
for all a ∈ A and m ∈ M . The reader can check that ψ is well defined, i.e. the
relations defining A ⊗B M are preserved by ψ. That proves the existence of ψ
To check uniqueness we just note that for all a ∈ A and m ∈ M , ψ must satisfy
the relation
ψ (a ⊗ m) = aψ (1 ⊗ m) = aϕ (m) .
40 2. MODULES WITH APPLICATIONS TO FINITE GROUPS
To prove the theorem we observe that by the above lemma the map ψ 7→ ϕ := ψ◦j
gives an isomorphism between HomA (A ⊗B M, N ) and HomB (M, N ).
Remark 4.5. For readers familiar with category theory the former theorem can
be reformulated as follows. Since any A-module M is automatically a B-module,
we have a natural functor Res from the category of A-modules to the category of B-
modules. This functor is usually called the restriction functor. The induction functor
Ind from the category of B-modules to the category of A-modules which sends M to
A ⊗B M is left adjoint of Res.
Example 4.6. Let k ⊂ F be a field extension. For any vector space M over k,
F ⊗k M is a vector space of the same dimension over F . If we have an exact sequence
of vector spaces
0 → N → M → L → 0,
then the sequence
0 → F ⊗k N → F ⊗k M → F ⊗k L → 0
is also exact. In other words the induction in this situation is an exact functor.
Exercise 4.7. Let A be a ring and B be a subring of A.
(a) Show that if a sequence of B-modules
N →M →L→0
is exact, then the sequence
A ⊗B N → A ⊗B M → A ⊗B L → 0
of induced modules is also exact. In other words the induction functor is right exact.
(b) Assume that A is a free right B-module, then the induction functor is exact.
In other words, if a sequence
0→N →M →L→0
of B-modules is exact, then the sequence
0 → A ⊗B N → A ⊗B M → A ⊗B L → 0
is also exact.
(c) Let A = Z[X]/(X 2 , 2X) and B = Z. Consider the exact sequence
ϕ
0→Z−
→ Z → Z/2Z → 0,
where ϕ is the multiplication by 2. Check that after applying induction we get a
sequence of A-modules
0 → A → A → A/2A → 0,
which is not exact.
Later we discuss general properties of induction but now we are going to study
induction for the case of groups.
5. INDUCED REPRESENTATIONS FOR GROUPS. 41
Moreover, for any g ∈ G, s ∈ S there exists a unique s′ ∈ S such that (s′ )−1 gs ∈ H.
Then the action of g on s ⊗ v for all v ∈ V is given by
(2.7) g (s ⊗ v) = s′ ⊗ ρ(s′ )−1 gs v.
Example 5.2. Let ρ be the trivial representation of H. Then IndG H ρ is the
permutation representation of G obtained from the natural left action of G on the
set of left cosets G/H, see Example 3 in Section 4.2 Chapter 1.
Lemma 5.3. We keep the notations of the previous lemma. Denote by IndG
Hχ
the character of the induced representation. Then one has for g ∈ G
X
(2.8) IndG
H χ (g) = χ s−1 gs .
s∈S,s−1 gs∈H
Proof. (2.6) and (2.7) imply
X
IndG
H χ (g) = δs,s′ tr ρ(s′ )−1 gs .
s∈S
Corollary 5.4. In the notations of Lemma 5.3 we have
X
IndGH χ (g) = χ u−1 gu .
u∈G,u−1 gu∈H
Proof. If s−1 gs ∈ H, then for all u ∈ sH we have χ(u−1 gu) = χ(s−1 gs). There-
fore
1 X
χ(s−1 gs) = χ u−1 gu .
|H| u∈sH
Hence the statement follows from (2.8).
42 2. MODULES WITH APPLICATIONS TO FINITE GROUPS
IndG
H χρ (12) = 0,
IndG
H χρ (123) = −1.
Therefore IndG H ρ is the irreducible 2-dimensional representation of S3 .
(b) Next, consider the 2-element subgroup K of G = S3 generated by the trans-
position (12), and let σ be the (unique) non-trivial one-dimensional representation of
K. Show that
IndG
K χσ (1) = 3,
IndG
K χσ (12) = −1,
IndG
H χρ (123) = 0.
Therefore IndGK σ is the direct sum of the sign representation and the 2-dimensional
irreducible representation.
Now we assume that k has characteristic zero. Let us recall that, in Section 4.2
Chapter 1, we defined a scalar product on the space C(G) of class functions by (1.2).
When we consider several groups at the same time we specify the group by the a
lower index.
Theorem 5.7. Consider two representations ρ : G → GL (V ) and σ : H →
GL (W ). Then we have the identity
(2.9) IndGH χσ , χρ G = (χσ , ResH χρ )H .
Proof. The statement follows from Frobenius reciprocity (Theorem 4.3) and
Corollary 4.7 in Chapter 1, since
dim HomG IndG H W, V = dim HomH (W, V ) .
Exercise 5.8. Prove Theorem 5.7 directly from Corollary 5.4. Define two maps
ResH : C (G) → C (H) , IndG
H : C (H) → C (G) ,
the former is the restriction on a subgroup, the latter is defined by (2.8). Then for
any ϕ ∈ C (G) , ψ ∈ C (H)
IndG
H ϕ, ψ G
= (ϕ, ResH ψ)H .
6. DOUBLE COSETS AND RESTRICTION TO A SUBGROUP 43
We define the set of double cosets by K\G/H. One can identify K\G/H with K-
orbits on S = G/H in the obvious way and with G-orbits on G/K × G/H by the
formula
KtH → G (K, tH) .
Example 6.1. Let F be a field. Let G = GL2 (F) be the group of all invertible
2 × 2 matrices with coefficients in F. Consider the natural action of G on F2 . Let B
be the subgroup of upper-triangular matrices in G. We denote by P1 the projective
line which is the set of all one-dimensional linear subspaces of F2 . Clearly, G acts on
P1 .
Exercise 6.2. Prove that G acts transitively on P1 and that the stabilizer of any
point in P1 is isomorphic to B.
By the above exercise one can identify G/B with the set of lines P1 . The set of
double cosets B\G/B can be identified with the set of G-orbits in P1 × P1 or with
the set of B-orbits in P1 .
Exercise 6.3. Check that G has only two orbits on P1 × P1 : the diagonal and
its complement. Thus, |B\G/B| = 2 and
G = B ∪ BsB,
where
0 1
s=
1 0
`
Theorem 6.4. Let T ⊂ G such that G = s∈T KsH. Then
ResK IndG K s
H ρ = ⊕s∈T IndK∩sHs−1 ρ ,
where
def
ρsh = ρs−1 hs ,
for any h ∈ sHs−1 .
Proof. Let s ∈ T and W s = k (K) (s ⊗ V ). Then by construction, W s is K-
invariant and
k (G) ⊗k(H) V = ⊕s∈T W s .
44 2. MODULES WITH APPLICATIONS TO FINITE GROUPS
7. Mackey’s criterion
In order to compute IndG H χ, Ind G
H χ , we use Frobenius reciprocity and Theo-
rem 6.4. One has:
X
IndG
H χ, Ind G
H χ G
= Res H Ind G
H χ, χ H
= IndH s
H∩sHs−1 χ , χ H =
s∈T
X X
= (χs , ResH∩sHs−1 χ)H∩sHs−1 = (χ, χ)H + (χs , ResH∩sHs−1 χ)H∩sHs−1 .
s∈T s∈T \{1}
We call two representation disjoint if they do not have any irreducible component
in common, or in other words if their characters are orthogonal.
Theorem 7.1. (Mackey’s criterion) The representation IndG H ρ is irreducible if
and only if ρ is irreducible and ρs and ρ are disjoint representations of H ∩ sHs−1
for all s ∈ T \ {1}.
Proof. Write the condition
IndG G
H χ, IndH χ G
=1
and use the above formula.
Corollary 7.2. Let H be a normal subgroup of G. Then IndG H ρ is irreducible
if and only if ρs is not isomorphic to ρ for any s ∈ G/H, s ∈
/ H.
8. HECKE ALGEBRAS, A FIRST GLIMPSE 45
Remark 7.3. Note that if H is normal, then G/H acts on the set of representa-
tions of H. In fact, this is a part of the action of the group Aut H of automorphisms
of H on the set of representation of H. Indeed, if ϕ ∈ Aut H and ρ : H → GL (V ) is
a representation, then ρϕ : H → GL (V ) defined by
ρϕt = ρϕ(t) ,
is a new representation of H.
We can identify the Hecke algebra with ΠH k(G)ΠH . Therefore a basis of the Hecke
algebra can be enumerated by the double cosets, i.e. elements of H\G/H.
Set, for g ∈ G,
ηg := ΠH gΠH .
it is clear that those functions are constant on double cosets and give a basis of the
Hecke algebra. Then, the multiplication is given by the formula
X 1
(2.10) ηg ηg ′ = |gHg ′ ∩ Hg ′′ H|ηg′′ .
g ′′ ∈G
|H|
Exercise 8.3. Consider the pair G = GL2 (Fq ), H = B the subgroup of upper
triangular matrices. Then by Exercise 6.3 we know that the Hecke algebra H(G, B)
is 2-dimensional. The identity element ηe corresponds to the double coset B. The
second element of the basis is ηs . Let us compute ηs2 using (2.10). We have
ηs2 = aηe + bηs ,
where
|sBs ∩ B| sBs ∩ BsB
a= , b= .
|B| |B|
46 2. MODULES WITH APPLICATIONS TO FINITE GROUPS
Since sBs is the subgroup of the lower triangular matrices in G, the intersection
subgroup sBs ∩ B is the subgroup of diagonal matrices. Therefore we have
1 q−1
|B| = (q − 1)2 q, |sBs ∩ B| = (q − 1)2 , a = , b = 1 − a = .
q q
Definition 8.4. We say that a G-module V is multiplicity free if any simple
G-module appears in V with multiplicity either 0 or 1.
Proposition 8.5. Assume that k is algebraically closed. The following conditions
on the pair H ⊂ G are equivalent
(1) The G-module IndG H triv is multiplicity free;
(2) For any G-module M the dimension of subspace M H of H-invariants is at
most one;
(3) The Hecke algebra H(G, H) is commutative.
Proof. (1) is equivalent to (2) by Frobenius reciprocity. Equivalence of (1) and
(3) follows from Lemma 1.11.
Lemma 8.6. Let G be a finite group and H ⊂ G be a subgroup. Let ϕ : G → G
be antiautomorphism of G such that for any g ∈ G we have ϕ(g) ⊂ HgH. Then
H(G, H) is commutative.
Proof. Extend ϕ to the whole group algebra k(G) by linearity. Then ϕ is an
antiautomorphism of k(G) and for all g ∈ G we have ϕ(ηg ) = ηg . Therefore for any
g, h ∈ H\G/H we have
X X
ηg ηh = cug,h ηu = cug,h ϕ(ηu ) = ϕ(ηg ηh ) = ϕ(ηh )ϕ(ηg ) = ηh ηg .
u∈H\G/H
Exercise 8.7. Let G be the symmetric group Sn and H = Sp × Sn−p . Prove that
H(G, H) is abelian. Hint: consider ϕ(g) = g −1 and apply Lemma 8.6.
9. Some examples
Let H be a subgroup of G of index 2. Then H is normal and G = H ∪ sH for
some s ∈ G\H. Suppose that ρ is an irreducible representation of H. There are two
possibilities
(1) ρs is isomorphic to ρ;
(2) ρs is not isomorphic to ρ.
Hence there are two possibilities for IndG
Hρ :
G ′ ′
(1) IndH ρ = σ ⊕ σ , where σ and σ are two non-isomorphic irreducible repre-
sentations of G;
(2) IndG H ρ is irreducible.
10. SOME GENERALITIES ABOUT FIELD EXTENSION 47
where sgn denotes the sign representation. Furthermore, the induced modules IndG
H ρ4
G
and IndH ρ5 are isomorphic and irreducible. Thus in dimensions 1, 4 and 5, S5 has
two non-isomorphic irreducible representations and only one in dimension 6.
Now let G be the subgroup of GL2 (Fq ) consisting of matrices of shape
a b
,
0 1
where a ∈ F∗q and b ∈ Fq . Let us classify complex irreducible representations of G.
One has |G| = q 2 − q. Furtheremore G has q conjugacy classes with the following
representatives
1 0 1 1 a 0
, , ,
0 1 0 1 0 1
(in the last case a 6= 1). Note that
1 b
H={ , b ∈ Fq }
0 1
is a normal subgroup of G and the quotient G/H is isomorphic to F∗q which is cyclic
of order q − 1.
Therefore G has q − 1 one-dimensional representations which can be lifted from
G/H. That leaves one more representation, its dimension must be q −1. Let us try to
obtain it using induction from H. Let σ be a non-trivial irreducible representation of
H, its dimension is automatically 1. Then the dimension the induced representation
IndGH σ is equal to q − 1 as required. We claim that it is irreducible. Indeed, if ρ is
a one-dimensional representation of G, then by Frobenius reciprocity, Theorem 5.7,
we have
IndG
H σ, ρ G = (σ, ResH ρ)H = 0,
1. Compact groups
Let G be a group which is also a topological space. We say that G is a topological
group if both the multiplication from G × G to G and the inverse from G to G are
continuous maps. Naturally, we say that G is compact (respectively, locally compact)
if it is a compact (resp., locally compact) topological space.
Examples.
• The circle
S 1 = {z ∈ C | |z| = 1} .
• The torus T n = S 1 × · · · × S 1 .
Note that in general, the direct product of two compact groups is com-
pact.
• The unitary group
Un = X ∈ GLn (C) | X̄ t X = 1n .
To see thatPUn is compact, note that a matrix X = (xij ) ∈ Un satisfies the
equations nj=1 |xij |2 = 1 for j = 1, . . . , n. Hence Un is a closed subset of
the product of n spheres of dimension (2n − 1).
• The special unitary group
SUn = {X ∈ Un | det X = 1} .
• The orthogonal group
On = x ∈ GLn (R) | X t X = 1n .
• The special orthogonal group
SOn = {X ∈ On | det X = 1} .
In the same way there exists a unique left-invariant measure d′ g such that
Z
d′ g = 1.
G
Moreover, dg = d′ g.
We do not give the proof of this theorem in general. In this sketch of proof, we
assume general knowledge of submanifolds and of the notion of vector bundle. All
examples we consider here are smooth submanifolds in GLk (R) or GLk (C).
Exercise 1.3. Assume that G is a subgroup of GLk (R) or GLk (C) and G is the
set of zeros of smooth functions f1 , . . . , fk . Then G is a smooth submanifold in GLk .
Hint: consider the map mg : G → G given by left multiplication by g ∈ G. Then its
differential (mg )∗ : Te G → Tg G is an isomorphism between tangent spaces at e and
g.
To define the invariant measure we just need to define a volume form on the
tangent space at identity Te G and then use right (left) multiplication to define it on
the whole group. More precisely, let γ ∈ Λtop Te∗ G. Then the map
g 7→ γg := m∗g (γ) ,
Remark 1.4. If G is locally compact but not compact, there are still left-invariant
and right-invariant measures on G, each is unique up to scalar multiplication, but the
left-invariant ones are not necessarily proportional to the right-invariant ones. We
speak of left-Haar measure or right-Haar measure.
1. COMPACT GROUPS 51
exists. Then L2 (G) is a Hilbert space with respect to the Hermitian form
Z
hϕ, ψi = ϕ̄ (g) ψ (g) dg.
G
Moreover, the representation R of G in L2 (G) given by
Rg ϕ (h) = ϕ (hg)
is continuous and the Hermitian form is G-invariant. This representation is called
the regular reprensentation of G.
1.4. Linear operators in a Hilbert space. We will recall certain facts about
linear operators in a Hilbert space. We only sketch the proofs hiding technical details
in exercises. The enthusiastic reader is encouraged to supply those details and the less
enthusiastic reader can find those details in textbooks on the subject, for instance,???.
Definition 1.7. A linear operator T in a Hilbert space is called bounded if there
exists C > 0 such that for any v ∈ V we have kT vk ≤ Ckvk.
Exercise 1.8. Let B(V ) denote the set of all bounded operators in a Hilbert
space V .
(a) Check that B(V ) is an algebra over C with multiplication given by composi-
tion.
(b) Show that T ∈ B(V ) if and only if the map T : V → V is continuous.
(c) Introduce the norm on B(V ) by setting
kT k = supkvk=1 kT vk.
Check that kT1 T2 k ≤ kT1 kkT2 k and kT1 + T2 k ≤ kT1 k + kT2 k for all T1 , T2 ∈ B(V )
and that B(V ) is complete in the topology defined by this norm. Thus, B(V ) is a
Banach algebra.
Theorem 1.9. Let T ∈ B(V ) be invertible. Then T −1 is also bounded.
Proof. Consider the unit ball
B := {x ∈ V | kxk < 1}.
For any k ∈ N denote by Sk the closure of T (kB) = kT (B) and let Uk = V \ Sk .
Note that [
V = kB.
k∈N
Since T is invertible, it is surjective, and therefore
[
Sk = V.
k∈N
We claim that there exists k such that Uk is not dense. Indeed, otherwise there exists
a sequence of embedded balls Bk ⊂ Uk , Bk+1 ⊂ Bk , which has a common point by
completeness of V . This contradicts to the fact that the intersection of all Uk is
1. COMPACT GROUPS 53
empty. Then Sk contains a ball x + εB for some x ∈ V and ε > 0. It is not hard to
see that for any r > kε + kxk, Sr contains B.
Now we will prove the inclusion B ⊂ T (2rB) for r as above. Indeed, let y ∈ B ⊂
Sr . There exists x1 ∈ rB such that ky − T x1 k < 21 . Note that y − T x1 ∈ 12 B ⊂ 21 Sr .
Then one can find x2 ∈ 2r B such that ky − T x1 − T x2 k < 41 . Proceeding in this way
1
we can construct a sequence {xn ∈ 2n−1 B} such that ky − T (x1 + · · · + xn )k < 21n .
X∞
Consider w = xi , which is well defined due to completeness of V . Then w ∈ 2rB
i=1
and T w = y. That implies B ⊂ T (2rB).
Now we have T −1 B ⊂ 2rB and hence kT −1 k ≤ 2r.
Bounded operators have a nice spectral theory, see ???.
Definition 1.10. Let T be bounded. The spectrum σ(T ) of T is the subset of
complex numbers λ such that T − λ Id is not invertible.
In a finite-dimensional Hilbert space σ(T ) is the set of eigenvalues of T . In the
infinite-dimensional case a point of the spectrum is not necessarily an eigenvalue. We
need the following fundamental result.
Theorem 1.11. If T is bounded, then σ(T ) is a non-empty closed bounded subset
of C.
Proof. The main idea is to consider the resolvent R(λ) = (T − λ Id)−1 as a
function of λ. If T is invertible, then we have the decomposition
R(λ) = T −1 (Id +T −1 λ + T −2 λ2 + . . . ),
which converges for |λ| < kT 1−1 k . Thus, R(λ) is analytic in a neighbourhood of 0.
Using shift R(λ) → R(λ + c) we obtain that R(λ) is analytic in its domain which is
C \ σ(T ). The domain of R(λ) is an open set. Hence σ(T ) is closed.
Furthermore, we can write the series for R(λ) at infinity:
(3.1) R(λ) = −λ−1 (Id +λ−1 T + λ−2 T 2 + . . . ).
This series converges for |λ| > kT k. Therefore σ(λ) is a subset of the circle |λ| ≤ kT k.
Hence σ(T ) is bounded.
Finally, (3.1) also implies lim R(λ) = 0. Suppose that σ(T ) = ∅, then R(λ) is
λ→∞
analytic and bounded. By Liouville’s theorem R(λ) is constant, which is impossible.
Definition 1.12. For any linear operator T in a Hilbert space V we denote by
T ∗ the adjoint operator. Since V ∗ ≃ V , we can consider T ∗ as a linear operator in V
such that for any x, y ∈ V
hx, T yi = hT ∗ x, yi .
54 3. REPRESENTATIONS OF COMPACT GROUPS
(b) Let F(V ) be the ideal in B(V ) of all operators with finite-dimensional image.
Prove that C(V ) is the closure of F(V ).
Lemma 1.19. Let A be a compact self-adjoint operator in V . Then
λ := sup hAu, ui
u∈S
and let Z
Q= ρg T ρ−1
g dg.
G
As follows from Schur’s lemma, since ρ is irreducible, Q is a scalar multiplication.
Since
tr Q = tr T = hv, v ′ i ,
we obtain
hv, v ′ i
Q= Id .
dim ρ
Hence
1
hw′ , Qwi = hv, v ′ i hw′ , wi .
dim ρ
On the other hand,
Z Z
′
′
−1 ′
hw , Qwi = w , v, ρg w ρg v dg = fw,v g −1 fv′ ,w′ (g) dg =
G G
Z
1
= f¯v,w (g) fv′ ,w′ (g) dg =
hfv,w , fv′ ,w′ i .
G dim ρ
If fv,w and fv′ ,w′ are matrix coefficients of two non-isomorphic representations, then
Q = 0, and the calculation is even simpler.
Corollary 2.2. Let ρ : G → GL(V ) and σ : G → GL(W ) be two irreducible
unitary representations, then hχρ , χσ i = 1 if ρ is isomorphic to σ and hχρ , χσ i = 0
otherwise.
Proof. Let v1 , . . . , vn be an orthonormal basis in V and w1 , . . . , wm be an or-
thonormal basis in W . Then
n X
X m
hχρ , χσ i = fvi ,vi , fwj ,wj .
i=1 j=1
For any such f there exists w ∈ V such that f (e, v) = hw, vi. The above condition
implies f (g, v) = hw, ρg vi, i.e. f = ϕw .
Theorem 2.4. (Peter–Weyl) Matrix coefficients of all irreducible unitary repre-
sentations span a dense subspace in L2 (G) for a compact group G.
Proof. We apply Lemma 1.30 to the regular representation of G. Let ρ ∈ G. b
Lemma 2.3 implies that Vρ ⊗ HomG (Vρ , L2 (G)) coincides with the space of all matrix
coefficients of ρ. Hence the span of matrix coefficients is the unique semisimple
G-submodule in L2 (G).
2.2. Convolution algebra. For a group G we define by L1 (G) the set of all
complex valued functions ϕ on G such that
Z
kf k1 := |ϕ(g)|dg
G
is finite.
Definition 2.5. The convolution product of two continuous complex valued func-
tions ϕ and ψ on G is defined by the formula:
Z
(3.4) (ϕ ∗ ψ)(g) := ϕ(h)ψ(h−1 g)dh.
G
3. Examples
3.1. The circle. Let T = S 1 = {z ∈ C | |z| = 1}, if z ∈ S 1 , one can write
dθ
z = eiθ with θ in R/2πZ. The Haar measure on S 1 is equal to 2π . All irreducible
1 1
representations of S are one-dimensional since S is abelian. They are given by the
characters χn : S 1 → C∗ , χn (θ) = einθ , n ∈ Z. Hence Sb1 = Z and
L2 S 1 = ⊕n∈Z Ceinθ .
This is a representation-theoretic explanation of the Parseval theorem, meaning that
every square integrable periodic function is the sum (with respect to the L2 norm)
of its Fourier series.
3. EXAMPLES 61
3.2. The group SU2 . Consider the compact group G = SU2 . Then G consists
of all matrices
a b
,
−b̄ ā
satisfying the relations |a|2 +|b|2 = 1. Thus, as a topological space, SU2 is isomorphic
to the 3-dimensional sphere S 3 .
Exercise 3.1. Check that SU2 is isomorphic to the multiplicative subgroup of
quaternions withnorm 1 by identifying
the quaternion a+bi+cj+dk = a+bi+j(c+di)
a + bi c + di
with the matrix .
−c + di a − di
To find the irreducible representations of SU2 , consider the polynomial ring
C [x, y], with the action of SU2 given by the formula
a b
ρg (x) = ax + by, ρg (y) = −b̄x + āy, if g = .
−b̄ ā
Let ρn be the representation of G in the space Cn [x, y] of homogeneous polynomi-
als of degree n. The monomials xn , xn−1 y, . . . , y n form a basis of Cn [x, y]. Therefore
dim ρn = n + 1. We claim that all ρn are irreducible and that every irreducible repre-
sentation of SU2 is isomorphic to ρn for some n ≥ 0. We will show this by checking
that the characters χn of ρn form an orthonormal basis in the Hilbert space of class
functions on G.
Recall that every unitary matrix is diagonal in some orthonormal basis, therefore
z 0
every conjugacy class of SU2 intersects the diagonal subgroup. Moreover,
0 z̄
z̄ 0
and are conjugate. Hence the set of conjugacy classes can be identified with
0 z
the quotient of S 1 by the equivalence relation z ∼ z̄. Let z = eiθ , then
z n+1 − z −n−1 sin (n + 1) θ
(3.5) χn (z) = z n + z n−2 + · · · + z −n = −1
= .
z−z sin θ
First, let us compute the Haar measure for G.
Exercise 3.2. Let G = SU2 .
(a) Using Exercise 3.1 show that the action of G × G given by the multiplication
on the right and on the left coincides with the standard action of SO(4) on S 3 . Use
it to prove that SO(4) is isomorphic to the quotient of G × G by the two element
subgroup {(1, 1), (−1, −1)}.
(b) Prove that the Haar measure on G is proportional to the standard volume
form on S 3 invariant under the action of the orthogonal group SO4 .
More generally: let us compute the volume form on the n-dimensional sphere S n ⊂
Rn+1 which is invariant under the action SOn+1 . We use the spherical coordinates in
62 3. REPRESENTATIONS OF COMPACT GROUPS
Rn+1 ,
x1 = r cos θ, x2 = r sin θ cos ϕ1 , x3 = r sin θ sin ϕ1 cos ϕ2 ,
...
xn−1 = r sin θ sin ϕ1 sin ϕ2 . . . sin ϕn−2 cos ϕn−1 ,
xn = r sin θ sin ϕ1 sin ϕ2 . . . sin ϕn−2 sin ϕn−1 ,
where r > 0, θ, ϕ1 , . . . , ϕn−2 vary in [0, π] and ϕn−1 ∈ [0, 2π]. The Jacobian relating
spherical and Euclidean coordinates is equal to
rn sinn−1 θ sinn−2 ϕ1 . . . sin ϕn−2 ,
thus when we restrict to the sphere r = 1 we obtain the volume
sinn−1 θ sinn−2 ϕ1 . . . sin ϕn−2 dθdϕ1 . . . dϕn−1 ,
which is SOn+1 -invariant. It is not normalized.
Let us return to the case G = SU2 ≃ S 3 . After normalization the invariant
volume form is
1
2
sin2 θ sin ϕ1 dθdϕ1 dϕ2 .
2π
The conjugacy class C (θ) of all matrices with eigenvalues eiθ , e−iθ (θ ∈ [0, π]) is
the set of points in S 3 with spherical coordinates (1, θ, ϕ1 , ϕ2 ): indeed, the minimal
polynomial on R of the quaternion with those coordinates is
t2 − 2t cos θ + 1
which is also the characteristic polynomial of the corresponding matrix in SU2 , so it
belongs to C(θ).
Hence, one gets that, for a class function ψ on G
Z Z π Z π Z 2π Z
1 2 2 π
ψ (g) dg = 2 ψ (θ) sin θdθ sin ϕ1 dϕ1 dϕ2 = ψ (θ) sin2 θdθ.
G 2π 0 0 0 π 0
Exercise 3.3. Prove that the functions χn form an orthonormal basis of the
space L2 ([0, π]) with the measure π2 sin2 θdθ and hence of the space of class functions
on G.
3.3. The orthogonal group G = SO3 . Recall that SU2 can be realized as
the set of quaternions with norm 1. Consider the representation γ of SU2 in the
space of quaternions H defined by the formula γg (α) = gαg −1 . One can see that
the 3-dimensional
space Him of pure imaginary quaternions is invariant and (α, β) =
Re αβ̄ is an invariant positive definite scalar product on Him . Therefore ρ defines
a homomorphism γ : SU2 → SO3 .
Exercise 3.4. Check that Ker γ = {1, −1} and that γ is surjective. Hence SO3 ∼ =
SU2 / {1, −1}. Thus, we can see that as a topological space SO3 is a 3-dimensional
sphere with opposite points identified, or the real 3-dimensional projective space.
3. EXAMPLES 63
where w runs the set of unit vectors orthogonal to v, and θ is the angular parameter
on the circle S 2 ∩ v ⊥ . Check that T ϕ (v) is the area of the cross section by the plane
v ⊥ . We have to prove that T is invertible.
Obviously T commutes with the SO3 -action. Therefore T can be diagonalized by
using Schur’s lemma and the decomposition
M
[
L2 (G)even = H2n .
n∈N
Indeed, T acts on H2n as the scalar operator λn Id. We have to check that λn 6= 0 for
all n. Consider again ϕ2n ∈ H2n . Then ϕ2n (1, 0, 0) = 1 and
Z Z
1 2π 2n (−1)n 2π 2n
T ϕ2n (1, 0, 0) = (iy) dθ = sin θdθ,
2 0 2 0
here we take the integral over the circle x22 +x23 = 1, and assume x2 = sin θ, x3 = cos θ.
Since T ϕ = λn ϕ, we obtain
Z
(−1)n 2π 2n
λn = sin θdθ 6= 0.
2 0
CHAPTER 4
Remark 1.2. (1) One checks that lim fˆ(ξ) = 0 and that fˆ is continuous on
ξ→∞
V ∗.
(2) Nevertheless, there is no reason for fˆ to belong to L1 (V ∗ ) (check on the
characteristic function of an interval in R).
(3) The Fourier transform of the convolution (cf Definition 2.5 Chapter III) of
two functions is the product of the Fourier transforms of the two factors.
(4) (Adjunction formula for Fourier transforms), let f ∈ L1 (V ) and ϕ ∈ L1 (V ∗ ),
then Z Z
f (x)ϕ̂(x)dx = fˆ(ξ)ϕ(ξ)dξ.
V V∗
Exercise 1.3. Let γ ∈ GL(V ), show that the Fourier transform of the function
γ.f defined by (γ.f )(x) = f (γ −1 (x)) is det(γ)t γ −1 .fˆ.
Let us consider the generalized Wiener algebra W(V ) consisting of integrable
functions on V whose Fourier transform is integrable on V ∗ .
Proposition 1.4. The subspace W(V ) ⊂ L1 (V ) is a dense subset (for the L1 -
norm).
Proof. Let Q be a positive definite quadratic form on V , denote by B its polar-
ization and by Q−1 the quadratic form on V ∗ whose polarization is B −1 . Let Disc(Q)
denote the discriminant of Q in a basis of V of volume 1.
Lemma 1.5. The Fourier transform of the function φ : x 7→ e−πQ(x) on V is the
−1
function ξ 7→ Disc(Q)−1/2 e−πQ (ξ) on V ∗ .
Proof. (of the lemma) One can reduce this lemma to the case n = 1 by using an
orthogonal basis for Q and Fubini’s theorem. We just need to compute the Fourier
2
transform of the function ǫ(x) := x 7→ e−πx on the line R.
One has Z Z
−πx2 −2iπξx −πξ 2 2
ǫ̂(ξ) = e dx = e e−π(x+iξ) dx.
R R
1. UNITARY REPRESENTATIONS OF Rn AND FOURIER TRANSFORM 69
If λ goes to 0, the function x 7→ f (λx) tends to f (0) and remains bounded by sup | f |.
By dominated convergence, we obtain the equality
ˆ ˆ
(4.2) f (0)ϕ̂(0) = fˆ(0)ϕ(0).
ˆ
We know that, if ϕ(ξ) = φ(ξ) (see Lemma 1.5) ϕ̂ˆ = ϕ, thus fˆ(0) = f (0).
We use the actions of the additive group V on W(V ) given by
τy (f ) : (x 7→ f (x − y))
and on W(V ∗ ) given by
µy (ϕ) : ξ 7→ e−2iπ<ξ,y> ϕ(ξ)
for all y ∈ V .
Exercise 1.8. Check that
(1) τ[ ˆ 1
y (f ) = µy (f ) for all f ∈ L (V ),
\
(2) µ 1
y (ϕ) = τ−y (ϕ̂) for all ϕ ∈ L (V ).
∗
[ ˆˆ
We apply τy to f , Exercise 1.8 shows that τ[
y (f ) = τ−y f , hence the result.
70 4. SOME RESULTS ABOUT UNITARY REPRESENTATIONS
ˆ¯
But, by Remark 1.9 (4.3), fˆ = f¯, QED.
Remark 1.13. This result amounts to saying that the Fourier transform in gen-
eralized Wiener algebras changes the usual product into the convolution product.
2. HEISENBERG GROUPS AND THE STONE-VON NEUMANN THEOREM 71
1.3. The link between Fourier series and Fourier transform on R. Let f
be a function over the interval [− 12 , 21 ]. The Fourier series of f is
X
(4.5) cn (f )e2iπnx
n∈Z
where Z 1
2
cn := f (t)e−2iπnt dt.
− 21
Now, for λ ∈ R>0 , if g is a function defined over the interval [− λ2 , λ2 ], changing the
variable by y := λx, the corresponding Fourier series is written
!
X Z λ2 1 u y
(4.6) g(u)e−2iπn λ du e2iπn λ .
n∈Z −λ2
λ
Remark 2.9. Note that the Fourier transform is defined on the dual V ∗ of V ,
and those spaces are identitified through the symplectic form ω. We will refer to this
characterisation of ϕ as the functional equation.
Proof. We use Lemma 2.7. Let v ∈ V . We compute
ZZ
Tϕ ρ(v)Tϕ = ϕ(x)ϕ(y)ρ(x)ρ(v)ρ(y)dxdy
V ×V
ZZ
= ϕ(x)ϕ(y)eiπ((x|v)+(x|y)+(v|y)) ρ(x + v + y)dxdy
V ×V
ZZ
= ϕ(x)ϕ(z − v − x)eiπ((x|v)+(x|z)+(v|z)) ρ(z)dxdz.
V ×V
For almost every value of z, this operator is Tψ for
Z
ψ(v, z) = ϕ(x)ϕ(z − v − x)eiπ((x|v)+(x|z)+(v|z)) dx
V
by Fubini’s theorem. The relation (4.11) is equivalent to the fact that for every v,
ψ = C(v)ϕ. So (4.11) is equivalent to:
Z
ϕ(x)ϕ(z − v − x)eiπ((x|v)+(x|z)+(v|z)) dx = C(v)ϕ(z).
V
In the left hand side, we set x = −y and use ϕ∗ = ϕ. Then we obtain
Z
ϕ(y)ϕ(v − z − y)e−iπ((y|v)+(y|z)+(z|v)) dy = C(v)ϕ(z) = C(z)ϕ(v).
V
Hence
ϕ(z) ϕ(v)
(4.12) =
C(z) C(v)
ϕ(z)
so that C(z)
does not depend on z, moreover it is equal to its complex conjugate hence
ϕ(z)
belongs to R. We set C = C(z) , and get C(z) = Cϕ(−z).
Finally,
Z
ϕ(x)ϕ(z − v − x)eiπ((x|v)+(x|z)+(v|z)) dx = Cϕ(−v)ϕ(z).
V
2.3. Fock representation. Let us continue with a lovely avatar of this repre-
sentation, the Fock representation. We would like to characterize the image τ ∗ (H)
inside L2 (V ).
Just before Lemma 2.10, we chose a quadratic form Q on V such that (Ω−1 B)2 =
−IdV , and this equips V with a structure of complex vector space of dimension g for
which Ω−1 B is the scalar multiplication by the imaginary unit i. We denote by J
this complex structure and by VJ the corresponding complex space.
Furthermore B + iΩ : VJ → VJ∗ is a sesiquilinear isomorphism, we denote by A
the corresponding Hermitian form on VJ .
In this context, for a given x ∈ V we have:
Q(y−x) Q(x)+Q(y)
(4.13) r(1,x) ϕ(y) = ϕ(y − x)eiπ(x|y) = e−π 2
+iπ(x|y)
= e−π( 2
−A(x,y))
Q(x)
which is the product of ϕ(y) with a holomorphic function of f (y) = e−π( 2 −A(x,y)) .
The Fock representation associated to the complex structure J is the subspace
FJ ⊂ L2 (V ) consisting of the products f ϕ where ϕ was defined before and f is
a holomorphic function on VJ . We have just proven that this space is stable un-
der the H-action. Moreover, it is closed in L2 (V ) since holomorphy is preserved
under uniform convergence on compact sets. Let us choose complex coordinates P
z = (z1 , . . . , zg ) in VJ so that the Hermitiam product has the form A(w, z) = w̄i zi .
The scalar product (·, ·)F in FJ is given by
Z
2
(f ϕ, gϕ)F = f¯(z)g(z)e−π|z| dz̄dz,
V
Pg
2
where |z| = i=1 |zi | . If m = (m1 , . . . , mg ) ∈ Zg we denote by z m the monomial
2
m
function z1m1 . . . zg g . Any analytic function f (z) can be represented by a convergent
series
X
(4.14) f (z) = am z m .
m∈Zg
Consider the Taylor series (4.14). By Exercise 2.13 we have for any w ∈ VJ
Z Pg 2
f (z)f¯(z)e i=1 wi zi e−π|z| dz̄dz = 0,
V
in particular,
Z
f (z)f¯(z)dz̄dz = 0,
V
in this case H1+ consists of all densities which converge to L2 -functions on the bound-
ary (real line).
Exercise 3.4. Check that this Hermitian product is invariant.
3. REPRESENTATIONS OF SL2 (R) 81
It is easy to check now that Hn+ is irreducible. Indeed, every invariant closed subspace
M in Hn+ has a topological basis consisting of eigenvectors of K, in other words
wk (dw)n/2 for some positive k must form a topological basis of M . Without loss of
generality assume that M contains (dw)n/2 , then by applying ρg one can get that
1
(bw+a)n
(dw)n/2 , and in Taylor series for (bw+a)
1
n all elements of the basis appear with
is invariant. The property of invariance justify the choice of weight for the density as
1+s 1+s̄
(dx) 2 (dx) 2 = dx, thus the integration is invariant. To check that the represen-
tation is irreducible one can move the real line to the unit circle as in the example
1+s
of discrete series and then use eikθ (dθ) 2 as an orthonormal basis in Ps+ . Note that
the eigen values of ρgθ in this case are e2kiθ for all integer k.
82 4. SOME RESULTS ABOUT UNITARY REPRESENTATIONS
The second principal series Ps− can be obtained if instead of densities we consider
the pseudo densities which are transformed by the law
1+s ax + b 1+s
ρg ϕ (x) (dx) 2 =ϕ |cx + d|−s−1 sgn (cx + d) dx 2 .
cx + d
3.4. Complementary series. Those are representations which do not appear
in the regular representation L2 (G). They can be realized as the representations in
1+s
Cs of all densities ϕ (x) (dx) 2 for real 0 < s < 1 and have an invariant Hermitian
product
Z ∞Z ∞
(4.18) hϕ, ψi = ϕ̄ (x) ψ (y) |x − y|s−1 dxdy.
−∞ −∞
CHAPTER 5
On algebraic methods
1. Introduction
Say a few words about infinite direct sums and products, talk about Zorn’s lemma.
Emphasize that we are now in full generality.
3. Wedderburn–Artin theorem
A ring R is called semisimple if every R-module is semisimple. For example, a
group algebra k (G), for a finite group G such that char k does not divide |G|, is
semisimple by Maschke’s Theorem 3.3.
Lemma 3.1. Let R be a semisimple ring. Then as a module over itself R is
isomorphic to a finite direct sum of minimal left ideals.
Proof. Consider R as an R-module. By definition the simple submodules of R
are exactly the minimal left ideals of R. Hence since R is semisimple we can write R
as a direct sum ⊕i∈I Li of minimal left ideals Li . It remains to show that this direct
sum is finite. Indeed, let li ∈ Li be the image of the identity element 1 under the
projection R → Li . But R as a module is generated by 1. Therefore li 6= 0 for all
i ∈ I. Hence I is finite.
Corollary 3.2. A direct product of finitely many semisimple rings is semisimple.
Exercise 3.3. Let D be a division ring, and R = Matn (D) be a matrix ring over
D.
(a) Let Li be the subset of R consisting of all matrices which have zeros everywhere
outside the i-th column. Check that Li is a minimal left ideal of R and that R =
L1 ⊕ · · · ⊕ Ln . Therefore R is semisimple.
(b) Show that Li and Lj are isomorphic R-modules and that any simple R-module
is isomorphic to Li .
(c) Using Corollary 2.12 show that F := EndR (Li ) is isomorphic to Dop , and that
R is isomorphic to EndF (Li ).
By the above exercise and Corollary 3.2 a direct product Matn1 (D1 ) × · · · ×
Matnk (Dk ) of finitely many matrix rings is semisimple. In fact any semisismple ring
is of this form.
Theorem 3.4. (Wedderburn-Artin) Let R be a semisimple ring. Then there exist
division rings D1 , . . . , Dk such that R is isomorphic to a finite product of matrix rings
Matn1 (D1 ) × · · · × Matnk (Dk ) .
Furthermore, D1 , . . . , Dk are unique up to isomorphism and this presentation of R is
unique up to permutation of the factors.
Proof. Take the decomposition of Lemma 3.1 and combine isomorphic factors
together. Then the following decomposition holds
R = L1⊕n1 ⊕ · · · ⊕ Lk⊕nk ,
where Li is not isomorphic to Lj if i 6= j. Set Ji = Li⊕ni . We claim that Ji is
actually a two-sided ideal. Indeed Lemma 1.10 and simplicity of Li imply that Li r
is isomorphic to Li for any r ∈ R such that Li r 6= 0. Thus, Li r ⊂ Ji .
4. JORDAN-HÖLDER THEOREM AND INDECOMPOSABLE MODULES 87
Now we will show that each Ji is isomorphic to a matrix ring. Let Fi := EndJi (Li ).
The natural homomorphism Ji → EndFi (Li ) is surjective by Corollary 2.12. This
homomorphism is also injective since rLi = 0 implies rJi = 0 for any r ∈ R. Then,
since Ji is a unital ring r = 0. On the other hand, Fi is a division ring by Schur’s
lemma. Threfore we have an isomorphism Ji ≃ EndFi (Li ). By Exercise 1.7 Li is a
free Fi -module. Moreover, Li is finitely generated over Fi as Ji is a sum of finitely
many left ideals. Thus, by Exercise 3.3 (c), Ji is isomorphic to Matni (Di ) where
Di = Fiop .
The uniqueness of presentation follows easily from Krull-Schmidt theorem (Theo-
rem 4.19) which we prove in the next section. Indeed, let S1 , . . . , Sk be a complete list
of non-isomorphic simple R-modules. Then both Di and ni are defined intrinsically,
since Diop ≃ EndR (Si ) and ni is the multiplicity of the indecomposable module Si in
R.
Proof. First note that if M is simple, then the statement is trivial. We will
prove that if the statement holds for any proper submodule of M then it is also true
for M . If M1 = N1 , then the statement is obvious. Otherwise, M1 + N1 = M , hence
we have two isomorphisms M/M1 ∼ = N1 / (M1 ∩ N1 ) and M/N1 ∼ = M1 / (M1 ∩ N1 ).
Like the second isomorphism theorem for groups. Now let
M1 ∩ N 1 ⊃ K 1 ⊃ · · · ⊃ K s = 0
be a Jordan-Hölder series for M1 ∩ N1 . This gives us two new Jordan-Hölder series
of M
M = M 0 ⊃ M 1 ⊃ M1 ∩ N 1 ⊃ K 1 ⊃ · · · ⊃ K s = 0
and
M = N0 ⊃ N1 ⊃ N1 ∩ M1 ⊃ K1 ⊃ · · · ⊃ Ks = 0.
These series are obviously equivalent. By our assumption on M1 and N1 the first
series is equivalent to M = M0 ⊃ M1 ⊃ · · · ⊃ Mk = 0, and the second one is
equivalent to M = N0 ⊃ N1 ⊃ · · · ⊃ Nl = {0}. Hence the original series are also
equivalent.
Thus, we can now give two definitions:
Definition 4.8. First, we define the length l (M ) of an R-module M which
satsfies ACC and DCC as the length of any Jordan-Hölder series of M . Note that
we can easily see that if N is a proper submodule of M , then l (N ) < l (M ).
Furthermore, this gives rise to a notion of finite length R-module.
Remark 4.9. Note that in the case of infinite series with simple quotients, we
may have many non-equivalent series. For example, consider Z as a Z-module. Then
the series
Z ⊃ 2Z ⊃ 4Z ⊃ . . .
is not equivalent to
Z ⊃ 3Z ⊃ 9Z ⊃ . . . .
4.3. Indecomposable modules and Krull–Schmidt theorem. A module M
is indecomposable if M = M1 ⊕ M2 implies M1 = 0 or M2 = 0.
Example 4.10. Every simple module is indecomposable. Furthermore, if a semisim-
ple module M is indecomposable then M is simple.
Definition 4.11. An element e ∈ R is called an idempotent if e2 = e.
Lemma 4.12. An R-module M is indecomposable if and only if every idempotent
in EndR (M ) is either 1 or 0.
Proof. If M is decomposable, then M = M1 ⊕ M2 for some proper submodules
M1 and M2 . Then the projection e : M → M1 with kernel M2 is an idempotent
in EndR M , which is neither 0 nor 1. Conversely, any non-trivial idempotent e ∈
EndR M gives rise to a decomposition M = Ker e ⊕ Im e.
90 5. ON ALGEBRAIC METHODS
hence
l
X (1) (2) (2) (1)
p1 ◦ qj ◦ pj ◦ q1 = idM1 .
j=1
(1) (2) (2) (1)
By Corollary 4.17 there exists j such that p1 ◦ qj ◦ pj ◦ q1 is an isomorphism.
After permuting indices we may assume that j = 1. Then Lemma 4.14 implies that
(2) (1)
p1 ◦ q1 is an isomorphism between M1 and N1 . Set
M ′ := M2 ⊕ · · · ⊕ Mk , N ′ := N2 ⊕ · · · ⊕ Nl .
(2)
Since M1 intersects trivially N ′ = Ker p1 we have M = M1 ⊕ N ′ . But we also
M = M1 ⊕ M ′ . Therefore M ′ is isomorphic to N ′ . By induction assumption the
statement holds for M ′ ≃ N ′ . Hence the statement holds for M .
Come up with examples of modules for which Krull–Schmidt does not hold.
n
X
δ(u ⊗ w) := dfj (u) ⊗ (ej ∧ w) for all u ∈ S(V ), w ∈ Λ(V ).
j=1
g f
Ci −−−→ Ci′ −−−→ Ci′′
dy dy
′ d y
′′
g ′ ′′ f
Ci−1 −−−→ Ci−1 −−−→ Ci−1
In this diagram δ = g −1 ◦ d′ ◦ f −1 goes from the upper right to the lower left corners.
Theorem 5.7. (Long exact sequence). The following sequence
δ g∗ f∗ δ g∗
− → Hi (C•′ ) −
→ Hi (C• ) − → Hi (C•′′ ) −
→ Hi−1 (C• ) −
→ ...
is actually an exact complex.
We skip the proof of this theorem. The enthusiastic reader might verify it as an
exercise or read the proof in Weibel, MacLane.
5.3. Homotopy.
Definition 5.8. Consider complexes (C• , d), (C•′ , d′ ) of R-modules and let f, g :
C• → C•′ be morphisms. We say that f and g are homotopically equivalent if there
exists a map h : C• → C•′ of degree 1 such that
f − g = h ◦ d + d′ ◦ h.
Lemma 5.9. If f and g are homotopically equivalent then f∗ = g∗ .
94 5. ON ALGEBRAIC METHODS
6. Projective modules
Let R be a unital ring.
f
P A_ _ _/ M
AA ψ
AA ϕ
AA
A
N
Example 6.1. A free R-module F is projective. Indeed, let {ei }i∈I be a set of
generators of F . Define f : F → M by f (ei ) = ϕ−1 (ψ (ei )).
Lemma 6.2. Let P be an R-module, the following conditions are equivalent
(1) P is projective;
(2) There exists a free module F such that F is isomorphic to P ⊕ P ′ ;
(3) Any exact sequence of R-modules
0→N →M →P →0
splits.
Proof. (1) ⇒ (3)
Consider the exact sequence
ϕ
0→N →M −
→ P → 0.
Set ψ = idP . Since ϕ is surjective and P is projective, there exists f : P → M such
that ψ = idP = ϕ ◦ f .
(3) ⇒ (2) Every module is a quotient of a free module. Therefore we just have to
apply (3) to the exact sequence
0→N →F →P →0
for a free module F .
(2) ⇒ (1) Choose a free module F such that F = P ⊕ P ′ . Let ϕ : M → N be a
surjective morphism of R-modules and ψ a morphism ψ : P → N . Now extend ψ to
ψ̃ : F → N such that the restriction of ψ̃ to P (respectively, P ′ ) is ψ (respectively,
zero). There exists f : F → M such that ϕ ◦ f = ψ̃. After restriction to P we get
ϕ ◦ f|P = ψ̃|P = ψ.
96 5. ON ALGEBRAIC METHODS
d p
/ P1 / P0 /M /0
f
0 id
p′
/ P′ d′ / P′ /M /0
1 0
then we have f (Ker p) ⊂ Ker p′ . We construct f1 : P1 → P1′ using the following
commutative diagram:
/ P1 d / Ker p /0
f f0
1
/ P′ d′ / Ker p′ / 0.
1
The existence of f1 follows from projectivity of P1 and surjectivity of d′ .
We repeat the procedure to construct fi : Pi → Pi for all i.
Suppose now that f and g are two morphisms satisfying the assumptions of the
lemma. Let us prove that f and g are homotopically equivalent. Let ϕ = f − g. We
have to prove the existence of maps hi : Pi → Pi+1 such that hi ◦ d = d′ ◦ hi+1 . Let
us explain how to construct h0 and h1 using the following diagram
d d p
/ P2 / P1 / P0 /M /0
~ ~
ϕ2 ~ ϕ1 ~ ϕ0 0
~ h1
~ h0
~~ ~~ p′
/ P′ d′ / P′ d′ / P′ /M / 0.
2 1 0
γ is not unique. If we choose another pair ψ ′ ∈ HomR (P0 , Q) and γ ′ ∈ HomR (P1 , N ),
then there exists θ ∈ HomR (P0 , N ) such that ψ ′ − ψ = α ◦ θ as in the diagram below
/ P2 d / P1 d / P0
} AA
θ }} AA ϕ
0 }} AA
AA
}
~}} α β
0 /N /Q /M / 0.
Proof. Let N be a left ideal. We prove the statement by induction on the degree
of nilpotency d(N ). The case d(N ) = 1 is trivial. Let d(N ) > 1. Set n = r2 − r, then
n belongs to N and rn = nr. Therefore we have
(r + n − 2rn)2 ≡ r2 + 2rn − 4r2 n mod N 2 .
We set s = r + n − 2rn. Then we have
s2 ≡ s mod N 2 , s≡r mod N.
Since d(N 2 ) < d(N ), the induction assumption ensures that there exists an idempo-
tent e ∈ R such that e ≡ s mod N 2 , hence e ≡ r mod N .
For an R-module M let
Ann M = {x ∈ R | xM = 0} .
Definition 7.2. The Jacobson radical rad R of a ring R is the intersection of
Ann M for all simple R-modules M .
Exercise 7.3. (a) Prove that rad R is the intersection of all maximal left ideals
of R as well as the intersection of all maximal right ideals.
(b) Show that x belongs rad R if and only if 1 + rx is invertible for any r ∈ R.
(c) Show that if N is a nilpotent left ideal of R, then N is contained in rad R.
Lemma 7.4. Let e ∈ rad R such that e2 = e. Then e = 0.
Proof. By Exercise 7.3 (b) we have that 1 − e is invertible. But e(1 − e) = 0
and therefore e = 0.
7.2. The Jacobson radical of an Artinian ring.
Definition 7.5. A ring R is artinian if it satisfying the descending chain condi-
tion for left ideals.
A typical example of artinian ring is a finite-dimensional algebra over a field. It
follows from the definition that any left ideal in an Artinian ring contains a minimal
(non-zero) ideal.
Lemma 7.6. Let R be an artinian ring, I ⊂ R be a left ideal. If I is not nilpotent,
then I contains a non-zero idempotent.
Proof. Since R is Artinian, one can can find a minimal left ideal J ⊂ I among
all non-nilpotent ideals of I. Then J 2 = J. We will prove that J contains a non-zero
idempotent.
Let L ⊂ J be some minimal left ideal such that JL 6= 0. Then there exists x ∈ L
such that Jx 6= 0. By minimality of L we have Jx = L. Therefore there exists r ∈ J
such that rx = x. Hence (r2 − r) x = 0. Let N = {y ∈ J | yx = 0}. Note that N is
a proper left ideal of J and therefore N is nilpotent. Thus, we have r2 ≡ r mod N .
By Lemma 7.1 there exists an idempotent e ∈ R such that e ≡ r mod N , and we
are done.
7. REPRESENTATIONS OF ARTINIAN RINGS 101
Theorem 7.7. If R is artinian then rad R is the unique maximal nilpotent ideal
of R.
Proof. By Exercise 7.3 every nilpotent ideal of R lies in rad R. It remains to
show that rad R is nilpotent. Indeed, otherwise by Lemma 7.6, rad R contains a
non-zero idempotent. This contradicts Lemma 7.4.
Lemma 7.8. An Artinian ring R is semisimple if and only if rad R = 0.
Proof. If R is semisimple and Artinian, then by Wedderburn-Artin theorem it
is a direct product of matrix rings, which does not have non-trivial nilpotent ideals.
If R is Artinian with trivial radical, then by Lemma 7.6 every minimal left ideal
L of R contains an idempotent e such that L = Re. Hence R is isomorphic to
L ⊕ R(1 − e). Therefore R is a direct sum of its minimal left ideals.
Corollary 7.9. If R is Artinian, then R/ rad R is semisimple.
Proof. By Theorem 7.7 the quotient ring R/ rad R does not have non-zero nilpo-
tent ideals. Hence it is semisimple by Lemma 7.8.
Recall that EndR (R) = Rop . Therefore the canonical projection on each compo-
nent Li is given by multiplication (on the right) by some idempotent element ei ∈ Li .
In other words R has a decomposition
(5.4) R = Re1 ⊕ · · · ⊕ Ren .
Moreover, ei ej = δij ei . Once more by Krull-Schmidt theorem this decomposition is
unique up to multiplication by some invertible element on the right.
Definition 7.13. An idempotent e ∈ R is called primitive if it can not be written
e = e′ + e′′ for some non-zero idempotents e′ , e′′ such that e′ e′′ = e′′ e′ = 0.
Exercise 7.14. Prove that the idempotent e ∈ R is primitive if and only if Re
is an indecomposable R-module.
In the decomposition (5.4) the idempotents e1 , . . . , en are primitive.
Lemma 7.15. Assume R is Artinian, N = rad R and e ∈ R is a primitive idem-
potent. Then N e is a unique maximal submodule of Re.
Proof. Due to Corollary 7.11 it is sufficient to show that Re/N e is a simple
R-module. Since e is primitive, the left ideal Re is an indecomposable R-module.
Assume that Re/N e is not simple. Then Re/N e = Ree1 ⊕ Ree2 for some non-zero
idempotent elements e1 and e2 in the quotient ring R/N . By Lemma 7.1 there exist
idempotents f1 , f2 ∈ R such that fi ≡ ei mod N . Then Re = Rf1 ⊕ Rf2 which
contradicts indcomposability of Re.
Theorem 7.16. Assume R is Artinian.
(1) Every simple R-module S has a projective cover which is isomorphic to Re
for some primitive idempotent e ∈ R.
(2) Let P be an indecomposable projective R-module. There exists a primitive
idempotent e ∈ R such that P is isomorphic to Re. Furthermore, P has a
unique simple quotient.
Proof. Let S be a simple R-module. There exists a surjective homomorphism
f : R → S. Consider the decomposition (5.4). There exists i ≤ n such that the
restriction of f on Rei is non-zero. By the simplicity of S the restriction f : Rei → S
is surjective. It follows from Lemma 7.15 that Rei is a projective cover of S.
Let P be an indecomposable projective module. By Lemma 7.10 the quotient
P/(rad RP ) is semisimple. Let S be a simple submodule of P/(rad RP ). Then we
have a surjection f : P → S. Let g : Q → S be a projective cover of S. There
exists a morphism ϕ : P → Q such that f = g ◦ ϕ. Since Q has a unique simple
quotient, the morphism ϕ is surjective. Then P is isomorphic to Q ⊕ Ker ϕ. The
indecomposability of P implies that P is isomorphic to Q.
Example 7.17. Consider the group algebra R = F3 (S3 ). First let us classify
simple and indecomposable projective R-modules.
7. REPRESENTATIONS OF ARTINIAN RINGS 103
simple modules L1 , . . . , Ln associated with those idempotents and that the dimension
of every Li over F is 1. Finally check that
(
F, if i = j, p = 0 or i = j + 1, p = 1
Extp (Li , Lj ) = .
0, otherwise
8. Abelian categories
An abelian category is a generalization of categories of modules over a ring.
Let us start with definition of an additive category.
Definition 8.1. A category C is called additive if for any two objects A and B,
(1) The set of morphisms HomC (A, B) is an abelian group.
(2) There exist an object A ⊕ B, called a direct sum, and a pair of morphisms
iA : A → A⊕B and iB : B → A⊕B such that for any morphisms ϕ : A → M
and ψ : B → M there exists a unique morphism τ : A ⊕ B → M such that
τ ◦ iA = ϕ and τ ◦ iB = ψ.
(3) There exist an object A × B called a direct product and a pair of morphisms
pA : A×B → A and iB : A×B → B such that for any morphisms α : M → A
and β : M → M there exists a unique morphism θ : M → A × B such that
pA ◦ θ = α and pB ◦ θ = β.
(4) The induced morphism A ⊕ B → A × B is an isomorphism.
Definition 8.2. An abelian category is an additive category C such that, for
every morphism ϕ ∈ HomC (A, B)
i
(1) There exist an object and a morphism Ker ϕ −
→ A such that for any mor-
phism γ : M → A such that, ϕ ◦ γ = 0, there exists a unique morphism
δ : M → Ker ϕ such that γ = i ◦ δ.
p
(2) There exist an object and morphism B −→ Coker ϕ such that for any mor-
phism τ : B → M such that, τ ◦ ϕ = 0, there exists a unique morphism
σ : Coker ϕ → M such that τ = σ ◦ p.
(3) There is an isomorphism Coker i → Ker p.
Exercise 8.3. Let R be a ring, show that the category of finitely generated R-
modules is abelian. Show that the category of projective R-modules is additive but
not abelian in general. Finally show that the category of projective R-modules is
abelian if and only if R is a semisimple ring.
In an abelian category we can define the image of a morphism, a quotient object,
exacts sequences, projective and injective objects. All the results of Sections 4, 5 and
6 can be generalized for abelian categories. If we want to define extension groups we
have to assume the existence of projective covers.
8. ABELIAN CATEGORIES 105
Example 1.10. Consider the partition λ = (1, . . . , 1) whose Young diagram con-
sists of one column with n boxes. Then Qt(λ) = Sn , Pt(λ) is trivial and
X
ct(λ) = bt(λ) = (−1)s s.
s∈Sn
Let E denote the permutation representation of Sn . Let us show that Q(Sn )ct(λ) is
the n − 1 dimensional simple submodule of E. Indeed, at(λ) ct(λ) = ct(λ) , therefore
the restriction of Vt(λ) to Pt(λ) contains the trivial representation of Pt(λ) . Recall
that the permutation representation can be obtained by induction from the trivial
representation of Sn−1 :
E = IndSPnλ triv .
By Frobenius reciprocity Q(Sn )ct(λ) is a non-trivial submodule of E. .
In the rest of this Section we prove Theorem 1.7.
First, let us note that Sn acts simply transitively on the set of Young tableaux of
the same shape by permuting the entries, and for any s ∈ Sn we have
ast(λ) = sat(λ) s−1 , bst(λ) = sbt(λ) s−1 , cst(λ) = sct(λ) s−1 .
Therefore if we have two tableaux t(λ) and t′ (λ) of the same shape, then
Q(Sn )ct(λ) = Q(Sn )ct′ (λ) s−1
for some s ∈ Sn . Hence Q(Sn )ct(λ) and Q(Sn )ct′ (λ) are isomorphic Q(Sn )-modules.
In what follows we denote by Vλ a fixed representative of the isomorphism class
of Q(Sn )ct(λ) for some tableau t(λ). As we have seen this does not depend on the
tableau but only on its shape.
Exercise 1.12. Let t(λ) be a Young tableau and s ∈ Sn . Show that if s does
not belong to the set Pt(λ) Qt(λ) , then there exist two entries i, j which lie in the same
row of t(λ) and in the same column of st(λ). In other words, the transposition (ij)
lies in the intersection Pt(λ) ∩ Qst(λ) . Hint: Assume the opposite, and check that one
can find s′ ∈ Pt(λ) and s′′ ∈ Qt(λ) such that s′ t(λ) = s′′ st(λ).
Next, observe that for any p ∈ Pλ and q ∈ Qλ we have
pct(λ) q = (−1)q ct(λ) .
110 6. SYMMETRIC GROUPS, SCHUR–WEYL DUALITY AND PSH ALGEBRAS
Lemma 1.13. Let t(λ) be a Young tableau and y ∈ Q(Sn ). Assume that for all
p ∈ Pt(λ) and q ∈ Qt(λ) we have
pyq = (−1)q y.
Then y = act(λ) for some a ∈ Q.
Proof. Let T be a set ofFrepresentatives of the double cosets Pt(λ) \Sn /Qt(λ) .
Then Sn is the disjoint union s∈T Pt(λ) sQt(λ) and we can write y in the form
X X X
ds (−1)q psq = ds at(λ) sbt(λ) .
s∈T p∈Pt(λ) ,q∈Qt(λ) s∈T
Proof. We have to check that ct(λ) sct′ (µ) s = 0 for any s ∈ Sn , which is equivalent
to ct(λ) cst′ (µ) = 0. Therefore it suffices to prove that ct(λ) ct′ (µ) = 0. By Exercise 1.17
there exists a transposition τ which belongs to the intersection Qt(λ) ∩ Pt′ (µ) . Then,
repeating the argument from the proof of Lemma 1.13, we obtain
ct(λ) ct′ (µ) = ct(λ) τ 2 ct′ (µ) = −ct(λ) ct′ (µ) .
Now we show the second statement of Theorem 1.7.
Lemma 1.19. Two irreducible representations Vλ and Vµ are isomorphic if and
only if λ = µ.
Proof. It suffices to show that if λ 6= µ, then Vλ and Vµ are not isomorphic.
Without loss of generality we may assume λ < µ and take some Young tableaux t(λ)
and t′ (µ). By Lemma 1.18 we obtain that ct(λ) acts by zero on Vµ . On the other
hand, by Corollary 1.16, ct(λ) does not annihilate Vλ . Hence the statement.
By Remark 1.8 the proof of Theorem 1.7 is complete.
Remark 1.20. Note that in fact we have proved that if λ 6= µ then ct(λ) Q(Sn )ct′ (µ) =
0 for any pair of tableaux t(λ), t′ (µ). Indeed, if ct(λ) Q(Sn )ct′ (µ) 6= 0, then
Q(Sn )ct(λ) Q(Sn )ct′ (µ) = Q(Sn )ct′ (µ) .
But this is impossible since Q(Sn )ct(λ) Q(Sn ) has only components isomorphic to Vλ .
Lemma 1.21. Let ρ : Sn → GL (V ) be a finite-dimensional representation of Sn .
Then the multiplicity of Vλ in V equals the rank of ρ ct(λ) .
Proof. The rank of ct(λ) in Vλ is 1 and ct(λ) Vµ = 0 for all µ 6= λ. Hence the
statement.
Exercise 1.22. Let λ be a partition and χλ denote the character of Vλ .
(1) Prove that χλ (s) ∈ Z for all s ∈ Sn .
(2) Prove that χλ (s) = χλ (s−1 ) for all s ∈ Sn and hence Vλ is self-dual.
(3) For a tableau t(λ) let c̄t(λ) = bt(λ) at(λ) . Prove that Q(Sn )ct(λ) and Q(Sn )c̄t(λ)
are isomorphic Q(Sn )-modules.
Exercise 1.23. Let λ be a partition. We define the conjugate partition λ⊥ by
setting λ⊥
i to be equal to the length of the i-th row in the Young diagram λ. For
example, if λ = , then λ⊥ = .
Prove that for any partition λ, the representation Vλ⊥ is isomorphic to the tensor
product of Vλ with the sign representation.
Since Q is a splitting field for Sn , Theorem 1.7 provides classification of irreducible
representations of Sn over any field of characteristic zero.
112 6. SYMMETRIC GROUPS, SCHUR–WEYL DUALITY AND PSH ALGEBRAS
2. Schur–Weyl duality.
2.1. Dual pairs. We will start the following general statement.
Theorem 2.1. Let G and H be two groups and ρ : G × H → GL (V ) be a
representation in a vector space V . Assume that V has a decomposition
m
M
V = Vi ⊗ HomG (Vi , V )
i=1
for all v1 , . . . , vn ∈ V and s ∈ Sn . One can easily check that the actions of GL(V )
and Sn in the space V ⊗n commute. We will show that GL(V ) and Sn form a dual
pair.
Theorem 2.4. (Schur–Weyl duality) Let m = dim V and Γn,m denote the set of
all Young diagrams with n boxes such that the number of rows of λ is not bigger
than m. Then M
V ⊗n = Vλ ⊗ Sλ (V ),
λ∈Γn,m
where Vλ is the irreducible representation of Sn associated to λ and
Sλ := HomSn (Vλ , V )
is an irreducible representation of GL(V ). Moreover, Sλ (V ) and Sµ (V ) are not iso-
morphic if λ 6= µ.
Remark 2.5. If λ ∈ Γn,m and t(λ) is arbitrary Young tableau of shape λ, then the
image of the Young symmetrizer ct(λ) in V ⊗n is a simple GL(V )-module isomorphic
to Sλ (V ).
Example 2.6. Let n = 2. Then we have a decomposition V ⊗V = S 2 (V )⊕Λ2 (V ).
Theorem 2.4 implies that S 2 (V ) = S(2) (V ) and Λ2 (V ) = S(1,1) (V ) are irreducible rep-
resentations of GL(V ). More generally, S(n) (V ) is isomorphic to S n (V ) and S(1,...,1) (V )
is isomorphic to Λn (V ).
Let us prove Theorem 2.4.
Lemma 2.7. Let σ : k(GL(V )) → Endk (V ⊗n ) be the homomorphism induced by
the action of GL(V ) on V ⊗n . Then
EndSn V ⊗n = σ(k(GL(V )).
Proof. Let E = Endk (V ). Then we have an isomorphism of algebras
Endk V ⊗n ≃ E ⊗n .
We define the action of Sn on E ⊗n by setting
s(X1 ⊗ · · · ⊗ Xn ) := Xs(1) ⊗ · · · ⊗ Xs(n)
for all s ∈ S and X1 , . . . , Xn ∈ E. Then EndSn (V ⊗n ) coincides with the subalgebra
of Sn -invariants in E ⊗n that is with the n-th symmetric power S n (E) of E. Therefore
it suffices to show that S n (E) is the linear span of σ(g) for all g ∈ GL(V ).
We will need the following
Exercise 2.8. Let W be a vector space. Prove that for all n ≥ 2 the following
identity holds in the symmetric algebra S (W )
X n
2n−1 n!x1 . . . xn = (−1)i2 +···+in x1 + (−1)i2 x2 + · · · + (−1)in xn .
i2 =0,1,...,in =0,1
114 6. SYMMETRIC GROUPS, SCHUR–WEYL DUALITY AND PSH ALGEBRAS
Let us choose a basis e1 , . . . , em2 of E such that all non-zero linear combinations
a1 e1 +· · ·+am2 em2 with coefficients ai ∈ {−n, . . . , n} belong to GL(V ). (The existence
of such basis follows from density of GL(V ) in E in Zariski topology.) By the above
exercise the set
{(a1 ei1 + · · · + an ein )⊗n | ai = ±1, i1 , . . . in ≤ N }
spans S n (E). On the other hand, every non-zero (a1 ei1 + · · · + an ein ) belongs to
GL(V ). Therefore we have
(a1 ei1 + · · · + an ein )⊗n = σ(a1 ei1 + · · · + an ein ).
Hence S n (E) is the linear span of σ(g) for g ∈ GL(V ).
Lemma 2.9. Let λ = (λ1 , . . . , λp ) be a partition of n. Then Sλ (V ) 6= 0 if and
only if λ ∈ Γn,m .
Proof. Consider the tableau t(λ) with entries 1, . . . , n placed in increasing order
from top to bottom of the Young diagram λ starting from the first column. For
1 3 5
instance, for λ = we consider the tableau t(λ) = 2 4 . By Remark 2.5
Sλ (V ) 6= 0 if and only if ct(λ) (V ⊗n ) 6= 0.
If λ⊥ = µ = (µ1 , . . . , µr ), then
bt(λ) (V ⊗n ) = ⊗ri=1 Λµi (V ).
If λ is not in Γn,m , then µ1 > m and bt(λ) (V ⊗n ) = 0. Hence ct(λ) (V ⊗n ) = 0.
Let λ ∈ Γn,m . Choose a basis v1 , . . . , vm in V , then
B := {vi1 ⊗ · · · ⊗ vin | 1 ≤ i1 , . . . , in ≤ m}
⊗n
is a basis of V . Consider the particular basis vector
u := v1 ⊗ . . . vµ1 ⊗ · · · ⊗ v1 ⊗ . . . vµr ∈ B.
One can easily see that in the decomposition of ct(λ) (u) in the basis B, u occurs with
p
Y
coefficient λi !. In particular, ct(λ) (u) 6= 0. Hence the statement.
i=1
Lemma 2.7, Lemma 2.9 and Theorem 2.1 imply Theorem 2.4. Furthermore, Theo-
rem 2.4 together with the Jacobson density theorem (Theorem 2.9 Chapter V) implies
the double centralizer property:
Corollary 2.10. Under assumptions of Theorem 2.4 we have
EndGL(V ) (V ⊗n ) = ρ(k(Sn )).
Definition 2.11. Let λ be a partition of n. The Schur functor Sλ is the functor
from the category of vector spaces to itself defined by
V 7→ Sλ (V ) = HomSn (Vλ , V ⊗n ).
2. SCHUR–WEYL DUALITY. 115
Proof. From the properties (3) and (4) we have that I = Ker e∗ . Write
m∗ (x) = y ⊗ 1 + 1 ⊗ z + m∗+ (x).
We have to check that y = z = x. But this immediately follows from the counit
axiom.
Proposition 3.9. Let A be a connected graded bialgebra and P be the set of
primitive elements of A. Assume that I 2 ∩ P = 0. Then A is commutative and has
the antipode.
Proof. Let us prove first that A is commutative. Assume the opposite. Let
x ∈ Ak , y ∈ Al be some homogeneous element of A such that [x, y] 6= 0 and k + l
minimal possible. Then m∗ ([x, y]) = [m∗ (x), m∗ (y)]. By minimality of k + l we have
that [m∗+ (x), m∗+ (y)] = 0, hence [x, y] is primitive. On the other hand, [x, y] ∈ I ⊗ I,
hence [x, y] = 0. A contradiction.
Next, let us prove the existence of antipode. For every x ∈ An we construct
S(x) ∈ An recursively. We set
S(x) := −x for n = 1, S(x) = −x − m ◦ (Id ⊗S) ◦ m∗+ (x) for n > 1.
Exercise 3.10. Check that S satisfies the antipode axiom.
4. The Hopf algebra associated to the representations of symmetric
groups
Let us consider the free Z-module A = ⊕n∈N A(Sn ) where A(Sn ) is freely gener-
ated by the characters of the irreducible representations (in C-vector spaces) of the
symmetric group Sn . (Note that since every Sn -module is semi-simple, A(Sn ) is the
Grothendieck group of the category Sn − mod of finite dimensional representations of
Sn ). It is a N-graded module, where the homogeneous component of degree n is equal
to A(Sn ) if n ≥ 1 and the homogeneous part of degree 0 is Z by convention. More-
over, we equip it with a Z-valued symmetric bilinear form, denoted h; i, for which
the given basis of characters is an orthonormal basis, and with the positive cone A+
generated over the non-negative integers by the orthonormal basis.
In order to define the Hopf algebra structure on A, we use the induction and
restriction functors:
Proof. Assume the equality doesn’t hold, then there exists an ω ∈ Ω which does
not belong to this sum. We choose such an ω with minimal degree k. Since ω is not
primitive, it is not orthogonal to I 2 and therefore belongs to the support of some
ηη ′ with η, η ′ belonging to Ω. Hence k = k ′ + k ′′ where k ′ (resp. k ′′ ) is degree of η
(resp. η ′ ). By minimality of k, η and η ′ lie in the direct sum, thus, by Lemma 5.2, a
contradiction.
Lemma 5.5. Let π α and π β be elements in M which are relatively prime. Then
the restriction of the multiplication induces an isomorphism Aα ⊗ Aβ ≃ Aα+β given
by a bijection between Supp(π α ) × Supp(π β ) and Supp(π α+β ).
122 6. SYMMETRIC GROUPS, SCHUR–WEYL DUALITY AND PSH ALGEBRAS
Proof. We will prove that the Gram matrix (see Exercise 4.8)
Gram((ωη)ω∈Supp(πα ),η∈Supp(πβ ) )
is the identity. This will be enough since it implies that the products ωη are distinct
elements of Ω (again, see Exercise 4.8), and they exhaust the support of π α+β which
belongs to their linear span.
Let ω1 , ω2 (resp. η1 , η2 ) be elements of Supp(π α ) (resp. Supp(π β )), one has
hω1 η1 , ω2 η2 i = hm∗ (ω1 η1 ), ω2 ⊗ η2 i = hm∗ (ω1 )m∗ (η1 ), ω2 ⊗ η2 i.
L ′ ′′ L ′ ′′
One has m∗ (ω1 ) ∈ α′ +α′′ =α Aα ⊗ Aα and m∗ (η1 ) ∈ β ′ +β ′′ =β Aβ ⊗ Aβ (this is
just a transposed version of Lemma 5.2), hence
X ′ ′ ′′ ′′
m∗ (ω1 )m∗ (η1 ) ∈ Aα +β ⊗ Aα +β .
α′ +α′′ =α,β ′ +β ′′ =β
We will now use the definitions and notations for partitions introduced in Section
1. For a partition λ = (λ1 , . . . , λn ), we denote by eλ the product eλ = eλ1 . . . eλn and
set a similar definition for hλ . Note that in general, the elements eλ , hλ do not belong
to Ω.
Definition 6.7. Let λ = (λ1 , . . . , λr ) and µ = (µ1 , . . . , µs ) be two partitions of
the same integer n. We say that λ is greater or equal than µ for the dominance order
and denote it λ µ if, for every k ≤ inf(r, s), λ1 + . . . + λk ≥ µ1 + . . . + µk .
Lemma 6.8. Let λ and µ be partitions of a given integer n, define Mλ,µ as the
number of matrices with entries belonging to {0, 1} such that the sum of the entries
in the i-th row (resp. column) is λi (resp. µi ). Then one has:
(1) heλ , hµ i = Mλ,µ ,
(2) Mλ,λ⊥ = 1,
(3) Mλ,µ 6= 0 implies λ µ⊥
Proof. (Sketch) We write λ = (λ1 , . . . , λr ) and µ = (µ1 . . . µs ). By Exercise 6.5,
we have X
e∗λ1 (hµ1 . . . hµs ) = hµ1 −ν1 . . . hµs −νs .
P
νi =0,1 νi =λ1
Next, we apply e∗λ2 to this sum, e∗λ3 to the result, and so on. We obtain:
X
heλ , hµ i = e∗λ (hµ ) = hµ1 −Pi ν1i . . . hµs −Pi νsi .
P
νij ∈{0,1}, j νij =λi
P
The terms in the sum of the right-hand side are equal to 0 except when µi = j νij
for all i, in which case the value is 1. The statement (1) follows.
For statement P(2), we see easily
P that the only matrix N = (νi,j ) with entries in
{0, 1} such that j νij = λi and i νij = λ⊥ j is the one such that the entries decrease
along both the rows and the columns, hence the result. P
Finally, consider a matrix N = (νij ) with entries in {0, 1} such that j νij = λi
P
and i νij = µj . The sum λ1 + . . . + λi is the sum of theP entries of the columns
⊥ ⊥
of index ≤ i of N . Furthermore, µ1 + . . . + µi is equal to j≤i jlj where lj is the
number of rows of N which have sum j. It is easy to check that statement (3) follows.
Corollary 6.9. The matrix (heλ , hµ⊥ i)λ,µ ⊢n is upper triangular with 1’s on the
diagonal. In particular, its determinant is equal to 1.
Proposition 6.10. When λ varies along the partitions of n, the collection of eλ ’s
is a basis of the homogeneous component of degree n, An , of A.
Proof. First we notice that every hi is a polynomial with integral coefficients in
the ej ’s. This follows immediately from Proposition 6.6. Therefore the base change
126 6. SYMMETRIC GROUPS, SCHUR–WEYL DUALITY AND PSH ALGEBRAS
matrix P from (hλ )λ⊢n to (eλ )λ⊢n has integral entries. Then the Gram matrix Ge of
(eλ )λ⊢n satisfies the equality
(heλ , hµ i)λ,µ ⊢n = P t Ge ,
where P t denotes the transposed P . The corollary 6.9 ensures that the left-hand
side has determinant ±1 (the corollary is stated for µ⊥ and µ 7→ µ⊥ is an involution
which can produce a sign). Hence Ge has determinant ±1: we refer to Exercise 4.8
to ensure that the Z-module generated by (eλ )λ⊢n has a basis contained in Ω. Since
the support of en1 is the set of all ω ∈ Ω of degree n, we conclude that (eλ )λ⊢n is a
basis of An .
We deduce from the results of this section:
Theorem 6.11. (Zelevinsky) Up to isomorphism, there is only one rank one PSH
algebra. It has only one non-trivial automorphism ι, which takes any homogeneous
element x of degree n to (−1)n S(x) where S is the antipode.
Remark 6.12. The sets of algebraically independent generators (en ) and (hn ) of
the Z-algebra A play symmetric roles, and they are exchanged by the automorphism
ι of the theorem.
7. Bases of PSH algebras of rank one
Let A be a PSH algebra of rank one, with basis Ω and scalar product h , i, we will
use the sets of generators (en ) and (hn ). We keep all the notations of the preceding
section.
We will first describe the primitive elements of A. We denote AQ := A ⊗ Q.
Exercise 7.1. Consider the algebra A[[t]] of formal power series with coefficients
in A. Let f ∈ A[[t]] such that m∗ (f ) = f ⊗ f and the constant term of f is 1. Show
′
that the logarithmic derivative g := ff satisfies m∗ (g) = g ⊗ 1 + 1 ⊗ g.
Proposition 7.2. (1) For every n ≥ 1, there is exactly one primitive el-
ement of degree n, pn , such that hpn , hn i = 1. Moreover, every primitive
element of degree n is a integral multiple of pn .
(2) In the formal power series ring AQ [[t]], we have the following equality:
!
X pn X
(6.6) exp tn = hn tn .
n≥1
n n≥0
Proof. We first show that the set of primitive elements of degree n is a sub-
group of rank 1. Indeed, we recall that the primitive elements form the orthogonal
complement of I 2 in I (see just below Exercise 5.3). Since all the elements (hλ )λ⊢n
except hn are in I 2 , the conclusion follows. Moreover, An is its own dual with respect
to the scalar product. Let denote by (hλ )λ⊢n the dual basis of (hλ )λ⊢n . Clearly, hn
can be chosen as pn . Hence statement (1).
7. BASES OF PSH ALGEBRAS OF RANK ONE 127
P
Consider the formal series H(t) := n≥0 hn tn ∈ A[[t]], it satisfies the relation
m( H) = H ⊗ H by Proposition 6.4, re-written in terms of h’s instead of e’s. Hence,
′ (t)
using Exercise 7.1, we get P (t) := HH(t) which satisfies m∗ (P ) = P ⊗1+1⊗P . Hence all
P
the coefficients of P are primitive elements in A. Write P (t) = i≥1 ̟i+1 ti . To prove
statement (2), it remains to check that h̟n , hn i = 1. We have P (t)H(t) = H ′ (t), so
when we compare the terms on both sides we get
(6.7) ̟n + h1 ̟n−1 + . . . + hn−1 ̟1 − nhn = 0.
By induction on n, this implies
X
̟n = (−1)n hn1 + cλ h λ ,
λ⊢n,λ6=(1,...,1)
where the cλ ’s are integers. Now, we compute the scalar product with en : we apply
Lemma 6.8 and find that there is no contribution from the terms indexed by λ if
λ 6= (1, . . . , 1). Therefore, h̟n , en i = (−1)n . Finally, we use the automorphism of
A to get the conclusion that pn = ̟n since ι(en ) = hn and ι(̟n ) = (−1)n ̟n by
Proposition 6.6.
For every partition λ = (λ1 , . . . , λr ), we set pλ = pλ1 . . . pλr . Let us compute their
Gram matrix:
Proposition 7.3. The family (pλ ) is an orthogonal basis of AQ and one has
Y Y
hpλ , pλ i = (λ⊥j − λ ⊥
j+1 )! λi
j i
Our first goal is to express hλ ’s in terms of ωµ ’s. First, we compute h∗i (ωλ ), and
for this, we need to introduce some notations.
Let λ be a partition, or equivalently a Young diagram. We denote by r(λ) (resp.
c(λ)) the number of rows (resp columns) of λ.
We denote by Rλi the set of all µ’s such that µ is obtained from λ by removing
exactly i boxes, at most one in every row of λ. Similarly, Cλi is the set of all µ’s such
that µ is obtained from λ by removing exactly i boxes, at most one in every column
of λ. In the specific case where i = r(λ), there is only one element in the set Rλi and
this element will be denoted by λ← , it is the diagram obtained by removing the first
column of λ, similarly, if i = c(λ) the unique element of Cλi will be denoted by λ↓
and is the diagram obtained by suppressing the first row of λ.
Remark 7.5. Note that if µ ∈ Cλi , then µ⊥ ∈ Ri (λ⊥ ).
Theorem 7.6. (Pieri’s rule) One has:
X
h∗i (ωλ ) = ωµ ,
µ∈Cλ
i
and X
e∗j (ωλ ) = ωµ .
µ∈Rλ
j
Proof. First, let us prove that if aiλ,µ 6= 0, then r(λ) = r(µ) or r(λ) = r(µ) + 1.
Assume r(µ) > r(λ) and aiλ,µ 6= 0: then we have ωµ ≤ h∗i (ωλ ), therefore, applying
Lemma 7.9, we get
ωµ← = e∗r(µ) (ωµ ) ≤ e∗r(µ) ◦ h∗i (ωλ ) = h∗i ◦ e∗r(µ) (ωλ ) = 0,
which gives a contradiction.
Assume r(µ) < r(λ) − 1 and aiλ,µ 6= 0: then applying the equation (6.8), we have
ωλ ≤ hi ωµ , therefore applying Lemma 7.9 and Lemma 7.7, we get
ωλ← = e∗r(λ) (ωλ ) ≤ e∗r(λ) ◦ hi (ωµ ) = hi ◦ e∗r(λ) (ωµ ) + hi−1 ◦ e∗r(λ)−1 (ωµ ) = 0,
which again gives a contradiction.
Next, we look at the case r(λ) = r(µ). We do a direct computation:
X X
h∗i (ωλ← ) = e∗r(λ) ◦ h∗i (ωλ ) = aiλ,µ e∗r(λ) (ωµ ) = aiλ,µ ωµ← .
µ r(λ)=r(µ)
For a partition λ, we introduce the notion of semistandard tableau of shape λ: the
Young diagram of shape λ is filled with entries wich are no longer distinct, with the
condition that the entries are non decreasing along the rows and increasing along the
columns of λ. For instance,
1 1 1
2 3
3
4
is a semistandard tableau.
To such a semistandard tableau, we associate its weight, which is the sequence
mi consisting of the numbers of occurences of the integer i in the tableau: in our
example, m1 = 2, m2 = 1, m3 = 2, m4 = 1 and all the other mi ’s are zero.
7. BASES OF PSH ALGEBRAS OF RANK ONE 131
and eventually
X X X
hhm1 . . . hmr , ωλ i = (hm1 . . . hmr−1 hmr )∗ (ωλ ) = ... 1,
µ µ
µ1 ∈Cλ
m1 µ2 ∈Cm12 µr ∈Cmr−1
r
and this assertion is clear since A the polynomial algebra Z[(hi )i∈N ].
132 6. SYMMETRIC GROUPS, SCHUR–WEYL DUALITY AND PSH ALGEBRAS
We notice that, in this determinant, every entry is a partial sum of the entry which
is just above it, we are led to substract the i + 1th row from the ith row for all i.
This doesn’t affect the value of the determinant, therefore we obtain the equality
X
H(̟λ ) = Det hki −i+j .
λ2 ≤k1 ≤λ1 ,...,λr ≤kr−1 ≤λr−1 ,kr ≤λr
1≤i,j≤r
Now each family of indices k1 , . . . , kr gives rise to a partition µ belonging to Cλm for
m = n − k1 − . . . − kr , from which we deduce the result.
8. Harvest
In the last four sections, we defined and classified PSH algebras and we obtained
precise results in the rank one case. Now it is time to see why this was useful. In this
section, we will meet two avatars of the rank one PSH algebra, namely A of section
4, and the Grothendieck group of polynomial representations of the group GL∞ :
this interpretation will give us precious information concerning the representation
theory in both cases. The final section of this chapter will be devoted to another
very important application of PSH algebras, in infinite rank case, associated to linear
groups over finite fields. We will only state the main results without proof and refer
the reader to Zelevinsky’s seminal book.
Proposition 8.1. The PSH algebra A is of rank one with basic primitive element
π, the class (in the Grothendieck group) of the trivial representation of the trivial
group S1 .
Proof. Our goal is to show that every irreducible representation of Sn (n ∈
N\{0}) appears in π n . It is clear that π n is the regular representation of Sn , hence
the result.
The choice of π gives us gives us two isomorphisms between A and A (one is
obtained from the other by application of the automorphism ι). We choose the
isomorphism which send h2 to the trivial representation of S2 (hence it sends e2 to
the sign representation of S2 ).
Let us give an interpretation of the different bases (eλ ), (hλ ), (ωλ ), (pλ ) in this
setting.
Exercise 8.2. (1) Check that ei corresponds to the sign representation of
Si and that hi corresponds to the trivial representation of Si .
(2) Show that ωλ corresponds to the class of the irreducible representation Vλ
defined in section 1.
Remark 8.3. In the case of symmetric groups, the Grothendieck group is also
the direct sum of Z-valued central functions on Sn when n varies. See Chapter 1,
associated with the fact that the characters of the symmetric groups take their values
in Z.
Exercise 8.4. (1) Show that the primitive element pii is the characteristic
function of the circular permutation of Si .
(2) Interpreting the induction functors involved, show that, for every partition
λ, pλ is the characteristic function of the conjugacy class cλ corresponding
to λ times |c|λ|!
λ|
.
The following Proposition is now clear:
Proposition 8.5. The character table of Sn is just the transfer matrix expressing
the pλ ’s in terms of ωλ ’s when λ varies along the partitions of n.
Our goal now is to prove the Hook formula:
Let λ be a partition, let a = (i, j) be any every box in the Young diagram λ, we
denote h(a) the number of boxes (i′ , j ′ ) of the Young diagram such that i′ = i and
j ′ ≥ j or i′ ≥ i and j ′ = j: h(a) is called the hook length of a.
Theorem 8.6. (Hook formula) For every partition λ of n, the dimension of the
Sn -module Vλ is equal to
n!
dim Vλ = Q
a∈λ h(a)
134 6. SYMMETRIC GROUPS, SCHUR–WEYL DUALITY AND PSH ALGEBRAS
Proof. For any Sn -module V , let us denote by rdimV the reduced dimension of
V that is the quotient dim n!
V
: this defines a ring homomorphism from A to Q, as one
can easily see computing the dimension of an induced module.
We write λ = (λ1 , . . . , λr ). Set Li = λi + r − i and consider the new partition
consisting of (L1 , . . . , Lr ) := L. We apply Theorem 7.15 and notice that rdim(hp ) =
1
p!
: therefore one has
!
1
rdim(ωλ ) = Det .
(Li − r + j)! 1≤i,j≤r
Since Li ! = (Li − r + j)!Pr−j (Li ) where Pk (X) is the polynomial X(X − 1) . . . (X −
k + 1), the right-hand side becomes
1
Det (Pr−j (Li ))1≤i,j≤r .
L1 ! . . . L r !
Now Pk is a polynomial of degree k with leadingQcoefficient 1, hence this determinant
is a Vandermonde determinant and is equal to 1≤i<j≤r (Li − Lj ) and we get
n! Y
dim Vλ = (Li − Lj ).
L1 ! . . . Lr ! 1≤i<j≤r
Li !
Noting that Q (L i −Lj )
is product of the hook lengths of boxes of the i-th row of λ,
i<j
we obtained the wanted Hook formula.
Theorem 8.7. For every partition λ of n, the restriction of Vλ to Sn−1 is the
direct sum ⊕µ Vµ where the Young diagram of µ is obtained from the Young diagram
of λ by deleting exactly one box.
Proof. This restriction is h∗1 (ωλ ). Hence the result.
Exercise 8.8. Compute the dimension of the S6 -module Vλ for λ = (3, 2, 1).
Calculate the restriction of Vλ to S5 .
8.2. Symmetric polynomials in infinitely many variables over Z. Let R
be a unital commutative ring, let us define the ring SR of symmetric polynomials in
a fixed infinite sequence (Xi )i∈N>0 of variables with coefficients in R. Recall that the
symmetric group Sn acts on the polynomial ring R[X1 , . . . , Xn ] by σ(Xi ) := Xσ(i) ,
the ring of invariants consits of the symmetric polynomials in n variables. There is a
surjective algebra homomorphism which preserves the degree
ψn : R[X1 , . . . , Xn+1 ]Sn+1 → R[X1 , . . . , Xn ]Sn
P (X1 , . . . , Xn+1 ) 7→ P (X1 , . . . , Xn , 0).
By definition, SR is the projective limit of the maps (ψn )n∈N>0 in the category of
graded rings.
8. HARVEST 135
In order to be more explicit, we need to introduce the ring of formal power series
R[[X1 , . . . , Xn , . . .]]
P consisting of (possibly infinite) formal linear combinations, with
α
coefficients in R, α aα X , where α runs along multi-indices (αi )i≥1 of integers with
finite support. There is no difficulty in defining the product since, for any multi-index
α, there are only finitely many ways of expressing α into a sum α1 + α2 . We set S∞ to
be the groups of permutations of all positive integers generated by the transpositions.
Then SR is the subring of R[[X1 , . . . , Xn , . . .]] whose elements are invariant under S∞
and such that the degrees of the monomials are bounded.
Let A be the PSH algebra of rank one.
Theorem 8.9. The map ψ : A → SZ , given by, for all a ∈ A,
X Y
(6.9) ψ(a) = ha, hαi iX α ,
α i
is an algebra isomorphism.
Remark 8.10. We deduce immediately from the formula for ψ the following
statements:
P P
(1) ψ(hn ) = |α|=n X α , where |α| := i αi if α = (α1 , . . . , αi . . .),
P
(2) ψ(en ) = α=(α1 ,...) X α , where every αi is either 0 or 1 and |α| = n,
P
(3) ψ(pn ) = i≥1 Xin .
Finally, if we denote by h♦λ the dual basis of hλ with respect to the scalar
product onP A, one has
(4) ψ(h♦λ ) = α X α
where α runs along the multi-indices whose non-increasing
rearrangement is λ.
Proof. We follow the proof given in Zelevinsky’s book, attributed to Bernstein.
Let us first define the homomorphism ψ: we iterate the comultiplication A → A ⊗ A
and obtain an algebra homomorphism µn : A → A⊗n for any n (one has µ2 = m∗ ).
Furthermore, the counit ε induces a map εn : A⊗n+1 → A⊗n such that the follow-
ing diagram is commutative:
µn+1
(6.10) A CC / A⊗n+1
CC µn εn uuu u
CC u
CC uu
! uz u
A⊗n
If B is a N-graded commutative ring and t is an indeterminate, we can define
a canonical homomorphistm βB : B → B[t] by setting, for any b ∈ B of degree k,
βB (b) := btk : thus we obtain a homomorphism βA⊗n : A⊗n → A[X1 , . . . , Xn ]. Note
that, in order to obtain a homogeneous homomorphism, we have to forget the grading
of A for the definition of the degree in A[X1 , . . . , Xn ]: in this algebra, the elements
of A have degree 0. Note also that the image of µn is always contained in (A⊗n )Sn .
136 6. SYMMETRIC GROUPS, SCHUR–WEYL DUALITY AND PSH ALGEBRAS
Exercise 8.11. Show that the following diagram is commutative (the symmetric
group acts both on the set of variables and on the factors of A⊗k ):
⊗n+1
βA
(6.11) (A⊗n+1 )Sn+1 / (A⊗n+1 [X1 , . . . , Xn+1 ])Sn+1
εn
(A⊗n )Sn ⊗n
/ (A⊗n [X1 , . . . , Xn ])Sn
βA
Lemma 8.12. There are exactly two positive algebra homomorphisms from A to
Z, conjugate up to ι (see Theorem 6.11) which transform the basic primitive element
π into 1. One of them, denoted by δ, is such that δ(hi ) = 1 for all i and δ(ωλ ) = 0
whenever ωλ is not one of the hi s.
Proof. Such a homomorphism maps π 2 onto 1, but π 2 = e2 + h2 and since it is
positive, either e2 or h2 is sent to one 1 (and the other to 0). Since ι exchanges e2 and
h2 , we can assume that h2 is sent to 1 (and e2 to 0). We denote this homomorphism
by δ. Let ω be a basic element of degree n in A, distinct from hn . By Lemma 6.3,
e∗2 (ω) 6= 0, hence ω e2 π n−2 and since δ(e2 ) = 0 and δ is positive, δ(ω) = 0. Then,
since δ(π n ) = 1, we obtain δ(hn ) = 1, hence the Lemma.
Set ψn = δ ⊗n ◦ βA⊗n ◦ µn : A → Z[X1 , . . . , Xn ]. Applying Lemma 8.12 and (6.12),
we obtain
X
(6.13) ψn (a) = ha, hi1 . . . hin iX1i1 . . . Xnin .
(i1 ,...,in )∈Nn
Taking the projective limit, we get the morphism ψ : A → SZ we are looking for and
the item (4) of Remark 8.10 ensures that ψ is an isomorphism.
We now compute ψ(ωλ ) for any partition λ, and more precisely ψn (λ) for any
n ≥ |λ|.
Corollary 8.13. For any partition µ = (µ1 , . . . , µk ) set
X
Xµ = Xiµ11 . . . Xiµkk .
i1 6=i2 6=...ik
8. HARVEST 137
Then X
ψ(ωλ ) = Kλµ X µ .
|µ|=|λ|
We first introduce the following notation for the generalized Vandermonde de-
terminant: let µ = (µ1 , . . . , µn ) be a decreasing sequence of non-negative integers,
µ
we set Vµ (X1 , . . . , Xn ) = det((Xi j )1≤i,j≤n ). Notice that V(n−1,n−2,...,1,0) (X1 , . . . , Xn )
is the usual Vandermonde determinant.
Proposition 8.14. One has
V(λ1 +n−1,λ2 +n−2,...λn ) (X1 , . . . , Xn )
(6.14) ψn (ωλ ) = .
V(n−1,n−2,...,1,0) (X1 , . . . , Xn )
Exercise 8.15. Prove Proposition 8.14. Hint: Let Sλ denote the right hand side
of (6.14). Prove that X
ψn (ek )Sλ = Sµ ,
µ∈Tk (λ)
where Tk (λ) is he set of all partitions obtained from λ be adding k boxes, at most
one box in each row, satisfying the additional restriction that the number of rows of
µ is not bigger than n. Check that it is consistent with dual Pieri formula. Then
show that for any µ one can find k > 0 and λ such that Tk (λ) contains only µ and
partition less that µ in lexicographic order. Then prove the statement by induction
on lexicographic order.
8.3. Complex general linear group for an infinite countable dimensional
vector space. Let V be an infinite countable dimensional complex vector space, we
consider the group G = GL(V ). Denote by T the full subcategory of the category of
G-modules whose objects are submodules of direct sums of tensor powers of V . We
saw in section 2 that T is a semisimple category. The simple modules are indexed
by partitions and we denote by Sλ (V ) the simple module associated to the partition
λ. We denote by K(T ) the Grothendieck group of T .
Our aim is to equip K(T ) with a structure of PSH algebra of rank one.
We define the multiplication: m([M ], [N ]) = [M ⊗ N ] for M and N in T (recall
that if M ∈ T , we denote by [M ] its class in the Grothendieck group).
We define the scalar product: h[M ], [N ]i = dim HomG (M, N ), and the grading:
by convention, the degree of V ⊗n is n.
Finally we proceed to define the comultiplication m∗ , and it is a trifle more tricky.
Since V is infinite dimensional, we can choose an isomorphism ϕ : V → V ⊕ V . By
composition with ϕ, we obtain a group morphism Φ : G × G to G,
A 0 −1 A 0
Φ =ϕ ◦ ◦ ϕ.
0 B 0 B
We have two canonical projectors of V ⊕ V and we denote by V1 (resp.V2 ) the image
of the first (resp. second) one.
138 6. SYMMETRIC GROUPS, SCHUR–WEYL DUALITY AND PSH ALGEBRAS
Exercise 8.16. Show that V1⊗p ⊗ V2⊗q is a semisimple G × G-module and that
its irreducible components are of the form Sλ (V1 ) ⊗ Sµ (V2 ), where λ is a partition of
p and µ is a partition of q.
Therefore, if we denote T̃ the full sucategory of the category of G × G-modules
whose objects are submodules of direct sums of V1⊗p ⊗ V2⊗q , then its Grothendieck
group is isomorphic to K(T ) ⊗ K(T ).
Hence, the restriction functor Res (with respect to the inclusion of G × G in
G) maps the category T to the category T̃ . Therefore it induces a linear map
m∗ : K(T ) → K(T ) ⊗ K(T ).
Theorem 8.17. The Grothendieck group K(T ), equipped the operations de-
scribed above and the basis given by the classes of simple modules, is a PSH algebra
of rank one, and the basic primitive element is the class of V , [V ].
Proof. The only axiom of the definition of PSH algebras which is not straight-
forward and needs to be checked is the self-adjointness, namely the fact that m and
m∗ are mutually adjoint with respect to the scalar product. For this, we have to find
a functorial bijective map HomG (M ⊗ N, P ) → HomG×G (M ⊗ N, Res(P )) (where
M , N , P are objects of T ). Since any G-module is the direct sum of its homoge-
neous components, we may asume that M , N , P are homogeneous of degree p, q, n
respectively, with n = p + q.
For any object W ∈ T homogeneous of degree r, set ΠW := HomG (V ⊗r , W ) which
is an Sr -module; Schur-Weyl duality (see Proposition 2.12) can be reformulated in
saying that there is a canonical isomorphism of G-modules W ≃ ΠW ⊗C(Sr ) V ⊗r . We
set M1 := ΠM ⊗C(Sp ) V1⊗p ֒→ M , N2 := ΠN ⊗C(Sq ) V2⊗q ֒→ N .
Then we have an inclusion M1 ⊗ N2 ⊂ M ⊗ N , and the restriction defines a map
HomG (M ⊗ N, P ) → HomG×G (M1 ⊗ N2 , Res(P )). This is the functorial map we
where looking for.
In order to show this map is bijective, it is enough (by the semisimplicity of the
categories T and T̃ ) to check it for M = V ⊗p , N = V ⊗q and P = V ⊗n with p + q = n
indeed, on one hand, one has:
dim HomG (V ⊗p ⊗ V ⊗q , V ⊗n ) = dim HomG×G (V1⊗p ⊗ V2⊗q , V ⊗n ) = n!
the first equality coming from the Schur-Weyl duality and the second equality comes
from the formula n!
⊕ r!(n−r)!
⊗(n−r)
V ⊗n ≃ ⊕nr=0 V1⊗r ⊗ V2 .
On the other hand, the map is injective because V1⊗p ⊗ V2⊗q spans the G-module V ⊗n .
Exercise 8.18. Let V have dimension n. Show Q that
i<j (λi − λj + j − i)
dim Sλ (V ) = ψn (ωλ )(1, . . . , 1) = Q .
i<j (j − i)
8. HARVEST 139
1. Representations of quivers
A quiver is an oriented graph. For example
γ
α ε
1 + 2 s
• k • k •3
β δ
is a quiver.
In this chapter we consider only finite quivers, namely quivers with finitely many
vertices and arrows.
The underlying graph of a quiver Q is the graph obtained from Q by forgetting
the orientation of the arrows.
If Q is a quiver, we denote by Q0 the set of vertices of Q and by Q1 the set of
arrows of Q. In the example above, Q0 = {1, 2, 3} and Q1 = {α, β, γ, δ, ε}.
A quiver Q′ is a subquiver of a quiver Q if Q′0 ⊂ Q0 and Q′1 ⊂ Q1 .
γ
For every arrow γ ∈ Q1 : i → − j we define s(γ) = i as the source or tail of γ and
t(γ) = j as the target or head of γ. In the example the vertex 1 is the source of α
and the target of β.
An oriented cycle is a subgraph with vertices C0 := {s1 , . . . , sr } ⊂ Q0 and arrows
C1 = {γ1 , . . . , γr } ⊂ Q1 such that γi goes from si to si+1 if i < r and γr goes from
α
+
sr to s1 . In our example •1 k •2 is a oriented cycle. A loop is an arrow with the
β
γ
same head and tail. In our example, there is only one loop •2 .
Definition 1.1. Fix a field k. Let Q be a quiver. Consider a k-vector space
M
V = Vi
i∈Q0
GL(d1 ) and GL(d2 ) act on this space by multiplication on the left and on the right
respectively. We would like to describe all the orbits for this action.
Consider a representation (V, ρ) of Q. Choose subspaces W1 ⊂ V1 and W2 ⊂ V2
such that V1 = Ker ργ ⊕ W1 and V2 = ργ (W1 ) ⊕ W2 . Note that ργ induces an iso-
morphism α : W1 → ργ (W1 ). Then (V, ρ) is the direct sum of the subrepresentations
0 0 α
Ker ργ →
− 0, 0 −→ W2 and W1 − → ργ (W1 ). It is clear that the first representation can
0
be written as a direct sum of several copies of k −→ 0, the second one is a direct sum
0
of several copies of 0 −
→ k. These decompositions are not unique, they depend on the
α
choice of basis in Ker ργ and W2 . Finally the representation W1 − → ργ (W1 ) can be
id
written as a direct sum of several copies of k −→ k.
Therefore there are three (up to isomorphism) indecomposable representations
of Q. Their dimensions are (1, 0), (0, 1) and (1, 1). Furthermore, in every dimen-
sion there are finitely many non-isomorphic representations. Quivers with the latter
property are called quivers of finite type.
Example 1.5. Consider the quiver Q with one vertex and one loop. Then a
finite-dimensional representation of Q is a pair (V, T ), where V is a finite-dimensional
vector space and T is a linear operator in V . Isomorphism classes of representations
of this quiver are the same as conjugacy classes of n × n matrices when n is the
dimension of V . If k is algebraically closed, this classification problem amounts to
describing Jordan canonical forms of n × n matrices. In particular, indecomposable
representations correspond to matrices with one Jordan block.
If k is not algebraically closed, the problem of classifying conjugacy classes of
matrices is more tricky. This example shows that representation theory of quivers
depends very much on the base field.
Example 1.7. Now let Q be the quiver with one vertex and two loops. Repre-
sentation theory of Q is equivalent to classifying pairs of linear operators (T, S) in a
vector space V up to conjugation. In contrast with all previous examples in this case
the number of variables parametrizing indecomposable representations of dimension
n grows as n2 . We call a pair (T, S) generic if T is diagonal in some basis e1 , . . . , en
with distinct eigenvalues and the matrix of S in this basis does not have any zero
entry.
144 7. INTRODUCTION TO REPRESENTATION THEORY OF QUIVERS
Exercise 1.8. Check that if (T, S) is generic and W ⊂ V is both T -stable and
S-stable, then W = 0 or W = V . Thus, the corresponding representation of Q is
irreducible.
Therefore every generic pair of operators (T, S) gives rise to an irreducible repre-
sentation of Q. The eigenvalues of T give n distinct parameters. If T is diagonalized,
we can conjugate S by linear operators diagonal in the eigenbasis of T . Thus, we
have n2 − n parameters for the choice of S.
The situation which appears in this example is refered to as wild. There is a
precise definition of wild quivers and we refer reader to ?? for further reading on this
subject.
2. Path algebra
As in the case of groups, we can reduce the representation theory of a quiver to
the representation theory of some associative ring. In the case of groups, this ring is
the group algebra, while in the case of quivers it is the path algebra.
Definition 2.1. Let Q be a quiver. A path p is a sequence γ1 , . . . , γk of arrows
such that s (γi ) = t (γi+1 ). Set s (p) = s (γk ), t (p) = t (γ1 ). The number k of arrows
is called the length of p.
Definition 2.2. Let p1 = γ1 , . . . , γk and p2 = δ1 , . . . , δl be two paths of Q. We
define the product of p1 and p2 to be the path δ1 , . . . , δl , γ1 , . . . γk if t(γ1 ) = s(δl ) and
zero otherwise.
Next we introduce elements ei for each vertex i ∈ Q0 and define the product of
ei and ej by the formula
ei ej = δij ei .
For a path p, we set
(
p, if i = t (p)
ei p = ,
0 otherwise
(
p, if i = s (p)
pei = .
0 otherwise
The path algebra k (Q) of Q is the vector space of k-linear combinations of all
paths of Q and elements {ei }i∈Q0 , with the multiplication law obtained by extending
of the product defined above by bilinearity. P
Note that every ei , i ∈ Q0 , is an idempotents in k(Q) and that i∈Q0 ei = 1.
Example 2.3. Let Q be the quiver with one vertex and n loops then k (Q) is the
free associative algebra with n generators.
2. PATH ALGEBRA 145
Exercise 2.4. Let Q be a quiver such that the underlying graph of Q does not
contain any cycle or loop. Let Q0 = {1, . . . , n}. Show that the path algebra k (Q) is
isomorphic to the subalgebra of the matrix algebra Matn (k) generated by the subset
of elementary matrices {Eii | i ∈ Q0 }, {Eji | γ ∈ Q1 , s(γ) = i, t(γ) = j}.
In particular, show that the path algebra of the quiver
• ← • ← ··· ← •
is isomorphic to the algebra Bn of upper triangular matrices, see Example 7.19 Chap-
ter V.
Lemma 2.5. Let Q be a quiver.
(1) The path algebra k (Q) is generated by the idempotents {ei | i ∈ Q0 } and
the paths {γ | γ ∈ Q1 } of length 1.
(2) The algebra k (Q) is finite-dimensional if and only if Q does not contain an
oriented cycle.
(3) If Q is the disjoint union of two quivers Q′ and Q′′ , then k (Q) is isomorphic
to the direct product k (Q′ ) × k (Q′′ ).
(4) The path algebra has a natural Z-grading
∞
M
k (Q) = k (Q)(n) ,
n=0
where k(Q)(0) is the span of the idempotents ei for all i ∈ Q0 and k(Q)(n) is
the span of all paths of length n.
(5) For every vertex i ∈ Q0 the element ei is a primitive idempotent of k (Q),
and hence k(Q)ei is an indecomposable projective k (Q)-module.
Proof. The first four assertions are straightforward and we leave them to the
reader as an exercise. Let us prove (5).
Let i ∈ Q0 . By Exercise 7.14 Chapter V, proving (5) amounts to checking that if
ε ∈ k(Q)ei is an idempotent such that ei ε = εei = ε, then ε = ei or ε = 0. We use
the grading of k(Q) defined in (4). By definition, the left ideal k(Q)ei inherits this
grading. Hence we can write
∞
M
k(Q)ei = k (Q)(n) ei ,
n=0
where k(Q)(0) ei = kei and, for n > 0, the graded component k(Q)(n) ei is spanned
by the paths of length n with sourse at i. We can write ε = ε0 + · · · + εl with
εn ∈ k(Q)(n) ei . Since ε is an idempotent, we have ε20 = ε0 , which implies ε0 = ei or
ε0 = 0. In the latter case let εp be the first non-zero term in the decomposition of ε.
Then the first non-zero term in the decomposition of ε2 has degree no less than 2p.
This implies ε = 0. If ε0 = ei , consider the idempotent ei − ε and apply the above
argument again.
146 7. INTRODUCTION TO REPRESENTATION THEORY OF QUIVERS
M
Given a representation (V, ρ) of a quiver Q, V = Vi one can equip V with a
i∈Q0
structure of k (Q)-module in the following way
(1) The idempotent ei acts on Vj by δij IdVj .
(2) For γ ∈ Q1 and v ∈ Vi we set γv = ργ (v) if i = s(γ) and zero otherwise.
(3) We extend this action for the whole k(Q) using Lemma 2.5 (1).
Conversly, every k (Q)-module V gives rise to a representation ρ of Q when one
sets Vi = ei V .
This implies the following Theorem. 1
Theorem 2.6. The category of representations of Q over a field k is equivalent
to the category of k (Q)-modules.
Exercise 2.7. Let Q be a quiver and J(Q) be the ideal of k(Q) generated by
all arrows γ ∈ Q1 . Then the quotient k(Q)/J is a semisimple commutative ring
isomorphic to k Q0 .
Exercise 2.8. Let Q′ be a subquiver of a quiver Q. Let I(Q′ ) be the ideal of
/ Q′0 and by all γ ∈
k(Q) generated by ei for all i ∈ / Q′1 . Prove that k(Q′ ) is isomorphic
to the quotient ring k(Q)/I(Q′ ).
∞
M
Lemma 2.9. Let A = A(i) be a graded algebra and R be the Jacobson radical
i=0
of A. Then
∞
M
(1) R is a graded ideal, i.e. R = R(i) , where R(i) = R ∩ A(i) ;
i=0
(2) If u ∈ R(p) for some p > 0, then u is nilpotent.
Proof. Assume first that the ground field k is infinite. Let t ∈ k ∗ . Consider
the automorphism ϕt of A such that ϕt (u) = tp u for all u ∈ A(p) . Observe that
ϕt (R) = R. Suppose that u belongs to R and write it as the sum of homogeneous
components u = u0 + · · · + un with uj ∈ A(j) . We have to show that ui ∈ R for all
i = 1, . . . , n. Indeed,
ϕt (u) = u0 + tu1 + · · · + tn un ∈ R
for all t ∈ k ∗ . Since k is infinite, this implies ui ∈ R for all i. If k is finite, consider
the algebra A ⊗k k̄ and use the fact that R ⊗k k̄ is included in the radical of A ⊗k k̄.
Let u ∈ R(p) . Then 1 − u is invertible. Hence there exists ai ∈ A(i) , for some
i = 1, . . . , n such that
(a0 + a1 + · · · + an )(1 − u) = 1.
This relation implies a0 = 1 and apj = uj for all j > 0. Thus uj = 0 for sufficiently
large j.
1Compare with the analogous result for groups in Chapter 2.
3. STANDARD RESOLUTION AND CONSEQUENCES 147
Let us call a path p of a quiver Q a one way path if there is no path from t(p) to
s(p).
Exercise 2.10. The span of all one way paths of Q is a two-sided nilpotent ideal
in k(Q).
Lemma 2.11. The Jacobson radical of the path algebra k (Q) is the span of all
one way paths of Q.
Proof. Let N be the span of all one way paths. By Exercise 2.10 R is contained
in the radical of k(Q).
Assume now that y belongs to the radical of k(Q). Exercise 2.7 implies that y ∈
J(Q) and moreover by Lemma 2.9(2) we may assume that y is a linear combination
of paths of the same length. We want to prove that y ∈ N . Note that ei yej belongs
to the radical for all i, j ∈ Q0 . Assume that the statement is false. Then there
exists i and j such that z := ei yej is not in N , in other words there exists a path u
with source j and target i. Furthermore zu is a linear combination of oriented cycles
u1 , . . . , ul of the same length. By Lemma 2.9(2) u must be nilpotent. But it is clearly
not nilpotent. Contradiction.
Lemma 2.11 implies the following
Proposition 2.12. Let Q be a quiver which does not contain oriented cycles.
Then k(Q)/ rad k(Q) ≃ k n , where n is the number of vertices. In particular, every
simple k(Q)-module is one dimensional.
Proof. The assumption on Q implies that every path is a one way path. Hence
the radical of k(Q) is equal to J(Q).
where
f aet(γ) ⊗ v = aet(γ) γ ⊗ v − aet(γ) ⊗ γv
for all γ ∈ Q1 , v ∈ Vs(γ) , and
g (aei ⊗ v) = av
for any i ∈ Q0 , v ∈ Vi , is exact. Hence it is a projective resolution of V .
148 7. INTRODUCTION TO REPRESENTATION THEORY OF QUIVERS
Since all Aei for i ∈ Q0 are homogeneous left ideals of A, there are induced gradings
X = ⊕p≥0 X(p) and Y = ⊕p≥0 Y(p) . Define f0 : X → Y and f1 : X → Y by
f1 aet(γ) ⊗ v = aet(γ) γ ⊗ v, f0 aet(γ) ⊗ v = aet(γ) ⊗ γv.
Note that for any p ≥ 0 we have f1 (X(p) ) ⊂ X(p+1) and f0 (X(p) ) ⊂ X(p) . Moreover, it
is clear from the definition that f1 is injective. Since f = f1 − f0 , we obtain that f
is injective by a simple argument on gradings.
It remains to prove that Im f = Ker g.
Exercise 3.3. Show that for any p > 0 and y ∈ Y(p) there exists y ′ ∈ Y(p−1) such
that y ′ ≡ y mod Im f . (Hint: it suffices to check the statement for x = u ⊗ v where
v ∈ Vi and u is a path of length p with source i).
The exercise implies that for any y ∈ Y there exists y0 ∈ Y(0) such that y ≡ y0
mod Im f . Let y ∈ Ker g, then y0 ∈ Ker g. But g restricted to Y 0 is injective. Thus,
y0 = 0 and y ∈ Im f .
by the formula
(7.2) dφ (x) = φ (γx) − γφ (x)
for any γ ∈ Q1 , x ∈ Xs(γ) and φ ∈ Homk Xs(γ) , Ys(γ) . Theorem 3.1 implies that
Ext1 (X, Y ) is isomorphic to the cokernel of the map d.
3. STANDARD RESOLUTION AND CONSEQUENCES 149
where the second equality follows from calculating Euler characteristic in (7.1). The
symmetric form
(x, y) := hx, yi + hy, xi
4. BRICKS 151
is called the Tits form of the quiver Q. We also consider the corresponding quadratic
form
q (x) := hx, xi .
4. Bricks
Here we discuss further properties of finite-dimensional representations of a path
algebra k (Q). In the rest of this chapter we assume that the ground field k
is algebraically closed and all representations are finite-dimensional.
Definition 4.1. A k(Q)-module X is a brick, if EndQ (X) = k.
Exercise 4.2. If X is a brick, then X is indecomposable. If X is indecomposable
and Ext1 (X, X) = 0, then X is a brick (by Lemma 3.6).
Example 4.3. Consider the quiver • → •. Then every indecomposable represen-
tation is a brick.
For the Kronecker quiver • ⇒ • the representation k 2 ⇒ k 2 with ργ1 = Id,
ργ2 = (01 01
00 ) is not a brick because ϕ = (ϕ1 , ϕ2 ) with ϕ1 = ϕ2 = (00 ) is a non-scalar
element in EndQ (X).
Lemma 4.4. Let X be an indecomposable k(Q)-module which is not a brick.
Then X contains a brick W such that Ext1 (W, W ) 6= 0.
Proof. We will prove the lemma by induction on the length l of X. The base
case l = 1 is trivial, since in this case X is irreducible and hence a brick by the Schur
lemma.
Recall that if X is indecomposable and has finite length, then ϕ ∈ EndQ (X)
is either isomorphism or nilpotent. Therefore, since k is algebraically closed and
X is not a brick, the algebra EndQ (X) contains a non-zero nilpotent element. Let
ϕ ∈ EndQ (X) be a non-zero operator of minimal rank. Then ϕ is nilpotent and
rk ϕ2 < rk ϕ, hence ϕ2 = 0.
Let Y := Im ϕ, Z := Ker ϕ. Clearly, Y ⊂ Z. Consider a decomposition
Z = Z1 ⊕ · · · ⊕ Zp
into a sum of indecomposable submodules. Denote by pi the projection Z → Zi .
Let i be such that pi (Y ) 6= 0. Set η := pi ◦ ϕ, Yi := pi (Y ) = η(Z). Note that by
our assumption rk η = rk ϕ, therefore Yi is isomorphic to Y . Let Yi = pi (Y ). Then
Ker η = Z and Im η = Yi .
Note that the exact sequence
η
0→Z→X−
→ Yi → 0
does not split since X is indecomposable. Let Xi be the quotient of X by the
submodule ⊕j6=i Zj and π : X → Xi be the canonical projection. Then we have the
exact sequence
η̄
0 → Zi → X i −
→ Yi → 0,
152 7. INTRODUCTION TO REPRESENTATION THEORY OF QUIVERS
where η̄ := η ◦ π −1 is well define since Ker π ⊂ Ker η. We claim that (7.6) does not
split. Indeed, if it splits, then Xi decomposes into a direct sum Zi ⊕ L for some
submodule L ⊂ Xi which is isomorphic to Yi . But then X = Zi ⊕ π −1 (L), which
contradicts indecomposability of X.
Therefore we have shown that Ext1 (Yi , Zi ) 6= 0. Recall that Yi is a submodule of
Zi . By Corollary 3.5 we have the surjection
Ext1 (Zi , Zi ) → Ext1 (Yi , Zi ) .
Hence Ext1 (Zi , Zi ) 6= 0.
The length of Zi is less than the length of X. If Zi is not a brick, then it contains
a brick W by the induction assumption.
Corollary 4.5. Assume that Q is a quiver such that its Tits form is positive
definite. Then every indecomposable representation X of Q is a brick with trivial
Ext1 (X, X). Moreover, if x = dim X, then q (x) = 1.
Proof. Assume that X is not a brick, then it contains a brick Y such that
Ext1 (Y, Y ) 6= 0. Then
q (y) = dim EndQ (Y ) − dim Ext1 (Y, Y ) = 1 − dim Ext1 (Y, Y ) ≤ 0,
but this is impossible. Therefore X is a brick. Then
q (x) = dim EndQ (X) − dim Ext1 (X, X) = 1 − dim Ext1 (X, X) ≥ 0.
By positivity of q we have q (x) = 1 and dim Ext1 (X, X) = 0.
Note that X X
dim Rep (x) = xs(γ) xt(γ) , dim G = x2i ,
γ∈Q1 i∈Q0
therefore
(7.6) dim Rep (x) − dim G = −q (x) .
Let us formulate without proof certain properties of G-action on Rep(x). They
follow from the general theory of algebraic groups, see for instance Humphreys We
work in Zariski topology.
• Each orbit is open in its closure;
• if O and O′ are two distinct orbits and O′ belongs to the closure of O, then
dim O′ < dim O;
• If (X, ρ) is a representation of Q, then dim OX = dim G − dim StabX , where
StabX denotes the stabilizer of ρ.
Lemma 5.1. For any representation (X, ρ) of dimension x we have
dim StabX = dim AutQ (X) = dim EndQ (X) .
Proof. The condition that φ ∈ EndQ (X) is not invertible is given by the poly-
nomial equations Y
det φi = 0.
i∈Q0
Since AutQ (X) is not empty and open in EndQ (X), we obtain that AutQ (X) and
EndQ (X) have the same dimension.
Corollary 5.2. If (X, ρ) is a representation of Q and dim X = x, then
codim OX = dim Rep (x)−dim G+dim StabX = −q (x)+dim EndQ (X) = dim Ext1 (X, X) .
Lemma 5.3. Let (Z, τ ) be a nontrivial extension of (Y, σ) by (X, ρ), i.e. there is
a non-split exact sequence
0 → X → Z → Y → 0.
Then OX⊕Y belongs to the closure of OZ and OX⊕Y 6= OZ .
Proof. Following Section 3.2 for every i ∈ Q0 consider a decomposition Zi =
Xi ⊕ Yi such that for every γ ∈ Q1 and (x, y) ∈ Xs(γ) ⊕ Ys(γ)
τγ (x, y) = (ργ (x) + ψγ (y), σγ (y))
for some ψγ ∈ Hom(Ys(γ) , Xt(γ) ).
Next, for every λ ∈ k \ 0 define g λ ∈ G by setting for every i ∈ Q0
giλ |Xi = IdXi , giλ |Yi = λ IdYj .
Then we have
g λ τγ (x, y) = (ργ (x) + λψγ (y), σγ (y)).
154 7. INTRODUCTION TO REPRESENTATION THEORY OF QUIVERS
The latter formula makes sence even for λ = 0 and g 0 τ lies in the closure of {g λ τ | λ ∈
k \ 0}. Furthermore g 0 τ is the direct sum X ⊕ Y . Hence OX⊕Y belongs to the closure
of OZ .
It remains to check that X ⊕ Y is not isomorphic to Z. This follows immediately
from the inequality
dim HomQ (Y, Z) < dim HomQ (Y, X ⊕ Y ) .
The following corollary is straightforward.
Corollary 5.4. If the orbit OX is closed in Rep(x), then X is semisimple.
m
M
Corollary 5.5. Let (X, ρ) be a representation of Q and X = Xj be a
j=1
decomposition into the direct sum of indecomposable submodules. If OX is an orbit
of maximal dimension in Rep(x), then Ext1 (Xi , Xj ) = 0 for all i 6= j.
Proof. If Ext1 (Xi , Xj ) 6= 0, then by Lemma 5.3 we can construct a representa-
tion (Z, τ ) such that OX is in the closure of OZ . Then dim OX < dim OZ .
6. Coxeter–Dynkin and affine graphs
6.1. Definition and properties. Let Γ be a connected non-oriented graph with
vertices Γ0 and edges Γ1 . We define the Tits form (·, ·) on ZΓ0 by
X X
(x, y) := (2 − 2l(i))xi yi − xi yj ,
i∈Γ0 (i,j)∈Γ1
where l(i) is the number of loops at i. If we equip all edges of Γ with orientation
then the symmetric form coincides with the introduced earlier symmetric form of the
corresponding quiver. We define the quadratic form q on ZΓ0 by
(x, x)
q(x) := .
2
By {ǫi | i ∈ Γ0 } we denote the standard basis in ZΓ0 . If Γ does not have loops,
then (ǫi , ǫi ) = 2 for all i ∈ Γ0 . If i, j ∈ Γ0 and i 6= j, then (ǫi , ǫj ) equals minus the
number of edges between i and j. The matrix of the form (·, ·) in the standard basis
is called the Cartan matrix of Γ.
2−1
Example 6.1. The Cartan matrix of • − • is −12 . The Cartan matrix of the
loop is (0).
Definition 6.2. A connected graph Γ is called Coxeter–Dynkin if its Tits form
(·, ·) is positive definite and affine if (·, ·) is positive semidefinite but not positive
definite. If Γ is neither Coxeter–Dynkin nor affine, then we say that it is of indefinite
type.
6. COXETER–DYNKIN AND AFFINE GRAPHS 155
Remark 6.3. For affine graph Γ the form (·, ·) is necessarily degenerate. Fur-
thermore
(7.7) Ker(·, ·) = {x ∈ ZQ0 | (x, x) = 0}.
Lemma 6.4. (a) If Γ is affine then the kernel of (·, ·) equals Zδ for some δ ∈ NΓ0
with all δi > 0.
(b) If Γ is of indefinite type, then there exists x ∈ NΓ0 such that (x, x) < 0.
Proof. Let x ∈ ZQ0 . We define supp x to be the set of vertices i ∈ Q0 such
that xi 6= 0. Let |x| be defined by the condition |x|i = |xi | for all i ∈ Q0 . Note that
supp x = supp |x| and by the definition of (·, ·) we have
(7.8) (|x|, |x|) ≤ (x, x).
To prove (b) we just notice that if Γ is of indefinite type then there exists x ∈ ZQ0
such that (x, x) < 0. But then (7.8) implies (|x|, |x|) < 0.
Now let us prove (a). Let δ ∈ Ker(·, ·) and δ 6= 0. Then (7.8) and (7.7) imply
that |δ| also lies in Ker(·, ·). Next we prove that supp δ = Q0 . Indeed, otherwise we
can choose i ∈ Q0 \ supp δ such that i is connected with at least one vertex in supp δ.
Then (ǫi , δ) < 0, therefore
(ǫi + 2δ, ǫi + 2δ) = 2 + 4(ǫi , δ) < 0
and Γ is not affine.
Finally let δ ′ , δ ∈ Ker(·, ·). Since supp δ = supp δ ′ = Q0 , one can find a, b ∈ Z
such that supp(aδ + bδ ′ ) 6= Q0 . Then by above aδ + bδ ′ = 0. Hence Ker(·, ·) is
one-dimensional and the proof of (a) is complete.
Note that (a) implies the following
Corollary 6.5. Let Γ be Coxeter–Dynkin or affine. Any proper connected
subgraph of Γ is Coxeter–Dynkin.
Definition 6.6. A non-zero vector x ∈ ZQ0 is called a root if q (x) ≤ 1. Note for
every i ∈ Q0 , ǫi is a root. It is called a simple root.
Exercise 6.7. Let Γ be a connected graph. Show that the number of roots is
finite if and only if Γ is a Coxeter–Dynkin graph.
Lemma 6.8. Let Γ be Coxeter–Dynkin or affine. If x is a root, then either all
xi ≥ 0 or all xi ≤ 0.
Proof. Assume that the statement is false. Let
X
I + := {i ∈ Q0 | xi > 0}, I − := {i ∈ Q0 | xi < 0}, x± = x i ǫi .
i∈I ±
We call a root x positive (resp. negative) if xi ≥ 0 (resp. xi ≤ 0) for all i ∈ Q0 .
6.2. Classification. The following are all Coxeter–Dynkin graphs (below n is
the number of vertices).
An r r r ... r
Dn r r r ... r
E6 r r r r r
E7 r r r r r r
E8 r r r r r r r
r
6. COXETER–DYNKIN AND AFFINE GRAPHS 157
The affine graphs, except the loop • , are obtained from the Coxeter–Dynkin
graphs by adding a vertex (see Corollary 6.5). Here they are.
Â1 r r
For n > 1, Ân is a cycle with n + 1 vertices. In this case δ = (1, . . . , 1).
In what follows the numbers are the coordinates of δ.
1 2 2 2 1
D̂n r r r ... r r
1r 1r
1 2 3 2 1
Ê6 r r r r r
2r
1r
1 2 3 4 3 2 1
Ê7 r r r r r r r
2r
2 4 6 5 4 3 2 1
Ê8 r r r r r r r r
3r
158 7. INTRODUCTION TO REPRESENTATION THEORY OF QUIVERS
The proof that the above classification is complete is presented below in the
exercises.
Exercise 6.9. Check that An , Dn , E6 , E7 , E8 are Coxeter–Dynkin using the Sylvester
criterion and the fact that every subgraph of a Coxeter–Dynkin graph is Coxeter–
Dynkin. One can calculate the determinant of a Cartan matrix inductively. It is
n + 1 for An , 4 for Dn , 3 for E6 , 2 for E7 and 1 for E8 .
Exercise 6.10. Check that the Cartan matrices of Ân , D̂n , Ê6 , Ê7 , Ê8 have corank
1 and every proper connected subgraph is Coxeter–Dynkin. Conclude that these
graphs are affine.
Exercise 6.11. Let Γ be a Coxeter–Dynkin graph. Using Corollary 6.5 prove
that Γ does not have loops, cycles and multiple edges. Prove that Γ has no vertices
of degree 4 and at most one vertex of degree 3.
Exercise 6.12. Let a Coxeter–Dynkin graph Γ have a vertex of degree 3. Let
p, q and r be the lengths of “legs” coming from this vertex. Prove that p1 + 1q + 1s > 1.
Use this to complete classification of Coxeter–Dynkin graphs.
Exercise 6.13. Complete classification of affine graphs using Corollary 6.5, Ex-
ercise 6.10 and Exercise 6.12.
7. Quivers of finite type and Gabriel’s theorem
Recall that a quiver is of finite type if it has finitely many isomorphism classes of
indecomposable representations.
Exercise 7.1. Prove that a quiver is of finite type if and only if all its connected
components are of finite type.
Theorem 7.2. (Gabriel) Let Q be a connected quiver and Γ be its underlying
graph. Then
(1) The quiver Q has finite type if and only if Γ is a Coxeter–Dynkin graph.
(2) Assume that Γ is a Coxeter–Dynkin graph and (X, ρ) is an indecomposable
representation of Q. Then dim X is a positive root.
(3) If Γ is a Coxeter–Dynkin graph, then for every positive root x ∈ ZQ0 there
is exactly one indecomposable representation of Q of dimension x.
Proof. Let us first prove that if Q is of finite type then Γ is a Coxeter–Dynkin
graph. Indeed, if Q is of finite type, then for every x ∈ NQ0 , Rep (x) has finitely
many G-orbits. Therefore Rep (x) must contain an open orbit. Assume that Q is
not Coxeter–Dynkin. Then there exists a non-zero x ∈ ZQ0 such that q (x) ≤ 0. Let
OX ⊂ Rep (x) be an open orbit. Then codim OX = 0. But by Corollary 5.2
(7.9) codim OX = dim EndQ (X) − q (x) > 0.
This is a contradiction.
7. QUIVERS OF FINITE TYPE AND GABRIEL’S THEOREM 159