The L 00 -Property of S n

In this subsection, we will see that S_nhas the L⁰⁰-property. but the following lemma should be introduced first. It will be used in the proof of 3.17.

Lemma 3.16. (The branching theorem)

Let µ be a partition of n, and let x₁, · · · , x_a be all removable nodes of D_µ. For all i, denote λ_ias the partition of n − 1 which corresponds to D_µ− x_i. Then

Res^S_Sⁿ

n−1(S^µ_F

0) ∼= (S^λ_F¹

0) ⊕ · · · ⊕ (S^λ_F^a

0)

Theorem 3.17. (cf. [J2], theorem A)

Let D be a diagram for S_n. Then

χ_D|

S^(p)n ≥ φ_D^r with multiplicity 1.

Moreover, if χ_D|

S^(p)n ≥ φ for some irreducible p-modular character φ with φ 6= φ_D^r, then φ = φ_Dfor some p-regular diagram D . D^r.

Proof.

For convenience, if E is a diagram for S_n and E₁ is a p-regular diagram for S_n, the notation

χ_E|

S^(p)_n = eφE1+ · · · menas χ_E|

S^(p)n ≥ φE1 with multiplicity e and all its other constituents belong to

{φE2| E₂. E₁}.

We will prove this theorem by induction. Assume that if E is a diagram for Sn−1, then

χ_E|

S^(p)_n−1 = φ_E^r+ · · · .

We claim that it will also hold on an arbitrary diagram for S_n. Note that it holds trivially on S₁.

Let D be an arbitrary diagram for S_n. Let

X_D= {x₁, · · · , x_a} be the set of removable nodes of D^r,

Y_D= {y₁, · · · , y_b} be the set of regular-removable nodes of D^r, Z_D= {z₁, · · · , z_c} be the set of removable nodes of D.

Put an equivalence relation ∼ on X_Dby x_i∼ x_j if and only if they are in the same ladder.

Then Y_Dprovides ∼-class representatives (observe that each y_imust lie in distinct ladders since a node is regular-removable only if it is the lowest rung in D of a ladder; on the other hand, if {x_a1, · · · x_aj} is a ∼-class, then they lie in the same ladder and the lowest x_ai is regular-removable). Note that the diagrams D^r− x_iare totally ordered byD (the higher x_iis, the smaller D^r− x_iis byD), so in particular, we may assume that

D^r− y_b. D^r− y_b−1. · · · . D^r− y₁.

Also note that under this assumption, y_i−1 is higher than y_i. Moreover, since y₁, · · · , y_b are in distinct ladders, if we denote l_i as the ladder in which y_i lies, then y_i−1 is either lefter than l_i or righter than l_i. However, since D^r is p-regular, y_i−1 can not lefter than l_i. So y_i−1 is righter than l_i, which means l_i−1 is longer than l_i. Thus l₁ is the longest ladder between l₁, · · · , l_b, and hence y₁ is the first shadow node of D^r. In this proof, we continuously adopt the notation l_i.

Denote r(y_i) be the size of the ∼-class containing y_i. Then l₁hits D^r in r(y₁) nodes, and so hits D in r(y₁) nodes. Since all ladders longer than l₁ miss D^r, and so miss D.

So the rungs of l₁ in D are removable nodes of D, and we may take them to be z_i for

i= 1, · · · , r(y₁). Then for i = 1, · · · , r(y₁), we have

(D − z_i)^r= D^r− y₁

(because y₁is the lowest rung in D^r of l₁), and for i > r(y₁), we have

(D − z_i)^r= D^r− y_j

for some j > 1 (Note that not every D^r− y_i need turn up as a (D − z_i)^r. For example, p= 3, D = {(1, 1), (2, 1), (3, 1)} has 1 removable node, but D^r = {(1, 1), (2, 1), (1, 2)}

has 2 regular-removable nodes.). So by induction hypothesis, we have

χ_D−zi|

S^(p)_n−1 = φ_(D−zi)^r+ · · · = φD^r−y₁+ · · · , if i = 1, · · · , r(y₁), χ_D−zi|

S^(p)_n−1 = φ_(D−zi)^r+ · · · = φ_D^r_−yj+ · · · , if i > r(y₁) ( j 6= 1).

Now observe that by lemma 3.16, we have Res^S_Sⁿ

n−1(χD) = ∑i(χD−zi). So

Res^S_Sⁿ

n−1(χD)|

S^(p)_n−1 =

∑

i

(χD−z_i)|

S^(p)_n−1= r(y₁)φD^r−y₁+ · · ·

(recall that D^r− y_j. D^r− y₁for j > 1). Moreover, since D^rr= D^r(so Y_D= Y_D^r) and D is arbitrary, we also have

Res^S_Sⁿ

n−1(χD^r)|

S^(p)_n−1 = r(y₁)φD^r−y1+ · · · .

On the other hand, if x_i∼ y_i, then we have (D^r− x_i)^r= D^r− y_i. Since by lemma 3.16,

we have Res^S_Sⁿ

Using these results, we claim the following: (we denote this claim as (∗))

If E is a p-regular diagram for S_n(so E = E^r), and v is a shadow node of E, then

Res^S_Sⁿ

n−1(φE) ⊇ r(v)φE−v,

where r(v) denotes the number of rungs in E of the ladder in which v lies (it is identical with the above definition). Moreover, if w is the first shadow node of E, then

Res^S_Sⁿ

n−1(φE) = r(w)φE−w+ · · · .

To prove this, observe that since E is p-regular, by proposition 2.24, we have

χ_E|

n−1(φE). Moreover, collecting this result and the previous results, we obtain (recall that a shadow node is regular-removable)

Res^S_Sⁿ

So, to prove (∗), it suffices to prove that there is no p-regular diagram eE for S_nsatisfying:

Ee6= E, χE|

Sn^(p)⊇ φ

Ee, and Res^S_Sⁿ

n−1(φ

Ee) ⊇ φE−v. Assume there is a such eE. First observe that since χ_E|

S^(p)n ⊇ φ

Ee, by proposition 2.24, we obtain

Ee. E.

Moreover, since eE is p-regular, we have χ

Ee|

Thus by induction hypothesis, eE has a removable node t such that

E− v D ( eE− t)^r.

Note that since eEis p-regular,

( eE− t)^r= eE− s

for some removable node s of eE. This two conditions

Ee. E and E − v D eE− s

imply clearly that s is higher than v, and it is not hard to see that E and eE must be of the form:

(1) There exist R_a,1, R_b,1, R_a,2, R_b,2, · · · , R_a,m, R_b,m ∈ N with

R_a,1< R_b,1< R_a,2< R_b,2< · · · < R_a,m−1< R_b,m−1< R_a,m< R_b,m

the 1-st row, the 2-nd row, · · · , R_a,1− 1-th row of E andEe

are equal, and the length of the R_a,1-th row of eE is one more than the length of the R_a,1-th row of E. Moreover, for i = 1, · · · , m − 1,

the R_a,i+ 1-th row, · · · , R_b,i− 1-th row of E andEe

are equal, and the length of the R_b,i-th row of E is one more than the length of the R_b,i-th row of eE, and

the R_b,i+ 1-th row, · · · , R_a,i+1− 1-th row of E andEe

are equal, and the length of the Ra,i+1-th row of eE is one more than the length of the R_a,i+1-th row of E. Finally,

the Ra,m+ 1-th row , · · · , the R_b,m− 1-th row of E andEe

are equal, and either

• The last row of E is exactly the R_b,m-th row and its length is 1, and the last row of eE is exactly the R_b,m− 1-th row.

or

• The number of rows of E andEeare equal. The length of the R_b,m-th row of E is one more than the length of the R_b,m-th row of eE, and rows of E and eE which below to the R_b,m-th row (if exist) are all equal.

(2) The node s is the rightest node of the Ra,1− 1-th row, or the R_a,1-th row of eE.

The node v is the rightest node of the Rb,m-th row of E.

(3) Note that for each i, the rightest node in the R_b,i-th row of E is removable.

The nodes which are marked in E, eEare v, s respectively.

Now since χ_E| p-block, and hence by proposition 3.10 (3), they are in the same block. Moreover, if we denote a_i as the rightest nodes in the Ra,i-th row of eE, and b_i as the rightest nodes in the R_b,i-th row of E for all i, then by using the above observation, we can see that if we raise b_iinto the R_a,i-th row of E for all i, then we can obtain eE from E. So

the set of the colors of b₁, · · · , b_m

is equal to

the set of the colors of a₁, · · · , a_m.

This implies that there exists one of b₁, · · · , b_mhas the same color as a₁, which contradicts to the assumption that v is a shadow node. Thus such eE does not exist, and hence we

Go back to this theorem. Recall that we have shown

Res^S_Sⁿ

n−1(χ_D)|

S^(p)_n−1 = r(y₁)φ_D^r_−y1+ · · ·

and we want to show

χ_D|

S^(p)n = φD^r+ · · · . Let E be a p-regular diagram with χ_D|

S^(p)n ≥ φE and let w be the first shadow node of E. Then by (∗), we have

Res^S_Sⁿ

n−1(χD)|

S^(p)_n−1 = Res^S_Sⁿ

n−1(χD|

S_n^(p)) ⊇ Res^S_Sⁿ

n−1(φE) = r(w)φE−w+ · · · .

So we obtain E − wD D^r− y₁. On the other hand, since

Res^S_Sⁿ

n−1(χD|

S^(p)n ) ⊇ φD^r−y1,

there is some such E, say E₁, such that

Res^S_Sⁿ

n−1(φE1) ⊇ φD^r−y₁.

Moreover, by (∗), since (here w₁denotes the first shadow node of E₁)

Res^S_Sⁿ

n−1(φE₁) = r(w₁)φE₁−w₁+ · · · ,

we obtain D^r− y₁D E1− w₁, and hence

D^r− y₁= E₁− w₁.

Now observe that since χ_D|

S^(p)n ≥ φE1 and χ_E1|

S^(p)n ≥ φE1, by proposition 3.10 (3), D and E₁are in the same block, and since D and D^r are in the same block by proposition 3.12 (2), D^r and E₁are in the same block. Thus by proposition 3.14 (2), since D^r and E₁are

and recall again that we have shown

Res^S_Sⁿ

where w is the first sahdow node of E. Let B be the set of nodes of E not lower than w.

Because D^r− y₁is p-regular and E − w . D^r− y₁,

(∗∗) the set of nodes of D^r− y₁not lower than w are contained in B.

Indeed, if it is false, then there is a node x, which is not lower than w, of D^r− y₁ such that it is to the right of the ladder in which w lies. Note that there is no node of E to the right of the ladder of w since w is the first shadow node of E. Say x lies in the i-th row of D^r− y₁. Then the fact D^r− y₁is p-regular implies that for all j ≤ i, the length of the j-th row of D^r− y₁is greater than the length of the j-th row of E (because by proposition 3.10 (2), if the k-th rung of a ladder l belongs to D^r− y₁, then so does the k − 1-th rung of l), and this contradicts to E − w . D^r− y₁.

Observe that there are only three possibility that y1may lie in D^r:

y₁belongs to B.

(i) (ii) y₁is lower than w. (iii) not (i) and (ii).

In the first case, by (∗∗) and E − w . D^r− y₁, we can see that E . D^r. In the second case, by E − w . D^r− y₁, we also can see that E . D^r. Now we show that the third case is impossible.

Assume the third case holds, then it means y₁ is not lower than w, and y₁ does not belong to B. Since D^r is p-regular, (∗∗), and no node in E is to the right of the ladder in which w lies, the node y₁ is either in the rightest node in the first row of D^r, or below the 1-st rung of the ladder in which w lies. In both cases, w and y₁have distinct colors.

Moreover, since y1belongs to D^r− y_jfor all j > 1, the diagram E − w 6D D^r− y_j. Recall

that we have shown

∑

i

χ_D−zi|

S^(p)_n−1= Res^S_Sⁿ

n−1(χ_D)|

S^(p)_n−1 ⊇ φE−w

χ_D−zi|

S^(p)_n−1 = φD^r−y₁+ · · · , if i = 1, · · · , r(y₁), χ_D−zi|

S^(p)_n−1 = φ_D^r_−yj+ · · · , if i > r(y₁) ( j 6= 1).

So the fact E − w 6D D^r− y_jimplies that there is a i ∈ {1, · · · , r(y₁)} such that

χ_D−zi|

S^(p)_n−1 ⊇ φE−w.

Moreover, since E − w is p-regular, we also have

χE−w|

S^(p)_n−1⊇ φE−w.

So D − z_iand E − w are in the same block by proposition 3.10 (3), and hence

(D − z_i)^r and E − w

are in the same block. Note that (D − z_i)^r= D^r− y₁. Now the fact y₁and w have distinct colors means D^r and E are not in the same block, and hence D and E are not in the same block. However, recall that

χ_D|

Sn^(p)≥ φE, and since E is p-regular, we also have

χ_E| (p)≥ φE.

So D and E are in the same block by proposition 3.12. This gets a contradiction.

Now we completes the proof of the theorem.

Corollary 3.17.1.

The group Snhas the L⁰⁰-property.

Proof.

It is obvious by theorem 3.17.

4 About A

n

Recall that (K₀, R, K_p) is a splitting p-modular system for G. But (K₀, R, K_p) may not be a splitting p-modular system for H. So in this section, we assume that (K0, R, K_p) is a splitting p-modular system for H and for G. For example, if m = lcm(ord(g) | g ∈ G) and K₀contains the m-th root of unity, then it is a splitting p-modular system for any subgroup H of G (cf., for example, [S1], theorem 24, page 94). In the case H = Anand G = Sn, if (K₀, R, K_p) is a splitting p-modular system for A_n, then it is also for S_n.

To investigate a minimal p-splitting field for G, let char(F) = p, ¯F be the algebraic closure of F, M be a ¯FG-module, S be a ¯F-basis of M, θ_g^M : M → M given by m 7→ g · m, and ρ_g^M be the matrix representation with respect to S. Then by [I], theorem 9.14, page 150, if F contains Tr(ρ_g^M) for all non-isomorphic simple ¯FG-module M and all g ∈ G, then F is the minimal p-splitting field for G. But it may not be ture when char(F) = 0.

To be an example, we list minimal s-splitting fields of A₅, where s = 0, 2, 3, and 5.

The field Q(√

5) is a minimal 0-splitting field of A₅. The field F4, F9, and F5are minimal 2, 3, and 5-splitting fields of A₅ respectively, where Fqdenotes the finite field of order q.

For more information about the p-splitting field of A_n, one may see [W], theorem 4.1.3.

It tells us a p-splitting field of A_nshould contain a special element.

Let f be an irreducible character of An. To show that f has the the (L⁰⁰, 2)-property, or (L⁰, p)-property for p > 2, observe that there exists an irreducible character χ of Sn

such that Res^S_Aⁿ discuss about Res^S_Aⁿ

n(χ) and Res^S_Aⁿ

n(φ ) for any irreducible character χ and any irreducible p-modular character φ of S_n, using Clifford’s Theorem. Also theorem 3.17 gives us im-portant informations about χ|

S^(p)n when showing that f has the (L⁰, p)-property for p > 2.

Note that by using Frobenius Formula (cf. for example, [F], 4.10, page 49), we can

obtain the character table of S_n for all n. Moreover, if χ is an irreducible character of S_n, then lemma 4.18 tell us when Res^S_Aⁿ

n(χ) splits, or be an irreducible character of A_n. So by investigating the character value of Res^S_Aⁿ

n(χ) when it splits, we can obtain the character table of A_nby the character table of S_n(cf. for example, [F], proposition 5.3, page 66).

For more informations about irreducible characters of A_n, one may see [F1] and [F2].

4.1 Tools

Proposition 4.1.

Let χ₁, · · · , χm be all distinct irreducible characters of G and let M_ibe a K₀G-module which affords χ_i. Then

dim_K0Hom_K0G(M_i, M_j) = δi j.

Proposition 4.2.

If V is a projective K_pG-module, then there is a unique (up to isomorphism) RG-lattice Msuch that its reduction mod m is V .

Proof.

Cf., for example, [S1], proposition 42, page 119. Note that the completion of K₀ is necessary.

Definition 4.3.

A FG-module homomorphism f : A → B is called essential if f (A) = B and f (A⁰) 6= B for all proper FG-submodule A⁰of A.

Let P be a projective FG-module. We say P is a projective envelope of an FG-module Mif there is an essential FG-homomorphism from P to M.

Proposition 4.4.

(1) Every FG-module M has a projective envelope which is unique up to isomorphism.

(2) Let M₁, · · · , M_k be FG-modules and let P_i be the projective envelope of M_i. Then

⊕k i=1

P_iis the projective envelope of ⊕^k

i=1

M_i.

(3) Let E1, E2 be simple KpG-modules, and let Pi be the projective envelope of Ei for i= 1, 2. Then P₁∼=_KpGP₂if and only if E₁∼=_KpGE₂.

Proof.

Cf., for example, [S1] proposition 14.1, page 117.

Proposition 4.5.

Let E₁, · · · , E_k be all distinct simple K_pG-modules, and let P_i be the projective enve-lope of E_i. Then each P_i is indecomposable among projective K_pG-modules. Moreover, if V is a projective K_pG-module, then

V = e₁P₁⊕ · · · ⊕ e_kP_k

for some ei∈ N ∪ {0}.

Proof.

Cf., for example, [S1], corollary 1, page 140.

Proposition 4.6.

Let E₁, · · · , E_k be all distinct simple K_pG-modules, and let P_i be the projective enve-lope of E_i. Then

dim_KpHom_KpG(P_i, E_j) = δi j.

Proof.

Cf., for example, [S1], page 121.

Proposition 4.7. (Frobenius Reciprocity)

Let M be an FH-module, and N be an FG-module. Then

Hom_FH(M, Res^G_H(N)) ∼=F Hom_FG(Ind^G_H(M), N).

Proof.

Cf., for example, [CR], theorem 10.8, page 232.

Definition 4.8.

Let M be a semisimple FG-module and write M = e₁M₁⊕ · · · ⊕ e_kM_k for some pair-wisely non-isomorphic FG-module M₁, · · · , M_k, and some e₁, · · · , e_k∈ N. Then eiM_i is called an FG-homomgeneous component of M.

Definition 4.9.

(1) Assume H E G. Let M be an FH-module. For any g ∈ G, define M^(g) as an FH-module which M^(g) agrees with M as abelian group, and for any h ∈ H, m ∈ M^(g),

h· m := (g⁻¹hg)m.

We say two FH-modules M, N are conjugate (under G) if N ∼=FH M^(g) for some g∈ G.

(2) If F = K₀and χ_M is the characters of M, then we denote

χ_M^(g)(h) = χM(g⁻¹hg).

It is the character of the K₀H-module M^(g). We say χ_Mand a character χ are conjugate (under G) if χ = χ_M^(g)for some g ∈ G.

(3) If F = K_pand φ_M is the p-modular characters of M, then we denote

φ_M^(g)(h) = φM(g⁻¹hg).

It is the p-modular character of the K_pH-module M^(g). We say φ_M and a p-modular character φ are conjugate (under G) if φ = φ_M^(g) for some g ∈ G.

Theorem 4.10. (Clifford’s Theorem)

Assume HE G. Let V be a simple FG-module, and M be a simple FH-submodule of Res^G_H(V ). Then Res^G_H(V ) is a semisimple FH-module. Moreover,

(1) Let

Mebe the FH-homogeneous components of Res^G_H(V ) containing M, He= {x ∈ G | x eM= eM} be the stabilizer ofMe(note that eHE G), g₁, · · · , g_k be a set of representative of G/ eH.

(g)

each appears with the same multiplicity t ∈ N in Res^G_H(V ), i.e.

Res^G_H(V ) ∼=FH

⊕k i=1

tM^(gi).

(2) The module eMis a simple F eH-module and we have

V ∼=FGInd^G

He( eM).

(3) If F = K₀or K_p, then t²≤ |He: H|.

Proof.

For (1) and (2), cf., for example, [CR], theorem 11.1, page 259. Now we are going to show (3).

Assume F = K₀. Since Res^G_H(V ) ∼=K₀H

⊕k i=1

tM^(gi), we have

dim_K0Hom_K0H(Res^G_H(V ), Res^G_H(V )) =

k

∑

i=1

t²dim_K0Hom_K0H(M^(gi), M^(gi)) = |G : eH|t².

On the other hand, by proposition 4.7, we have

dim_K0Hom_K0H(Res^G_H(V ), Res^G_H(V )) = dim_K0Hom_K0G(Ind^G_H(Res^G_H(V )),V ).

Note that since V occurs in composition factors of Ind^G_H(Res^G_H(V )) at most |G : H| times, we have

dim_K0Hom_K0G(Ind^G_H(Res^G_H(V )),V ) ≤ |G : H|.

Therefore, we obtain |G : eH|t²≤ |G : H|, and hence t²≤ |He: H|.

Assume F = K_p. Denote

P_i: the KpH projective envelope of the KpH-module M^(gi), P_ResG

H(V ): the K_pH projective envelope of the K_pH-module Res^G_H(V ), P_V : the K_pGprojective envelope of the K_pG-module V .

First observe that by proposition 4.4 (2), we have

P_ResG

H(V )= ⊕^k

i=1

tP_i.

On the other hand, let f be an essential K_pG-homomorphism from P_V to V . Then f is a surjective K_pH-homomorphism from Res^G_H(P_V) to Res^G_H(V ). Let h be an essential K_pH-homomorphism from P_ResG

H(V ) to Res^G_H(V ). Then since Res^G_H(P_V) is a projective K_pH-module, we have the following commutative diagram

P_ResG (Note that ker(g) is also a projective K_pH-module). So

Res^G_H(P_V) ∼=_KpH

 k

⊕tP_i



⊕ ker(g),

and hence by proposition 4.6, we obtain

dimKpHomKpH(Res^G_H(P_V), Res^G_H(V )) ≥ kt²= |G : eH|t².

On the other hand, by proposition 4.7, we have

dim_KpHom_KpG(Ind^G_H(Res^G_H(P_V)),V ) = dim_KpHom_KpH(Res^G_H(P_V), Res^G_H(V )).

Since Ind^G_H(Res^G_H(P_V)) = K_pG⊗_KpHRes^G_H(P_V), it is a projective K_pG-module, and its dimension is

|G : H| dim_Kp(P_V).

So if we write Ind^G_H(Res^G_H(P_V)) as the direct sum of some indecomposable projective K_pG-modules, then P_V occured in it at most |G : H| times (recall that P_V is an indecom-posable projective K_pG-modules since V is a simple K_pG-module). So

|G : H| ≥ dim_KpHom_KpG(Ind^G_H(Res^G_H(P_V)),V ).

Now combine these inequalities, we obtain |G : H| ≥ |G : eH|t², and hence | eH: H| ≥ t².

By theorem 4.10 (3), we immediately have the following two lemmas.

Lemma 4.11.

Assume HE G with |G : H| = 2. Let χ be an irreducible character of G. Then

either Res^G_H(χ) is irreducible or Res^G_H(χ) = χ1+ χ2

for some distinct irreducible characters χ1, χ2of H, where χ1, χ2are conjugate.

Lemma 4.12.

Assume HE G with |G : H| = 2. Let φ be an irreducible p-modular character of G.

Then

either Res^G_H(φ ) is irreducible or Res^G_H(φ ) = φ1+ φ2

for some distinct irreducible p-modular characters φ1, φ2of H, where φ1, φ2are conjugate.

Lemma 4.13.

Let f be an irreducible character of H. Then there is an irreducible character χ of G such that Res^G_H(χ) ≥ f .

Proof.

Let M be a simple K₀H-module which affords f . Since K₀H is a K₀H-submodule of Res^G_H(K₀G), we have

dim_K0Hom_K0H(Res^G_H(K₀G), M) 6= 0.

On the other hand, let V₁, · · · ,V_mbe all non-isomorphic simple K₀G-modules of G. Ob-serve that

K₀G ∼=_K0GdimK₀(V₁)V₁⊕ · · · ⊕ dim_K0(V_m)V_m.

So

Res^G_H(K₀G) ∼=_K0Gdim_K0(V₁) Res^G_H(V₁) ⊕ · · · ⊕ dim_K0(V_m) Res^G_H(V_m).

Hence

0 6= dimK₀HomK₀H(Res^G_H(K₀G), M) =

m i=1

∑

dimK₀(V_i) dim_K0HomK₀H(Res^G_H(V_i), M).

This means there exists a j ∈ {1, 2, · · · , m} such that dim_K0Hom_K0H(Res^G_H(V_j), M) 6= 0, i.e. Res^G_H(χ_j) ≥ f .

在文檔中 在對稱群上的特徵標 (頁 50-72)