Representations of Sd

4 Pages·2015·0.076 MB·English

by  Sam Evens

REPRESENTATIONS OF S
d
SAM EVENS
1. Actors
Let λ be a partition of d. Let Y be the collection of numberings T of shape λ. By
definition, a numbering of λ is a way to assign the numbers {1,...,d} to the boxes
of the Young diagram of λ so each number goes in precisely one box. The symmetric
group S acts on Y by (σ,T) 7→ σ·T. Here if T has the number a in a given box, σ·T
d
has σ(a) in the given box. The S -action on Y is transitive, and the stabilizer of any
d
numbering is trivial.
For a given numbering T, the row group R(T) is the elements σ ∈ S such that for
d
each row A of T, σ permutes the elements of the row A . The column group C(T) is
i i
the elements of S which permute the elements of each column of T.
d
For numberings T ,T of shape λ, we say T ∼ T if there is σ ∈ R(T ) such that
1 2 1 2 1
T = σ·T . Then the row group R(T ) = R(T ) by equation (1) on page 84 of Fulton.
2 1 2 1
It follows easily that ∼ is an equivalence relation.
ForanumberingT ofshapeλ, let[T]denotetheequivalenceclassofT. LetZ bethe
set of all equivalence classes. It is routine to check that if T ∼ T , then σ·T ∼ σ·T
1 2 1 2
for σ ∈ S , so S acts on Z by the formula σ·[T] = [σ·T].
d d
Let Mλ be the complex vector space with basis given by [T] for [T] in Z. Let T be
0
a fixed numbering. Then S acts transitively on Z and the stabilizer of T is R(T ).
d 0 0
Hence, there is a bijection S /R(T ) → Z, so |Z| = |S /R(T )|. If λ is the partition
d 0 d 0
d = Pk λ , then R(T ) ∼= Qk S , so
i=1 i 0 i=1 λi
d!
|Z| = .
k
Q ((λ )!)
i+1 i
Note that S acts on Mλ by σ · Pa [T] = Pa [σ · T], and this makes Mλ into a
d T T
representation of S .
d
There is a bijection iSRd(T)(Ctriv) = C[Sd]⊗C[R(T0)]Ctriv → Mλ. Indeed, let g1,...,gl
be representatives in the distinct cosets of S /R(T ). Let a ,...,a ∈ C. The bijection
d 0 1 l
takes an element Pl g ⊗a of iSd (C ) to Pa [g ·T ], and it is easily seen to
i=1 i i R(T) triv i i 0
be S -equivariant. We can do something like this for any transitive action of a finite
d
group on a set.
1
2 SAMEVENS
For a numbering T of shape λ, let b = P sgn(q)q ∈ C[S ]. Let v = b ·[T] ∈
T q∈C(T) d T T
Mλ. We let Vλ = PCv ⊂ Mλ, where the sum is over all numberings of T of shape
T
λ. Remark 3.4 asserts that v is nonzero, so Vλ is nonzero. By Lemma 3.5, if σ ∈ S ,
T d
then σ·vT = vσ·T. It follows that Vλ is a subrepresentation of Mλ.
Recall also the order relation on partitions of d. Let λ = (λ ,...,λ ) and µ =
1 k
(µ ,...,µ ) be partitions of d with the convention that λ ≥ λ ≥ ··· ≥ λ ≥ 0 and
1 k 1 2 k
µ ≥ µ ≥ ··· ≥ µ ≥ 0. If instead µ = (µ ,...,µ ) as above, but with k > j, we may
1 2 k 1 j
add µ = ··· = µ = 0 to the partition, and thereby assume j = k (and similarly if
j+1 k
j > k). For λ,µ as above, we say λ D µ if for all i, Pi λ ≥ Pi µ . If λ D µ and
j=1 j j=1 j
λ 6= µ, we write λ ⊲ µ. This is a partial order on partitions of d.
These are the basic actors in our story, Mλ,Vλ,[T],v and b , but one can also
T T
include the row and column groups R(T) and C(T), and the relation D.
2. Main Results
Proposition 2.1. (see Diaconis, Lemma 1) Let λ and λ′ be partitions of d and let T
(resp. T′) be numberings of shape λ (resp. λ′). Assume that b ·[T′] 6= 0.
T
(1) If b ·[T′] 6= 0, then λ D λ′.
T
(2) If λ = λ′, then b ·[T′] ∈ Cv .
T T
Proof. First suppose there exists i 6= j such that {i,j} are in the same row of T′ and
the same column of T. Then there is a transposition q = (ij) in R(T′) and C(T).
Hence,
′ ′ ′ ′
b ·[T ] = b ·[q·T ] = b ·q·[T ] = −b ·[T ],
T T T T
using Lemma 3.3 for the last step. Thus b ·[T′] = 0. Hence, by assumption, there is
T
no such pair (i,j), so by Lemma 3.2, λ D λ′. This proves (1). To prove (2), we saw in
the proof of (1) that if there is a pair (i,j) in the same row of T′ and the same column
of T, then b ·[T′] = 0. If such a pair does not exist, then by the second part of Lemma
T
3.2, there is p′ ∈ R(T′) and q ∈ C(T) so that p′·T′ = q·T. Hence, by Lemma 3.3,
′ ′ ′
b ·[T ] = b ·[p ·T ] = b ·[q·T] = sgn(q)b ·[T] = ±v .
T T T T T
This proves (2). (cid:3)
Proposition 2.2. (see Fulton, (5a) and (5b), p. 87) Let λ and λ′ be partitions of d,
and let T be a numbering of shape λ.
(1) b (Mλ) = b (Vλ) = Cv is nonzero.
T T T
(2) Assume λ 6D λ′. Then b (Mλ′) = b (Vλ′) = 0.
T T
Proof. For (1), note Vλ ⊂ Mλ implies that b (Vλ) ⊂ b (Mλ). By the definition of
T T
Mλ, b (Mλ) is the span of elements b [T′], as T′ runs over numberings of shape λ. By
T T
REPRESENTATIONS OF Sd 3
part (2) of Proposition 2.1, each b [T′] ∈ C·v . But by Lemma 3.3,
T T
|C(T)|v = |C(T)|b [T] = b ·b ·[T] = b ·v ∈ b (Vλ),
T T T T T T T
so
Cv ⊂ b (Vλ) ⊂ b (Mλ) ⊂ Cv ,
T T T T
and this establishes (1). For part (2), note that Mλ is spanned by elements [T′] where
T′ is a numbering of shape λ′. The assertion follows from Proposition 2.1 since λ 6D λ′
implies b [T′] = 0. (cid:3)
T
Theorem 2.3. (i) If λ is a partition of d, then Vλ is an irreducible representation of
Sd.
(ii) If λ and µ are partitions of d, then Vλ 6∼= Vµ.
(iii)ThedistinctirreduciblerepresentationsofS aretheVλ asλrangesoverpartitions
d
of d.
Proof. (i) Let V be a nonzero subrepresentation of Vλ. By complete reducibility, there
is a subrepresentation W of Vλ so that Vλ = V ⊕W. Let T be a numbering of shape
λ. Since b ∈ C[S ], b (V) ⊂ V and b (W) ⊂ W. Hence, b (Vλ) = b (V)⊕b (W).
T d T T T T T
By Proposition 2.2,
Cv = b (V)⊕b (W),
T T T
so 1 = dim(b (V))+dim(b (W)), and either b (V) = Cv and b (W) = 0, or else
T T T T T
b (W) = Cv and b (V) = 0. In the first case, V = C[S ]·v ⊂ V, and V = V so
T T T λ d T λ
W = 0. In the second case, V = W and V = 0, which is a contradiction. Hence, V is
λ λ
irreducible.
For (ii), since λ 6= µ, either λ 6⊲ µ or µ 6⊲ λ. In the first case, for T a numbering of
shape λ, b (Vµ) = 0 by Proposition 2.2, while b (Vλ) 6= 0 by the same proposition, so
T T
Vλ 6∼= Vµ. In the second case, simple reasoning shows Vλ 6∼= Vµ.
(iii) is an easy consequence of (i) and (ii) since the number of irreducible representa-
tionsofS isthenumberofpartitionsofd, asthenumberofirreduciblerepresentations
d
equals the number of conjugacy classes. (cid:3)
Theorem 2.4. Let λ be a partition of d, and write the representation
Mλ = (Vλ)⊕kλ ⊕⊕ (Vµ)⊕kµ,λ.
µ
Then k = 1 and k = 0 unless µ D λ.
λ µ,λ
Proof. Let V ⊂ Mλ be a subrepresentation such that V ∼= Vλ. We show that V = Vλ,
and it follows that k = 1. First, since V ∼= Vλ and b (Vλ) 6= 0 by Proposition
λ T
2.2, it follows that b (V) 6= 0. But b (V) ⊂ b (Mλ) = Cv by Proposition 2.2,
T T T T
so b (V) = Cv . Since b (V) ⊂ V, it follows that v ∈ V, so by Remark 3.6,
T T T T
V = C[S ]·v ⊂ V, andV = Vλ. Nowsupposeµ 6= λandsupposethereisaninjective
λ d T
S -module homomorphism ψ : Vµ → Mλ. Let T be a numbering of shape µ and
d
4 SAMEVENS
suppose µ 6D λ. Then b (Vµ) 6= 0 by Proposition 2.2, so b (ψ(Vµ)) = ψ(b (Vµ)) 6= 0.
T T T
But then b (Mλ) 6= 0, so by Proposition 2.2, µ D λ. This proves the assertion. (cid:3)
T
This last result says that Vλ occurs in a unique way in Mλ.
3. Lemmas and Remarks needed for Main Results
Lemma 3.1. Let T be a numbering of shape λ where λ is a partition of d and σ ∈ S .
d
Then σR(T)σ−1 = R(σ·T) and σC(T)σ−1 = C(σ·T).
Lemma 3.2. (See Lemma 0 in Diaconis) Let λ and λ′ be partitions of d, and let T
(resp. T′) be a numbering of shape λ (resp. λ′). Suppose for every pair i,j in any row
of T′, the elements i and j are in different columns of T.
Then λ D λ′.
If λ = λ′, there are p′ ∈ R(T′) and q ∈ C(T) so that p′·T′ = q·T.
Lemma 3.3. Let T be a numbering of λ with λ a partition of d.
(i) If q ∈ C(T), then b ·q = sgn(q)b = q·b .
T T T
(ii) b ·b = |C(T)|b .
T T T
Remark 3.4. For a numbering T of λ with λ a partition of d, the element v 6= 0. This
T
follows from the easy fact that R(T)∩C(T) = {e}.
Lemma 3.5. For a numbering T of λ with λ a partition of d,
(i) if σ ∈ Sd, then σ·vT = vσ·T.
(ii) If σ ∈ C(T), then σ·V = sgn(σ)v .
T T
Remark 3.6. For T and λ as in the previous Lemma, Vλ is a S -subrepresentation of
d
Mλ. Further, Vλ = C[S ]·v .
d T
Department of Mathematics, University of Notre Dame, Notre Dame, IN 46556
E-mail address: [email protected]