text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
, where ϕ� are some algebraic conjugates of ϕi.
i
m
Proof. It suffices to prove this for two summands. If ϕi are eigenvalues of rational matrices Ai of
smallest size (i.e., their characteristic polynomials are the minimal polynomials of ϕ i), then ϕ1 + ϕ2
is an eigenvalue of A := A1
A2. Therefore, so is any algebrai... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
consider the Galois conjugates of the
representation V , i.e. representations obtained from V by the action of the Galois group of K over
Q on the matrices of group elements in the basis from part (a)). Then derive a contradiction.
2 for g =
νV (g)
|
|
Remark. Here is a modification of this argument, which does not ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
for all i.
C h. Then P is a central element of Z[G], so it
Proof. Let C be a conjugacy class in G, and P =
�
acts on V by some scalar ∂, which is an algebraic integer (indeed, since Z[G] is a finitely generated
⎨
Z-module, any element of Z[G] is integral over Z, i.e., satisfies a monic polynomial equation with
integ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
is an irreducible representation, (νV , νV ) = 1, so
∂iνV (gCi ) =
G
|
|
dim V
.
�
Ci
G
Since dim|
|V �
Q and
∂iνV (gCi )
Ci
⎨
G
A, by Proposition 4.13 dim|
|V �
�
Z.
4.5 Burnside’s Theorem
Definition 4.18. A group G is called solvable if there exists a series of nested normal subgroups
{
where Gi+1/Gi is ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
, dim(V )) = 1. Then for any g
|
C
�
52
The proof will be based on the following lemma.
Lemma 4.22. If π1, π2 . . . πn are roots of unity such that
integer, then either π1 = . . . = πn or π1 + . . . + πn = 0.
1
n
(π1 + π2 + . . . + πn) is an algebraic
n
Proof. Let a = 1 (π1 + . . . + πn). If not all πi... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
1
n
is an algebraic integer. Thus, by Lemma 4.22, we get that either π1 = . . . = πn or π1 + . . . + πn =
νV (g) = 0. In the first case, since δV (g) is diagonalizable, it must be scalar. In the second case,
νV (g) = 0. The theorem is proved.
Theorem 4.23. Let G be a finite group, and let C be a conjugacy class in G... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
(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 nonzero.
53
⇒
⇒
Now pick V
N such that νV (g) = 0; it exists by Lemma 4.24. Theorem 4.21 implies that g
�
(and hence any element of C) acts by a scalar in V... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
orem 4.25. Let G, H be finite groups,
field k (of any characteristic), and
irreducible representations of G
Wj
}
H over k are
{
{
Vi}
Vi
×
.
Wj}
�
{
be the irreducible representations of G over a
be the irreducible representations of H over k. Then the
Proof. This follows from Theorem 2.26.
4.7 Virtual repr... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
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 representation of V
by the vector space V and the action δResG V = δV H .
|
There is also a natura... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
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
→ →
K V .
G are groups and V a representation of K then IndG
H IndH
K V
Exercise. Let K G be finite groups, and ν : K
cor... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
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 for any right H-coset ε, νε(g) = 0 if ε = εg... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
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 defined and inverse to each oth... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
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 induced
repre... | 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
Re... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
]2 , V ).
(d) Let W be a representation of H. 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 ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
[H] V
IndG
×
and write IndG
1 for every g
H (V ⊕)
G.]
⊃
�
⎩
⊃
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... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
(a) Z2
(b) Z3
(c) Z5
(d) A4
(e) Z2
Z2
×
4.12 Representations 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+... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
Define the Young projectors:
a� :=
1
P�|
g,
�
P�
g
�
|
1
Q�|
|
1)g denotes the sign of the permutation g. 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... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
C3
+.
−
Corollary 4.38. All irreducible representations of Sn can be given by matrices with rational entries.
Problem 4.39. Find the sum of dimensions of all irreducible representations of the symmetric
group Sn.
Hint. Show that all irreducible representations of Sn are real, i.e., admit a nondegenerate
invariant... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
same column in the tableau T �
same shape as T obtained by permuting the entries of T by the permutation g). Thus, it suffices to
1
show that if such a pair does not exist, then g
1, q = g−
such that pT = q�T �
�
1q�g
Q��
Any two elements in the first row of T must be in different columns of T �, so there exists q1� �... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
previous lemma, it suffices to show that for any g
= gTµ. We claim that there are
transposition t
two integers which are in the same row of T and the same column of T �. Indeed, if ∂1 > µ1, this is
clear by the pigeonhole principle (already for the first row). Otherwise, if ∂1 = µ1, like in the proof
have the
of t... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
left A-module M ,
one has HomA(Ae, M ) ∪= eM (namely, x
M given by fx(a) = ax,
a
eM corresponds to fx : Ae
Ae).
⊃
�
�
Proof. Note that 1
the fact that HomA(A, M ) ∪= M and the decomposition A = Ae
−
e is also an idempotent in A. Thus the statement immediately follows from
A(1
−
e).
�
Now we are ready t... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
1.
Proposition 4.44. Hom(U�, Vµ) = 0 for µ < ∂, and dim Hom(U�, V�) = 1. Thus, U� =
�µ
∧
Definition 4.45. The integers Kµ� are called the Kostka numbers.
Proof. By Lemmas 4.42 and 4.43,
Hom(U�, Vµ) = Hom(C[Sn]a�, C[Sn]aµbµ) = a�C[Sn]aµbµ,
and the result follows from Lemmas 4.40 and 4.41.
Now let us compute the cha... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
Ci), divided by
elements x
�
|
P� is the number of elements in P�
P� is
). Thus,
conjugate to g (i.e.
P�|
) times the order of the centralizer Zg of g (which is n!/
Ci
|
�
∂i!, and the number of elements x such that xgx−
Sn such that xgx−
P�|
Ci
⎛
�
∈
�
1
1
i
|
|
Now, it is easy to see that the centralizer ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
!
im
rjm!
j
� �
r m
⎛
(x1 + ... + xN )im
�
1
m
∧
(rjm is the number of times we take xj ).
m
m
m
4.15 The Frobenius character formula
Let �(x) =
to Frobenius, gives a character formula for the Specht modules V�.
xj ). Let δ = (N
2, ..., 0)
1, N
(xi
i<j
−
−
−
�
N
→
→
⎛
1
CN . The following theorem, due... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
�
−
1)ενU�+�−�(�) ,
−
where if the vector ∂ + δ
ε(δ) has a negative entry, the corresponding term is dropped, and if
it has nonnegative entries which fail to be nonincreasing, then the entries should be reordered in
(i.e., we agree that
the nonincreasing order, making a partition that we’ll denote ∂ + δ
is obt... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
m, we get
�
(
m
j xj )im (
m)im
k yk
⎨
⎨
mim im!
(
m
m
j,k xj yk /m)im
im!
.
=
⎨
� �
i m
S(x, y) =
exp(
�
m
�
j,k
x m
j yk
m/m) = exp(
log(1
−
xj yk)) =
−
�
j,k
�
j,k
(1
xjyk)−
1
−
Thus,
R(x, y) =
Now we need the following lemma.
Lemma 4.48.
i<j
⎛
(xi
i<j
⎛
i,j
⎛
xj )(yi
−
(1
−
xiyj)
... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
) = det(
1
−
1
xiyj
) =
1
N (1
j=1
−
xjyε(j))
�
SN
ε
�
⎛
.
62
Corollary 4.49 easily implies that the coefficient of x�+χy�+χ is 1. Indeed, if ε = 1 is a permu
tation in SN , the coefficient of this monomial in
1
j y�(j)) is obviously zero.
x
Q
(1
−
Remark. For partitions ∂ and µ of n, ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
1
A(µ)V�.
��
�
Problem 4.51. The content c(∂) of a Young diagram ∂ is the sum
Let C =
C[Sn] be the sum of all transpositions. Show that C acts on the Specht module V� by
j
⎨ ⎨
j).
−
i<j (ij)
�j
i=1(i
�
multiplication by c(∂).
⎨
Problem 4.52. (a) Let V be any finite dimensional representation of Sn. Show that... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
the determinant as a sum over
permutations, we get
−
dim V� =
SN :lj
�
N
∧
s
�
(
s(j)
−
1)s
−
⎛
(lj
j
−
n!
N + s(j))!
=
n!
l1!...lN !
1)s
(
�
SN
s
�
−
�
j
lj (lj
−
1)...(lj
−
N +s(j)+1) =
n!
j lj!
det(lj (lj
1)...(lj
−
−
N + i + 1)).
Using column reduction and the Vandermonde determinant formul... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
) One has
dim V� =
n!
.
h(i, j)
i �j
→
⎛
Proof. The formula follows from formula (5). Namely, note that
l1!
1<j N
→
⎛
(l1
−
lj )
=
k.
�
1 k l1,k=l1
→ →
−
lj
It is easy to see that the factors in this product are exactly the hooklengths h(i, 1). Now delete the
first row of the diagram and proceed by induction... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
Wi, where Wi :=
�i End Vi. Therefore, by Schur’s lemma, B = EndA(E) is naturally identi
�i End(Wi). This implies all the statements of the theorem.
I Vi
�
�i
�
We will now apply Theorem 4.54 to the following situation: E = V �
n , where V is a finite
dimensional vector space over a field of characteristic zero, and ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
elements of
the form u
�
u, u
...
�
�
U .
(ii) For any algebra A over k, the algebra S nA is generated by elements �n(a), a
A.
�
Proof. (i) The space SnU is an irreducible representation of GL(U ) (Problem 3.19). The subspace
u is a nonzero subrepresentation, so it must be everything.
spanned by u
...
... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
L� are some distinct irreducible
representations of gl(V ) or zero.
��V�
�
�
n =
4.19 Schur-Weyl duality for GL(V )
The Schur-Weyl duality for the Lie algebra gl(V ) implies a similar statement for the group GL(V ).
Proposition 4.58. The image of GL(V ) in End(V �
n) spans B.
Proof. Denote the span of g�
n , g
... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
4.20 Schur polynomials
Let ∂ = (∂1, ..., ∂p) be a partition of n, and N
p. Let
⊂
D�(x) =
(
�
SN
s
�
−
Define the polynomials
N
1)s
�
j=1
x
�j +N
s(j)
j
−
�j +N
= det(x
i
j
−
).
S�(x) :=
D�(x)
D0(x)
(clearly D0(x) is just �(x)). It is easy to see that these are indeed polynomials, as D� is an
tisymmetric... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
The first identity is obtained from the definition using the Vandermonde determinant. The
second identity follows from the first one by setting z = 1.
4.21 The characters of L�
Proposition 4.61 allows us to calculate the characters of the representations L�.
Namely, let dim V = N , g
the character νL� (g), let us cal... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
of squares but at most N = dim V rows.
n for some n are labeled
Proposition 4.64. The representation L�+1N (where 1N = (1, 1, ..., 1)
L�
N V .
� √
ZN ) is isomorphic to
�
N V
Proof. Indeed, L�
the same character as L�
� √
n
N V
V �
N V is L�+1N . This implies the statement.
n+N , and the only component of... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
(
Theorem 4.66. (i) Every finite dimensional polynomial representation of GL(V ) is completely
reducible, and decomposes into summands of the form L� (which are pairwise non-isomorphic).
(ii) (the Peter-Weyl theorem for GL(V )). Let R be the algebra of polynomial functions on
1xh),
GL(V ) (with action (δ(g, h)θ)(x)... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
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 representation of GL(V ), and let us regard R as a representation
of GL(V ) via (δ(h)θ)(x) = θ(xh). Then HomGL(V )(Y, R) is the space of polyno... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
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 the irreducible representations are parametrized by integer sequences ∂ 1
up to a simultaneous shift by a constant.
...
⊂
⊂
In particular, one can show that any finite ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
. 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 conjugation. An example of an element of Rk,N
is the function Tw := ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
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
�
F2
F×q , π
q (char... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
in Fq. 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 =... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
to show that 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 ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
�1,�2 = IndG
B C�,
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 = ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
B
a
�
≥
�
If
∂(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 ca... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
(q
1) = 2
G
|
|
,
−
−
so
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
⇒
... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
−
To prove the third assertion of the theorem, we look at the characters on hyperbolic 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... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
map.
73
⇒
∞
∞
∞
∞
We thus have
K Cξ
because they have the same character. 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 ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
imply
that it is the character of an irreducible representation of G.
Lemma 4.72. Let ν be the character of the ”virtual representation” defined above. Then
and
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 un... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
F×
λ
q
�
2(q
−
−
1) =
since F× is cyclic of order q
q
1. Therefore,
−
ν, ν
�
◦
=
(q
−
1)2
1
(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
... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
groups of a finite group G. Then
two conditions are equivalent:
(i) Any element of G belongs to a subgroup H
X.
�
(ii) The character of any irreducible representation of G belongs to the Q-span of characters of
induced representations IndG
H V , where H
X and V
is an irreducible representation of H.
�
Remark. ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
Proof that (i) implies (ii). Let U be a virtual representation of G over C (i.e., a linear combina
tion of irreducible representations with nonzero integer coefficients) such that (νU , νIndG V ) = 0 for
all H, V . So by Frobenius reciprocity, (νU
H , νV ) = 0. This means that νU vanishes on H for any
H
X. Hence by ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
and Gx the stabilizer of x in G. Let U be an irreducible representation of Gx. Then we define a
representation V(O,U ) of G ∼ A as follows.
�
As a representation of G, we set
V(O,x,U ) = IndG U =
Gx
f : G
U
f (hg) = hf (g), h
|
{
⊃
Next, we introduce an additional action of A on this space by (af )(g) = x(g(a))f... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
of V = V(O,U ) is given by the Mackey-type formula
νV (a, g) =
1
Gx|
|
�
G:hgh−1
h
�
Gx
�
x(h(a))νU (hgh−
1).
76
Proof. (i) Let us decompose V = V(O,U ) as an A-module. Then we get
V =
OVy,
�y
�
}
av = (y, a)v, a
v
{
). So if W
V(O,U )|
�
V is a subrepresentation, then W =
→
v(g) = 0 unless gy =
whe... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
O
O
G/Gx||
||
Gx|
|
=
G
|
|
�
O
=
O
|
|
G
|
A∗
||
=
G ∼ A
|
|
.
|
(iv) The proof is essentially the same as that of the Mackey formula.
Exercise. Redo Problems 3.17(a), 3.18, 3.22 using Theorem 4.75.
Exercise. Deduce parts (i)-(iii) of Theorem 4.75 from part (iv).
77
5 Quiver Representations
5.... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
x1, ..., xN be linear coordinates on V . Let us say that a subset X of V is Zariski dense
if any polynomial f (x1, ..., xN ) which vanishes on X is zero (coefficientwise). Show that if G has
finitely many orbits on V then G has at least one Zariski dense orbit on V .
(b) Use (a) to construct a field embedding k(x1, ...,... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
= 2I
R�, where I is the identity matrix.
×
−
Main definition: � is said to be a Dynkin diagram if the quadratic from on RN with matrix
A� is positive definite.
Dynkin diagrams appear in many areas of mathematics (singularity theory, Lie algebras, rep
resentation theory, algebraic geometry, mathematical physics, etc... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
a)).
Show they are Dynkin diagrams.
(c) Show that if � is a Dynkin diagram, it cannot have cycles. For this, show that det(A�) = 0
for a graph � below 9
1
1
1
1
1
(show that the sum of rows is 0). Thus � has to be a tree.
(d) Show that if � is a Dynkin diagram, it cannot have vertices with 4 or more incoming ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
laced) affine Dynkin diagram is a connected graph without self-loops such that the
quadratic form defined by A� is positive semidefinite. Classify affine Dynkin diagrams. (Show that
they are exactly the forbidden diagrams from (c)-(e)).
Problem 5.4. Let Q be a quiver with set of vertices D. We say that Q is of finite type ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
: xi = ni. First assume that ni
⊂
the space W of representations V of Q such that dimVi = ni. Show that the group
with finitely many orbits on W
result in the case when ni are arbitrary integers.
0, and consider
i GLni (k) acts
k, and use Problem 5.2 to derive the inequality. Then deduce the
⎛
�
(b) Deduce that... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
but not definite, and use
For this, show that the matrix aij = 2ζij
Problem 5.3.
−
Hint. Let f =
0.
When is it equal to 0? Next, show that M (G) has no self-loops, by using that if G is not cyclic
⎨
then G contains the central element
xiνVi , where νVi be the characters of Vi. Show directly that ((2
νV )f, f )
... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
the type 1
�0 we mean a map from a one-dimensional vector
space to the zero space. Similarly, an object of the type 0
�1 is a map from the zero space into
a one-dimensional space. The object 1
�1 means an isomorphism from a one-dimensional to
another one-dimensional space. The numbers in such diagrams always mean... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
1
�1 , the
third of 0
�1 . We see that the quiver A2 has three indecomposable representations, namely
1
�0 ,
1
�1
and
0
�1 .
Example 5.9 (A3). The quiver A3 consists of three vertices and two connections between them.
So we have to choose between two possible orientations.
� �
•
� �
•
or
•
� �
� �
•
•
•
1... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
B
A
.
� �
•W
� �
� �
•
Y
be a complement of X in V . Let W �
�
be a complement
�
A
•
V
� �
•
W
B
=
� �
� �
•
Y
•
X
A
� �
A(•X)
B
� �
•
0
�
A
�
•
X
�
� �
•
W
�
B
� �
� �
•
Y
0 . Looking at the second summand,
The first of these summands is a multiple of 1
we now have a situation where A is injective, B is sur... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
0
�1
�0 .
1
�0
�0 ,
0
�0
�1 ,
1 � �
�1
�0 ,
1 � �
�1 � �
�1 ,
0
�1 � �
�1 ,
0
�1
�0
2. Now we look at the orientation
� �
� �
•
.
•
•
Very similarly to the other orientation, we can split away objects of the type
which results in a situation where both A and B are injective:
1
�0 �
0 ,
0
�0 �
1
... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
1 � �
first summand. Again, to simplify notation, we let
�1 �
� �
�
Y
V
� �
•
∈
� � 1 . We go on decomposing the
� �
Y
•
∈
•
∈
V
V
Y
.
�
V = V �, W = W �, Y = Y �.
We can now assume that V
we get
∈
Y = 0. Next, let W �
be a complement of V
Y in W . Then
�
�
�
•
V
•
W
�
•
Y
=
�
•
V
Y
•V
�
�
•
Y
�
... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
Example 5.10 (D4). We restrict ourselves to the orientation
.
•
•
� �
� �
� �•
•
So a representation of this quiver looks like
•
V1
A1 V A3
� �
� �•
� �
A2
•
V3
•
V2
83
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
The first thing we can do is - as usual... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
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 next step is to split away a multiple of
to reach a situation where
•
0
0 •
� �
1
� �•
� �
•
0
V1 + V2 + V3 = V.
By letting Y = V1
∈
i = 1, 2, 3, we can... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
•
∈
∈
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 such that V3
V
�
,
→
•
0
�
Y
� �
•
•
Y
�
�
•
V2
The second summand is a multiple of the indecomposable object
.
•0
�
•
1
1
� �
•... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
� �
•
•
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 split away
� �
•
0
1
� �
•
� �
�
� �
•
0
•
0
and
�
1 �
� �
� �•
... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
= 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)
x1,
�⊃
V1 = 0, V3
Since V3
then define the isomorphism
∈
∈
B2 : ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
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 other possible orientations, one would also find 12
indecomposa... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
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 in Zn .
So for the inner product B, a root 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 |
i
⎨
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 ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
0, . . . , 0, 1, 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.... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
the dimension vector of this representation.
d(V ) = (dim V1, . . . , dim Vn)
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 indeco... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
of Q. Then we
�
+F
i
: RepQ
RepQi
⊃
by the rule
F +(V )k = Vk
i
if 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. ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
Vj
Vi
⊃
is surjective.
2. Let i
�
Q 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 summan... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
always mean by i
quiver Q. We only establish the first statement and we also restrict 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 app... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
0,
j = i. In the last case
+F
V = 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 ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
= ker �. Then
� :
�
i
j
⊥
Vj
Vi
⊃
Therefore we get
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 =... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
−
· · ·
as operators on Rn . This implies what we need, since α has at least one strictly positive coefficient,
so 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 + ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
This is possible, since we can assign the highest label to any
sink, remove this sink from the quiver, assign the next highest label to 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... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
i)
k
�
= skd
V (i)
.
�
�
1) are surjective at the appropriate vertices, then
−
V (i)
d
�
�
= . . . sn
1snd(V ).
−
By Lemma 5.33 this cannot continue 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
⎩
5.3... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
(i
−
V (i) = F + . . . F + F +V (0)
1 n
n
k
−
+ . . . F n
(i) = Fk
+
V �
1Fn
−
(0) have to be surjective at the appropriate vertices.
1), . . . , V �
−
+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) = ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
. . .
and
α = ϕi
(sq . . . sn
−
1sn)ϕ = ϕi.
We let C(i) be the representation having dimension vector ϕi. Then we define
This is an indecomposable representation and
1 . . . Fq−C(i).
V = Fn−F n−
−
d(V ) = ϕ.
Example 5.38. Let us demonstrate by example how reflection functors work. Consider the quiver
D4 with the... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
up to
isomorphism. Namely:
W and B : W
⊃
⊃
(a) Consider the following pairs (for n
1):
⊂
1) En,�: V = W = Cn , A is the Jordan block of size n with eigenvalue ∂, B = 1 (∂
C).
�
2) En,
≤
: is obtained from En,0 by exchanging V with W and A with B.
96
3) Hn: V = Cn with basis vi, W = Cn
1 with basis wi, Avi ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
u, Xu, X 2u, ...X l
W ). Deduce that (1)-(4) are the only indecomposable
sition (i.e., for every chain u
representations of Q2.
V or u
�
�
�
−
(d)(harder!) generalize this classification to the Kronecker quiver, which has two vertices 1 and
2 and two edges both going from 1 to 2.
(e)(still harder!) can you gener... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
are equal.
(d) Show that E6, E7, E8 have 72,126,240 roots, respectively (enumerate types of roots in terms
of the presentations in the basis ei, and count the roots of each type).
Problem 5.41. Let V� be the indecomposable representation of a Dynkin quiver Q which corre
sponds to a positive root ϕ. For instance, if... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
indispensable in many areas of mathematics, such as algebraic geometry, topology, representation
theory, and many others.
We will now give a very short introduction to Category theory, highlighting its relevance to the
topics in representation theory we have discussed. For a serious acquaintance with category theory... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
. The category Vectk of vector spaces over a field k (morphisms are linear maps).
4. The category Rep(A) of representations of an algebra A (morphisms are homomorphisms of
representations).
5. The category of topological spaces (morphisms are continuous maps).
6. The homotopy category of topological spaces (morphism... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
the category of abelian groups or vector spaces with the tensor product operation). This means that
for each X, Y
� C
Hom(X, Z) is a morphism in
composition is bilinear, i.e. gives rise to a linear map Hom(Y, Z)
a more detailed discussion of this, we refer the reader to [McL].
⊃
is the category of vector spaces, ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
F (Y )) which preserves
⊃
Note that functors can be composed in an obvious way. Also, any category has the identity
functor.
Example 6.5. 1. A (locally small) category
A functor between such categories is a homomorphism of monoids.
C
with one object X is the same thing as a monoid.
2. Forgetful functors Groups
... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
The functors of direct
n ,
sum and tensor product are then functors Vectk
⊃
nV are functors on Vectk. More generally, if β is a representation of Sn, we
V
√
n). Such functors (for irreducible β) are called the Schur functors.
HomSn (β, V �
have functors V
�⊃
They are labeled by Young diagrams.
Vectk. Also the... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
C
A morphism a : F
a−
1 and a−
G is an isomorphism if there is another morphism a−
F such that
a are the identities. The set of morphisms from F to G is denoted by Hom(F, G).
⊃
⊃
1
1 : G
∞
a
∞
Example 6.7. 1. Let FVectk be the category of finite dimensional vector spaces over k. Then the
on this category are... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
as a representation of GL(V ).
F (V ) = V ⊕
⊃
3. Let A be an algebra over a field k, and F : A
mod
Then as follows from Problem 1.22, EndF = Hom(F, F ) = A.
−
Vectk be the forgetful functor.
⊃
4. The set of endomorphisms of the identity functor on the category A
mod is the center of
−
A (check it!).
6.4 Equi... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
useful. We might suspect so since we have
used the word “equal” for objects of a category (namely, functors) which we are not supposed to
do. And in fact here is an example of two categories which are “the same for all practical purposes”
but are not isomorphic; it demonstrates the deficiency of our definition.
such... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
categories. This is accomplished by the definition of an equivalence of categories.
Definition 6.8. A functor F :
F �
such that F
and F �
C ⊃ D
∞
∞
F are isomorphic to the identity functors.
is an equivalence of categories if there exists F �
:
D ⊃ C
In this situation, F �
In particular, the above categories
... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf |
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