text stringlengths 16 3.88k | source stringlengths 60 201 |
|---|---|
). We first describe the physical motivation and then state the mathematical problem.
Imagine a material which occupies a certain region U in the physical space V = R3 (a space
with a positive definite inner product). Suppose the material is deformed. This means, we have
U �. The question in elasticity theory
applie... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
part on the opposite side. It is easy to deduce
from Newton’s laws that Fv is linear in v, so there exists a linear operator SP : V
V such that
Fv = SP v. It is called the stress tensor.
⊃
||
||
v
An elasticity law is an equation SP = f (dP ), where f is a function. The simplest such law is a
linear law (Hooke’s ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
and for x
real numbers K, µ.
�
�
In fact, it is clear from physics that K, µ are positive. Physically, the compression modulus K
characterises resistance of the material to compression or dilation, while the shearing modulus µ
characterizes its resistance to changing the shape of the object without changing its vol... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
cases, let B be the invariant bilinear form on V , and
(, ) the invariant positive Hermitian form (they are defined up to a nonzero complex scalar and a
V such that B(v, w) = (v, jw).
positive real scalar, respectively), and define the operator j : V
ij), and j 2 = ∂ Id, where ∂ is a real number, positive in
Show t... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
, all five irreducible representations
, C2 , C3
tions C+, C
−
3 , C3 , C4 , C−5 are of real type. As for Q8, its one-dimensional representations are of
of A5 – C, C+
real type, and the two-dimensional one is of quaternionic type.
⊃
−
−
−
Definition 4.3. The Frobenius-Schur indicator F S(V ) of an irreducible represen... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
V (
−
G
|
|
g 2) = νS2V (P )
�
G
g
�
ν
−
2V (P ) = dim(S2V )G
−
√
dim(
√
2V )G
=
Finally, the number of involutions in G equals
1,
1,
0,
−
if V is of real type
if V is of quaternionic type
if V is of complex type
⎞
⎟
⎠
1
G
|
�
V
|
dim V νV (
g 2) =
dim V
�
G
g
�
�
real V
dim V.
−
�
quat V
Corollary 4.5. Assu... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
by Frobenius, became the starting
point for creation of representation theory (see [Cu]).
Theorem 4.7.
r
det XG =
Pj (x)deg Pj
�
j=1
for some pairwise non-proportional irreducible polynomials Pj (x), where r is the number of conju
gacy classes of G.
We will need the following simple lemma.
Lemma 4.8. Let Y be a... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
ke’s theorem, we have
r
detV L(x) =
(detVi L(x))dim Vi ,
�
i=1
where Vi are the irreducible representations of G. We set Pi = detVi L(x). Let
and Ei,jk
End Vi be the matrix units in these bases. Then
{
is a basis of C[G] and
eim}
be bases of Vi
Ei,jk}
{
�
L(x) Vi =
|
�
j,k
yi,jkEi,jk,
where yi,jk are new ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
coefficients).
To show (4.9)
(4.10), suppose z is a root of
≥
(4.9), notice that z is a root of the characteristic polynomial of the matrix
≥
p(x) = x n + a1x n
−
1 + . . . + an
1x + an.
−
Then the characteristic polynomial of the following matrix (called the companion matrix) is
p(x):
49
0 0 0
1 0 ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
� B
The corresponding eigenvector is in both cases v
w. This shows that both A and Q are rings.
To show that the latter is a field, it suffices to note that if ϕ = 0 is a root of a polynomial p(x) of
1 is a root of x p(1/x). The last statement is easy, since a number ϕ is algebraic
degree d, then ϕ−
if and only if it... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
conjugates of ϕ; they are roots of any polynomial q with rational coefficients such that q(ϕ) = 0.
50
⇒
⇒
Note that any algebraic conjugate of an algebraic integer is obviously also an algebraic inte
ger. Therefore, by the Vieta theorem, the minimal polynomial of an algebraic integer has integer
coefficients.
... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
Hint. Consider the representations of G over the field Q of algebraic numbers.
(b) Show that if V is an irreducible complex representation of a finite group G of dimension
> 1 then there exists g
G such that νV (g) = 0.
�
Hint: Assume the contrary. Use orthonormality of characters to show that the arithmetic mean
1... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
and derive a contradiction.
g=1
|
|
�
�⊃
4.4 Frobenius divisibility
Theorem 4.16. Let G be a finite group, and let V be an irreducible representation of G over C.
Then
Proof. Let C1, C2, . . . , Cn be the conjugacy classes of G. Set
dim V divides G .
|
|
∂i = νV (gCi ) |
Ci|
dim V
,
where gCi is a representative o... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
gC|
|
i
= e in G, the eigenvalues of δ(gCi ) are roots of unity), and
(iii) A is a ring (Proposition 4.12).
On the other hand, from the definition of ∂i,
∂iνV (gCi ) =
�
Ci
�
i
Ci|
|
νV (gCi )νV (gCi )
.
dim V
Recalling that νV is a class function, this is equal to
νV (g)νV (g)
dim V
�
G
g
�
= |
G
(νV , ν... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
mathematician William Burnside
in the early 20-th century, using representation theory (see [Cu]). Here is this proof, presented in
modern language.
Before proving Burnside’s theorem we will prove several other results which are of independent
interest.
Theorem 4.21. Let V be an irreducible representation of a fini... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
of unity, so
νV (g) is an algebraic integer. Since
νV (g) is an algebraic integer. Also, by Proposition 4.17, n |
gcd(n, C ) = 1, there exist integers a, b such that a C + bn = 1. This implies that
C
1
|
|
|
|
|
νV (g)
n
= (π1 + . . . + πn).
1
n
is an algebraic integer. Thus, by Lemma 4.22, we get that either π1... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
N such that νV (g) = 0.
�
Proof. If V
�
D, the number 1 dim(V )νV (g) is an algebraic integer, so
p
a =
1
p
�
D
V
�
dim(V )νV (g)
is an algebraic integer.
Now, by (4), we have
0 = νC(g) +
dim V νV (g) +
dim V νV (g) = 1 + pa +
dim V νV (g).
�
V
D
�
�
V
N
�
�
N
V
�
This means that the last summand is... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
qb, we see that there has
to be more than one conjugacy class consisting just of one element. So G has a nontrivial center,
which gives a contradiction.
⊂
a
4.6 Representations of products
Theorem 4.25. Let G, H be finite groups,
field k (of any characteristic), and
irreducible representations of G
Wj
}
H over k... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
, so ni = +1 and we are done.
= 1, meaning that ni =
2
i . So
2
i ni
i n
⎨
⎨
⎨
±
niVi. Then by
1 for exactly
4.8
Induced Representations
Given a representation V of a group G and a subgroup H G, there is a natural way to construct
H V is the representation given
a representation of H. The restricted represent... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
�)(f )(x) for any g, g�, x
G and h
H.
�
�
Remark 4.30. Notice that if we choose a representative xε from every right H-coset ε of G, then
any f
is uniquely determined by
IndG
.
f (xε)
H V
�
{
}
Because of this,
dim(IndG
H V ) = dim V
.
G
|
H
|
|
|
·
Problem 4.31. Check that if K H
is isomorphic to IndG
→... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
�
�
G:xgx−1
x
�
Proof. For a right H-coset ε of G, let us define
Vε
=
f
{
IndG
H V
|
�
f (g) = 0
g
⊕
ε
.
}
⇒�
Then one has
and so
IndG =
H V
Vε,
�
ε
ν(g) =
νε(g),
�
ε
55
where νε(g) is the trace of the diagonal block of δ(g) corresponding to Vε.
Since g(ε) = εg is a right H-coset... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
representation of G and W a representation of H. Then
→G
H W ) is naturally isomorphic to HomH (ResG
H V, W ).
Proof. Let E = HomG(V, IndG
H W ) and E�
E as follows: F (ϕ)v = (ϕv)(e) for any ϕ
= HomH (ResG
⊃
E and (F �(α)v)(x) = α(xv) for any α
H V, W ). Define F : E
E
and F �
�
E �.
: E
�
⊃
�
�
are well de... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
(g(F �(α)v))(x).
(d) F
∞
F �
= IdE⊗ .
This holds since F (F �(α))v = (F �(α)v)(e) = α(v).
(e) F �
∞
F = IdE , i.e., (F �(F (ϕ))v)(x) = (ϕv)(x).
Indeed, (F �(F (ϕ))v)(x) = F (ϕxv) = (ϕxv)(e) = (xϕv)(e) = (ϕv)(x), and we are done.
Exercise. The purpose of this exercise is to understand the notions of restricted and i... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
Then, V is a left k [G]-module, thus a (k [G] , k)-bimodule.
Thus, the tensor product k [G]1 �k[G] V is a (k [H] , k)-bimodule, i. e., a left k [H]-module. Prove
that this tensor product is isomorphic to ResG
⊃
k [G]1 �k[G] V
H V as a left k [H]-module. The isomorphism ResG
�k[G] v for every v
is given by v
ResG... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
Then, W is a left k [H]-module, thus a (k [H] , k)
, W ), but also isomorphic to
bimodule. Show that IndG
k [G]2�k[H]W is given by f
k [G]2�k[H]W . The isomorphism Homk[H] (k [G]1 , W )
f (g) for every f
H-cosets in G. (This isomorphism is independent of the choice of representatives.)
H W is not only isomorphi... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
�
�
⎩
⊃
k given by
IndG
⊕
as representations of
H (V ⊕) as Homk[H] (k [G]1 , V ⊕). Prove
(f (Sx)) (v) is
k [G] is the linear map defined by
H V
�
⎩
x
f,
�k[H] v
⎩
�⊃
��
⎩
�
4.11 Examples
Here are some examples of induced representations (we use the notation for representations from
the character tables).
... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
of Sn
In this subsection we give a description of the representations of the symmetric group S n for any
n.
Definition 4.35. A partition ∂ of n is a representation of n in the form n = ∂1 + ∂2 + ... + ∂p,
where ∂i are positive integers, and ∂i
∂i+1.
⊂
∗
∗
To such ∂ we will attach a Young diagram Y�, which is the ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
Set c� = a�b�. Since P�
�
Q�
g
�
b� :=
1)gg,
−
(
where (
element is nonzero.
−
Q� =
, this
1
}
{
∈
The irreducible representations of Sn are described by the following theorem.
Theorem 4.36. The subspace V� := C[Sn]c� of C[Sn] is an irreducible representation of Sn under
left multiplication. Every irreducible re... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
representations of Sn are real, i.e., admit a nondegenerate
invariant symmetric form. Then use the Frobenius-Schur theorem.
4.13 Proof of Theorem 4.36
Lemma 4.40. Let x
�
C[Sn]. Then a�xb� = σ�(x)c�, where σ� is a linear function.
P�Q�, then g has a unique representation as pq, p
1)qc�.
Proof. If g
Thus, to p... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
ent columns of T �, so there exists q1� �
have
which moves all these elements to the first row. So there is p1
P� such that p1T and q1� T �
the same first row. Now do the same procedure with the second row, finding elements p2, q2� such
have the same first two rows. Continuing so, we will construct the desired
that ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
��nd elements p1
same first row, and repeat the argument for the second row, and so on. Eventually, having done
1 such steps, we’ll have ∂i > µi, which means that some two elements of the i-th row of the first
i
−
tableau are in the same column of the second tableau, completing the proof.
1 such that p1T and q1� T ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
to prove Theorem 4.36. Let ∂
µ. Then by Lemmas 4.42, 4.43
⊂
HomSn (V�, Vµ) = HomSn (C[Sn]c�, C[Sn]cµ) = c�C[Sn]cµ.
The latter space is zero for ∂ > µ by Lemma 4.41, and 1-dimensional if ∂ = µ by Lemmas 4.40
and 4.42. Therefore, V� are irreducible, and V� is not isomorphic to Vµ if ∂ = µ. Since the number
of parti... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
the character of U�. Let Ci be the conjugacy class in Sn having il cycles
1 (here i is a shorthand notation for (i1, ..., il, ...)). Also let x1, ..., xN be
of length l for all l
variables, and let
⊂
be the power sum polynomials.
Hm(x) =
m
xi
�
i
Theorem 4.46. Let N
�j in the polynomial
of x� :=
xj
⊂
p (wh... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
=
Zg|
|
�
m
m im im!,
νU� (Ci) = m
⎛
mim im!
j ∂j!
|
Ci
.
P�|
∈
m Sim ∼ (Z/mZ)im , so
⎛
Now, since P� =
j S�j , we have
⎛
⎛
Ci
P�|
∈
|
=
r
� �
1
j
∧
∂j !
1 mrjm rjm!
,
m
∧
⎛
where r = (rjm) runs over all collections of nonnegative integers such that
mrjm = ∂j ,
rjm = im.
�
m
�
j
Indeed, an element ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
�
p. Then νV� (Ci) is the coefficient of x�+χ :=
�j +N
xj
j
−
in the polyno
�(x)
Hm(x)im .
�
1
m
∧
⎛
Remark. Here is an equivalent formulation of Theorem 4.47: νV� (Ci) is the coefficient of x�
in the (Laurent) polynomial
1
−
��
i<j
Hm(x)im .
i � �
1
m
x
j
x
∧
61
Proof. Denote... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
µ > ∂ or µ = ∂, and the case µ = ∂ arises only if
of the form ei
−
ε = 1, as desired.
ε(δ)
�
∂ + δ
ε(χ)
−
−
−
−
◦
◦
�
�
Therefore, to show that χ� = ν�, by Lemma 4.27, it suffices to show that (χ�, χ�) = 1.
We have
Using that
(χ�, χ�) =
1
!
n
�
i
χ�(Ci)2 .
Ci
|
|
=
Ci
|
|
n!
,
m mim im!
we conclude that (χ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
xiyj)
−
yj )
.
(zj
zi)(yi
−
i,j(zi
−
yj)
−
yj )
= det(
zi
1
−
).
yj
⎛
(zi−
Proof. Multiply both sides by
yj). Then the right hand side must vanish on the hyperplanes
zi = zj and yi = yj (i.e., be divisible by �(z)�(y)), and is a homogeneous polynomial of degree
1). This implies that the right hand s... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
This is a partial order, and µ
�
⇔
−
∂ is a sum of
∂ implies
⇔
∂. It follows from Theorem 4.47 and its proof that
vectors of the form ei
µ
−
⊂
This implies that the Kostka numbers Kµ� vanish unless µ
�
∂.
⇔
ν� =
�µ
≥
�K
µ�νUµ .
4.16 Problems
In the following problems, we do not make a distinction between Young... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
1
and n
1.
−
−
Hint. Represent E as Cn
Cn
−
−
1, where Cn = C is the element from Problem 4.51.
(b) Show that the element (12) + ... + (1n) acts on V� by a scalar if and only if ∂ is a rectangular
Young diagram, and compute this scalar.
4.17 The hook length formula
Let us use the Frobenius character formula t... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
)
�
1 i<j N
→
→
(5)
⇒
(where N
p).
⊂
In this formula, there are many cancelations. After making some of these cancelations, we
∂ j),
j.
1, i
obtain the hook length formula. Namely, for a square (i, j) in a Young diagram ∂ (i, j
⊂
∗
= i, j�
= j or i�
define the hook of (i, j) to be the set of a... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
, and B = EndA E. Then:
(i) A = EndB E (i.e., the centralizer of the centralizer of A is A);
(ii) B is semisimple;
(iii) as a representation of A
Wi, where Vi are all the
B, E decomposes as E =
irreducible representations of A, and Wi are all the irreducible representations of B. In particular,
we have a natur... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
�
1 + 1
�
b
�
...
�
1 + ... + 1
�
1
�
...
�
b,
gl(V ).
b
�
Proof. Clearly, the image of U(gl(V )) is contained in B, so we just need to show that any element
of B is contained in the image of U(gl(V )). By definition, B = S n End V , so the result follows from
part (ii) of the following lemma.
64
... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
which goes under the name “Schur-Weyl duality”.
Theorem 4.57. (i) The image A of C[Sn] and the image B of U(gl(V )) in End(V �
izers of each other.
n) are central
(ii) Both A and B are semisimple. In particular, V �
n is a semisimple gl(V )-module.
(iii) We have a decomposition of A
L�, where the summation
B-mo... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
ducible representations of GL(V ) or zero.
GL(V ), V �
n decomposes as
��V�
L�, where
�
Example 4.60. If ∂ = (n) then L� = SnV , and if ∂ = (1n) (n copies of 1) then L� =
shown in Problem 3.19 that these representations are indeed irreducible (except that
if n > dim V ).
√
√
nV . It was
nV is zero
65
4.... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
Proposition 4.62.
Therefore,
S�(1, z, z 2 , ..., z N
1) =
−
S�(1, ..., 1) =
�
1 i<j N
→
→
∂i
z�i
i
−
i
z−
j
z�j
−
j
z−
−
−
−
∂j + j
i
j
i
−
−
�
1 i<j N
→
→
Proof. The first identity is obtained from the definition using the Vandermonde determinant. The
second identity follows from the first one by set... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
of ∂. If N
Therefore, the dimension of L� is given by the formula
⊂
dim L� =
∂i
�
1 i<j N
→
→
i
−
−
∂j + j
i
j
−
This shows that irreducible representations of GL(V ) which occur in V �
by Young diagrams with any number of squares but at most N = dim V rows.
n for some n are labeled
Proposition 4.64. The ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
�1, ..., ∂N ) of integers (not necessarily positive) such that ∂1
∂N . This sequence is called
the highest weight of L�.
r 1N := L�
·
N V ⊕)�
...
�
⊂
⊂
√
(
Theorem 4.66. (i) Every finite dimensional polynomial representation of GL(V ) is completely
reducible, and decomposes into summands of the form L� (which ar... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
we may assume that Y is contained in a quotient of a (finite)
form Sr(V
(
√
N V ⊕, Y is contained in a direct sum of
direct sum of such representations. As V ⊕
N V ⊕)�
representations of the form V �
s, and we are done.
N V ⊕)�
V ⊕)
=
� √
1V
Y
⊃
�
→
�
�
�
√
(
n
N
−
�
√
(ii) Let Y be a polynomial representa... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
by
scalars, the above results extend in a straightforward manner to representations of the Lie algebra
sl(V ). Similarly, the results for GL(V ) extend to the case of the group SL(V ) of operators with
determinant 1. The only difference is that in this case the representations L� and L�+1m are
∂N
isomorphic, so th... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
�ecting the Young diagram of ∂.
�
�
−
1)ss for any permutation s.
. Show also that θ(C[Sn]a) = C[Sn]θ(a),
is the conjugate partition to ∂,
−
�
−
= V�� , where ∂⊕
C
Problem 4.69. Let Rk,N be the algebra of polynomials on the space of k-tuples of complex N by N
matrices X1, ..., Xk, invariant under simultaneous c... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
G
|
GL2(Fq)
|
|
= q(q + 1)(q
1)2 .
−
68
The goal of this section is to describe the irreducible representations of G.
To begin, let us find the conjugacy classes in GL2(Fq).
Representatives
Scalar
0
x
0 x
�
⎩
Parabolic
1
x
0 x
�
⎩
Hyperbolic
0
x
0 y
, y = x
�
⎩
x ζy
y x
Elliptic
Fq,
, x
�
F... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
Let A be such
a matrix, and consider a quadratic extension of Fq,
Over this field, A will have eigenvalues
and
with corresponding eigenvectors
Fq(∀π), π
Fq
�
F2 .q
\
ϕ = ϕ1 + ∀πϕ2
ϕ = ϕ1
−
∀πϕ2,
v, v
(Av = ϕv, Av = ϕv).
Choose a basis
In this basis, the matrix A will have the form
e1 = v + v, e2 = ∀π(v
... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
only the first two matrices are commutators. But it is easy to see that the matrix
is the commutator of the matrices
1 1
0 1
�
�
while the matrix
A =
1 1/2
1
0
�
�
, B =
1
0
�
0
1
�
−
,
a
0
1
0 a−
�
�
is the commutator of the matrices
A =
a 0
0 1
�
�
, B =
0 1
1 0
�
,
�
This completes the proof.
There... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
where C� is the 1-dimensional representation of B in which B acts by ∂. We have
Wµ, where Wµ is a q-dimensional irreducible representation of
if and only if
∂1, ∂2}
{
=
⊗
∂ ⊗
1, ∂ 2}
{
(in the
∂(aga−
1).
dim(V�1,�2 ) = |
|
G
B
|
|
= q + 1.
V�1,�2 is irreducible.
Theorem 4.71.
2. ∂1 = ∂2 = µ
G.
≥
1. ∂1 =... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
∂(g) 1,
·
1
aga−
B
a
�
≥
�
B.
g =
x 0
0 y
�
�
,
∂1(x)∂2(y) + ∂1(y)∂2(x)
1,
·
�
⎩
B or a is an element of B multiplied by the transposition matrix.
g =
x πy
x
y
�
�
, x = y
the expression on the right evaluates to 0 because matrices of this type don’t have eigenvalues over
Fq (and thus cannot be conjugate... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
Clearly,
since
νV�1,�2
, νV�1,�2 �
◦
= 2.
IndG
B Cµ,µ,
Cµ
∧
HomG(Cµ, IndG
BCµ,µ) = HomB(Cµ, Cµ) = C (Theorem 4.33).
Therefore, IndG
BCµ,µ = Cµ
values of µ, proving that Wµ are distinct.
Wµ; Wµ
�
is irreducible; and the character of Wµ is different for distinct
72
⇒
∞
If ∂1 = ∂2, let z = xy... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
bolic elements
and note that the function
∂1(x)∂2(y) + ∂1(y)∂2(x)
determines ∂1, ∂2 up to permutation.
4.24.4 Complementary series representations
∩
Fq be a quadratic extension Fq(∀π), π
Let Fq2
q. We regard this as a 2-dimensional vector
�
space over Fq; then G is the group of linear transformations of Fq2 ov... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
Therefore, for λ q =
K Cξq ∪
IndG
= IndG
λ we get 2
1 q(q
1) representations.
−
Next, we look at the following tensor product:
�
where 1 is the trivial character and W1 is defined as in the previous section. The character of this
representation is
W1
V�,1,
x 0
0 x
�
ν
�
= q(q + 1)ϕ(x);
ν(A) = 0 for A para... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
Proof.
ν, ν
�
◦
= 1
ν(1) > 0.
We now compute the inner product ν, ν
�
◦
ν(1) = q(q + 1)
(q + 1)
q(q
1) = q
1 > 0.
−
−
. Since ϕ is a root of unity, this will be equal to
−
−
1
(q
1)2q(q + 1)
�
(q
−
1) (q
·
−
−
1)2 1+(q
·
2
1) 1 (q
·
·
−
−
q(q
1)+
−
2
1)
·
(λ(ψ)+λq(ψ))(λ(ψ) + λ q(ψ))
�
λ elliptic... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
(q
q(q + 1)
⎩
1) (q
·
−
1)2 1+(q
·
−
2
1) 1 (q
·
·
−
−
1)+
q(q
−
2
1)
·
2
(2(q
q)
−
−
2(q
−
1))
= 1.
�
We have now shown that for any λ with λ q = λ the representation Yξ with the same character
as
W1
�
V�,1
−
V�,1
−
IndG
K Cξ
⊃
We have thus found q
λq), so there are q(q
−
2
exists and is irred... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
of characters of
induced representations IndG
H V , where H
X and V
is an irreducible representation of H.
�
Remark. Statement (ii) of Theorem 4.73 is equivalent to the same statement with Q-span
replaced by C-span. Indeed, consider the matrix whose columns consist of the coefficients of the
decomposition of IndG ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
νU vanishes on H for any
H
X. Hence by (i), νU is identically zero. This implies (ii) (because of the above remark).
H
|
�
Corollary 4.74. Any irreducible character of a finite group is a rational linear combination of
induced characters from its cyclic subgroups.
4.26 Representations of semidirect products
Let ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
{
⊃
Next, we introduce an additional action of A on this space by (af )(g) = x(g(a))f (g). Then it’s
easy to check that these two actions combine into an action of G ∼ A. Also, it is clear that this
O,
representation does not really depend on the choice of x, in the following sense. Let x, y
G be such that gx = y... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
W
V(O,U )|
�
V is a subrepresentation, then W =
→
v(g) = 0 unless gy =
where Vy =
Vy. Now, Vy is a
x
representation of Gy, which goes to U under any isomorphism Gy
G
mapping x to y. Hence, Vy is irreducible over Gy, so Wy = 0 or Wy = Vy for each y. Also, if hy = z
then hWy = Wz, so either Wy = 0 for all y o... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
. Field embeddings. Recall that k(y1, ..., ym) denotes the field of rational functions
of y1, ..., ym over a field k. Let f : k[x1, ..., xn]
k(y1, ..., ym) be an injective k-algebra homomor
n. (Look at the growth of dimensions of the spaces WN of polynomials of
phism. Show that m
⊂
degree N in xi and their images ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
5.1.
⊃
(c) generalize the result of this problem to the case when G = GLn1 (k)
×
...
×
GLnm (k).
Problem 5.3. Dynkin diagrams.
Let � be a graph, i.e., a finite set of points (vertices) connected with a certain number of edges
(we allow multiple edges). We assume that � is connected (any vertex can be connected t... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
diagram if and only if it is one on the following graphs:
An :
•
Dn:
•
E6 :
•
�−−�
· · ·
�−−�
�−−�
· · ·
�−−�
|
�
�−−�−−�−−�−−�
|
�
78
E7 :
E8 :
•
•
�−−�−−�−−�−−�−−�
|
�
�−−�−−�−−�−−�−−�−−�
|
�
(a) Compute the determinant of A� where � = AN , DN . (Use the row decomposition rule, and
write dow... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
�) = 0 for all graphs � below:
1
2
1
2
3
2
1
2
1
2
3
4
3
2
1
7Recall the Sylvester criterion: a symmetric real matrix is positive definite if and only if all its upper left corner
principal minors are positive.
8The Sylvester criterion says that a symmetric bilinear form (, ) on RN is positive definite if and o... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
iver Q is of finite type if and only if the corresponding unoriented
graph (i.e., with directions of arrows forgotten) is a Dynkin diagram.
In this problem you will prove the “only if” direction of this theorem (i.e., why other quivers
are NOT of finite type).
(a) Show that if Q is of finite type then for any rational... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
��nite subgroup of SU (2), and V be the 2-dimensional representation
of G coming from its embedding into SU (2). Let Vi, i
I, be all the irreducible representations of
G. Let rij be the multiplicity of Vi in V
Vj.
�
�
(a) Show that rij = rji.
(b) The McKay graph of G, M (G), is the graph whose vertices are labe... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
numbers labeling the vertices of the affine Dynkin
diagrams on our pictures). Compare with the results on subgroups of SO(3) we obtained in
Problem 3.24.
→
5.2
Indecomposable representations of the quivers A1, A2, A3
We have seen that a central question about representations of quivers is whether a certain connected ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
W .
⊃
•
� �
•
A
� �
W•
V•
be a complement to the kernel of A in V and
be a complement to the image of A in W . Then we can decompose the representation as
To decompose this representation, we first let V �
let W �
follows
A
V•
W•
=
0
ker• A
•0 �
V•�
A
� �
Im•A � •0
0
W•
�
The first summand is a multiple of t... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
split away
�0
�0 . Next, let Y �
be a complement of ImB in Y .
0
� �
0
•
0
� �
•
Y
�
•
0
which is a multiple of the object 0
A is injective and the map B is surjective (we rename the spaces to simplify notation):
�1 . This results in a situation where the map
�0
Next, let X = ker(B
of A(X) in W such that A(X... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
•
W
� =
•
Y
•
V
A
�
B
� �
•
X
�
•
0
A(•V )
� �
•
W
�
B
� �
� �
•
X
�
Here, the first summand is a multiple of 1 � �
�1 � �
the second summand can be decomposed into multiples of 0
So, on the whole, this quiver has six indecomposable representations:
�1 � �
�1 . By splitting away the kernel of B,
�1 and 0
�1
�0... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
��rst choose a complement W �
Y .
Then we can decompose the representation as follows:
Y in W , and set V �
= W �
= W �
V , Y �
of V
∈
∈
∈
�
•
V
� �
� �
•
W
�
•
Y
=
•
V
�
� �
•
W
� �
�
•
Y
The second summand is a multiple of the object 1 � �
first summand. Again, to simplify notation, we let
�1 �
� �
�
Y
... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
other orientation - we get 6 indecomposable representations of A3:
1
0
�0 �
�1�
0 ,
0
0 ,
1
�0 �
�1 �
1 ,
1
�
�1 �
�
1 ,
0 ,
0
�1 �
1
5.3
Indecomposable representations of the quiver D4
As a last - slightly more complicated - example we consider the quiver D4.
Example 5.10 (D4). We restrict ourselves ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
a situation where all of the maps A1, A2, A3 are injective.
� A1 V A3
� �
� �•
� �
�
•
V1
�
•
V3
•
0
0
� �
•
•
0
0
•
1
A2
•
V2
As in 2, we can then identify the spaces V1, V2, V3 with subspaces of V . So we get to the triple of
subspaces problem of classifying a triple of subspaces of a given space V .
The ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
to simplify notation - we are in a
situation where
As a next step, we let Y = V1
and set V
= V �
1� = V �
�
•
V1
∈
V1, V2�
V
� �
•
∈
∈
V1
V2
V = V1 + V2 + V3,
V3 = 0.
V2 and we choose a complement V
�
∈
V2. This yields the decomposition
∈
�
•
V3
�
•
V3
V
��
� �
•
•
Y
•
1�V
�
=
�
•
2�V
of Y in V su... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
2 = V1
V3 = V2
V3 = 0.
∈
∈
∈
(V2
�
•
Y
•
0
�
�
•
V3
V
��
� �
•
•
V2
�
•
V1
V
� �•
�
V3 •
� V
1�
� �
•
•
V
1�
=
�
•
V2
The first of these summands is a multiple of
�
•
1
•
0
1
� �•
•
0
•
0
∧
By splitting these away we get to a situation where V1
objects of the type
V2
V3. Similarly, we can spl... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
implies that
V3 = 0.
∈
V3
V1
∧
�
V2.
So we get
and
V1
�
V2 = V1
�
V3 = V2
�
V3 = V.
dim V1 = dim V2 = dim V3 = n
dim V = 2n.
Since V3
V1
∧
�
V2 we can write every element of V3 in the form
V3,
x
�
x = (x1, x2), x1
V1, x2
V2.
�
�
We then can define the projections
B1 : V3
⊃
V1,
(x1, x2)... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
: V1
decomposition
�
�
V•
1
V
•�
�
V•
2
�
�
C(1•, 0)
�
�
V•
3
=
n
�
j=1
These correspond to the indecomposable object
•
1
� �
2
� �
•
•
1
•
1
�
C(0•
, 1)
Thus the quiver D4 with the selected orientation has 12 indecomposable objects. If one were to
explicitly decompose representations for the oth... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
2. By the definition of the Cartan matrix we get
B(x, x) = x T A�x =
xi aij xj = 2
�
i,j
�
i
2
xi +
�
i,j, i=j
xi aij xj = 2
�
i
2
xi + 2
·
�
i<j
aij xixj
which is even.
Definition 5.13. A root with respect to a certain positive inner product is a shortest (with respect
to this inner product), nonzero vector i... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
Next, let φ be the edge connecting i with the next vertex towards j and i�
end of φ. We then let �1, �2 be the graphs obtained from � by removing φ. Since � is supposed
to be a Dynkin diagram - and therefore has no cycles or loops - both �1 and �2 will be connected
graphs, which are not connected to each other.
•
•
... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
⎨
kiϕi a positive root if all ki
0. A root for which ki
0
∗
⊂
Remark 5.18. Lemma 5.16 states that every root is either positive or negative.
Example 5.19.
1 can be realized as
1. Then the lattice L = ZN
1. Let � be of the type AN
−
a subgroup of the lattice ZN by letting L
ZN be the subgroup of all vectors (x1... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
0, . . . , 0) =
−
(ϕi + ϕi+1 +
+ ϕj
1)
−
· · ·
−
are the roots of L. Therefore the number of positive roots in L equals
N (N
2
−
1)
.
2. As a fact we also state the number of positive roots in the other Dynkin diagrams:
DN
E6
E7
E8
1)
−
N (N
36 roots
63 roots
120 roots
Definition 5.20. Let ϕ
�
Zn be ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
We are now able to formulate Gabriel’s theorem using roots.
Theorem 5.23 (Gabriel’s theorem). Let Q be a quiver of type An, Dn, E6, E7, E8. Then Q has
finitely many indecomposable representations. Namely, the dimension vector of any indecomposable
representation is a positive root (with respect to B�) and for any pos... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
k = i
+F
i
(V )i = ker
⎧
� :
Vi
�
.
Vj
⊃
⎝
�
i
j
⊥
�
Also, all maps stay the same but those now pointing out of i; these are replaced by compositions
of the inclusion of ker � into
Vj with the projections
Vk.
Vj
�
�
⊃
Definition 5.27. Let Q be a quiver, i
the canonical map
�
Q be a source. Let V be a repr... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
be a source. Then either dim Vi = 1, dim Vj = 0 for j = i or
ξ : Vi
Vj
⊃
�
j
i
⊥
is injective.
Proof.
1. Choose a complement W of Im�. Then we get
•
V = 0
W
� �
� �
� �
•
•
0
•
0
V �
�
Since V is indecomposable, one of these summands has to be zero. If the first summand is
zero, then � has to be surjective. I... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
ourselves to showing that the
spaces of V and Fi−F +V are the same. It is enough to do so for the i-th space. Let
⊃
i
� :
�
i
j
⊥
Vj
Vi
⊃
91
⇒
⇒
be surjective and let
K = ker �.
When applying F
+
i
, the space Vi gets replaced by K. Furthermore, let
After applying Fi−, ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
0.
i
� :
�
i
j
⊥
Vj
Vi
⊃
So we can assume that � is surjective. In this case, assume that F
V is decomposable as
+
i
with X, Y = 0. But F
V is injective at i, since the maps are canonical projections, whose direct
sum is the tautological embedding. Therefore X and Y also have to be injective at i and hence (by
... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
dim K =
dim Vj
�
i
j
⊥
dim Vi.
−
d(F +V )
i
−
d(V )
i
�
⎩
=
dim Vj
�
i
j
⊥
2 dim Vi =
−
−
B (d(V ), ϕi)
and
This implies
⎩
d(Fi
+V )
−
j = 0,
j = i.
d(V )
�
+V )
d(Fi
d(F +V ) = d(V )
−
i
−
⊆
d(V ) =
B (d(V ), ϕi) ϕi
−
B (d(V ), ϕi) ϕi = si (d(V )) .
5.7 Coxeter elements
Definition 5.32. Let Q ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
one of the elements
must have at least one strictly negative one. Furthermore, it is enough to show that 1 is not an
eigenvalue for c, since
cα, c2α, . . . , cM
1α
−
≥
cw = c
⎩
1 + c + c +
�
(1 + c + c 2 +
2
+ c M
· · ·
1
−
· · ·
+ c M
1)v = w = 0
−
2
v = (c + c + c +
3
· · ·
+ c M
1 + 1)v = w.
−
Assume th... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
a sink of the remaining
quiver and so on. This way we create a labeling of the desired kind.
We now consider the sequence
V (0)
= V, V (1)
= F +
n V, V (2)
= F + F +
n
−
1 n V, . . .
This sequence is well defined because of the selected labeling: n has to be a sink of Q, n
1 has
to be a sink of Qn (where Qn ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
indefinitely - since d
may not have any negative entries.
Let i be smallest number such that V (i) is not surjective at the appropriate vertex. By Proposition
⎩
5.30 it is indecomposable. So, by Proposition 5.28, we get
�
V (i)
for some p.
d(V (i)) = ϕp
We are now able to prove Gabriel’s theorem. Namely, we get th... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
V �
(0).
F −F − . . . F −V i =
n
1
n
k
−
Fn−F − . . . F −F + . . . F + Fn
+
1
n
n
1
k
k
−
−
+
+ . . . F n
+V �
1 . . . F k−Fk
Fn−F n−
1Fn
−
−
V (0) = V (0)
(0) = V �
= V
(0) = V �
�
95
These two corollaries show that there are only finitely many indecomposable representations
(since there a... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
functors work. Consider the quiver
D4 with the orientation of all arrows towards the node (which is labeled by 4). Start with the
1-dimensional representation V�4 sitting at the 4-th vertex. Apply to V�4 the functor F3−F2−F1−.
This yields
F1−F2−F3−V�4 = V�1+�2+�3+�4 .
Now applying F4− we get
F4−F1−F2−F3−V�4 = V�1... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
by exchanging V with W and A with B.
96
3) Hn: V = Cn with basis vi, W = Cn
1 with basis wi, Avi = wi, Bwi = vi+1 for i < n, and
−
Avn = 0.
4) Kn is obtained from Hn by exchanging V with W and A with B.
Show that these are indecomposable and pairwise nonisomorphic.
(b) Show that if E is a representation of Q2 su... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
classification to Qn, n > 2, with any orientation?
Problem 5.40. Let L
or all half-integers (but not integers), and the sum of all coordinates is an even integer.
1 Z8 be the lattice of vectors where the coordinates are either all integers
2
→
(a) Let ϕi = ei
of L (over Z).
−
ei+1, i = 1, ..., 6, ϕ7 = e6 + e7, ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
else.
(a) Show that if i is a source then Ext1(V, V�i ) = 0 for any representation V of Q, and if i is
a sink, then Ext1(V�i , V ) = 0.
(b) Given an orientation of the quiver, find a Jordan-H¨older series of V� for that orientation.
97
⇒
6
Introduction to categories
6.1 The definition of a category
We have now ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
write f : X
C
Y );
⊃
(iii) For any objects X, Y, Z
(f, g)
f
∞
�⊃
g,
Ob(
C
�
), a composition map Hom(Y, Z)
×
Hom(X, Y )
⊃
Hom(X, Z),
which satisfy the following axioms:
1. The composition is associative, i.e., (f
g)
∞
), there is a morphism 1X
∞
h = f
(g
∞
∞
h);
2. For each X
Ob(
f = f and g ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
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