Part III Representation Theory
Lectured by Stuart Martin, Lent Term 2013
Please send any comments and corrections to [email protected].
Contents
0 Preliminaries
2
1 Semisimple algebras
4
2 Irreducible modules for the symmetric group
8
3 Standard basis of Specht modules
12
4 Character formula
14
5 The hook length formula
21
6 Multilinear algebra and algebraic geometry
24
Interlude: some reminders about affine varieties
29
7 Schur–Weyl duality
31
8 Tensor decomposition
34
9 Polynomial and rational representations of GL(V )
37
Interlude: some reminders about characters of GLn (C)
40
10 Weyl character formula
41
11 Introduction to invariant theory and first examples
44
12 First fundamental theorem of invariant theory
50
Example sheet 1
53
Example sheet 2
57
Last updated Thursday 21 March 2013
1
0
Preliminaries
In this course we will study representations of
(a) (finite) symmetric groups
(b) (infinite) general linear groups
all over C, and apply the theory to “classical” invariant theory.
We require no previous knowledge, except for some commutative algebra, ordinary representation
theory and algebraic geometry, as outlined below.
Commutative algebra
References: Part III course, Atiyah–Macdonald.
It is assumed that you will know about rings, modules, homomorphisms, quotients and chain conditions
(the ascending chain condition, ACC, and descending chain condition, DCC).
Recall: a chain, or filtration, of submodules of a module M is a sequence (Mj : 0 ≤ j ≤ n) such
that
M = M0 > M1 > · · · > Mn > 0
The length of the chain is n, the number of links. A composition series is a maximal chain;
equivalently, each quotient Mj−1 /Mj is irreducible.
Results:
1. Suppose M has a composition series of length n. Then every composition series of M has length
n, and every chain in M can be extended to a composition series.
2. M has a composition series if and only if it satisfies ACC and DCC.
Modules satisfying both ACC and DCC are called modules of finite length. By (1), all composition
series of M have the same length, which we denote by `(M ) and call the length of M .
3. (Jordan–Hölder) If (Mi ) and (Mi0 ) are composition series then there exists a one-to-one cor0
respondence between the set of quotients (Mi−1 /Mi ) and the set of quotients (Mi−1
/Mi0 ) such
that the corresponding quotients are isomorphic.
Remark. For a vector space V over a field K, the following are equivalent:
(i) finite dimension
(ii) finite length
(iii) ACC
(iv) DCC
and if these conditions are satisfied then `(V ) = dim(V ).
(Ordinary) representation theory (of finite groups)
References: Part II course, James–Liebeck, Curtis–Riener Vol. 1, §1 below.
Work over C or any field K where char(K) - |G|. Let V be a finite-dimensional vector space.
2
ˆ GL(V ) is the general linear group of K-linear automorphisms of V .
ˆ If dimK (V ) = n, choose a basis e1 , . . . , en of V over K to identify it with K n . Then θ ∈
GL(V ) ↔ Aθ = (aij ) ∈ Mn (K), where
θ(ej ) =
n
X
aij ei
i=1
and, in fact, Aθ ∈ GLn (K). This gives rise to a group isomorphism GL(V ) ∼
= GLn (K).
ˆ Given G and V , a representation of G on V is a homomorphism
ρ = ρV : G → GL(V )
ˆ G acts linearly on V if there is a linear action G × V → V , given by (g, v) 7→ g · v, where
(a) (action) (g1 g2 ) · v = g1 · (g2 · v) and e · v = v
(b) (linearity) g · (v1 + v2 ) = g · v1 + g · v2 , g · (λv) = λ(g · v)
ˆ If G acts linearly on V then the map G → GL(V ) given by g 7→ ρg , where ρg (v) = g · v, defines
a representation of G on V . Conversely, given a representation ρ : G → GL(V ) we have a linear
action of G on V given by g · v = ρ(g)v.
In this case, we say V is a G-space or G-module. By linear extension, V is a KG-module,
where KG is the group algebra – more to come later.
For finite groups you should know
ˆ If S is an irreducible KG-module then S is a composition factor of KG
ˆ (Maschke’s theorem) If G is finite and char(K) - G, then ever KG-module is completely reducible;
or equivalently, every submodule is a summand.
ˆ The number of inequivalent (ordinary) irreducible representations of G is equal to the number
of conjugacy classes of G.
ˆ If S is an irreducible CG-module and M is any CG-module, then the number of composition
factors of M isomorphic to S is equal to dim HomCG (S, M ).
Algebraic geometry
References: Part II course, Reid, §6 below.
You should know about affine varieties, polynomial functions, the Zariski topology (in particular,
Zariski denseness) and the Nullstellensatz.
3
1
Semisimple algebras
Conventions All rings are associative and have an identity, which we denote by 1 or 1R . Modules
are always left R-modules.
(1.1) Definition A C-algebra is a ring R which is also a C-vector space, whose notions of addition
coincide, and for which
λ(rr0 ) = (λr)r0 = r(λr0 )
for all r, r0 ∈ R and λ ∈ C. There are obvious notions of subalgebras and algebra homomorphisms.
Usually, our algebras will be finite-dimensional over C.
(1.2) Examples and remarks
(a) C is a C-algebra, and Mn (C) is a C-algebra.
P
If G is a finite group thenPthe group
P algebraPCG = { g αg g : αg ∈ C} is a C-algebra with
ˆ pointwise addition:
g αg g +
g βg g =
g (αg + βg )g
P P
0
ˆ multiplication:
g
g
hh0 =g αh βh
Recall the correspondence between representations of G over C and finite-dimensional CG-modules.
(b) If R, S are C-algebras then the tensor product R ⊗ S = R ⊗C S is a C-algebra. [More on ⊗ can
be found in Atiyah–MacDonald p. 30.]
(c) If 1R = 0R then R = {0R } is the zero ring. Otherwise, for λ, µ ∈ C, we have λ1R = µ1R if and
only if λ = µ, so we can identify λ ∈ C with λ1R ∈ R.
With the identification C ,→ R we can view C as a subalgebra of R.
(d) An R-module M becomes a C-space via λm = (λ1R )m for λ ∈ C, m ∈ M .
Given any R-module N , HomR (M, N ) is a C-subspace of HomC (M, N ). In particular, it is a
C-space, with structure
(λϕ)(m) = λϕ(m) = ϕ(λm) ∀λ ∈ C, m ∈ M, φ ∈ HomR (M, N )
(e) If M be an R-module. Then EndR (M ) = HomR (M, M ) is a C-algebra with multiplication given
by composition of maps:
(ϕψ)(m) = ϕ(ψ(m))
∀m ∈ M, φ, ψ ∈ EndR (M )
In particular, if V is a C-space then EndC (V ) is an algebra. Recall that EndC (V ) ∼
= Mn (C), where
n = dimC (V ).
(f) Let R be a C-algebra and X ⊆ R be a subset. Define the centraliser of X in R by
cR (X) = {r ∈ R : rx = xr ∀x ∈ X}
It is a subalgebra of R. The centre of R is cR (R) = Z(R).
(g) Let R be C-algebra and M be a R-module. Then the map αR → EndC (M ) given by α(r)(m) = rm
for r ∈ R, m ∈ M is a map of C-algebras.
Now, EndR (M ) = cEndC (M ) (αR), and so αR ⊆ cEndC (M ) (EndR (M )).
(h) An R-module M is itself naturally an EndR (M )-module, where the action is evaluation, and
the action commutes with that of R. The inclusion α as in (g) says that elements of R act as
EndR (M )-module endomorphisms of M .
Given an R-module N , HomR (M, N ) is also an EndR (M )-module, where the action is composition
of maps.
4
(1.3) Lemma Let R be a finite-dimensional C-algebra. Then there are only finitely many isomorphism classes of irreducible R-modules. Moreover, all the irreducible R-modules are finite-dimensional.
Proof. Let S be an irreducible R-module and pick 0 6= x ∈ S. Define a map R → S by r 7→ rx. This
map has nonzero image, so its image is S by irreducibility. In particular,
dimC (S) ≤ dimC (R) < ∞
But S must occur in any composition series of R, so by the Jordan–Hölder theorem, there are only
finitely many isomorphism classes of irreducible modules.
(1.4) Lemma (Schur’s lemma)
Let R be a finite-dimensional C-algebra. Then
(a) If S T are nonisomorphic irreducible R-modules then HomR (S, T ) = 0.
(b) If S is an irreducible R-module then EndR (S) ∼
= C.
Proof
(a) If f : R → S is a map of algebras then ker(f ) ≤ R and im(f ) ≤ S. By irreducibility, ker(f ) = 0
or R and im(f ) = 0 or S. If ker(f ) = 0 then R ∼
= im(f ) = S, contradicting nonisomorphism; so
we must have ker(f ) = R and im(f ) = 0, i.e. f = 0.
(b) Clearly D = EndR (S) is a division ring. Since S is finite-dimensional, so is D. Thus, if d ∈ D,
then the elements 1, d, d2 , . . . are linearly dependent. Let p ∈ C[X] be a nonzero polynomial with
p(d) = 0.
Since C is algebraically closed, p factors as
p(X) = c(X − a1 ) · · · (X − an )
for some 0 6= c ∈ C and ai ∈ C. Hence
(d − a1 1D ) · · · (d − an 1D ) = 0
But, being a division ring, D has no zero divisors, so one of the terms must be zero. Thus
d = aj 1D ∈ C1D . But d ∈ D was arbitrary, so D = C1D ∼
= C.
(1.5) Definition An R-module is semisimple (or completely reducible) if it is a direct sum of
irreducible submodules. A finite-dimensional C-algebra R is semisimple if it is so as an R-module.
(1.6) Proposition
(a) Every submodule of a semisimple module is semisimple, and every such submodule is a direct
summand.
(b) Every quotient of a semisimple module is semisimple.
(c) Direct sums of semisimple modules are semisimple.
In fact, M is semisimple if and only if every submodule of M is a direct summand. The proofs of this
and of (1.6) are left as an exercise.
Remarks
1. R a semisimple C-algebra ⇒ any R-module is semisimple.
2. G a finite group ⇒ CG is semisimple. (This is precisely Maschke’s theorem.)
(1.7) Definition If R and S are C-algebras, M is an R-module and N is an S-module, then M ⊗ N
has the structure of an R-module via
r(m ⊗ n) = rm ⊗ n,
r ∈ R, m ∈ M, n ∈ N
5
and the structure of an S-module via
s(m ⊗ n) = m ⊗ sn,
s ∈ S, m ∈ M, n ∈ N
and these actions commute:
r(s(m ⊗ n)) = r(m ⊗ sn) = rm ⊗ sn = s(rm ⊗ n) = s(r(m ⊗ n))
hence the images of R, S in EndC (M ⊗ N ) commute.
If N has basis {e1 , . . . , ek } then the map M ⊕k → M × N given by
(m1 , . . . , mk ) 7→ m1 ⊗ e1 + · · · + mk ⊗ ek
is an isomorphism of R-modules.
Similarly, if M has basis {f1 , . . . , f` } then the map N ⊕` → M ⊗ N given by
(n1 , . . . , n` ) 7→ n1 ⊗ f1 + · · · + n` ⊗ f`
is an isomorphism of S-modules.
(1.8) Lemma
Let M be a semisimple R-module. Then the evaluation map
M
S ⊗ HomR (S, M ) → M
S
is an isomorphism of R-modules and of EndR (M )-modules. Here S runs over a complete set of nonisomorphic irreducible R-modules and we use the action of R on S and of EndR (M ) on HomR (S, M ).
Proof Check that the map is a map of R-modules and of EndR (M )-modules. It’s an isomorphism
of vector spaces, by observing that we can reduce to the case where M is irreducible and apply Schur’s
lemma (1.4).
(1.9) Lemma
Let M be a finite-dimensional semisimple R-module. Then
Y
EndC HomR (S, M )
EndR (M ) ∼
=
S
where S runs over a complete set of nonisomorphic irreducible R-modules.
Q
Proof Observe that the product E = S EndC HomR (S, M ) acts naturally on S ⊗ HomR (S, M )
and, since this action commutes with that of R, there exists a homomorphism E → EndR (M ), which
is injective.
Now count dimensions by (1.8), M is isomorphic to a direct sum of dimC HomR (S, M ) copies of each
irreducible module S, so by (1.4),
P
dimC EndR (S, M ) = S (dimC HomR (S, M ))2 = dimC E
(1.10) Theorem (Artin–Wedderburn Theorem) Any finite-dimensional semisimple C-algebra is
isomorphic to a product
k
Y
R=
EndC (Vi )
i=1
where Vi are finite-dimensional vector spaces. Conversely, if R has this form then it is semisimple, the
nonzero Vi s form a complete set of nonisomorphic irreducible R-modules and, as an R-module, R is
isomorphic to a direct sum of dimC Vi copies of each Vi .
Proof
We’ll take this in steps.
Step 1 (Deal with the product) Without loss of generality, assume each Vi is nonzero. The Vi are
naturally R-modules, with the factors other than EndC (Vi ) acting as 0.
6
Now EndC , and hence R, has transitively on Vi − {0}, thus the Vi are irreducible R-modules.
If Vi has basis {ei1 , . . . , eimi } then the map
R → V1 ⊕ · · · ⊕ V1 ⊕ · · · ⊕ Vk ⊕ · · · ⊕ Vk
|
{z
}
|
{z
}
m1 copies
given by
mk copies
(f1 , . . . , fk ) 7→ (f1 (e11 ), . . . , f1 (e1m1 ), . . . , fk (ek1 ), . . . , fk (ekmk ))
is an injective map of R-modules, and hence is an isomorphism by counting dimensions. Thus R is
semisimple.
The Vi form a complete set of irreducible R-modules (by Jordan–Höldeer, as in (1.3)), and they are
nonisomorphic: if i 6= j then (0, . . . , 0, 1, 0, . . . , 0) annihilates Vj but not Vi .
Step 2
Let R be any ring. Now the natural map
α : R → EndEndR (R) (R)
sending r to the map x 7→ rx is an isomorphism. It’s injective, and if θ ∈ EndEndR (R) (R) then it
commutes with the endomorphisms αr ∈ EndR (R), where αr (x) = xr. Now if r ∈ R then
θ(r) = θ(1r) = θ(αr (1)) = αr (θ(1)) = θ(1)r
so θ acts by left-multiplication. Hence θ = α(θ(1)) ∈ α(R), so α is surjective.
Assume R is a semisimple C-algebra. Then EndR (R) is semisimple by (1.9), and so R is a semisimple
EndR (R)-module. By (1.9) and the isomorphism α, R has the required form.
(1.11) Lemma
If R is semisimple and M is a finite-dimensional R-module then the natural map
α : R → EndEndR (M ) (M )
sending r to the map m 7→ rm is surjective.
Proof Let I = M ◦ = {r ∈ R : rM = 0} be the annihilator of M ; then ker α = I. Now, R/I is
semisimple and M is an R/I-module, and EndR (M ) = EndR/I (M ). So we can replace R by R/I and
suppose that M is faithful (and α is injective).
Q
By (1.9), EndR (M ) = S EndC HomR (S, M ) and, since M is faithful, all the spaces HomR (S, M )
are nonzero because they L
are precisely the simple EndR (M )-modules. By (1.8), M is isomorphic (as
an EndR (M )-module) to S S ⊗ HomR (S, M ), so it’s the direct sum of dimC S copies of the simple
module HomR (S, M ) for each S.
As in (1.9), this implies that
dimC EndEndR (M ) (M ) =
X
(dimC S)2
S
But this is equal to dimC R, so α is an isomorphism.
Finally, it’s worth noting:
(1.12) Lemma Let R be a C-algebra and let e ∈ R be ‘almost idempotent’ in the sense that e2 = λe
for some 0 6= λ ∈ C. Then for any R-module M , we have an isomorphism of EndR (M )-modules
HomR (Re, M ) ∼
= eM
Proof
eM .
eM is an EndR (M )-submodule of M : if ϕ ∈ EndR (M ) and em ∈ eM then ϕ(em) = eϕ(m) ∈
Replacing e by
e
λ,
we may suppose that e is idempotent. Now ther exists a map of EndR (M )-modules
Hom(Re, M ) → eM
given by ϕ 7→ ϕ(e), whose inverse sends m to the map r 7→ rm.
7
Chapter I: Representation theory of the symmetric group
2
Irreducible modules for the symmetric group
Recall the correspondence between representations ρ : G → GL(V ) and CG-modules given by setting
g · v = ρ(g)v for g ∈ G and v ∈ V . The trivial representation G → C× = GL(C) is given by g 7→ 1,
and the corresponding CG-module is C, sometimes written CG to emphasise the context.
If X ⊆ {1, . . . , n}, write SX for the subgroup of Sn fixing every number outside of X.
Let ε : Sn → {±1} be the sign representation, and write
Y
εσ =
1≤i<j≤n
σ(i) − σ(j)
i−j
Throughout this section, write A = CSn ; this is a finite-dimensional semisimple C-algebra.
(2.1) Definition For λ
i ∈ N0 , we say λ = (λ1 , λ2 , λ3 , . . . ) is a partition of n, and write λ ` n, if
P
λ1 ≥ λ2 ≥ λ3 ≥ · · · and i λi = n. Write (λa1 1 , λa2 2 , . . . , λakk ) to denote
(λ1 , . . . , λ1 , λ2 , . . . , λ2 , . . . , λk , . . . , λk , 0, 0, . . . )
| {z } | {z }
| {z }
a1
with λ1 > λ2 > · · · > λk .
a2
ak
Partitions of n correspond with conjugacy classes of Sn , e.g.
(5, 22 , 1) ↔ (• • • • •)(• •)(•)
If λ, µ ` n, write λ < µ if and only if there is some i ∈ N with λj = µj for all j < i and λi < µi . This
is the lexicographic (or dictionary) order on partitions. It is a total order on the set of partitions
of n. E.g.
(5) > (4, 1) > (3, 2) > (3, 12 ) > (22 , 1) > (2, 13 ) > (15 )
(2.2) Definition
Partitions λ ` n may be geometrically represented by their Young diagram
[λ] = {(i, j) : i ≥ 1, 1 ≤ j ≤ λi } ⊆ N × N
If (i, j) ∈ [λ] then it is called a node of [λ]. The k th row (resp. column) of a diagram consists of
those nodes whose first (resp. second) coordinate is k.
Example
If λ = (4, 22 , 1) then
[λ]
=
•
•
•
•
•
•
•
•
•
← λ1
← λ2
← λ3
The height (or length) of λ is the length of the first column of [λ], i.e. max{j : λj 6= 0}, and is
denoted by ht(λ).
(2.3) Definition A λ-tableau tλ is one of the n! arrays of integers obtained by replacing each node
in [λ] by one of the integers 1, . . . , n, with no repeats. E.g. the following are (4, 3, 1)-tableaux
t1
=
1
3
8
2
6
4
7
5
,
t2
=
4
2
6
5
1
7
8
3
Equivalently, a λ-tableau can be regarded as a bijection [λ] → {1, 2, . . . , n}.
8
Sn acts on the set of [λ]-tableaux, e.g. (1 4 7 8 6)(2 5 3) sends t1 to t2 . Formally, given a tableau tλ and
π ∈ Sn , the composition of the functions π and tλ gives a new λ-tableau πtλ , where (πtλ )(x) = π(tλ (x)).
Given a partition λ ` n, we want to pick one representative of all the corresponding λ-tableaux. Let
t◦λ be the standard λ-tableau, i.e. the tableau numbered in order from left to right and from top to
bottom, e.g.
1 2 3 4 5
6 7
◦
t(5,22 ,1) =
8 9
10
Define the row stabilizer Rt and column stabilizer Ct of a tableau t by
σ ∈ Rt
σ ∈ Ct
⇔
⇔
each i ∈ {1, . . . , n} is in the same row of t and σt
each i ∈ {1, . . . , n} is in the same column of t and σt
For instance, if t = t1 as in the examples after (2.3) above, then
Rt1
Ct1
(2.4) Definition
=
=
S{1,2,4,5} × S{3,6,7} × S{8}
S{1,3,8} × S{2,6} × S{4,7} × S{1}
Let tλ be a λ-tableau. The Young symmetrizer is
X
h(tλ ) =
εc rc ∈ A
r∈Rtλ , c∈Ctλ
Write hλ = h(t◦λ ).
Examples
(a) If λ = (n), then h = h(t(n) ) =
P
σ∈Sn
σ, called the symmetrizer in A.
Since ρ · h = h for all ρ ∈ Sn , Ah = Ch is the trivial representation.
P
(b) If λ = (1n ), then h = h(t(1n ) ) = σ∈Sn εσ σ, called the alternizer in A.
Since ρ · h = ερ h, Ah = Ch is the sign representation.
Goal The left-ideals in A of the form Ahλ for λ ` n (known as Specht modules) form a complete
set of nonisomorphic irreducible A-modules: see (2.14) below.
(2.5) Lemma
If λ ` n, t = tλ is a tableau, σ ∈ Sn .
(a) Rt ∩ Ct = 1;
(b) The coefficient of 1 in h(tλ ) is 1;
(c) Rσ(t) = σRt σ −1 and Cσ(t) = σCt σ −1 ;
(d) h(σtλ ) = σh(tλ )σ −1 ;
(e) Ah(tλ ) ∼
= Ahλ as A-modules.
Proof
(a)–(d) are clear.
For (e), tλ = σt◦λ for some permutation σ ∈ Sn . Postmultiplication by σ then gives an isomorphism
Ah(tλ ) ∼
= Ahλ .
The following lemma is sometimes called the basic combinatorial lemma.
(2.6) Lemma
true:
Let λ, µ ` n with λ ≥ µ, and let tλ and tµ be tableaux. Then one of the following is
9
(A) There exist distinct integers i and j occurring in the same row of tλ and the same column of tµ ;
(B) λ = µ and tµ = rctλ for some r ∈ Rtλ and c ∈ Ctλ .
Proof Suppose (A) is false. If λ1 6= µ1 then λ1 > µ1 , so [µ] has fewer columns than [λ]. Hence two
of the numbers in the first row of tλ are in the same column of tµ , so (A) holds, contradicting our
assumption. So λ1 = µ1 .
Since (A) fails, some c1 ∈ Ctµ forces c1 tµ to have the same members in the first row of tλ . Now ignore
the first rows of tλ and c1 tµ . The same argument implies that λ2 = µ2 and there is c2 ∈ Ctµ such that
tλ and c2 c1 tµ have the same numbers in each of their first two rows.
Continuing in this way we see that λ = µ and there exists c0 ∈ Ctλ such that tλ and c0 tµ have the
same numbers in each row. Then rtλ = c0 tµ for some r ∈ Rtλ .
Now define
c = r−1 (c0 )−1 r ∈ r−1 c0 Ctµ (c0 )−1 r = r−1 Cc0 tµ r = Cc0 tµ = Ctλ
Then tµ = rctλ .
(2.7) Lemma If σ ∈ Sn cannot be written as rc for any r ∈ Rtλ and c ∈ Ctλ then there exist
transpositions u ∈ Rtλ and v ∈ Ctλ such that uσ = σv.
Proof Since (2.6)(B) fails for tλ and σtλ , there exist integers i 6= j in the same row of tλ and the
same column of σtλ . Take u = (i j) and v = σ −1 uσ.
(2.8) Lemma
Given a tableau tλ and a ∈ A, the following are equivalent:
(1) rac = εc a for all r ∈ Rtλ and c ∈ Ctλ ;
(2) a = kh(tλ ) for some k ∈ C.
Proof
(2)⇒(1) As r0 runs through Rtλ so does rr0 , and as c runs through Ctλ so does c0 c with εc0 c = εc0 εc .
Calculation reveals that rh(tλ )c = εc h(tλ ).
P
(1)⇒(2) Let a = σ∈Sn aσ σ. If σ is not of the form rc then by (2.7) there exist transpositions u ∈ Rtλ
and v ∈ Ctλ such that uσv = σ.
By assumption, uav = εv a, and the coefficient of σ gives auσv = εv aσ , so aσ = −aσ , and hence
aσ = 0. The coefficient of rc in (1) gives a1 = εc arc ∈ C. Thus
X
X
a=
arc rc =
εc a1 rc = a1 h(tλ )
r,c
rc
as required
(2.9) Lemma
Proof
If a ∈ A then h(tλ )ah(tλ ) = kh(tλ ) for some k ∈ C.
x = h(tλ )ah(tλ ) satisfies (2.8)(1).
(2.10) Proposition
(a) h(tλ )2 =
Define fλ = dimC Ahλ . Then
n!
h(tλ )
fλ
(b) fλ | n!
(c) h(tλ )Ah(tλ ) = Ch(tλ )
In particular,
fλ
h(tλ ) is idempotent.
n!
Proof
10
(a) Let h = h(tλ ). We already know that h2 = kh for some k ∈ C. Right-multiplication by h induces
a linear map ĥ : A → A. For a ∈ A, (ah)h = k(ah), so k̂|Ah is multiplication by k.
Extend a basis of Ah to a basis of A. With respect to this basis,
kIfλ ∗
ĥ ∼
0
0
and so tr(ĥ) = kfλ .
With respect to the basis Sn of A, ĥ has matrix H whose entries are
Hστ = coefficient of σ in τ h
But Hστ = 1, so tr(ĥ) = n!.
n!
Hence k =
.
fλ
(b) The coefficient of 1 in h2 = kh is
c1 , c2 ∈ Ctλ such that r1 c1 r2 c2 = 1.
P
εc1 εc2 ∈ Z where the sum runs over all r1 , r2 ∈ Rtλ and
(c) By (2.9), the only other possibility is hAh = 0; but h2 6= 0.
(2.11) Proposition
Ah(tλ ) is an irreducible A-module.
Proof Put h = h(tλ ). Then Ah 6= 0 and A is semisimple, so it is enough to show that Ah is
indecomposable.
Suppose Ah = U ⊕ V . Then Ch = hAh = hU ⊕ hV , so one of hU and hV is nonzero; without loss of
generality, hU 6= 0. Then hU = Ch, so Ah = AhU ⊆ U , so U = Ah and V = 0.
(2.12) Lemma
If λ > µ are partitions and tλ , tµ are tableaux then h(tµ )Ah(tλ ) = 0.
Proof By (2.6) there exist integers in the same row of tλ and the same column of tµ . Let τ ∈
Rtλ ∩ Ctµ be the corresponding transposition. Then
h(tµ )h(tλ ) = h(tµ )τ τ h(tλ ) = −h(tµ )h(tλ )
so h(tµ )h(tλ ) = 0.
Applying this to (σtλ , tµ ) for σ ∈ Sn gives
0 = h(tµ )h(tλ ) = h(tµ )σh(tλ )σ −1
and so h(tµ )σh(tλ ) = 0.
Thus h(tµ )Ah(tλ ) = 0.
(2.13) Lemma
If λ 6= µ and tλ , tµ are tableaux, then Ah(tλ ) Ah(tµ ).
Proof Assume without loss of generality that λ > µ. If there exists an A-module isomorphism
f : Ah(tλ ) ∼
= Ah(tµ ) then
(2.10)
(2.12)
f (Ch(tλ )) = f (h(tλ )Ah(tλ )) = h(tλ )f (Ah(tλ )) = h(tλ )Ah(tµ ) =
which is a contradiction.
(2.14) Theorem
A-modules.
Proof
0
The left-ideals {Ahλ : λ ` n} form a complete set of nonisomorphic irreducible
They are irreducible by (2.11) and nonisomorphic by (2.13) and, since
|{λ : λ ` n}| = |{ccls of Sn }| = |{irreducible CSn -modules}|
they form a complete set.
Further reading Chapter 4 of James’s lecture notes (Springer) – alternative proof using ‘polytabloids’ and certain permutation modules.
11
3
Standard basis of Specht modules
We first redefine what we mean by a ‘standard’ tableau.
(3.1) Definition t is a standard tableau if the numbers increase along the rows (left to right) and
columns (top to bottom) of t.
The standard tableaux of shape λ are ordered such that tλ < t0λ if tλ is smaller than t0λ in the first
place they differ when you read [λ] ‘like a book’. E.g. the standard (3, 2)-tableaux are
1
4
2
5
3
<
1
3
2
5
4
1
3
<
2
4
5
<
1
2
3
5
4
<
1
2
3
4
5
Let Fλ be the number of standard λ-tableaux.
Fλ = fλ ; recall from (2.10) that fλ = dimC Ahλ .
X
First we show
Fλ2 = n!.
Goal
λ`n
Notation
λ/µ means there is some m with λ ` m, µ ` m − 1 and [µ] ⊆ [λ] ⊆ N × N.
This idea will be used in induction arguments.
X
(3.2) Lemma Let λ ` m. Then Fλ =
Fµ .
µ : λ/µ
Proof If tλ is standard then t−1
λ ({1, 2, . . . , m − 1}) is the diagram of a partition µ ` m − 1; and
tλ |[µ] must be standard. The converse is similar.
(3.3) Lemma
If λ 6= π are partitions of m then
|{ν : ν/λ and ν/π}| = |{τ : λ/τ and π/τ }| ∈ {0, 1}
Proof If ν/λ and ν/π then [ν] ⊇ [λ] ∪ [π]. Since [λ] and [π] differ in at least one place, we must in
fact have [ν] ⊇ [λ] ∪ [π].
Likewise, if λ/τ and µ/τ then [τ ] ⊇ [λ] ∩ [π] and we must have [τ ] = [λ] ∩ [π].
Now [λ] ∪ [π] and [λ] ∩ [π] are partitions, so
ν exists ⇔ |[λ] ∪ [π]| = m + 1 ⇔ |[λ] ∩ [π]| = m − 1 ⇔ τ exists
This completes the proof.
(3.4) Lemma
If λ ` m then (m + 1)Fλ =
X
Fν .
ν : ν/λ
Proof by induction on m.
The case m = 1 is clear. Suppose the statement holds true for partitions of m − 1. Then


X
X
X
(3.2) X

Fν =
Fπ  = |{ν : ν/λ}|Fλ +
Fπ
ν : ν/λ
ν : ν/λ
π : ν/π
ν,π : ν/λ,π6=λ
Using the fact that |{ν : ν/λ}| = |{τ : λ/τ }| + 1 together with (3.3) gives that our expression is
equal to
X
(|{τ : λ/τ }| + 1)Fλ +
Fπ
τ,π : λ/τ,π/τ,π6=λ
= Fλ +
X
τ,π : λ/τ,π/τ
Fπ = Fλ +
X
τ : λ/τ
12
mFτ = Fλ + mFλ = (m + 1)Fλ
as required.
X
(3.5) Lemma
Fλ2 = m!
λ`m
Proof by induction on m.
The case m = 1 is clear. Suppose the statement holds true for partitions of m − 1. Then
X
λ`m
Fλ2
(3.2)
=
X
Fλ Fτ
X
(3.4)
=
τ `m−1
λ`m,λ/τ
mFτ2 = m · (m − 1)! = m!
as required.
(3.6) Lemma
If tλ > t0λ are standard then h(tλ )h(t0λ ) = 0.
Proof It suffices to show that there are integers i 6= j lying in the same row of t0λ and the same
column of tλ . For then the corresponding transposition τ = (i j) ∈ Rt0λ ∩ Ctλ gives the result as in the
proof of (2.12).
It remains to find such integers i and j. Let x ∈ [λ] be the first place where tλ and t0λ first differ. Let
t0λ = i and y = (t0λ )−1 (i) ∈ [λ]. Then y must be below and to the left of x since tλ is standard. In
particular, x cannot lie in the first column or the last row. Let z ∈ [λ] be in the same row as x and
the same column of y, and let j = tλ (z) = t0λ (z) be the common value of tλ and t0λ at z.
It is now elementary to verify that i and j satisfy the above assumption.
L
(3.7) Theorem CSn =
CSn h(tλ ), where the direct sum runs over all standard tableaux tλ for
all partitions λ ` n.
P
Proof The sum is direct: suppose we have a nontrivial relation
a(tλ )h(tλ ) with a(tλ ) ∈ CSn
are not all zero. Choose µ maximal (in the ordering on partitions) such that some a(tµ )h(tλ ) 6= 0,
and for this µ pick t0µ minimal (in the ordering on standard µ-tableaux) such that a(t0µ )h(t0µ ) 6= 0.
Multiplying the relation on the right by h(t0µ ) gives a(t0µ )h(t0µ )2 = 0 by (3.6) and (2.12), and hence
a(t0µ )h(t0µ ) = 0, contradicting our assumption.
L
L
fλ CSn hλ as CSn -modules,
It is clear that the
CSn h(tλ ) ≤ CSn . By Artin–Wedderburn, CSn ∼
=
while in our
direct
sum
there
are
F
copies
of
Ch
.
By
Jordan–Hölder,
F
≤
λ
λ
λ
P 2
P
Lfλ . But fλ ≤ Fλ since,
by (3.5),
Fλ = n! = fλ2 . Hence Fλ = fλ for each λ ` n, and so CSn ≤
CSn h(tλ ).
(3.8) Corollary
The number of standard λ-tableaux is equal to dimC (CSn hλ ).
L
h(tλ )M , where the direct
(3.9) Lemma Let M be a finite-dimensional CSn -module. Then M ∼
=
sum runs over all standard tableaux tλ for all partitions λ ` n.
P
Proof If
m(tλ ) = 0 is a nontrivial relation with m(tλ ) ∈ h(tλ )M then choose µ minimal and t0µ
maximal with m(t0µ ) 6= 0. Premultiply the relation by h(t0µ ) to contradict (3.6) and (2.12). Finally,
Proof
Hλ = hλ as in the proof of (3.7).
M
∼
= HomCSn
(1.12)
h(tλ )M
M
∼
=
M
HomCSn (CSn h(tλ ), M )
CSn h(tλ ), M ∼
= HomCSn (CSn , M ) ∼
=M
13
4
Character formula
Given a finite-dimensional CG-algebra M , its character is defined by
M →M
χM (g) = tr
m 7→ gm
and χM is a class function G → C, i.e. a function that is constant on the conjugacy classes of G. If
α is a conjugacy class in G, write χM (α) for the common value of χM on α.
Given λ ` n, denote the character of the irreducible CSn -module CSn hλ by χλ . We will calculate
χλ (α) and later use the formula to derive Weyl’s character formula for GLn (C).
Now Sn -class α comprises all permutations with fixed cycle type nαn · · · 2α2 1α1 , i.e. having αn n-cycles,
. . . , α2 2-cycles and α1 1-cycles. Let nα = |α|.
(4.1) Lemma
Proof
nα =
n!
1α1 2α2 . . . nαn · α1 !α2 ! . . . αn !
Any permutation in α is one of the n! of the form
n
n
n
z }| { z }| {
z }| {
(∗)(∗) · · · (∗) (∗∗)(∗∗) · · · (∗∗) · · · (∗ · · · ∗)(∗ · · · ∗) · · · (∗ · · · ∗)
|
{z
}|
{z
} |
{z
}
α1
α2
αn
αn
α1
Each such permutation can be represented in 1 . . . n · α1 ! . . . αn ! ways: each of the αk k-cycles can
be represented in k ways by cycling their components, and each block of αk k-cycles can be represented
in αk ! ways by permuting the position of the factors.
(4.2) Theorem (orthogonality
Then
(
X
n!
λ
µ
(a)
nα χ (α)χ (β) =
0
ccls α
(
X
n!/nα
(b)
χλ (α)χλ (β) =
0
λ`n
Proof
relations)
Let λ, µ ` n and let α and β be conjugacy classes in Sn .
if λ = µ
if λ =
6 µ
if α = β
if α =
6 β
Every element of Sn is real, i.e. conjugate to its inverse, and so
χλ (σ) = χλ (σ −1 ) = χλ (σ)
so χλ (σ) ∈ R for all σ ∈ Sn . These relations are then just the usual ones for general finite groups. Notation
Given x = (x1 , . . . , xn ) ∈ Cm , `1 , . . . , `m ∈ Z, define
`
|x`1 , . . . , x`m | = det(xi j )
Usually `j ≥ 0, in which case this is a homogeneous polynomial of degree
m
X
`j in x1 , . . . , xm .
j=1
Example The Vandermonde determinant is given by
V (x) = |xm−1 , . . . , 1|
(4.3) Lemma
Proof
V = V (x) =
Exercise.
Q
i<j (xi
− xj )
In fact, if `i ≥ 0 for each i then |x`1 , . . . , x`m | is divisible by V . We define the Schur polynomial s`
by
|x`1 , . . . , x`m |
s` (x) = m−1
|x
, . . . , 1|
14
This is indeed a polynomial in x1 , . . . , xm , provided that `i ≥ 0.
Example
s(1,2,...,n) (x) = x1 · · · xm .
If xi , yj ∈ C for 1 ≤ i, j ≤ m and xi yj 6= 1 for all i, j, then
(4.4) Lemma (Cauchy’s lemma)
det
1
1 − xi yj
Proof by induction on m.
m
Y
= xm−1 , . . . , 1 y m−1 , . . . , 1
1
1
−
xi yj
i,j=1
The case m = 1 is clear. Let m > 1.
Now
1
xi − x1
yj
1
−
=
·
1 − xi yj
1 − x1 yj
1 − x1 yj 1 − xi yj
So subtracting the first row from each other row in the determinant, one can remove the factor (xi −x1 )
1
from each row i 6= 1 and
from each column. So
1 − x1 yj
det
1
1 − xi yj
=
!
Y
m
Y
1
(xi − x1 ) 
1
−
x1 yj
i>1
j=1


1
1
y1
 1−x
 y12 y1
 det 
 1−x3 y1
|


y2
1−x2 y2
y2
1−x3 y2
..
.
..
.
y2
1−xn y2
y1
1−xn y1
{z
···
···
···
..
.
···
1
yn
1−x2 yn
yn
1−x3 yn
..
.
yn
1−xn yn
=∆







}
Now subtract the first column in the matrix above from the other columns and use the fact that
y1
yj − y1
1
yj
−
=
·
1 − xi yj
1 − xi y1
1 − xi y1 1 − xi yj
This gives

1

∗
!
m

Y
Y
1

∆ =  (yj − y1 )
det ∗
.
1
−
x
y
i
1
j>1
i=2
 ..
∗


0
0
1
1−x2 y2
1
1−x3 y2
1
1−x2 y3
1
1−x3 y3
1
1−xn y2
1
1−xn y3
..
.
..
.
···
···
···
..
.
···
0
1
1−x2 yn
1
1−x3 yn
..
.
1
1−xn yn
The result now follows by applying the induction hypothesis to this last determinant.
(4.5) Lemma
Choose xi , yj ∈ C for 1 ≤ i, j ≤ n with modulus < 1. Then
X
`
1
x 1 , . . . , x `m y `1 , . . . , y `m det
=
1 − xi yj
`1 >`2 >···>`m ≥0
Proof
We use the expansion
X
1
=
(xi yj )`
1 − xi yj
`≥0
The determinant is thus
X
σ∈Sm
εσ
m Y
i=1
1 + xi yσ(i) + (xi yσ(i) )2 + · · ·
and the monomial x`11 · · · x`mm (`i ∈ N) occurs with coefficient
m
X Y
σ∈Sm i=1
`i
yσ(i)
= y `1 , . . . , y `m 15







In particular, it this term is zero unless the `i are all distinct. So
X
1
det
=
x`11 . . . x`mm y `1 , . . . , y `m 1 − xi yj
`1 ,...,`m distinct
X
=
X
`1 >···>`m ≥0
π∈Sm
X
X
=
`1 >···>`m ≥0
X
=
`1 >···>`m ≥0
which is what we needed.
`
y π(1) , . . . , y `π(m) `
x1π(1)
`
. . . xmπ(m)
`
x1π(1)
`
. . . xmπ(m) επ
π∈Sm
`
y 1 , . . . , y `m !
!
`
x 1 , . . . , x`m y `1 , . . . , y `m Notation If λ = (λ1 , . . . , λm ) ∈ Zm , let `i = λi + m − i, i.e. `1 = λ1 + m − 1, . . . , `m = λm .
Sometimes in the literature, the `i are denoted by βi and are thus called ‘β-numbers’. Note that
(1) λ1 ≥ · · · ≥ λm if and only if `1 > · · · > `m
`
x 1 , . . . , x `m P
(2) If i λi = n and λi ≥ 0 then m−1
is a polynomial of degree m in the xi s.
|x
, . . . , 1|
Write Λ+ (m, n) for the set of all partitions λ of n into ≤ m parts.
(4.6) Definition
Given x1 , . . . , xm , y1 , . . . , ym ∈ C, then for any i ∈ N, set
si = xi1 + · · · + xim ,
ti = y1i + · · · + tim
These are called the power sums (sometimes referred to in the literature as Newton sums).
`
x 1 , . . . , x `m y `1 , . . . , y `m X
1 X
αn α1
αn
1
(4.7) Lemma
=
nα sα
1 . . . sn t1 . . . tn
|xm−1 , . . . , 1| |y m−1 , . . . , 1|
n!
+
ccls α
λ∈Λ (m,n)
Note that both quotients on the left-hand side genuinely are polynomials (in fact, Schur polynomials),
so they make sense even if the xi , yi are not distinct.
Proof Both sides are polynomials, so without loss of generality we may take all the xi , yi to have
modulus < 1. Then
n
n Y
X
xi yj
(xi yj )2
(xi yj )3
1
=
+
+
+ ···
log
1 − xi yj
1
2
3
i,j=1
i,j=1
=
s1 t1
s2 t2
s3 t3
+
+
+ ···
1
2
3
and hence
n
Y
s1 t1
s2 t2
s3 t3
+
+
+ ···
1
2
3
n
∞
X
1 s1 t1
s2 t2
s3 t3
=
+
+
+ ···
n!
1
2
3
n=0
α α
X 1
n!
s1 t1 1 s2 t2 2
=
·
···
n! α1 !α2 ! · · ·
1
2
1
= exp
1
−
xi yj
i,j=1
(∗)
where (∗) denotes the sum over all n and all sequences α1 , α2 , . . . of nonnegative integers with only
finitely many nonzero terms and α1 + α2 + · · · = n. But then
n
Y
X sα1 sα2 · · · tα1 tα2 · · ·
1
1 2
1 2
=
α1 2α2 · · · α !α ! · · ·
1
−
x
y
1
i
j
1 2
i,j=1
(∗)
16
By (4.4)(Cauchy) and (4.5), we have
`
m
x 1 , . . . , x`m y `1 , . . . , y `m X sα1 sα2 · · · tα1 tα2 · · ·
X
Y
1
1 2
1 2
=
=
α1 2α2 · · · α !α ! · · ·
|xm−1 , . . . , 1| |y m−1 , . . . , 1|
1
−
x
y
1
1 2
i
j
i,j=1
`1 >`2 >···>`m ≥0
(∗)
so we can equate terms which are of degree n in the xi to obtain the desired result.
(4.8) Definition For λ = (λ1 , . . . , λm ) ∈ Zm and α a conjugacy class in Sn , let ψλ (α) be the
αn
1
coefficient of the monomial xλ1 1 . . . xλmm in sα
1 . . . sn , i.e.
X
αn
1
ψλ (α)xλ1 1 . . . xλmm
sα
1 . . . sn =
λ∈Zm
Remarks
(1) ψλ is a class function for Sn .
(2) ψλ (α) = 0 if any λi is negative or if
m
X
i=1
λi 6= n.
(3) ψλ is a symmetric function of the λi s.
Notation
Set
ωλ (α) =
X
επ ψ(`π(1) +1−m,
`π(2) +2−m, ... , `π(m) ) (α)
π∈Sm
The idea now is to show that ωλ = χλ whenever λ ` n.
(4.9) Lemma (Character formula)
αn m−1
1
sα
x
, . . . , 1 =
1 . . . sn
X
λ∈Λ+ (m,n)
ωλ (α) x`1 , . . . , x`m Proof The left-hand side is equal to
X
X
0
ψλ (α)xλ1 1 . . . xλmm
ετ xm−1
τ (1) . . . xτ (m)
λ∈Zm
τ ∈Sm
which, by combining terms, is equal to
X X
λ
τ (1)
ετ ψλ (α)xτ (1)
+m−1
λ
τ (m)
. . . xτ (m)
λ∈Zm τ ∈Sm
Let `i = λτ (i) +m−i—note that this is not our usual convention—then since the ψλ (α) are symmetric,
our expression is equal to
X X
ψ(`1 +1−m, ... , `m ) (α)x`τ1(1) . . . x`τm(m)
λ∈Zm τ ∈Sm
=
X
ψ(`1 +1−m,
... , `m ) (α)
λ∈Zm
`
x 1 , . . . , x`m Whenever the `i are not distinct, the corresponding term gives zero. So setting λi = `i + i − m and
using the fact that `1 > · · · > `m if and only if λ1 ≥ · · · ≥ λm , our expression is equal to
X
X
ψ(`π(1) +1−m, ... , `π(m) ) επ x`1 , . . . , x`m λ1 ≥···≥λm π∈Sm
The terms for which λ is not a partition of n are zero, for if λm < 0 then `m + π −1 (m) − m < 0. Thus
we have the desired equality.
17
Let λ, λ0 ` n be partitions of ≤ m parts. Then
(
X
n! if λ = λ0
nα ωλ (α)ωλ0 (α) =
0 if λ 6= λ0
ccls α
(4.10) Lemma (Orthogonality of ω)
Proof
X
By (4.7) we have
λ∈Λ+ (m,n)
X
`
αn α1
αn m−1
1
x 1 , . . . , x`m y `1 , . . . , y `m = 1
x
, . . . , 1 y m−1 , . . . , 1
nα sα
1 . . . sn t1 . . . tn
n!
ccls α
By (4.9) this is equal to

X
1

n!
ccls α
X
λ,λ0 ∈Λ+ (m,n)

0
`
0 nα ωλ (α)ωλ0 (α) x 1 , . . . , x`m y `1 , . . . , y `m 
where `i come from λi and `0i come from λi in the usual way.
0
0 As λ, λ0 vary, the polynomials x`1 , . . . , x`m and y `1 , . . . , y `m are linearly independent in the polynomial ring C[x1 , . . . , xm , y1 , . . . , ym ], so the result follows.
Now that we have trudged through lots of combinatorics we will reintroduce the symmetric group.
(4.11) Lemma
If λ ∈ Λ+ (n, m) then ψλ (α) is the character of the CSn -module CSn rλ , where
X
rλ =
σ
σ∈Rt◦
λ
Proof (X-rated) Take the character χ and of CSn rλ and suppose σ ∈ α, where α is a conjugacy
class in Sn . Write R = Rt◦λ , and find a coset decomposition
Sn =
N
[
gi R
i=1
for the transversal {1 = g1 , g2 , . . . , gN }. (Recall: a transversal is a minimal set of coset representatives.) Then CSn rλ has basis {gi rλ }1≤i≤N , which we now use to calculate the traces. The CSn -action
is given by σgi rλ = gj rλ for σ ∈ Sn , where σgi ∈ gj R. Thus
χ(α) = {1 ≤ i ≤ N : gi−1 σgi ∈ R}
We will now take this and bash it very hard lots of times until enough juice comes out to give us what
we want.
Qm
Now g −1 σg ∈ R if and only if g ∈ gi R, where gi−1 σgi ∈ R, and |R| = 1 λj !, so
1
χ(α) = Qm
1
λj !
{g ∈ Sn : g −1 σg ∈ R}
Now since g −1 σg = h−1 σh if and only if gh−1 ∈ CSn (σ), and each value taken by g −1 σg for g ∈ Sn is
taken by |CSn | elements. By (4.1) we have
1
χ(α) = Qm
1α1 2α2 . . . nαn α1 !α2 ! . . . αn ! |α ∩ R|
1 λj !
since |CSn (σ)| = n!/nα . Now a permutation τ ∈ α ∩ R restricts to a permutation of the numbers in
the ith row of t◦λ . If this restriction involves, say, αij j-cycles, then the αij satisfy
αi1 + 2αi2 + · · · + nαin = λi (1 ≤ i ≤ m)
(∗)
α1j + α2j + · · · + αmj = αj (1 ≤ j ≤ n)
18
The number of permutations in R of this type is
λ2 !
λm !
λ1 !
·
·
·
1α11 . . . nα1n α11 ! . . . α1n !
1α21 . . . nα2n α21 ! . . . α2n !
1αm1 . . . nαmn αm1 ! . . . αmn !
and so
X χ(α) =
α1 !
1α11 . . . nα1n α11 ! . . . α1n !
···
αm !
1αm1 . . . nαmn αm1 ! . . . αmn !
(†)
where the sum runs over all αij satisfying (∗). By the multinomial theorem,
α1m
11
X nm
. . . xm
α1 !xα
αm !x1αn1 . . . xα
m
αn
1
1
.
.
.
s
=
sα
·
·
·
n
1
1α11 . . . nα1n α11 ! . . . α1n !
1αm1 . . . nαmn αm1 ! . . . αmn !
where the sum runs over all αij ∈ N satisfying
α1j + α2j + · · · + αmj = αj
for all 1 ≤ j ≤ n. So ψλ (α), which is the coefficient of xλ1 1 . . . xλmm , is equal to the right-hand side of
(†), and hence it equals χ(α).
(4.12) Lemma Let λ, µ ` n with µ ≤ λ. The irreducible module CSn hµ is isomorphic to a
submodule of CSn rλ if and only if λ = µ.
Proof With µ < λ and σ ∈ Sn , by (2.6) there exist two integers in the same row of t◦λ and the same
column of σ −1 t◦µ , so if τ is their transposition then στ σ −1 ∈ Ct◦µ . And
hµ σrλ = hµ στ σ −1 στ rλ = −hµ σrλ
⇒
hµ σrλ = 0
Thus 0 = hµ CSn rλ ∼
= HomCSn (CSn hµ , CSn rλ ) by (1.12).
Conversely suppose µ = λ. Then
hλ rλ
X
σ∈Ct◦
εσ σ = h2λ 6= 0
λ
so 0 6= hλ CSn rλ ∼
= HomCSn (CSn hλ , CSn rλ ).
Note that, in general, CSn hλ is not a submodule of CSn rλ .
(4.13) Lemma
Proof
Step 1
If λ ∈ Λ+ (n, m) then ωλ = χλ .
We’ll take this in steps.
ωλ is a Z-linear combination of ψν with ν ≥ λ and with coefficient of ψλ equal to 1.
If π ∈ Sn and µπ is the partition with parts
`π(1) + 1 − m, . . . , `π(m)
i.e. with parts λi + π −1 (i) − 1. Since ψλ (α) is symmetric in the λi , we can write
X
ωλ =
επ ψµπ
π∈Sn
If π = 1 then µπ = λ, so we’re done.
If π 6= 1, then λ1 +π −1 (1)−1 ≥ λ (with equality if and only if π −1 (1) = 1; and then λ2 +π −1 (2)−2 ≥ λ2
with equality if and only if π −1 (2) = 2; and so on. Thus µπ > λ.
/
X
Step 2 ωλ =
kλν χν with kλν ∈ Z, kλλ > 0 and kλν = 0 if ν < λ.
ν`n
By (4.11) and (4.12), ψλ is an N-linear combination χµ s with µ ≥ λ and with nonzero coefficient of
χλ . Thus ωλ is a Z-linear combination of χν s with ν ≥ λ and with positive coefficient of χλ .
/
19
Step 3
The result follows.
We know by (4.10) that
(
n! if λ = µ
nα ωλ (α)ωµ (α) =
0 if λ 6= µ
ccls α
X
2
In the case λ = µ, orthogonality of χλ implies
kλν
= 1, and thus kλν = 0 if λ 6= µ and kλλ = 1. /
X
ν`n
So we’re done.
The kλν s in the above proof are sometimes referred to as Kostka numbers in the literature.
At long last, take m ∈ N, x1 , . . . , xm ∈ C arbitrary, `i = λi + m − i for λ ` n, and si = xi1 + · · · + xim .
X
αn m−1
1
x
, . . . , 1 =
χλ (α) x`1 , . . . , x`n (4.14) Theorem sα
1 . . . sn
λ∈Λ+ (m,n)
Proof
(4.9) and (4.13).
Remarks
(1) With m ≥ n we can ensure that the right-hand side of the equation in (4.14) involves all partitions
of n.
(2) If λ ∈ Λ+ (m, n) then χλ (α) is the coefficient of the monomial x`11 . . . x`mm in the expansion of
n
s1α1 . . . sα
n .
20
5
The hook length formula
We already know by (3.8) that the dimension of an irreducible CSn -module is the number of standard
tableaux. We give some results which allow for easy calculation of this dimension.
If λ ∈ Λ+ (n, m) then
Q
1≤i<j≤n (`i − `j )
Qn
fλ = deg χλ = n! ·
i=1 `i !
(5.1) Theorem (Young–Frobenius formula)
Proof
deg χλ = χλ (1) is the coefficient of x`11 . . . x`mm in the expansion of
X
τ (1)−1
(x1 + · · · + xm )n ετ x1
. . . xτm(m)−m
(x1 + · · · + xm )n xm−1 , . . . , 1 =
τ ∈Sn
By the multinomial theorem, this coefficient equals
X
n!
(`1 + 1 − τ (1))! . . . (`m + 1 − τ (m))!
τ ∈Sn
1
1
1
= n! ,...,
, (` − m + 1)!
(` − 1)! `!
1
= n! · Qm
|. . . , `(` − 1), `, 1|
j=1 `j !
m−1
1
`
, . . . , `2 , `, 1
= n! · Qm
j=1 `j !
This determinant is a Vandermonde determinant, so applying (4.3) gives the result.
(5.2) Definitions
(a) Let λ ` n. The (i, j)-hook of λ is the intersection of an infinite Γ–shape having (i, j) as its corner
with [λ]. For example, the (2, 2)–hook of (42 , 3) is shown as stars (?) in the following tableau:
•
•
•
• • •
? ? ?
? •
(b) The (i, j)-hook of [µ] comprises the (i, j)-node along with the (µi − j) nodes to the right of it
(the arm) and the (µ0j − i) nodes below it (the leg). [Here, [µ] has column lengths µ01 , µ02 , . . . .]
The length of the (i, j)-hook is hij = µi + µ0j + 1 − i − j.
(c) Replacing the (i, j)-node of [µ] by hij for each
hook graph of (42 , 3) is
6 5
5 4
3 2
node gives the hook graph. For example, the
4
3
1
2
1
(d) A skew hook is a connected part of the rim of [µ] which can be removed to leave a proper
diagram. For example, the skew 4-hooks of (42 , 3) are
•
•
•
• • •
• ? ?
? ?
and
•
•
•
•
•
•
•
?
?
?
?
There is a natural one-to-one correspondence between hooks and skew hooks in a given tableau.
For each node (i, j) there is a unique skew hook starting in the ith row and ending in the j th
column, and it corresponds with the (i, j)-hook.
21
Theorem (5.3) (hook length formula) [Frame, Robinson & Thrall, 1954]
n!
fλ = Q
(i,j)∈[λ]
Proof
hij
Suppose λ has m parts. By (5.1) we have to show that
m
Y
(`i − `k )
k=i+1
λi
Y
hij = `i !
for each i
j=1
The (combined) product on the left-hand side is a product of λi + m − 1 = `i terms, so it suffices to
show that they are precisely 1, 2, . . . , `i in some order. Now
`i − `m > `i − `m−1 > · · · > `i − `i+1
λ01
hi1 > hi2 > · · · > hiλi
Since λ has m parts and
is the length of the first column, λ01 = m, and hi1 = λi . So each term is
≤ `i . Thus it is enough to show that hij 6= `i − `k for any j, k.
However if r = λ0j then λr ≥ j and λr+1 < j, so
hij − `i + `r+1 = (λi + r − 1 − j + 1) − (λi + m − i) + (λr + m − r) = λr + 1 − j > 0
hij − `i + `r+1 = (λi + r − 1 − j + 1) − (λi + m − i) + (λr+1 + m − r − 1) = λr+1 − j < 0
So `i − `r < hij < `i − `r+1 .
This theorem is often referred to as the FRT theorem in the literature.
Example
When n = 11 and λ = (6, 3, 2), the hook graph is given by
8
4
2
and so
fλ =
7
3
1
6
1
5
3
2
1
11!
= 990
8.7.5.4.32 .22 .13
Remarks
(1) The determinantal form of the Young–Frobenius formula is given by
1
fλ = n! (λi − i + j)! where λ ∈ Λ+ (n, m) and the determinant on the right is of an m × m matrix and we adopt the
1
convention
= 0 if m < 0. For a proof see e.g. Sagan or James §19.
m!
X
(2) The equation
fλ2 = n! can be proved as a purely combinatorial statement without reference to
λ`n
representations, using the so-called ‘RSK algorithm’ (for Robinson–Schensted–Knuth).
(3) The Murnaghan–Nakayama rule (e.g. example sheet 1, question 9) states that if πρ ∈ Sn with
ρ an r-cycle and π a permutation of the remaining n − r numbers then
X
χλ (πρ) =
(−)i χν (π)
ν
where ν runs over partitions for which [λ] − [ν] is a skew r-hook of leg length i. (Note: leg length
refers to the leg of the hook with which the skew hook corresponds.)
22
For example,
conj. class
2
χ(5,4
z }| {
2
)
((5, 4, 3, 1)) = −χ(5,3) + χ(3 ,2)
=
(22 )
(3,1)
− (χ
χ
|
| {z }
from (5,3)
(1)
= −χ
+0+0
on (4, 3, 1)
+χ
{z
(2,12 )
from (32 ,2)
= −1
23
)
}
on (3, 1)
on (1)
Chapter II: Representation theory of general linear groups
6
Multilinear algebra and algebraic geometry
Let V and W be finite-dimensional CG-modules, where G is any (possibly infinite) group. Recall the
following:
ˆ Tensor products: V ⊗C W is a CG-module with the diagonal action g(v ⊗ w) = gv ⊗ gw.
Basic properties:
– V ⊗C∼
=C
– V ⊗W ∼
=W ⊗V
∼ V ⊗ (W ⊗ Z)
– (V ⊗ W ) ⊗ Z =
– If θ : V → V 0 and ϕ : W → W 0 are CG-maps then there is a CG-map defined by
θ⊗ϕ : V ⊗W
v⊗w
→
7→
V 0 ⊗ W0
θ(v) ⊗ ϕ(w)
ˆ Hom spaces: HomC (V, W ) is a CG-module via
!
X
X
αi gi (f )(v) =
αi gi (f (gi−1 v))
i
i
ˆ Dual space: V ∗ = HomC (V, C).
Basic properties:
– If V is 1-dimensional then V ∗ ⊗ V ∼
= C.
– If θ : V → W is a CG-map then the dual θ∗ : W ∗ → V ∗ is too.
– The map
V∗⊗W
f ⊗w
is a CG-isomorphism.
→ HomC (V, W )
7→ (v 7→ f (v)w)
ˆ Tensor powers: The nth tensor power of V is defined by
T n V = V ⊗n = V ⊗ · · · ⊗ V ,
|
{z
}
T 0V = C
n copies
Basic properties:
– If V has basis {ei : 1 ≤ i ≤ m} then V ⊗n has basis
{ei1 ⊗ · · · ⊗ ein : 1 ≤ i1 , . . . , in ≤ m}
and hence dim(V ⊗n ) = mn .
– V ⊗n is naturally a left CSn -module via
σ(v1 ⊗ · · · ⊗ vn ) = vσ−1 (1) ⊗ · · · vσ−1 (n)
– The actions of CSn and CG commute, i.e.
σgx = gσx for all g ∈ G, σ ∈ Sn , x ∈ V ⊗n
(6.1) Definition
The nth exterior power of V is Λn V = V ⊗n /X, where
X = spanC {x − εσ σx : x ∈ V ⊗n , σ ∈ Sn }
The image of v1 ⊗ · · · ⊗ vn in Λn V under the quotient map will be denoted by v1 ∧ · · · ∧ vn .
Define the antisymmetric tensors by
T n Vanti = {x ∈ V ⊗n : σx = εσ x for all σ ∈ Sn }
(6.2) Lemma
24
(i) v1 ∧ · · · ∧ vn = εσ vσ−1 (1) ∧ · · · ∧ vσ−1 (n) for all σ ∈ Sn .
This tells us, by considering transpositions, that v1 ∧ · · · ∧ vn = 0 whenever vi = vj for some
i 6= j.
(ii) T n Vanti and Λn V are CG-modules.
(iii) Λn V has basis {ei1 ∧ · · · ∧ ein : 1 ≤ i1 < · · · < in ≤ n}, and so dim Λn V =
In particular, Λm V is one-dimensional and Λm+1 V = Λm+2 V = · · · = 0.
Proof
m
.
n
An easy exercise.
Remark
If V is m-dimensional over a field k then usually we define Λn V = V ⊗n /X, where
X = spanC {v1 ⊗ · · · ⊗ vn : vi = vj for some i 6= j}
If char k 6= 2 then this definition reduces to (6.2).
X
εσ σ is the alternizer [cf. (2.4)], and the natural
(6.3) Lemma T n Vanti = aV ⊗n , where a =
σ∈Sn
map T n Vanti → Λn V is an isomorphism of CG-modules.
Proof
If x is antisymmetric then
X
X
X
εσ σx =
εσ (εσ x) =
x = n! x
ax =
σ∈Sn
σ∈Sn
σ∈Sn
and so T n Vanti ⊆ aV ⊗n .
Conversely, since σa = εσ a for all σ ∈ Sn , any element of aV ⊗n must be antisymmetric.
We have a map
aV ⊗n ,→ V ⊗n → Λn V
which is a CG-homomorphism and whose kernel is X ∩ aV ⊗n ≤ aX, since x =
metric. But for y ∈ V ⊗n ,
1
ax for x antisymn!
a(y − εσ σy) = ay − εσ (εσ a)y = ay − ay = 0
so ax = 0 (since x ∈ X). Hence the map is injective. And it is surjective: any x ∈ Λn V is the image
1
of some y ∈ V ⊗n , but then x is the image of ay ∈ aV ⊗n . So the map is an isomorphism.
n!
(6.4) Lemma
Proof
Λn (V ∗ ) ∼
= (Λn V )∗ .
The quotient map V ⊗n Λn V is surjective, so induces an inclusion
(Λn V )∗ ,→ (T n V )∗ = T n (V ∗ )
Now by the universal property of Λn V (i.e. any multilinear map V × · · · × V → C factors through
|
{z
}
n copies
Λn V ), the image of the inclusion is T n (V ∗ )anti , which is isomorphic to Λn (V ∗ ).
Now prepare for some déjà vu in (6.5)–(6.8). . .
(6.5) Definition
The nth symmetric power of V is Sn V = V ⊗n /Y , where
Y = spanC {x − σx : x ∈ V ⊗n , σ ∈ Sn }
The image of v1 ⊗ · · · ⊗ vn in Sn V under the quotient map will be denoted by v1 ∨ · · · ∨ vn .
Define the symmetric tensors by
T n Vsym = {x ∈ V ⊗n : σx = x for all σ ∈ Sn }
(6.6) Lemma
25
(i) v1 ∨ · · · ∨ vn = vσ−1 (1) ∨ · · · ∨ vσ−1 (n) for all σ ∈ Sn .
(ii) T n Vsym and Sn V are CG-modules.
(iii) Sn V has basis {ei1 ∨ · · · ∨ ein : 1 ≤ i1 ≤ · · · ≤ in ≤ n}, and so dim Sn V =
Proof
An easy exercise.
T n Vsym = sV ⊗n , where s =
(6.7) Lemma
m+n−1
.
n
X
σ is the symmetrizer [cf. (2.4)], and the natural
σ∈Sn
map T n Vsym → Sn V is an isomorphism of CG-modules.
Proof
If x is symmetric then
sx =
X
σx =
σ∈Sn
and so T n Vsym ⊆ sV ⊗n .
X
x = n! x
σ∈Sn
Conversely, since σs = s for all σ ∈ Sn , any element of sV ⊗n must be symmetric.
We have a map
sV ⊗n ,→ V ⊗n → Sn V
which is a CG-homomorphism and whose kernel is Y ∩ aV ⊗n ≤ sY , since x =
But for y ∈ V ⊗n ,
1
sx for x symmetric.
n!
s(y − σy) = sy − sy = 0
so sx = 0 (since x ∈ Y ). Hence the map is injective. And it is surjective: any x ∈ Sn V is the image
1
of some y ∈ V ⊗n , but then x is the image of sy ∈ sV ⊗n . So the map is an isomorphism.
n!
(6.8) Lemma
Proof
Sn (V ∗ ) ∼
= (Sn V )∗ .
The quotient map V ⊗n Sn V is surjective, so induces an inclusion
(Sn V )∗ ,→ (T n V )∗ = T n (V ∗ )
Now by the universal property of Sn V (i.e. any multilinear map V × · · · × V → C factors through
|
{z
}
n copies
Sn V ), the image of the inclusion is T n (V ∗ )sym , which is isomorphic to Sn (V ∗ ).
Polynomial maps between vector spaces
(6.9) Definition Let V and W be finite-dimensional C-spaces with bases {ei : 1 ≤ i ≤ m} and
{fr : 1 ≤ r ≤ h}, respectively. A function ϕ : V → W is a polynomial map (resp. n-homogeneous
map) if, for all X1 , . . . , Xm ∈ C,
~ 1 + · · · + ϕh (X)f
~ h
ϕ(X1 e1 + · · · + Xm em ) = ϕ1 (X)f
~ abbreviates
for some polynomials (resp. homogeneous polynomials of degree n) ϕ1 , . . . , ϕh . [Here X
the m-tuple (X1 , . . . , Xm ).]
(6.10) Lemma The definitions in (6.9) do not depend on the choice of bases. That is, if there exist
such functions ϕi with respect to some bases {ei }, {fr }, then there exist such functions with respect
to any bases {e0i }, {fr0 }.
Proof
Write
e0i =
m
X
pji ej
and
fr =
h
X
s=1
j=1
26
qsr fs0
Then
m
X
ϕ
Xi e0i
i=1
!

= ϕ
m
X
i,j=1
=
h
X
Xi pji ej 
m
X
ϕr
r=1
=

h
X
ϕr
r,s=1
and the
h
X
r=1
Xi1 p1i1 , . . . ,
i1 =1
m
X
Xim pmim
im =1
m
X
Xi1 p1i1 , . . . ,
i1 =1
m
X
!
Xim pmim
im =1
fr
!
qsr fs0
ϕr · · · qsr are polynomials (resp. n-homogeneous polynomials) like the ϕr .
Let PC (V, W ) and HC,n (V, W ) denote the spaces of polynomial maps and n-homogeneous maps, respectively, from V to W . They are C-spaces, as can be easily verified.
(6.11) Lemma The composite of polynomial maps X → W → Z is a polynomial map X → Z.
Likewise, the composite of an n-homogeneous map X → W and a n0 -homogeneous map W → Z is an
nn0 -homogeneous map X → Z.
Proof
An easy exercise.
Basic examples
(1) HC,0 (V, W ) ∼
=W
(2) HC,1 (V, W ) = HomC (V, W )
(3) ∆ ∈ HC,n (V, Sn V ), where ∆ : v 7→ v ∨ · · · ∨ v. For
!
X
X
∆
Xi ei =
Xi1 . . . Xin ei1 ∨ · · · ∨ ein =
i
~i
X
i1 ≤···≤in
λ~i Xi1 . . . Xin ei1 ∨ · · · ∨ ein
for suitable λ~i ∈ C. [Here ~i abbreviates the n-tuple (i1 , . . . , in ).]
(6.12) Theorem
For C-spaces V and W , the map ψ 7→ ψ ◦ ∆ induces an isomorphism
∼
=
HomC (Sn V, W ) −→ HC,n (V, W )
called restitution.
Proof It’s injective: suppose ψ ◦ ∆ = 0. We use descending induction on i to show that ψ(v1 ∨ · · · ∨
vn ) = 0 whenever i of the terms are equal. When i = n this is precisely our assumption. Suppose it’s
true for i + 1, then for any α ∈ C we have
0 = ψ((x + αy) ∨ · · · ∨ (x + αy) ∨vi+2 ∨ · · · ∨ vn )
{z
}
|
i+1
=
i+1
X
j=0
αj
i+1
ψ(x
· · ∨ x} ∨ y ∨ · · · ∨ y ∨vi+2 ∨ · · · ∨ vn )
| ∨ ·{z
| {z }
j
i+1−j
j
Since this holds true for any α ∈ C, each term must be zero. In particular
i+1
ψ(x
· · ∨ x} ∨y ∨ vi+2 ∨ · · · ∨ vn ) = 0
| ∨ ·{z
1
i
If dim V = m and dim W = h, say, then
m+n−1
dim HomC (Sn V, W ) = h
= dim HC,n (V, W )
n
27
So the map is an isomorphism.
(6.13) Corollary
Sn V = spanC {v ∨ · · · ∨ v : v ∈ V }
Proof Take W = C in (6.12). If the elements v ∨ · · · ∨ v do not span then there exists a nonzero
map Sn V → C whose composition with ∆ is zero, contradicting (6.12).
Remark
We can construct the inverse explicitly. Let ϕ ∈ HC,n (V, W ), so
ϕ(X1 e1 + · · · + Xm em ) = ϕ1 (X1 , . . . , Xm )f1 + · · · + ϕk (X1 , . . . , Xm )fh
with ϕi homogeneous. The total polarisation P ϕ ∈ HomC (Sn V, W ) of ϕ is
P ϕ(ei1 ∨ · · · ∨ ein ) =
h
X
∂ n ϕj
fj
∂Xi1 . . . ∂Xin
j=1 |
{z
}
∈C
(The partial derivative is formal, is symmetric in the Xik , and lies in C since ϕj has degree N .) Now
X
(P ϕ)(∆(X1 e1 + · · · + Xm em )) =
Xi1 . . . Xin (P ϕ)(ei1 ∨ · · · ∨ ein )
i1 ,...,in
=
X X
j
=
X
Xi1 . . . Xin
i1 ,...,in
∂ n ϕj
fj
∂Xi1 . . . ∂Xin
n! ϕj (X1 , . . . , Xm )fj
j
= n!ϕ(X1 e1 + · · · + Xm em )
So P ϕ ◦ ∆ = n! ϕ. TheP
penultimate equality follows from Euler’s theorem, which states that if F
∂F
Xi = rF .
is r-homogeneous then i ∂X
i
Example
Let ϕ : V → C be a quadratic form:
!
m
X
X
ϕ
Xi ei =
aij Xi Xj ,
i=1
aij = aji
i,j
hP
P
i
P
Then (P ϕ) ( i Xi ei ) ∨
Y
e
= 2 i,j aij Xi Yj is twice the corresponding symmetric bilinear
j j J
form.
28
Interlude: some reminders about affine varieties
Let K be an infinite field. Then R = K[X1 , . . . , Xn ] a Noetherian ring by Hilbert’s basis theorem.
Define An K = K n , the n-dimensional affine space over K.
If f ∈ R and p = (a1 , . . . , an ) ∈ An then the element f (p) = f (a1 , . . . , an ) ∈ K is the evaluation of
f at p.
Define a map
{ideals I of R}
I
←→
7−→
{subsets of An }
V (I)
where V (I) = {p ∈ An : f (p) = 0 for all f ∈ I}.
A subset X ⊆ An is an algebraic set if X = V (I) for some ideal I. But I is finitely generated, say
I = (f1 , . . . , fm ), so
V (I) = {p ∈ An : fj (p) = 0 for each 1 ≤ j ≤ m}
is a locus of points satisfying a finite number of polynomial equations.
Algebraic subsets of An form the closed sets of a topology on An , called the Zariski topology. The
Zariski topology on An induces a topology on any algebraic set X ⊆ An : the closed subsets of X are
the algebraic subsets of X.
Example Zariski-closed subsets of A1 are A1 and finite subsets of A1 ; that is, the Zariski topology
on A1 is the cofinite topology.
Returning to our correspondence, we have a map in the other direction
{ideals I of R}
I(X)
←→
←−
{subsets of An }
X
where I(X) = {f ∈ R : f (p) = 0 for all p ∈ X}.
An algebraic set X is irreducible if there does not exist a decomposition X = X1 ∪ X2 with X1 , X2 (
X proper algebraic subsets. Note
X is irreducible ⇔ I(X) is a prime ideal
If I E R then the radical
√
I of I is given by
√
I = {f ∈ R : f m ∈ I for somem}
√
and I is a radical ideal if I = I. Note that prime ideals are automatically radical.
Theorem (Nullstellensatz)
Let K be algebraically closed. Then
(a) Maximal ideals of R are of the form
mj = (X1 − a1 , . . . , Xn − an )
for some p = (a1 , . . . , an ) ∈ An .
(b) If (1) 6= I E R then V (I) 6= ∅.
√
(c) I(V (I)) = I for all ideals I.
Proof
Omitted.
Let W a finite-dimensional K-space. A map f : W → K is regular (or polynomial) if it is defined
by a polynomial in the coordinates with respect to a given basis of W . (Check: it doesn’t matter
which basis we pick.)
29
Define the coordinate ring (or ring of regular functions) of W , denoted K[W ], to be the Kalgebra of polynomial functions on W . If e1 , . . . , em is a basis of W and x1 , . . . , xm is the corresponding
dual basis of W ∗ then
K[W ] = K[x1 , . . . , xm ]
We say f ∈ K[W ] is d-homogeneous if f (tw) = td f (w) for all d ∈ C and w ∈ W . Write K[W ]d for
the subspace of K[W ] of all d-homogeneous polynomials. Then
M
K[W ] =
K[W ]d
d≥0
is a graded K-algebra, i.e. if f, g are homogeneous of degree df , dg , respectively, then f g is homogeneous
of degree df + dg .
The monomials xd11 · · · xdmm with d1 + · · · + dm = d form a basis of K[W ]d . In particular, K[W ]1 = W ∗ .
This extends to a canonical identification of K[W ]d with Sd (W ∗ ), so that
M
M
K=
K[W ]d =
Sd (W ∗ ) = S(W ∗ )
d≥0
d≥0
We call S(W ∗ ) the symmetric algebra of W ∗ .
Zariski-dense subsets
A subset X ⊆ W , with W finite-dimensional, is Zariski-dense if every function f ∈ K[W ] vanishing
on X is the zero function. So every polynomial function f ∈ K[W ] is completely determined by its
restriction f |X to a Zariski-dense subset X ⊆ W .
Now let X be an arbitrary subset of W . As above, the ideal of X
I(X) = {f ∈ K[W ] : f (a) = 0 for all a ∈ X} =
\
a∈X
where ma = I({a}) is the maximal ideal of functions vanishing at a, i.e.
eva : K[W ] →
C
ma = ker
f
7→ f (a)
We say X ⊆ Y is Zariski-dense in Y if I(X) = I(Y ).
Further reading
Chapter 4 of Reid’s book.
30
ma
7
Schur–Weyl duality
Let V be an m-dimensional C-space and let n ∈ N.
V ⊗n is a right CSn -module via the action1
(v1 ⊗ · · · ⊗ vn )σ = vσ(1) ⊗ · · · ⊗ vσ(n)
So there is a map
CSn
σ
→ EndC V ⊗n
7→ (v 7→ vσ)
—(1)
Regarding V as a representation of GL(V ) in the natural way, V ⊗n is a (left) CGL(V )-module via
the action
g(v1 ⊗ · · · ⊗ vn ) = gv1 ⊗ · · · ⊗ gvn
So there is a map
CGL(V ) → EndC V ⊗n
g
7→ (v 7→ gv)
—(2)
The actions (1) and (2) commute, so by abusing notation slightly, (1) induces a map
ϕ : CSn → EndCGL(V ) (V ⊗n )
and (2) induces a map
ψ : CGL(V ) → EndCSn (V ⊗n )
Schur–Weyl duality asserts that the images of CSn and CGL(V ) in EndC (V ⊗n ) are each other’s
centralizers.
(7.1) Theorem (Schur–Weyl duality) [Schur, 1927]
ψ(CGL(V )) = EndCSn (V ⊗n )
and ϕ(CSn ) = EndCGL(V ) (V ⊗n )
Such an assertion looks innocuous, but it is fundamental in explaining why the symmetric group and
the general linear group are so closely related.
(7.2) Definition The C-algebra SC (m, n) is the subalgebra of EndC (V ⊗n ) consisting of endomorphisms which commute with the image of CSn , i.e.
SC (m, n) = EndCSn (V ⊗n )
It is called the Schur algebra. [For a definition over arbitrary fields, see J.A. Green’s book Polynomial
representations of GLn , SLN 830.]
CSn is semisimple and V ⊗n is a finite-dimensional CSn -module, so by (1.9), SC (m, n) is a semisimple
C-algebra.
Put W = EndC (V ). This is a CGL(V )-module under the conjugation action (check this).
(7.3) Lemma
There is a map α : W ⊗n → EndC (V ⊗n ) given by
f1 ⊗ · · · ⊗ fn 7→ v1 ⊗ · · · ⊗ vn 7→ f1 (v1 ) ⊗ · · · ⊗ fn (vn )
which is an isomorphism both of CSn -modules and of CGL(V )-modules.
1 Note
that, because this is a right-action, σ1 σ2 denotes the effect of first applying σ1 and then applying σ2 , which is
‘backwards’ compared to normal.
31
Proof
W ⊗n = W ⊗ · · · ⊗ W ∼
= (V ⊗ V ∗ ) ⊗ · · · ⊗ (V ⊗ V ∗ )
∼ (V ⊗ · · · ⊗ V ) ⊗ (V ∗ ⊗ · · · ⊗ V ∗ )
=
∼ V ⊗n ⊗ (V ⊗n )∗
=
∼
= EndC (V ⊗n )
It is left as an exercise to check that this isomorphism is the map we think it is, that W ⊗n has
the natural structure of a CSn -module and EndC (V ⊗n ) inherits structure from V ⊗n via θ · σ =
ϕ(σ)θϕ(σ −1 ), and finally that α is a CSn -map.
(7.4) Lemma
SC (m, n) = α(T n Wsym )
Proof SC (m, n) is the set of all x ∈ EndC (V ⊗n ) fixed under the action of Sn . By (6.5), T n Wsym is
the set of all y ∈ W ⊗n fixed under the action of Sn (which is induced by α, i.e. y ·σ ∈ W ⊗n corresponds
with ϕ(σ)α(y)ϕ(σ −1 ) ∈ EndC (V ⊗n ).
Recall affine n-space An from §6.
(7.5) Lemma An is irreducible; i.e. if An = X ∪Y with X, Y Zariski-closed then X = An or Y = An .
Proof The ring of regular functions on An is R = C[X1 , . . . , Xn ]. If X and Y are zero-sets of ideals
I, J E R then, by assumption, any maximal ideal contains either I or J. If I, J are both nonzero then
pick 0 6= f ∈ I and 0 6= g ∈ J. Any maximal ideal contains f g, so
By Nullstellensatz, f g ∈
p
for all a1 , . . . , an ∈ C
f g(a1 , . . . , an ) = 0
{0} = {0}, contradicting the fact that R is an integral domain.
(7.6) Lemma Let Y ≤ Cd be a subspace. Then Y is Zariski-closed, where Cd is identified with Ad
in the canonical way.
Proof Chose a basis ϕ1 , . . . , ϕh of HomC (Cd /Y, C) and view them as maps Cd → C. Then Y is the
zero set of the ϕi .
n
(7.7) Lemma
Proof
n
T WSym = spanC
z
}|
{
{ϕ ⊗ · · · ⊗ ϕ : ϕ ∈ GL(V )}
Define a subspace X of T n Wsym by
n
X = spanC
z
}|
{
{ϕ ⊗ · · · ⊗ ϕ : ϕ ∈ GL(V )}
n
z
}|
{
There is a map α : W → W
given by α(ϕ) = ϕ ⊗ · · · ⊗ ϕ, which is a regular map of the affine
2
2n
spaces Am ∼
= W and Am ∼
= W ⊗n .
⊗n
Now X is a subspace, so is Zariski-closed by (7.6), and hence α−1 (X) is Zariski-closed. Thus,
W = α−1 (X) ∪ {endomorphisms with zero determinant}
is a union of Zariski-closed subsets.
2
Since Am is irreducible (by (7.5)), α−1 (X) = W . Thus X contains all maps of the form ϕ ⊗ · · · ⊗ ϕ
for ϕ ∈ W . But they span T n Wsym since the images ϕ ∨ · · · ∨ ϕ span Sn W by (6.13).
In other words. . .
n
(7.8) Theorem
z
}|
{
SC (m, n) = spanC {ϕ ⊗ · · · ⊗ ϕ : ϕ ∈ GL(V )}
The assertion that the image of CGL(V ) is the centralizer of CSn is a reformulation
n
z
}|
{
of the assertion that SC (m, n) = spanC {ϕ ⊗ · · · ⊗ ϕ : ϕ ∈ GL(V )} , which is (7.8).
Proof of (7.1)
32
Conversely, SC (m, n) is a semisimple C-algebra. By (1.11), CSn maps onto EndSC (m,n) (V ⊗n ), and
since the image of GL(V ) spans SC (m, n), we must have
EndSC (m,n) (V ⊗n ) = EndCGL(V ) (V ⊗n )
Thus CSn maps onto EndCGL(V ) (V ⊗n ), i.e. the image of CSn in EndC (V ⊗n ) is the centralizer of the
image of CGL(V ).
33
8
Tensor decomposition
Let V be an m-dimensional C-space. We know from (3.9) that
L
V ⊗n =
h(tλ )V ⊗n
—(1)
where the direct sum runs over standard tableaux tλ with λ ` n.
Examples
If λ = (1n ) then hλ V ⊗n ∼
= T n Vanti ∼
= Λn V .
If λ = (n) then hλ V ⊗n ∼
= T n VSsym ∼
= Sn V .
In particular,when n = 2, this tells us that
V ⊗V ∼
= T 2 Vanti ⊕ T 2 Vsym ∼
= Λ2 V ⊕ Sn V
Now the actions of Sn and GL(V ) on V ⊗n commute. So if λ ` n and tλ is a λ-tableau, then h(tλ )V ⊗n
is a CGL(V )-submodule of V ⊗n . Observe that
h(tλ )V ⊗n ∼
= hλ V ⊗n
as CGL(V )-modules
since σh(tλ )σ −1 = hλ for some σ ∈ Sn . So premultiplication by σ induces an isomorphism
∼
=
h(tλ )V ⊗n −→ hλ V ⊗n
(8.1) Lemma The nonzero hλ V ⊗n with λ ` n are nonisomorphic irreducible CGL(V )-modules.
Moreover, if M is an SC (m, n)-module and M is viewed as a CGL(V )-module by restriction via the
natural map CGL(V ) → SC (m, n) then M is isomorphic to a direct sum of copies of the hλ V ⊗n .
Proof
Recall that SC (m, n) = EndCSn (V ⊗n ) and
hλ V ⊗n
(1.12)
∼
=
HomCSn (CSn hλ , V ⊗n )
By (1.9) and (1.10)(Artin–Wedderburn), the nonzero spaces hλ V ⊗n are a complete set of nonisomorphic irreducible SC (m, n)-modules.
Now SC (m, n) is semisimple, so the result follows from (2) and (3) below, both of which derive from
the fact (at the end of §7) that the map CGL(V ) → SC (m, n) is surjective.
(2) If M is an SC (m, n)-module and N is a CGL(V )-module of M then N is an SC (m, n)-submodule
of M .
(3) If M and N are SC (m, n)-modules and θ : M → N is a CGL(V )-module map then it is an
SC (m, n)-map.
This implies that (1) is a decomposition of V ⊗n into CGL(V )-submodules which are either zero or
irreducible. Which of these submodules are nonzero? We’ll prove:
(8.2) Theorem
Let λ ` n and m = dimC V . Then


if λm+1 6= 0
0
⊗n
dimC hλ V
= 1
if λm+1 = 0 andλ1 = · · · = λm


≥ 2 otherwise
First we have a slightly technical result. Take a partition λ ` n such that [λ] is partitioned into two
nonempty parts, say of sizes i and j = n − i, by a vertical wall. For example, if λ = (5, 3, 2, 1) ` 11
we could have
• • • • •
• • •
←− j = 4
i = 7 −→
• •
•
34
For λ ` n, let tλ be a tableau whose entries in the left-hand part are {1, . . . , i}. Let µ ` i be the
corresponding partition, so that tµ = tλ ↓[µ] . Let ν ` j correspond to the right-hand part, with
tableau tν . This is a map
tν : [ν] → {10 , . . . , j 0 }
(8.3) Lemma
where k 0 = k + i for 1 ≤ k ≤ j
There exists a surjective CGL(V )-module map
((h(tµ )V ⊗i ) ⊗ (h(tν )V ⊗j ) → h(tλ )V ⊗n
Proof
—(∗)
Let Si = Aut({1, . . . , i}) and Sj = Aut({10 , . . . , j 0 }).
Si and Sj are embedded in Sn and are disjoint, so we regard CSi and CSj as subsets of CSn which
commute. Also
Ctλ = Ctµ × Ctν and H = Rtµ × Rtν ≤ Rtλ
S
and H permutes each side of the wall. Take a transversal Rtλ = i ri H. Then
(2.4)
h(tλ ) =
X
X
r∈Rtλ c∈Ctλ
=
X X
i
=
X
εc rc
X
X
X
εc0 εc00 ri r0 r00 c0 c00
r 0 ∈Rtµ r 00 ∈Rtν c0 ∈Ctµ c00 ∈Ctν
ri h(tµ )h(tν )
i
So
h(tλ )h(tµ )h(tλ ) =
X
ri h(tµ )2 h(tν )2
(2.10)
=
kh(tλ )
i
where k =
i!j!
. We thus have a CGL(V )-map
fµ fν
V ⊗i ⊗ V ⊗j → h(tλ )V ⊗n
given by premultiplication by h(tλ ). Restriction of this map to the left-hand side of (∗) is surjective,
since
h(tλ )(x ⊗ y) = h(tλ ) ·
1
(h(tµ )x ⊗ h(tν )y)
k
(8.4) Lemma
(1) If λm+1 = 0 then hλ V ⊗n 6= 0
(2) If n > 0 and λm = 0 then dimC hλ V ⊗n ≥ 2
Proof
(1) Let ij be the row in which j occurs in t◦λ and let x = ei1 ⊗ · · · ⊗ ein . For σ ∈ Sn
σx = x
⇔
ij = iσ−1 (j) ∀j
⇔
j, σ −1 (j) are in the same row
⇔
σ ∈ Rt◦λ
⊗n
So
is
the
coefficient of x in the decomposition of hλ x with respect to the standard basis of V
Rt◦ 6= 0. So hλ x 6= 0.
λ
(2) Let y = e1+i1 ⊗ · · · ⊗ e1+in and adapt the argument above, to see that hλ x and hλ y are linearly
independent.
(8.5) Lemma
If λm+1 = 0 and λm > 0 then
hλ V ⊗n ∼
= Λm V ⊗ h(λ1 −1,...,λm −1) V ⊗(n−m)
35
Proof Put a wall in [λ] between the first column and the rest. Let tλ be a tableau whose first
column consists of {1, . . . , m}. By (8.3) there exists a surjection
Λm V ⊗ h(tν )V ⊗(n−m) h(tλ )V ⊗n
where ν = (λ1 − 1, . . . , λm − 1). Hence, with the usual identifications, there exists a map
Λm V ⊗ hν V ⊗(n−m) hλ V ⊗n
Both hν V ⊗(n−m) and hλ V ⊗n are nonzero, hence they are irreducible CGL(V )-modules by (8.1). Since
Λm V is one-dimensional, both sides are irreducible and the map is an isomorphism.
Proof of (8.2)
If λm+1 6= 0 then [λ] has i > m rows and as in (8.5) there exists a surjection
Λi V ⊗ hν V ⊗(n−i) hλ V ⊗n
but Λi V = 0, so hλ V ⊗n = 0.
If λm+1 = 0 then iterating (8.5) we have
dimC hλ V ⊗n = dimC h(λ1 −λm ,...,λm−1 −λm ) V ⊗(n−mλm )
(
1
if λ1 = · · · = λm
=
≥ 2 if not, by (8.4)
36
9
Polynomial and rational representations of GL(V )
Let V be an m-dimensional C-space with basis e1 , . . . , em .
(9.1) Definition A finite-dimensional CGL(V )-module M , with basis w1 , . . . , wn , is rational (resp.
polynomial, n-homogeneous) if there exist rational (resp. polynomial, n-homogeneous) functions
fij (Xrs ) for 1 ≤ i, j ≤ h in m2 variables Xrs for 1 ≤ r, s ≤ m such that the map
representation
basis e
basis wj
i
GL(V ) −−−−−−−−−→ EndC (M ) −−−−−→ Mh (C)
GLm (C) −−−−−→
sends (Ars )1≤r,s≤m 7→ (fij (Ars ))1≤i,j≤h .
Exercise
and M .
Check that the definitions in (9.1) are independent of the choice of bases ei and wj of V
Examples
(0) The natural representation of GL(V ) on V ⊗n is polynomial. In fact. . .
(1) V ⊗n is n-homogeneous; but. . .
(2) C ⊕ V is polynomial but not homogeneous; and. . .
(3) V ∗ is rational but not polynomial (or homogeneous).
(9.2) Lemma
(a) Submodules, quotients and direct sums of rational (resp. polynomial, n-homogeneous) modules
are rational (resp. polynomial, n-homogeneous).
(b) If M is rational then so is M ∗ .
(c) If M and N are rational (resp. polynomial, n- and n0 -homogeneous) then M ⊗ N is rational (resp.
polynomial, (n + n0 )-homogeneous).
(d) If n 6= n0 and M is both n- and n0 -homogeneous then M = 0.
(e) if M is rational (resp. polynomial) then Λn M and Sn M are rational (resp. polynomial).
Proof
Easy exercise.
(9.3) Definition
For λ1 ≥ · · · ≥ λm ≥ 0 with
P
i
λi = n, set
Dλ1 ,...,λm (V ) = h(λ1 ,...,λm ) (V ⊗n )
This is sometimes referred to in the literature as the Weyl module or Schur module.
(9.4) Theorem Every n-homogeneous CGL(V )-module is a direct sum of irreducible submodules.
Moreover, the modules of (9.3) form a complete set of nonisomorphic irreducible n-homogeneous
CGL(V )-modules.
Proof By (8.1) it suffices to show that any n-homogeneous CGL(V )-module is obtained from some
S(m, n)-module (by restriction).
Let M be an n-homogeneous CGL(V )-module. M corresponds to a map
ρ0 : GL(V ) → EndC (M )
Consider the following diagram
37
GL(V )
ρ0
EndC (M )
,→
Sn (EndC V )
EndC (V )
ρ1
∼
=
T n (EndC V )sym
ρ2
EndC (M )
EndC (M )
∼
=
S(m, n)
ρ4
ρ3
EndC (M )
EndC (M )
In the top row all the maps are natural, and the composite γ : GL(V ) → S(m, n) is the natural map
we use for restricting S(m, n)-modules to CGL(V )-modules.
Claim
There exist ρ1 , ρ2 , ρ3 , ρ4 as above making the diagram commute.
Since M is n-homogeneous, we can extend the domain of definition of ρ0 to obtain an n-homogeneous
map ρ1 . For ρ2 , use the universal property of symmetric powers. Then ρ3 , ρ4 are transported by the
isomorphisms.
We check that ρ4 is a map of C-algebras. Well
ρ4 (1) = ρ4 (γ(1)) = ρ0 (1) = 1
and
ρ4 (γ(gg 0 )) = ρ0 (gg 0 ) = ρ0 (g)ρ0 (g 0 ) = ρ4 (γ(g))ρ4 (γ(g 0 ))
for g, g 0 ∈ GL(V ). But γ(GL(V )) spans S(m, n) by (7.8).
So M is a CS(m, n)-module and the restriction to GL(V ) is the module we started with.
(9.5) Lemma Every polynomial module for GL(C) = C× decomposes as a direct sum of submodules
on whih g ∈ C× acts as multiplication by g r for some r.
Proof If ρ : C× → GL(M ) ∼
= GLh (C) is a polynomial representation then each ρ(g)ij is polynomial
in g. Choose N ∈ N such that each ρ(g)ij has degree < N . Then M is a CG-module by restriction,
where
ij
G = {e2π N : 0 ≤ j ≤ N } ⊆ C×
is a cyclic group of order N . Now
M = M1 ⊕ · · · ⊕ Mh
as a CG-module with each Mj one-dimensional, and g ∈ G acts as multiplication by g nj on Mj for
0 ≤ j < N . (These are the possible irreducible CG-modules.)
Choosing nonzero elements of the Mj s gives a basis for M , and if (ρ(g)ij ) is the matrix of ρ(g) with
respect to this basis then
(
g ni if i = j
ρ(g)ij =
0
if i 6= j
ij
for g ∈ e2π N (0 ≤ j < N ), and hence for all g ∈ C× since the ρ(g)ij are polynomials of degree < N
in g.
(9.6) Lemma
Every polynomial decomposes as a direct sum of n-homogeneous modules.
Proof Tke a representation ρ : GL(V ) → GL(U ) be a representation of GL(V ) on a C-space U . We
have an inclusion C× ,→ GL(V ), so that one can regard U as a polynomial representation of C× , so
by (9.5) U will split as
U = U0 ⊕ · · · ⊕ UN
with α1 ∈ GL(V ) acting as multiplication by αn on Un .
Let u ∈ Un and let g ∈ GL(V ). Write
gu = u0 + · · · + uN
38
where ui ∈ Ui for 0 ≤ i ≤ N . Now (α1)gu = g(α1)u for α ∈ C× , so
u0 + αu1 + · · · + αN uN = αn u0 + · · · + αn uN
and hence ui = 0 for i 6= n.
Thus gu ∈ Un and the Un are CGL(V )-submodules of U . Since α1 acts as multiplication by αn on
Un , it follows that Un is an n-homogeneous CGL(V )-module.
(9.7) Theorem Every polynomial CGL(V )-module is a direct sum of irreducible submodules. The
modules Dλ1 ,...,λm (V ) with λ1 ≥ · · · ≥ λm ≥ 0 are a complete set of nonisomorphic irreducible
polynomial CGL(V )-modules.
(9.8) Definition
sentation
If n ∈ Z then the one-dimensional CGL(V )-module corresponding to the repreGL(V ) →
C⊗
g
7→ (det g)n
is denoted by detn . Note that
det1 ∼
= Λm V,
(9.9) Definition
detn ∼
= (det1 )⊗n if n ≥ 0,
detn ∼
= (det−n )∗ if n ≤ 0
If λ1 ≥ · · · ≥ λm but λm < 0 then define
Dλ1 ,...,λm = Dλ1 −λm ,...,λm−1 −λm ,0 (V ) ⊗ detλm
Remark
If λ1 ≥ · · · ≥ λm > 0 then by iterating (8.5) (as in the proof of (8.2)) we know that
Dλ1 ,...,λm (V ) ∼
= Dλ1 −λm ,...,λm−1 −λm ,0 (V ) ⊗ detλm
(9.10) Theorem Every rational CGL(V )-module is a direct sum of irreducible submodules. The
modules Dλ1 ,...,λm (V ) for λ1 ≥ · · · ≥ λm form a complete set of nonisomorphic irreducible (rational)
CGL(V )-modules.
p
with p a polynomial function.
deti
Thus if M is a rational CGL(V )-module then M1 = M ⊗ detN is a polynomial CGL(V )-module for
some N . Since M1 decomposes as a direct sum of irreducibles, so does M . If M is irreducible then so
is M1 , since detN is one-dimensional, and thus M1 is of the form
Proof
The rational functions f : GL(V ) → C are of the form f =
M1 ∼
= Dµ1 ,...,µm (V ) for some µ1 ≥ · · · ≥ µm
Finally, M = Dµ1 −N,...,µm −N (V ) by the above remark.
(9.11) Theorem
Proof
The one-dimensional rational CGL(V )-modules are precisely the detn for n ∈ Z.
As above, pass to polynomial modules and then apply (8.2).
Remark
The characterisation of irreducible CGL(V )-modules can be summarised by:
n-homogeneous
polynomial
rational
λ1 ≥ · · · ≥ λm > 0 with
λ1 ≥ · · · ≥ λm ≥ 0
λ1 ≥ · · · ≥ λm
39
P
i
λi = n
(9.4)
(9.7)
(9.10)
Interlude: some reminders about characters of GLn (C)
Let Tn ≤ GLn (C) be the n-dimensional torus, i.e. the subgroup of GLn (C) consisting of diagonal
matrices.
If ρ : GLn (C) → GL(W ) is a rational representation then the character of ρ is the function


λ1


..
χρ : (x1 , . . . , xn ) 7→ tr ρ 

.
λn
sometimes denoted by χW .
Basic properties
Let ρ, ρ0 be rational representations of GLn (C) associated with W, W 0 .
−1
(a) χρ ∈ Z[x1 , x−1
1 , . . . , xn , xn ], and if ρ is polynomial then χρ ∈ Z[x1 , . . . , xn ].
(b) χρ is a symmetric function under the conjugation action of Sn on Tn , i.e. under permutations of
the diagonal entries.
(c) ρ ∼
= ρ0 ⇒ χρ = χρ0 ; ⇐ will be (10.6).
(d) χW ⊕W 0 = χW + χW 0 and χW ⊗W 0 = χW χW 0 .
(e) If W is an irreducible polynomial representation of degree m then χW is a homogeneous polynomial
of degree m.
−1
(f) χW ∗ (x1 , . . . , xn ) = χW (x−1
1 , . . . xn ).
(g) When V = Cn we have
χS2 V
χV ⊗n = (x1 + · · · + xn )m ,
X
X
=
xi xj ,
χΛ2 V =
xi xj ,
i<j
i≤j
χdet = x1 . . . xn ,
χΛn−1 V =
n
X
i=1
x1 . . . xbi . . . xn ,
χV ∗ =
x−1
1
+ · · · + x−1
n ,
χSj V = hj (x1 , . . . , xn ) =
X
xi1 . . . xij
i1 ≤···≤ij
The hj are called complete symmetric functions. We could (but won’t) study the theory of
symmetric functions and apply it here; see R. Stanley’s book for more.
40
10
Weyl character formula
Let V be an m-dimensional C-space. If λ = (λ1 , . . . , λm ) ∈ Zm and P
λ1 ≥ · · · ≥ λm then the character
of the CGL(V )-module Dλ1 ,...,λm (V ) (= hλ V ⊗n if λm ≥ 0 and n = i λi ) will be denoted by φλ .
(10.1) Lemma
Proof
If ξ ∈ EndC (V ) then φλ (ξ) is a symmetric rational function of the eigenvalues of ξ.
The function
P (x1 , . . . , xm ) = φλ (diag(x1 , . . . , xm ))
is a rational function of the x1 , . . . , xm , and it is symmetric since diag(x1 , . . . , xm ) is conjugate to
diag(xτ (1) , . . . , xτ (m) ) for any τ ∈ Sm .
Now choose a basis for V such that the matrix A1 of ξ is in Jordan normal form, and for t ∈ C let At
be the matrix obtained from A1 by replacing the upperdiagonal 1s by ts. Let ξt be the endomorphism
corresponding to At . For t 6= 0, At is conjugate to A1 , and so φλ (ξt ) = φλ (ξ). Since φλ is a rational
function it is continuous (wherever it is defined), and so
φλ (ξ) = lim φλ (ξt ) = φλ (lim ξt ) = φλ (ξ0 ) = P (x1 , . . . , xm )
t→0
t→0
which is a symmetric function of the eigenvalues of ξ.
(10.2) Lemma Let α be an Sn -conjugacy class with cycle type nαn , . . . , 1α1 , and let ξ be an
endomorphism of V with eigenvalues x1 , . . . , xm . If si = xi1 + · · · + xim then
X
αn
1
sα
χλ (α)φλ (ξ)
1 . . . sn =
λ∈Λ+ (m,n)
(Compare this with (4.14).)
Proof Pick g ∈ α. Suppose ξ has matrix diag(x1 , . . . , xm ) with respect to the standard basis
e1 , . . . , em of V . We compute the trace of the endomorphism of V ⊗n sending x 7→ (gξ)x = ξgx in two
ways.
Way 1
V ⊗n is a CSn -module, so by (1.8),
M
CSn hλ ⊗ HomCSn (CSn hλ , V ⊗n )
V ⊗n ∼
=
λ`n
which, by (1.12), becomes
V ⊗n ∼
=
M
(CSn hλ ⊗ hλ V ⊗n )
λ`n
This is an isomorphism both of CSn - and EndCSn (V ⊗n )-modules. Since the action of GL(V ) on V ⊗n
commutes with that of Sn , the corresponding action of gξ on the right-hand side is given by the action
of g on CSn hλ and of ξ on hλ V ⊗n . So the trace of the action is
P
λ
—(1)
λ∈Λ+ (n,m) ξ (α)φλ (ξ)
Way 2 (direct computation)
gξ(ei1 ⊗ · · · ⊗ ein ) = xi1 . . . xin eig−1 (1) ⊗ · · · ⊗ eig−1 (n)
and so
tr(gξ) =
X
xi1 . . . xin
D
where D = {(i1 , . . . , in ) : 1 ≤ i1 , . . . , in ≤ m and ig−1 (j) = ij ∀j}.
The condition ig−1 (j) = ij for all j is equivalent to requiring that the function j 7→ ij be constant on
the cycles involved in g. Hence
αn
1
tr(gξ) = sα
1 . . . sn
41
—(2)
Equating (1) and (2) gives the result.
(10.3) Theorem Let λ ∈ Λ+ (m, n) and let ξ ∈ EndC (V ) with eigenvalues x1 , . . . , xm . If si =
xi1 + · · · + xim , then
s αn
X χλ (α) s1 α1
n
···
φλ (ξ) =
α1 ! . . . αn ! 1
n
ccls α
Proof Take the formula of (10.2), multiply it by nα χµ (α) and sum over α; then use orthogonality
of the χs.
(10.4) Theorem (Weyl’s character formula for GLn ) Let ξ ∈ GL(V ) with eigenvalues x1 , . . . , xm .
Then
`
x 1 , . . . , x `m φλ (ξ) = m−1
|x
, . . . , 1|
where `i = λi + m − i.
Proof
(a) Suppose λi ≥ 0 for all i, so that λ ∈ Λ+ (n, m). By (10.2) and the character formula for the
symmetric group (4.14) we know that
`
x 1 , . . . , x`m X
X
α1
αn
λ
s1 . . . sn =
χ (α) m−1
=
χλ (α)φλ (α)
|x
,
.
.
.
,
1|
+
+
λ∈Λ (n,m)
λ∈Λ (n,m)
Then the orthogonality of χλ s allows us to equate coefficients.
(b) For general λ, since
(9.9)
Dλ ∼
= Dλ1 −λm ,...,λm −λm (V ) ⊗ detλm
it follows that
φλ (ξ) =
` −λ
x 1 m , . . . , x`m −λm |xm−1 , . . . , 1|
(x1 . . . xm )
λm
`
x 1 , . . . , x`m = m−1
|x
, . . . , 1|
Remark This short proof uses the character formula of Sn from §4. One can do this in reverse,
using integration to compute the Weyl formula for the compact subgroup Un of unitary matrices in
GLm (C), and then translate that to GLm (C). [See also Telemann’s notes for the 2005 Part II course.]
You then use (10.2) to pass to Sn . See also Weyl’s book.
(10.5) Theorem (Weyl’s dimension formula)
Q
1≤i<j≤m
deg φλ = dim Dλ1 ,...,λm (V ) = Q
Proof
(`i − `j )
1≤i<j≤m (j
− i)
For t ∈ C set xm = 1, xm−1 = et , . . . , x1 = e(m−1)t . Then
Y
`
x 1 , . . . , x`m =
(e`i t − e`j t )
1≤i<j≤m
since it’s the transpose of a Vandermonde matrix. The term of lowest degree is
Q
i<j ((`i
− `j )t).
Q
Q
Also xm−1 , . . . , 1 = i<j (e(m−i)t − e(m−j)t ) and the term of lowest degree in t is i<j ((j − i)t).
If ξt = diag(x1 , . . . , xm ) then by (10.4),
deg φλ = φλ (1) = limt→0 φλ (ξt ) =
42
Q
i<j (`i −`j )
Q
i<j (j−i)
(10.6) Theorem
phic.
If two rational CGL(V )-modules have the same characters then they are isomor-
Proof As λ varies, the rational functions in Weyl’s character formula are linearly independent
elements of C(x1 , . . . , xm ).
(10.7) Lemma
Dλ1 ,...,λm (V )∗ ∼
= D−λm ,...,−λ1 (V )
Let ψ be the character of Dλ1 ,...,λm (V ). Then
−`
−`
x m , . . . , x−`1 x 1 , . . . , x−`m −1
= ψ(ξ) = φλ (ξ ) = −(m−1)
x
, . . . , 1
1, . . . , x−(m−1) m−1−`
m−1−`
m
1
x
,...,x
=
= φµ (ξ)
|xm−1 , . . . , 1|
Proof
where µ = (−λm , . . . , −λ1 ). Hence we have the required isomorphism by (10.6).
Finally we have the famous Clebsch–Gordan formula.
(10.8) Theorem (Clebsch–Gordan formula)
Let V = C2 and p, q ∈ N. Then
min{p,q}
Dp,0 (V ) ⊗ Dq,0 (V ) ∼
=
Proof
M
Dp+q−r,r (V )
r=0
Exercise; write down the characters and use (10.6).
(10.9) Examples of rational CGL(V )-modules
ˆ C∼
= D0,0,...,0,0 (V )
ˆ deti ∼
= Di,i,...,i,i (V )
ˆ V ∼
= D1,0,...,0,0 (V )
ˆ Sn V ∼
= Dn,0,...,0,0 (V )
ˆ Λn V ∼
= D1,...,1,0,...,0 (V ) where n ‘1’s and m − n ‘0’s appear, and n ≤ m
ˆ V∗ ∼
= D0,0,...,0,−1 (V )
ˆ (Sn V )∗ ∼
= D0,0,...,0,−n (V )
ˆ (Λn V )∗ ∼
= D0,...,0,−1,...,−1 (V ) where m − n ‘0’s and n ‘−1’s appear, and n ≤ m
43
Chapter III: Invariant theory
11
Introduction to invariant theory and first examples
Let W be a finite-dimensional C-space and C[W ] be its coordinate ring (cf. §6).
Facts about C[W ]
ˆ C[W ] is a commutative C-algebra, which is infinite-dimensional if W 6= 0, with
(λf )(w) = λf (w),
(f + g)(w) = f (w) + g(w),
(f g)(w) = f (w)g(w),
1C[W ] (w) = 1
for all λ ∈ C, f, g ∈ C[W ] and w ∈ W .
ˆ C[W ] is the ring of regular functions of the affine variety Adim W ∼
= W.
C[W ] ∼
=
ˆ
M
(6.12)
HC,n (W, C)
∼
=
n≥0
M
Sn (W ∗ ) = S(W ∗ )
n≥0
This latter isomorphism is known as polarisation.
ˆ If W ∗ has basis w1∗ , . . . , wh∗ then the map
C[X1 , . . . , Xh ] →
Xi
7→
C[W ]
wi∗
is a C-algebra homomorphism.
ˆ If W is a CG-module then C[W ] is a CG-module via the action
gf (w) = f (g −1 w)
for g ∈ G, f ∈ C[W ] and w ∈ W .
ˆ For g ∈ G, the map
C[W ] → C[W ]
is a C-algebra automorphism, since
f
7→
gf
g(f f 0 )(w) = (f f 0 )(g −1 w) = f (g −1 w)f 0 (g −1 w) = (gf )(w)(gf 0 )(w) = ((gf )(gf 0 ))(w)
and
g1C[W ] (w) = 1C[W ] (g −1 w) = 1 = 1C[W ] (w)
(11.1) Definition
If V, W are CG-modules then f : W → V is concomitant if
f (gw) = gf (w)
for all w ∈ W, g ∈ G
A G-invariant for the CG-module W is a concomitant map f : W → C.
Remarks
(1) f is invariant if and only if it is a fixed point under the G-action, i.e. gf = f for all g ∈ G.
(2) linear concomitant ≡ CG-homomorphism
(3) The map ∆ :
V
w
→
Sn W
is an n-homogeneous concomitant.
7
→
w ∨ ··· ∨ w
(4) C[W ]G = {f ∈ C[W ] : gf = f for all g ∈ G} is a C-subalgebra of C[W ], called the polynomial
invariant ring. It is a graded C-algebra:
M
C[W ]G =
C[W ]G
d
d≥0
with grading inherited by that of C[W ].
44
(11.2) Problem
Compute C[W ]G , describing its generators and relations.
What follows are some classical results which we will state but not prove.
ˆ [Noether] If G is finite then C[W ]G is finitely generated as a C-algebra.
ˆ [Noether, 1916] If char K = 0 then for any representation W of the finite group G, the invariant
ring K[W ]G is generated by the invariants of degree ≤ |G|.
ˆ [Hilbert] If W is a rational CGL(V )-module then C[W ]GL(V ) and C[W ]SL(V ) are finitely-generated
C-algebras. In fact [Hochster–Roberts, 1974] they are also Cohen–Macaulay rings. Also, C[W ]SL(V )
is a unique factorisation domain, and hence it is a Gorenstein ring (i.e. it has finite injective
dimension).
ˆ [Popov, Astérisque, 1988] There is an explicit bound on the number of generators for C[W ]SL(V ) .
Examples
(11.3) Symmetric polynomials W = Cn with basis e1 , . . . , en , is naturally a CSn -module. If
ξ1 , . . . , ξn is the corresponding dual basis for W ∗ , then the isomorphism
∼
=
→ C[W ]
C[X1 , . . . , Xn ] −
Xi
7→
ξi
allows an identification C[W ] ↔ C[X1 , . . . , Xn ]. The polynomial invariant ring of the CSn -module W
is
C[X1 , . . . , Xn ]Sn = {p ∈ C[X1 , . . . , Xn ] : p symmetric in the Xi s}
We have elementary symmetric functions ej (X1 , . . . , Xn ) ∈ C[X1 , . . . , Xn ]Sn such tha
~ n−1 + · · · + en (X)
~
(t + X1 ) · · · (t + Xn ) = tn + e1 (X)t
~ = (X1 , . . . , Xn ). That is,
where X
~ = X1 + · · · + Xn ,
e1 (X)
~ =
ei (X)
X
Xj1 Xj2 . . . Xji ,
~ = X1 . . . Xn
en (X)
j1 <j2 <···<ji
The fundamental theorem of symmetric functions computes the polynomial invariants for W ;
namely, it shows that the map
C[Y1 , . . . , Yn ] → C[X1 , . . . , Xn ]Sn
~
Yi
7→
ei (X)
is a C-algebra isomorphism. The standard way to prove this is by showing that C[X1 , . . . , Xn ]Sn =
C[e1 , . . . , en ] by induction on n.
Recall for p(t) ∈ C[t] monic of degree n, say
p(t) = tn + a1 tn−1 + · · · + an =
its discriminant is
disc(p) =
Y
i<j
n
Y
(t + λi )
i=1
(λi − λj )2
2
Q
Now, since the polynomial X n−1 , . . . , 1 = i<j (Xi − Xj )2 is symmetric in the Xi s, it may be
represented as a polynomial in the ei s, say
Y
~ . . . , en (X))
~
(Xi − Xj )2 = D(e1 (X),
i<j
45
Also, disc(p) is a polynomial in the coefficients of p, for example
disc(t2 + bt + c) = b2 − 4c,
disc(t3 + bt + c) = −4b3 − 27b2
(11.4) Alternating groups
Restriction in (11.3) enables consideration of W as a representation of An (n ≥ 2). The ring of
invariants is C[X1 , . . . , Xn ]An . Consider the Vandermonde determinant
n−1
Y
X
, . . . , 1 =
(Xi − Xj )
i<j
n−1
2
X
, . . . , 1 is Sn -invariant since it is equal to D(e1 , . . . , en ) as above.
Claim
The C-algebra map
θ : C[Y1 , . . . , Yn , Z]
Yi
Z
is surjective, and ker(θ) = (Z 2 − D(Y1 , . . . , Yn )).
→ C[X1 , . . . , Xn ]An
7→
n−1ei
X
7→
, . . . , 1
Proof We prove the claim in three steps.
Step 1 θ is surjective. Take a transposition τ ∈ Sn . Let f ∈ C[X1 , . . . , Xn ]An . Then fsym = f + τ f
is a symmetric polynomial, since Sn is generated by τ and An , and
τ fsym = τ f + τ 2 f = τ f + f = fsym
and if σ ∈ An then
σfsym = σf + τ (τ −1 στ )f = f + τ f = fsym
Similarly, falt = f − τ f is an alternating polynomial.
Observe that falt (X1 , . . . Xn ) = 0 whenever Xi = Xj for i 6= j. Hence by Nullstellensatz
q
falt ∈ (Xi − Xj ) = (Xi − Xj )
and so (Xi − Xj ) | falt for all i 6= j. In particular,
falt = X n−1 , . . . , 1 g
for some polynomial g. Now g is symmetric, and hence
f=
1
1
(fsym + falt ) = (fsym + X n−1 , . . . , 1 g)
2
2
Since fsym and g lie in im(θ), so does f .
~ )) ⊆ ker(θ). This is clear since
Step 2 (Z 2 − D(Y
~ )) = X n−1 , . . . , 12 − D(e1 , . . . , en ) = D(e1 , . . . , en ) − D(e1 , . . . , en ) = 0
θ(Z 2 − D(Y
~ )).
Step 3 ker(θ) ⊆ (Z 2 − D(Y
~ , Z) is of the form
Any polynomial P (Y
~ , Z) = Q(Y
~ , Z)(Z 2 − D(Y
~ )) + A(Y
~ ) + B(Y
~ )Z
P (Y
| {z }
|
{z
}
quotient
remainder
~ ) + B(Y
~ )Z then P = 0.
We show that if P ∈ ker(θ) has the form A(Y
Suppose a1 , . . . , an ∈ C, and let
f (t) = tn + a1 tn−1 + · · · + an = (t + λ1 ) · · · (t + λn )
46
Then
P (a1 , . . . , an , λn−1 , . . . , 1) = θ(P )(λ1 , . . . , λn ) = 0
since P ∈ ker(θ). Exchanging λ1 and λ2 changes the sign of the Vandermonde determinant, so
1
P (a1 , . . . , an , ±δ) = 0, where δ = disc(tn + a1 tn−1 + · · · + an ) 2 . Using the given form for P ,
A(a1 , . . . , an ) ± B(a1 , . . . , an )δ = 0
and hence A = B = 0 on the Zariski-dense open subset
{(a1 , . . . , an ) ∈ Cn : disc(tn + a1 tn−1 + · · · + an ) 6= 0}
Then A = B = 0 everywhere on Cn , so P = 0.
(11.5) Characteristic polynomial
Let V be an m-dimensional C-space. GL(V ) acts naturally on U = EndC (V ) by conjugation, i.e.
(g · θ)(v) = g · θ(g −1 · v)
(or, in shorthand, g · θ = gθg −1 ).
Suppose χθ (t) = det(t1V + θ) is the characteristic polynomial of θ and cn (θ) is the coefficient of tm−n
in χθ (t). Then
χg·θ (t) = det(t1V + gθg −1 ) = det(g(t1V + θ)g −1 ) = χθ (t)
so cn : U → C is an (n-homogeneous) invariant.
If θ has eigenvalues lambda1 , . . . , λm , then putting θ in Jordan normal form gives
χθ (t) =
m
Y
(t + λi )
i=1
so that cn (θ) = en (λ1 , . . . , λm ); in particular
c1 (θ) = λ1 + · · · + λm = tr(θ)
Claim
The C-algebra map
and cm (θ) = λ1 . . . λm = det(θ)
C[Y1 , . . . , Ym ] → C[U ]GL(V )
is an isomorphism.
Yi
7→
ci
Sketch proof If f is a polynomial invariant for U then f (θ) is a symmetric polynomial function of
the eigenvalues of θ. Mimic the proof that the characters of rational CGL(V )-modules are symmetric
rational functions of the eigenvalues of g ∈ GL(V ) as done in (10.1).
(11.6) Discriminant of quadratic forms
Let V be an m-dimensional C-space and let
U = HC,2 (V, C) = {quadratic forms on V }
By polarisation (see end of §6) identify U with the symmetric m × m matrices via f 7→ A, where
f (x) = xt Ax
for x ∈ Cm a column vector.
Define disc(f ) = det(A).
GLm (C) acts on U by
t
(gf )(x) = f (g −1 x) = xt (g −1 Ag −1 )x,
so
t
f ∈ U, g ∈ GLm (C), x ∈ Cm
disc(gf ) = det(g −1 Ag −1 ) = (det g)−2 disc(f )
47
and therefore disc : U → C is an SLm (C)-invariant (but not a GLm (C)-invariant).
Claim
The C-algebra map
C[X] →
X
7→
C[U ]SLm (C)
is an isomorphism.
disc
Proof It is injective: if there is a polynomial P such that P (disc f ) = 0 for all f ∈ U then P (λ) = 0
for all λ ∈ C, since the quadratic form
fλ (x1 , . . . , xm ) = x21 + · · · + x2m−1 + λx2m
has discriminant λ, so P = 0.
Let θ : U → C be a polynomial SLn (C)-invariant and define
F :
C
λ
→
C
7
→
θ(fλ )
This is a polynomial function since θ is a polynomial map. We prove that θ(f ) = F (disc f ). It is
enough to show this for f with disc f 6= 0. For, if this is case is known, then
U = {f : disc f = 0} ∪ {f : θ(f ) = F (disc f )}
is a union of Zariski-closed subsets, but U is irreducible, so that U = {f : θ(f ) = F (disc f )}.
Recall that any symmetric matrix is congruent (over C) to a matrix of the form


1


..


.




1


diag(1, . . . , 1, 0, . . . , 0) = 

0




.
..


0
Thus if f ∈ U corresponds to the matrix A and λ = disc f = det A 6= 0 then there exists B ∈ GLm (C)
such that B t AB = I. Now if C = diag(1, . . . , 1, (det B)−1 ), then BC ∈ SLm (C), and
(BC)t A(BC) = C t B t ABC = diag(1, . . . , 1, (det B)−2 ) = diag(1, . . . , 1, λ)
and so θ(f ) = θ(fλ ) = F (λ), as required.
(11.7) Discriminant of binary forms
Define
U = HC,n (C2 , C) =
Take f ∈ U . Then f has the form
homogeneous polynomials of degree n
in two variables X1 , X2
f (X1 , X2 ) = a0 X1n + a1 X1n−1 X2 + · · · + an X2n
= b(λ1 X1 + µ1 X2 )(λ2 X1 + µ2 X2 ) · · · (λn X1 + µn X2 )
and define
disc(f ) = b2n−2
Y
i<j
(λi µj − λj µi )2
This is well-defined since it is unchanged if two terms are exchanged or if one term is enlarged and
another is reduced by the same factor.
Example (n = 3) disc(f ) = −27a20 a23 + 18a0 a1 a2 a3 − 4a0 a32 − 4a31 a3 + a21 a22
Claim
disc : U → C is a (2n − 2)-homogeneous SL2 (C)-invariant.
Proof
Exercise for the daring.
(11.8) Definition)
U ⊕ V → C.
A covariant for a CSL(V )-module U is a polynomial invariant of the form
Examples
48
ˆ Every invariant θ for U gives rise to a covariant
U ⊕V
(u, v)
→
C
7
→
θ(u)
ˆ If U = HC,n (V, C) then there is an ‘evaluation covariant’ ev :
U ⊕V
(f, v)
→
C
.
7→ f (v)
(11.9) The Hessian
If f is a function of X1 , . . . , Xm then the Hessian of f is
∂2f
H(f ) = det
∂Xi ∂Xj
It is itself a function of X1 , . . . , Xm .
Let U = HC,n (Cm , C), the n-homogeneous polynomials in X1 , . . . , Xm . The Hessian defines a polynomial map
U ⊕ Cm →
C
(f, v)
7→ H(f )(v)
Now U is naturally a CGLm (C)-module via the action
(gf )(v) = f (g −1 v),
f ∈ U, g ∈ GLm (C), v ∈ Cm
By the chain rule,
H(gf )(gv) = (det g)−2 H(f )(v)
and so H is an SL2 (C)-covariant.
49
12
First fundamental theorem of invariant theory
The first fundamental theorem of invariant theory gives generators for the set of polynomial invariants
when the module is the direct sum of copies of V and V ∗ . The hope was then in principle to compute
invariants of an arbtirary rational module, but in practice this proved to be impossible. We will state
and prove the first fundamental theorem for GL(V ) and state it for SL(V ).
As usual, V is an m-dimensional C-space with basis e1 , . . . , em .
(12.1) Theorem (multilinear first fundamental theorem for the general linear group)
For n, r ∈ N,
(i) HomCGL(V ) (T n (V ∗ ) ⊗ T r (V ), C) = 0 if n 6= r;
(ii) HomCGL(V ) (T n (V ∗ ) ⊗ T n (V ), C) = spanC {µσ : σ ∈ Sn }, where
µσ (ϕ1 ⊗ · · · ⊗ ϕn ⊗ v1 ⊗ · · · ⊗ vn ) = ϕσ(1) (v1 ) · · · ϕσ(n) (vn )
Remark
Proof
A proof in arbitrary characteristic can be found in de Cocini and Procesi (1976).
Let X, Y be CG-modules, where G is some group. We know that
HomC (X ∗ ⊗ Y, C) = (X ∗ ⊗ Y )∗ ∼
= X ∗∗ ⊗ Y ∗ ∼
=X ⊗Y∗ ∼
= HomC (Y, X)
Taking G-fixed points, we obtain
HomCG (X ∗ ⊗ Y, C) ∼
= HomCG (Y, X)
With this and the isomorphism T n (V ∗ ) ∼
= (T n V )∗ we have an isomorphism
∼
=
π :
HomCGL(V ) (T r V, T n V ) −→
f
7−→
HomCGL(V ) (T n (V ∗ ) ⊗ T r (V ), C)
ϕ1 ⊗ · · · ⊗ ϕn ⊗ ~v 7→ (ϕ1 ⊗ · · · ⊗ ϕn )(f (v))
Now
(i) If n 6= r then HomCGL(V ) (T r V, T n V ) = 0 because HomCGL(V ) (S 0 , S) = 0 if S 0 (resp. S) is an
irreducible r-homogeneous (resp. n-homogeneous) CGL(V )-module.
(ii) is just a restatement of Schur–Weyl duality. To see this, recall that EndCGL(V ) (T n V ) is spanned
by λσ for σ ∈ Sn , where
λσ (v1 ⊗ · · · ⊗ vn ) = vσ−1 (1) ⊗ · · · ⊗ vσ−1 (n)
and now π(λσ ) = µσ .
Notation For CG-modules U and W , let HomCG,n (U, W ) denote the vector space of all n-homogeneous
concomitants U → W (cf. (11.1)).
Note that HomCG,n (U, W ) is a CG-module with conjugation action
(gf )(u) = gf (g −1 u),
g ∈ G, u ∈ U, f ∈ HomCG,n (U, W )
Note also that HomCG,n (U, W ) = HC,n (U, W )G .
(12.2) Lemma
If U is a CG-module then
C[U ]G ∼
=
M
HomCG (U, C)
n≥0
Proof We know that
C[U ] ∼
=
M
HC,n (U, C)
n≥0
50
is an isomorphism of CG-modules; now take G-fixed points.
(12.3) Lemma If U and W are CCG-modules then there exist invrse isomorphisms of vector spaces
HomCG (Sn U, W ) →
ψ
7→
where ∆ :
U
u
HomCG,n (U, W ) →
HomCG,n (U, W )
and
ψ◦∆
f
7→
HomCG (Sn U, W )
1
n! P f
7
→
Sn U
and P f is the total polarisation of f (see after (6.13)).
7→ u ∨ · · · ∨ u
Clearly the maps induce isomorphisms between HomC (Sn U, W ) and HC,n (U, W ), so we’ll
Proof
prove:
(i) ψ concomitant ⇒ ψ ◦ ∆ concomitant;
(ii) f concomitant ⇒
1
n! P f
concomitant.
So. . .
(i) is clear since ∆ is a concomitant.
(ii) To prove this, there are (at least) three possible approaches.
Use the formula for P f (not recommended).
1
P f ◦ ∆, so
Approach 2 If f ∈ HomCG,n (U, W ) then f = n!
1
1
P f (g(u ∨ · · · ∨ u)) =
P f (g∆(u))
n!
n!
1
P f ∆(g(u))
=
n!
Approach 1
= f (g(u))
= gf (u)
1
=g
P f (u ∨ · · · ∨ u)
n!
Approach 3 HomC (Sn U, W ) and HC,n (U, W ) are CG-modules, with G acting by conjugation,
1
and ψ 7→ ψ ◦ ∆ is a CG-map and an isomorphism. So its inverse f 7→ n!
P f is also a CG-map.
These maps restrict to isomorphisms between sets of G-fixed points.
Remark
δ
natural
∆ factors as U −
→ T n U −−−−→ Sn U , where
δ(u) = u ⊗ · · · ⊗ u
| {z }
n
so composition with δ induces a surjection
HomCG (T n U, W ) −→ HomCG (U, W )
We’ll use this reformulation of (12.3) to prove
(12.4) Theorem (first fundamental theorem for the general linear group)
Let U = V ⊕ · · · ⊕ V ⊕ V ∗ ⊕ · · · ⊕ V ∗ . Then C[U ]GL(V ) is generated as a C-algebra by elements ρij
|
{z
} |
{z
}
p
q
for 1 ≤ i ≤ p, 1 ≤ j ≤ q, where
ρij (v1 , . . . , vp , ϕ1 , . . . , ϕq ) = ϕj (vi )
Proof
Clearly the ρij are polynomial invariants, since
gϕj (gvi ) = ϕj (g −1 gvi ) = ϕj (vi )
51
By (12.2) we must show that any n-homogeneous invariant f is a linear combination of products of
the ρij . By (12.3) there exists a surjection
HomCGL(V ) (T n U, C) −→ HomCGL(V ),n (U, C)
Lp+q
Put V1 = · · · = Vp = V and Vp+1 = · · · = Vp+q = V ∗ , so that U = i=1 Vi .
This induces a decomposition of T n U as
T nU =
p+q M
p+q
M
i1 =1 i2 =1
Thus
HomCGL(V ) (T n U, C) =
···
p+q
M
in =1
M
i1 ,...,in
Vi1 ⊗ Vi2 ⊗ · · · ⊗ Vin
HomCGL(V ) (Vi1 ⊗ · · · ⊗ Vin , C)
Focus on one of the summands. By (12.1), either HomCGL(V ) (Vi1 ⊗ · · · ⊗ Vin , C) = 0, or n = 2k, k of
the ij are ≤ p and k of the ij are > p.
In the latter case, say ij ≤ p for j = α1 , . . . , αk and ij > p for j = β1 , . . . , βk . In this case,
HomCGL(V ) (Vi1 ⊗ · · · ⊗ Vin , C) is spanned by the k! maps τσ , for σ ∈ Sk , given by
τσ (v1 ⊗ · · · ⊗ vn ) = ϕβσ(1) (vα1 ) · · · ϕβσ(k) (vαk )
The n-homogeneous invariant of U corresponding to τσ is just
ρiα1 ,iβσ(1) −p · · · ρiαk ,iβσ(k) −p
and we’re done.
(12.5) Theorem
If r ∈ N and U = EndC V ⊕ · · · ⊕ EndC V then C[U ]GL(V ) is generated as a
{z
}
|
r
C-algebra by invariants
ti1 ,...,ik :
where k ≥ 1 and 1 ≤ i1 < · · · < ik ≤ r.
U
→
C
(θ1 , . . . , θr ) 7→ tr(θi1 · · · θik
Remarks
(1) [Procesi, 1976] In fact, C[U ]GL(V ) is generated by the ti1 ,...,ik with 1 ≤ k ≤ 2m − 1.
(2) Procesi also shows that (12.5) is equivalent to the first fundamental theorem. This links invariant
theory to the representation theory of bouquets, a certain type of quiver.
∼
Proof By (12.2) and (12.3) we must compute HomCGL(V ) (T n (EndC V, C) which, since EndC V =
V ∗ ⊗V , is isomorphic to HomC (T n V ∗ ⊗T n V, C). By the multilinear first fundamental theorem (12.1),
this is spanned by the µσ for σ ∈ Sn , where
µσ (ϕ1 ⊗ ϕn ⊗ v1 ⊗ · · · ⊗ vn ) = ϕσ(1) (v1 ) · · · ϕσ(n) (vn )
We now compute the corresponding map
νσ : T n (EndC V ) −→ C
Let V have basis e1 , . . . , em and V ∗ have dual basis η1 , . . . , ηm . Let (Aij ) be the matrix of θ ∈ EndC V
with respect to these bases, i.e.
X
θ(v) =
Aij ηj (v)ei
i,j
∗
so that the corresponding element of V ⊗ V is
P
i,j
Aij ηj ⊗ ei .
If θk has matrix Akij then θ1 ⊗ · · · ⊗ θn corresponds with
X
X
A1a1 b1 · · · Anan bn ηb1 ⊗ · · · ⊗ ηbn ⊗ ea1 ⊗ · · · ⊗ ean ∈ T n V ∗ ⊗ T n V
a1 ,...,an b1 ,...,bn
52
so we have
νσ (θ1 ⊗ · · · ⊗ θn ) =
=
X
~
a,~b
X
~b
A1a1 b1 · · · Anan bn ηbσ(1) (ea1 ) · · · ηbσ(n) (ean )
A1bσ(1) b1 · · · Anbσ(n) bn
[The latter equality comes from the definition of dual basis.] Writing σ = (i1 · · · ik )(j1 · · · j` ) · · · as a
product of disjoint cycles, we reorder this to get
X
νσ (θ1 ⊗ · · · ⊗ θn ) =
Aib1i bi Aib2i bi · · · Aibki bi Ajb1j bj · · · Ajb`j bj · · ·
2
1
3
2
1
k
2
1
1
`
= tr(θik · · · θi1 ) tr(θj` · · · θj1 ) · · ·
which is what we required.
Theorem (first fundamental theorem for the special linear group)
Let V be an m-dimensional C-space with basis e1 , . . . , em and let V ∗ have dual basis η1 , . . . , ηm . If
U = V ⊕ ··· ⊕ V ⊕V ∗ ⊕ ··· ⊕ V ∗
{z
} |
{z
}
|
p
q
then C[U ]SL(V ) is generated as a C-algebra by the polynomial invariants which send
ˆ (v1 , · · · , vp , ϕ1 , · · · , ϕq ) 7→ ϕj (vi ) for 1 ≤ i ≤ p, 1 ≤ j ≤ q;
ˆ (v1 , · · · , vp , ϕ1 , · · · , ϕq ) 7→ |vi1 , . . . , vim | for 1 ≤ i1 < · · · < im ≤ p
ˆ (v1 , · · · , vp , ϕ1 , · · · , ϕq ) 7→ |ϕj1 , . . . , ϕjm | for 1 ≤ j1 < · · · < jm ≤ q
where |v| is the determinant of the matrix whose nth column is the coordinate of the vin with respect
to e1 , . . . , em and |ϕ| likewise with respect to ηj1 , . . . , ηjm .
End of course.
53
MATHEMATICAL TRIPOS PART III, 2013
REPRESENTATION THEORY
This sheet deals with the representation theory and combinatorics of Sn and of GLn .
Work over C unless told otherwise.
1
Let H be a subgroup of Sn ; define the element H − =
algebra CSn . Prove that
and
P
h∈H
εh h which lies in the group
πH − = H − π = επ H − for any π ∈ H,
πH − π −1 = (πHπ −1 )− for any π ∈ Sn .
2
Let r be a natural number. Let λ = (λ1 , . . . λ` ) be a sequence of positive integers. We
P
call λ a composition of r into ` parts if `i=1 λi = r. The elements λi are called the parts of
λ.
(a) Find a formula for the number of compositions of r.
(b) Find a formula for the number of compositions of r with at most k parts.
3
Let r, s be non-negative integers. A partition of the form λ = (r, 1s ) is called a hook
partition; we will write H(n) for the set of all hook partitions of n. Let f λ be the number of
standard tableaux of shape λ.
(a) Find a formula for the number of standard tableaux of shape λ = (r, 1s ) where λ is
a partition of n. Justify
P your answer.
(b) Prove that λ∈H(n) f λ = 2n−1 .
P
(c) Prove that λ∈H(n) (f λ )2 = 2(n−1)
.
n−1
4
Let n ∈ N, and let λ = (λi )i∈N and µ = (µj )j∈N be partitions of n. The conjugate
partition λ0 is defined to be the partition λ0 = (λ01 , λ02 , . . .) where λ0i = |{j|λj > i}|. The
dominance order on the set of partitions of n is defined as follows:
λ µ if and only if for all i ∈ N:
X
X
λj 6
µj .
j6i
j6i
0
0
(a) Prove that λ µ if and only if µ λ .
(b) We say that a partition λ is self-conjugate if λ = λ0 . Show that there are as many
self-conjugate partitions of n as there are partitions of n with distinct odd parts (i.e. all λi
are odd and distinct).
(a) Assuming k > 3, prove that then f (3 ) > 3k.
(b) Let λ = (l, . . . , l) = (lk ) be a partition of n = kl and λ̃ = (l + 1, . . . , l + 1) be a
partition of n + k. Prove (by using the Young-Frobenius formula (5.1)) that
5
k
f λ̃
(n + k)!
l!
=
·
.
λ
f
n!
(l + k)!
(c) Let k, l > 3 with kl = n. Prove that then f (l ) > n.
k
1
6
Let Ck be the set of all standard tableaux of shape (k, k) whose entry in box (2,1) is
some odd number.
(a) Find |Ck |.
(b) For a partition λ = (r, s) with two parts, write down f (r,s) (e.g. from the YoungFrobenius formula). Can you find limk→∞ |Ck |/f (k,k) ?
7
In this question λ = (λ1 λ2 , . . . , λk ) is a partition of n, and M λ is the permutation module
with basis the λ-tabloids (a λ-tabloid is an equivalence class of numberings of a Young diagram
(with distinct numbers 1,...,n), two being equivalent if the corresponding rows containing the
same entries.)
(a) Show that M λ is isomorphic to the permutation module C∆ where ∆ = cos(Sn : Sλ ),
(the coset space with Sλ the Young subgroup).
(b) Show that
n!
dim M λ = Qk
.
i=1 λi !
(c) In the case λ = (n − 2, 2) has two parts, show that M λ ∼
= CΩ(2) where Ω(2) is the set
of 2-element subsets of Ω = {1, . . . , n} (with the natural action).
P
8
For a λ-tableau T with n boxes, define the signed column sum κT = σ∈CT εσ σ . Let
eT = κT {T } be the λ-polytabloid. Let 1 denote the identity element of Sn .
(a) List the columns of T as C1 , C2 , . . . , Cs . Prove that
P κT = κC1 · · · κCs .
−
(b) Let H 6 Sn and let (a b) ∈ H. Define H = h∈H εh h, as in Q.1. Prove that
H − = s · (1 − (a b)) for some s ∈ CSn .
(c) Let π ∈ Sn and σ ∈ CT . Show that
κπT = πκT π −1 ,
κT σ = σκT = εσ κT
and
eπT = πeT ; ; σeT = εσ eT .
9
*[The Murnaghan-Nakayama Rule.]
Part (a) below is an optional task for enthusiasts only. You will need to read about
James’ method (lecture notes pp 79–83) using the Littlewood-Richardson Rule. A good
account appears in Sagan’s Springer GTM, p.182. This beautiful result has generalisations
to the irreducible characters of exotic structures called Iwahori-Hecke algebras of type An−1
due to Ram (who appealed to Schur-Weyl duality), and it appears to have been known to
Young in his work on hyperoctahedral groups (the so-called Weyl groups of type Bn ).
(a) Consider the character χλ (w) =: χλµ where w ∈ Sn has cycle type (µ1 , µ2 , . . . , µ` ).
Suppose there are precisely t partitions λ(1), . . . , λ(t) which are obtained by removing a rim
λ(i)
hook of length µ1 from [λ]. Find an expression for χλµ in terms of χ(µ2 ,µ3 ...,µ` ) .
(b) Let n > 6. Prove that
(n−3,3)
χ(2,1n−2 ) = (n − 3)(n − 4)(n − 5)/6.
10 Given some CSn -module M , with action written as π · m (π ∈ Sn , m ∈ M ). Define a
new CSn -module M ε as the module whose underlying vector space is M and with new action
π ∗ m := επ π · m (π ∈ Sn , m ∈ M ).
(a) Verify that M ε is a CSn -module.
(b) Prove that M is irreducible if and only if M ε is irreducible.
n
(c) Let χ be the character of M , and let χ(1 ) be the character of the Specht module
n
n
S (1 ) . Prove that the character χε of M ε is given by χε (π) = χ(1 ) (π) for π ∈ Sn .
n
[Why do you now know that the trivial module S (n) and the sign module S (1 ) are the only
1-dimensional CSn -modules.]
11 Let ρ :GL(V ) →GLn (C) be an irreducible polynomial representation. Show that the
matrix entries ρij ∈ C[GL(V )] are homogeneous polynomials of the same degree. More
generall, show that every polynomial representation ρ is a direct sum of representations ρ(i)
whose matrix entries are homogeneous polynomials of degree i.
12 Give a proof of the result proved in lectures that affine n-space An is irreducible without
appealing to the Nullstellensatz.
13 Prove that the 1-dimensional rational representations of C∗ =GL1 (C) are power maps
of the form u 7→ ur where r ∈ Z.
14 (a) Show that the set of diagonalisable matrices is Zariski dense in Mn (C) (hence every
invariant function on Mn (C) is completely determined by its restriction to Tn , the diagonal
matrices. [Actually this is nothing more than Jordan decomposition; if you work a bit harder
you can show this is true over any field, not necessarily algebraically closed].
(b) Let ρ :GLn →GL(W ) be a rational representation. Show that ρ is polynomial if and
only if ρ|Tn is polynomial. Show also that ρ is polynomial if and only if its character χρ is a
polynomial.
(c) Deduce that every 1-dimensional rational representation of GL(V ) is of the form
r
det :GL(V ) →GL1 , r ∈ Z.
15 Consider the linear action of GL(V ) on End(V ) by conjugation.
(a) What are the orbits of thisP
action? Any element α ∈End(V ) has a characteristic
n
polynomial of the form pα (t) = t + ni=1 (−1)i ei (α)tn−i , where n = dim V , showing that the
ei are invariant polynomial functions on End(V ).
(b) [HARD] Find a proof of the fact that the ring of invariants for the conjugation action
is generated by the si without using properties of symmetric functions.
(c) Show that the functions Tr1 . . . ,Trn generate the invariant ring C[End(V )]GL(V ) . Here
Trk :End(V ) → C is the map A 7→TrAk , (k = 1, 2, . . .).
SM, Lent Term 2013
Comments on and corrections to this sheet may be emailed to [email protected]
MATHEMATICAL TRIPOS PART III, 2013
REPRESENTATION THEORY
This sheet deals mostly with polynomial invariant theory. Work over C.
1
Let W be finite-dimensional over C. Check that the coordinate functions x1 , . . . , xm ∈
C[W ] are algebraically independent (equivalently, if f (a1 , . . . , am ) = 0 for a polynomial f and
all a = (a1 , . . . , am ) ∈ Cm then f is the zero polynomial). Hint: use induction on the number
of variables and the fact that a non-zero polynomial in one variable has only finitely many
zeroes.
2
(a) Show that the natural representation of SL2 (C) on C2 is self-dual (i.e. equivalent to
its dual representation). Hint: find a non-singular matrix P such that (A−1 )t = P AP −1 .
(b) Show that this representation has two orbits.
3
Consider GL2 acting on C[X, Y ] (induced by the natural representation of GL2 on C2 ).
(a) What is the image of X and Y under the action of any g ∈GL2 ?
(b) Calculate C[X, Y ]GL2 and C[X, Y ]SL2 .
(c) Calculate C[X, Y ]U where U is the subgroup of upper triangular unipotent matrices.
4
Consider the representation of SLn (C) in its action on Mn (C), as discussed in lectures.
Prove that the invariant ring is generated by the determinant. You may need the fact that
GLn is Zariski-dense in Mn .
5
Determine the invariant rings of C[M2 (C)]U and C[M2 (C)]T under left multiplication by
the subgroup U of upper triangular unipotent matrices and the 2-torus T =diag(t, t−1 ) of
diagonal matrices of SL2 .
6
Let Tn be the subgroup of GLn of non-singular diagonal matrices. Choose the standard
basis in W = Cn and the dual basis in W ∗ and thus identify the coordinate ring C[W ⊕ W ∗ ]
with C[x1 , . . . xn , ζ1 , . . . , ζn ]. Show that
C[W ⊕ W ∗ ]Tn = C[x1 ζ1 , . . . , xn ζn ].
What happens if you replace Tn by the subgroup Tn0 of diagonal matrices with determinant
1?
1
7
Recall that for each j > 1 the symmetric functions nj (x) = xj1 + xj2 + · · · xjn are known as
power sums (or Newton functions). As before the ej are the elementary symmetric functions.
(a) Show that
(−1)j+1 jej = nj − e1 nj−1 + e2 nj−2 − · · · + (−1)j−1 ej−1 n1
for all jQ= 1 . . . , n. This is known
P as the Newton-Girard identity. [Hint: let j = n, consider
f (t) = i (t − xi ) and calculate i f (xi ). For j < n what can you say about the right-hand
side?]
(b) Here is Weyl’s original proof1 of the identity in (a). Define the polynomial
n
Y
ψ(t) =
(1 − txi ) = 1 − e1 t + e2 t2 − · · · + (−1)n en tn ,
i=1
where the ei are the elementary symmetric functions. Determine its logarithmic derivative
0 (t)
− ψψ(t)
as a formal power series. Deduce (a).
(c) Show that (in characteristic 0) the power sums generate the symmetric functions.
8
(a) Let A be a commutative algebra and let G be a group algebra of automorphisms of
A. Assume the representation of G on A is completely reducible. Show that the subalgebra
AG of invariants has a canonical G-stable complement and the corresponding G-equivariant
projection π : A → AG satisfies π(hf ) = hπ(f ) for h ∈ AG , f ∈ A. This projection is
sometimes called the Reynolds operator.
(b) Let A = ⊕j>0 Aj be a graded K-algbbra (meaning Ai Aj ⊂ Ai+j ). Assume the
idea A+ = ⊕j>0 Aj is finitely-generated. Show that A is finitely-generated as an algebra
over A0 . (In fact if the ideal A+ is generated by the homogeneous elements a1 , . . . , an , then
A = A0 [a1 , . . . , an ].)
(c) Deduce Hilbert’s theorem: if W is a G-module and the representation of G on K[W ]
is completely reducible then the invariant ring K[W ]G is finitely-generated. [Hint: you will
need to use Hilbert’s Basis Theorem.] This shows also that K[W ]G is noetherian.
(d) Hilbert’s Theorem is applicable to finite groups provided Maschke’s Theorem holds
(i.e we require that |G| is invertible in K). In the special case of characteristic 0, show that
for any representation W of G, the ring of invariants K[W ]G is generated by invariants of
degree less than or equal to |G| (Noether’s Theorem).
Schmidt introduced a numerical invariant β(G) for every finite group G. It is defined
to be the minimal number m such that for every representation W of G, the invariant ring
K[W ]G is generated by the invariants of degree 6 m. With this definition, Noether’s Theorem
asserts that β(G) 6 |G|. In fact Schmidt showed that β(G) = |G| if and only if G is cyclic2.
It is also a fact that if G is actually abelain then β(G) coincides with the so-called Davenport
constant.
In the next few questions we’ll write V = C2 , G =SL(V ) =SL2 (C), and
Cn = HC,n (V, C) ∼
= (S n V )∗ ∼
= S n (V ∗ ),
which can be identified with the set of homogeneous polynomials of degree n in variables
X1 , X2 . Note that {Cn : n ∈ N} are the non-isomorphic simple CG-modules. For a fixed n,
recall that a covariant for Cn is a polynomial CSL(V )-invariant Cn ⊕ V → C. The following
questions sketch a classification of the generators for C[Cn ⊗ V )]G .
1Weyl,
The Classical Groups, II.A.3
Finite groups and invariant theory in Séminaire d’Algèbre Paul Dubreil et M.-P. Malliavin (Springer
Lecture Notes 1478, Springer 1991)
2see
Suppose f, g are functions of X1 , X2 and r ∈ N. Define the rth transvectant of f, g as
r
X
(−1)j
∂rf
∂rg
τr (f, g) =
r−j
j
j
r−j .
j!(r
−
j)!
∂X
∂X
∂X
∂X
1
2
1
2
j=0
r
1
∂ ∂
∂ ∂
=
−
(f (X1 , X2 )g(Y1 , Y2 ) |Y1 =X1 ,Y2 =X2 .
r! ∂X1 ∂Y2 ∂X2 ∂Y1
You should observe τ0 (f, g) = f g, τ1 (f, g) is the Jacobian of f, g and τ2 (f, f ) is the Hessian
of f . Show that
(a) if f, g are homogeneous polynomials of degrees p, q then τr (f, g) = 0 unless r 6
min{p, q}, in which case it is a homogeneous polynomial of degree p + q − 2r;
(b) if r 6 min{p, q} then the map τr : Cp ⊗ Cq → Cp+q−2r sending f ⊗ g 7→ τr (f, g) is a
non-zero map of CG-modules;
(c) any CG-module map Cp ⊗ Cq → Cp+q−2r is a multiple of τr . [Hint: use ClebschGordan];
(d) the τr give an isomorphism of CG-modules
9
min{p,q}
Cp ⊗ Cq →
M
Cp+q−2r .
r=0
10 Let p, q, n, r ∈ N and r 6 min{q, n}. Set N = max{0, r − p}. Show that there are scalars
λN , . . . , λr ∈ C, with λr 6= 0, such that
r
X
λj τj (τr−j (f, g), h),
f τr (g, h) =
j=N
for all f ∈ Cp , g ∈ Cq , h ∈ Cn . [HINT: Schur’s Lemma.] Thus the τj are trying hard to be
associative.
11 Fix n ∈ N. Let R = C[Cn ⊕ V ], so that RG is the set of covariants of Cn . If ϕ ∈ R and
f ∈ Cn , define ϕ(f ) ∈ C[V ] by ϕ(f )(x) = ϕ(f, x). If r ∈ N and φ, ψ ∈ R define τr (φ, ψ) as
the map Cn ⊕ V → C by (f, x) 7→ τr (φ(f ), ψ(f ))(x). Finally if d, i ∈ N, let Rdi be the set of
all φ ∈ R which are homogeneous, of degree d in Cn and degree i in V . Show that
G
(a) Rdi are CG-submodules of R, and R is graded, R = ⊕∞
d,i=0 Rdi , hence so too is R ;
(b) Rdi Rej ⊆ Rd+e,i+j and τr (Rdi , Rej ) ⊆ Rd+e,i+j−2r for r 6 min{i, j};
(c) the assignment φ 7→ (f 7→ φ(f )) induces an isomorphism
RG ∼
= homCG,d (Cn , Ci ),
di
and deduce that
G
(d) any covariant θ ∈ Rdi
(d > 1) can be expressed as a linear combination
min{n,i}
θ=
X
τn−r (φr , E)
r=0
G
with φr ∈ Rd−1,n+i−2r
and E is the evaluation map E(f, v) = f (v) [you should note that (c)
G
says that R1,i = C.E when i = n and 0 if i 6= n.]
12 [Gordan’s Theorem3 - a weak form] If S is a C-subalgebra of RG with the property that
τr (φ, E) ∈ S whenever r ∈ N and φ ∈ S, then S = RG . [Remark: after the last question the
proof should reduce to a two- line argument.]
13 [REALLY HARD] This uses Gordan’s Theorem to find a set of generators of RG in the
case n = 3. Given n = 3, show that RG is generated by the covariants
G
;
E ∈ R13
G
;
H = τ2 (E, E), the Hessian, ∈ R22
G
t = τ1 (H, E) ∈ R33 ;
G
D = τ3 (t, E), the discriminant (×48), ∈ R40
14 [REALLY REALLY HARD4 ] Take n = 4 and repeat the last question to generate RG
using quartic forms.
Classically, the (still unsolved) problem of computing all covariants of binary forms of
degree n was tackled by something called the symbolic method (using polarisation to reduce
it to the FFT). Associated to the symbolic method is the symbolic notation, designed to
make the calculations easier, but still non-trivial. You can consult any old text on Invariant
Theory, such as Grace and Young if interested.
SM, Lent Term 2013
Comments on and corrections to this sheet may be emailed to [email protected]
3Paul
Gordan, dubbed the ‘King of Invariant Theory’ is perhaps better known for being Emmy Noether’s
thesis advisor. The result appeared in 1868 in Crelle’s Journal as Beweis, dass jede Covariante und Invariante einer binären Form eine ganze Funktion mit numerischen Coeffizienten einer endlichen Anzahl solcher
Formen ist.
4J.J. Sylvester’s collected works (four volumes, edited by H.F. Baker) include tables with details about
higher degree forms. However his published tables for forms of degree larger than 6 appear to be totally wrong.
Corrections appear in von Gall (1888) and Dixmier/Lazard (1986), Shioda (1967) and Brouwer/Popoviciu
(2010).