Induction of P -schemes

Chapter VI: Constructing new P -schemes from old ones

6.2 Induction of P -schemes

For the second claim, assume to the contrary that it does not hold. Choosei ∈ [k]

such all strongly antisymmetric P|_Gi−1-schemes are inhomogeneous on G_i. Let C = {C_H : H ∈ P} be a strongly antisymmetricP-scheme that is homogeneous on G_k. By compatibility, we know C is homogeneous on G_i. Then C|_Gi−1 is also homogeneous onG_i. It is also strongly antisymmetric by Lemma 6.2, which contradicts the assumption.

The proof for antisymmetry is the same.

Now we are ready to prove Lemma 4.12. For convenience, we restate the lemma.

Lemma 6.6. Let k ∈ N⁺ and G_k ⊆ Gk−1 ⊆ · · · ⊆ G₁ ⊆ G₀ be a chain of finite groups. For i ∈ [k], let N_i be a subgroup of G_i that is normal in Gi−1, π_i : Gi−1 → Gi−1/N_i be the corresponding quotient map, andP_i be a subgroup system overGi−1/N_i that containsG_i/N_i. Define

P ={gπ_i⁻¹(H)g⁻¹ : 1≤i≤k, H ∈ P_i, g ∈G₀},

which is a subgroup system overG₀ and containsπ_i⁻¹(G_i/N_i) =G_i for alli∈[k].

Then we have

1. If for alli∈[k], all strongly antisymmetricP_i-schemes are discrete onG_i/N_i, then all strongly antisymmetricP-schemes are discrete onG_k.

2. If for somei∈[k], all strongly antisymmetricP_i-schemes are inhomogeneous onG_i/N_i, then all strongly antisymmetricP-schemes are inhomogeneous on G_k.

The same holds if strong antisymmetry is replaced by antisymmetry.

Proof. Fixi ∈[k]. By Lemma 6.4 and the definition ofP, if all strongly antisym- metricP_i-schemes are discrete (resp. inhomogeneous) onG_i/N_i, then all strongly antisymmetric P|_Gi−1-schemes are discrete (resp. inhomogeneous) on G_i. The same holds if strong antisymmetry is replaced by antisymmetry. The lemma now follows from Lemma 6.5.

which is a subgroup system overG. In this section, we show that everyP -scheme induces aP-scheme in a way that preserves antisymmetry and strong antisymmetry.

To achieve it, we need the following lemma.

Lemma 6.7. Given g₁, . . . , g_k ∈ Gsuch that {g⁻¹₁ , . . . , g_k⁻¹} is a complete set of representatives ofH\G/G⁰, there exists a bijection

φ:

k

a

i=1

(G⁰ ∩g_iHg_i⁻¹)\G⁰ →H\G

defined as follows: Forg ∈G, define the map

φ_H,g : (G⁰∩gHg⁻¹)\G⁰ →H\G

sending(G⁰∩gHg⁻¹)htoHg⁻¹hforh∈G⁰. The mapsφ_H,g are well defined. For i∈[k], the restriction ofφto(G⁰∩giHg_i⁻¹)\G⁰isφH,gi.

Proof. Consider the action ofG⁰onH\Gby inverse right translation. Fori∈[k], let O_i ={Hg_i⁻¹g⁻¹ :g ∈ G⁰}be theG⁰-orbits ofHg_i⁻¹ ∈ H\G. Then{O₁, . . . , O_k} is the partition ofH\Ginto theG⁰-orbits, i.e.,H\G= `k

i=1Oi. Fix i∈ [k]. The stabilizer ofHg_i⁻¹ is G⁰ ∩g_iHg_i⁻¹. So by Lemma 2.1, we have an equivalence of actions ofG⁰

λ_Hg⁻¹

i :O_i →(G⁰∩g_iHg_i⁻¹)\G⁰

sending^h(Hg_i⁻¹) =Hg⁻¹_i h⁻¹ to(G⁰∩g_iHg⁻¹_i )h⁻¹forh∈G⁰. The inverse of this map is exactlyφ_H,gi. As we are allowed to chooseg_ito be anyg ∈G, all the maps φ_H,g are well defined.

For eachH ∈ P, the subgroupsG⁰∩gHg⁻¹are inP⁰for allg ∈G. By Lemma 6.7, we can combine partitions of G⁰ ∩g_iHg_i⁻¹\G⁰, i = 1, . . . , k, into a partition of H\G. This leads to the following definition.

Definition 6.4(induction). LetG, G⁰,P andP⁰ be as above. LetC⁰ ={C_H⁰ :H ∈ P⁰}be aP⁰-scheme. ForH ∈ P, chooseg₁, . . . , g_k ∈Gsuch that{g₁⁻¹, . . . , g_k⁻¹} is a complete set of representatives ofH\G/G⁰. Define the partitionC_H ofH\Gby

C_H =n

φ_H,gi(B) :i∈[k], B ∈C_G⁰0∩g_iHg⁻¹_i

o ,

where the maps φ_H,gi are as in Lemma 6.7, i.e., each φ_H,gi sends(G⁰∩g_iHg_i⁻¹)h to Hg_i⁻¹h for h ∈ G⁰. Define the P-collection C = {C_H : H ∈ P}, called the inductionofC⁰ toP.

TheP-collectionC constructed as above is indeed aP-scheme:

Theorem 6.1. TheP-collectionCin Definition 6.4 is a well definedP-scheme, which does not depend on the choices of the elementsg_i. Moreover, ifC⁰ is antisymmetric (resp. strongly antisymmetric), so isC.

Proof. FixH ∈ P. It follows from Lemma 6.7 that CH is indeed a partition of H\G. We need to show that CH is independent of the choices of the elements g₁, . . . , g_k. So considerg₁⁰, . . . , g_k⁰ ∈ Gsuch that{g⁰⁻¹₁ , . . . , g_k⁰⁻¹}is a complete set of representatives ofH\G/G⁰as well. We want to show

C_H =n φ_H,g⁰

i(B) :i∈[k], B ∈C_G⁰ 0∩g_i⁰Hg_i⁰⁻¹

o .

As the right hand side is also a partition ofH\G, it suffices to show thatφ_H,g⁰

i(B)∈ C_H for i ∈ [k] and B ∈ C_G⁰_0∩g0

iHg⁰⁻¹_i . Fix i and B. Choose j ∈ [k] such that Hg_j⁻¹G⁰ = Hg_i⁰⁻¹G⁰. And chooseg ∈ G⁰ such thatHg_j⁻¹ = Hg_i⁰⁻¹g⁻¹. We have the conjugation

c_G0∩g_i⁰Hg_i⁰⁻¹,g : (G⁰ ∩g⁰_iHg_i⁰⁻¹)\G⁰ →(G⁰∩g_jHg_j⁻¹)\G⁰

sending(G⁰∩g_i⁰Hg_i⁰⁻¹)hto(G⁰ ∩g_jHg_j⁻¹)ghforh∈ G⁰. By invariance ofC⁰, the setc_G0∩g_i⁰Hg⁰⁻¹_i ,g(B)is a block ofC⁰

G⁰∩g_jHg⁻¹_j . Soφ_H,gj ◦c_G0∩g⁰_iHg⁰⁻¹_i ,g(B)is a block ofC_H. We claim

φ_H,gj◦c_G0∩g⁰_iHg_i⁰⁻¹,g=φ_H,g⁰

i, which holds since

φ_H,gj ◦c_G0∩g⁰_iHg⁰⁻¹_i ,g((G⁰∩g_i⁰Hg⁰⁻¹_i )h) =φ_H,gj((G⁰∩g_jHg_j⁻¹)gh) =Hg⁻¹_j gh

=Hg_i⁰⁻¹h =φ_H,g⁰

i((G⁰∩g_i⁰Hg⁰⁻¹_i )h) forh ∈ G⁰. It follows thatφ_H,g⁰

i(B) ∈C_H, as desired. SoC_H does not depend on the choices ofg₁, . . . , g_k.

Next we prove that C is a P-scheme. To prove compatibility, considerH, H⁰ ∈ P satisfyingH ⊆H⁰. Forg ∈G, the following diagram commutes:

(G⁰∩gHg⁻¹)\G⁰ (G⁰∩gH⁰g⁻¹)\G⁰

H\G H⁰\G.

π_G0∩gHg−1,G0∩gH0g−1

φH,g φ_H0,g

π_H,H0

ForB ∈C_H, we want to show thatπ_H,H⁰(B)is contained in a block ofC_H⁰. Choose g ∈ G and B˜ ∈ C_G⁰0∩gHg⁻¹ such that B = φ_H,g( ˜B). Such g and B˜ exist by the definition ofC_H and the fact that this definition does not depend on the choices of g₁, . . . , g_k(in particular we may chooseg₁ =g). Then

π_H,H⁰(B) =π_H,H⁰ ◦φ_H,g( ˜B) = φ_H⁰_,g(y)◦π_G⁰_∩gHg⁻¹_,G⁰_∩gH⁰_g⁻¹( ˜B).

Hereπ_G⁰_∩gHg⁻¹_,G⁰_∩gH⁰_g⁻¹( ˜B)is contained in a block ofC_G⁰ 0∩gH⁰g⁻¹ by compatibility of C⁰, and hence π_H,H⁰(B) is contained in a block of C_H⁰. It follows that C is compatible.

For regularity, considerH, H⁰as above andB ∈C_H. ChooseB⁰ ∈C_H⁰ containing π_H,H⁰(B). We claim thatπ_H,H⁰|_B :B →B⁰has constant degree, i.e., the number of preimages|(π_H,H⁰|_B)⁻¹(y)|is independent of the choices ofy∈B⁰. Chooseg ∈G and B˜ ∈ C_G⁰0∩gHg⁻¹ such that B = φ_H,g( ˜B). Let B˜⁰ = π_G⁰∩gHg⁻¹,G⁰∩gH⁰g⁻¹( ˜B). ThenB⁰ =φ_H⁰_,g( ˜B⁰). By regularity ofC⁰, the mapπ_G⁰∩gHg⁻¹,G⁰∩gH⁰g⁻¹|B˜ : ˜B →B˜⁰ has constant degree. The claim follows by noting that φ_H,g|B˜ : ˜B → B and φH⁰,g|B˜⁰ : ˜B⁰ →B⁰ are bijective. SoC is regular.

For invariance, considerH, H⁰ ∈ Pandh ∈GsatisfyingH⁰ =hHh⁻¹. Forg ∈G, we haveG⁰∩gHg⁻¹ =G⁰∩gh⁻¹H⁰(gh⁻¹)⁻¹, and the following diagram commutes

(G⁰∩gHg⁻¹)\G⁰ (G⁰∩gHg⁻¹)\G⁰

H\G H⁰\G,

id

φH,g φ_H0,gh−1

cH,h

whereiddenotes the identity map. It follows thatcH,h maps blocks ofCH to blocks ofC_H⁰. SoC is invariant.

Now assumeC is not strongly antisymmetric and we prove thatC⁰ is not either. By definition, there exists a nontrivial permutationτ =σ_k◦ · · · ◦σ₁of a blockB ∈C_H for someH ∈ P such that eachσ_i :Bi−1 → B_i is a map of the formπ_Hi−1,Hi|_Bi−1, (πHi,Hi−1|Bi)⁻¹, orcHi−1,h|Bi−1, andBi ∈CHi,Hi ∈ P,B =B0 =Bk,H =H0 = H_k(see Definition 2.7).

By the two diagrams above, we can choose g_i ∈ G and B˜_i ∈ C⁰

G⁰∩g_iHg_i⁻¹ for 0≤i ≤k, and chooseσ˜_i : ˜B_i−1 →B˜_iof the formπ_G0∩gi−1Hi−1g⁻¹_i−1,G⁰∩giHig_i⁻¹|B˜i−1, (π_G0∩giHig_i⁻¹,G⁰∩gi−1Hi−1g⁻¹_i−1|B˜i)⁻¹, or the identity map onB˜_i for i ∈ [k], such that φ_Hi,gi( ˜B_i) = B_i and σ_i ◦φ_{Hi−1,gi−1}|B˜i−1 = φ_Hi,gi|B˜i ◦ σ˜_i for i ∈ [k].1 Define

1For the case thatσ_i= (π_Hi,Hi−1|Bi)⁻¹, we chooseσ˜_i= (π_G0∩giH_ig⁻¹_i ,G⁰∩gi−1Hi−1g⁻¹_i−1|B˜_i)⁻¹, which is well defined sinceφHi−1,gi−1andφH_i,g_i are bijective.

˜

τ := ˜σ_k ◦ · · · ◦σ˜₁ which is a map from B˜₀ to B˜_k. Then the following diagram commutes.

B˜₀ B˜_k

B B

˜ τ

φH,g0|B˜0 φ_H,gk|_˜

Bk

τ

We haveHg₀⁻¹G⁰ = Hg_k⁻¹G⁰, since otherwise the image ofφ_H,g0 and that ofφ_H,gk would be disjoint (see Lemma 6.7). SoHg⁻¹₀ = Hg⁻¹_k g⁻¹ for someg ∈ G⁰. The first part of the proof shows that φ_H,g0 ◦c_G0∩g_kHg⁻¹_k ,g = φ_H,gk. By composing τ˜ withc_G0∩g_kHg⁻¹_k ,g, we may assumeg_k =g₀ andB˜_k = ˜B₀. Then asτ is a nontrivial permutation of B and φ_H,g0|B˜0 : ˜B₀ → B is bijective, we know τ˜is a nontrivial permutation ofB˜0. SoC is not strongly antisymmetric.

The proof for antisymmetry is the same except that we only consider maps τ that are conjugations.

Corollary 6.1. LetG, G⁰,P,P⁰be as above and letH be a subgroup inP.

1. Suppose all antisymmetricP-schemes are discrete onH. Then all antisym- metricP⁰-schemes are discrete onG⁰∩gHg⁻¹ for allg ∈G.

2. Suppose all antisymmetricP-schemes are inhomogeneous onH, andG⁰acts transitively on H\G by inverse right translation. Then all antisymmetric P⁰-schemes are inhomogeneous onG⁰∩gHg⁻¹ for allg ∈G.

The same claims hold if antisymmetry is replaced with strong antisymmetry.

Proof. We prove the claims by contrapositive. For the first claim, suppose C⁰ = {C_H⁰ 0 :H⁰ ∈ P⁰}is an antisymmetricP⁰-scheme that is not discrete onG⁰∩gHg⁻¹ for someg ∈ G. Choose B ∈C_G⁰ 0∩gHg⁻¹ that is not a singleton. By Theorem 6.4, the inducedP-schemeC ={C_H⁰ :H⁰ ∈ P}is antisymmetric. Moreover, we know C is not discrete onH since the blockφ_H,g(B)∈C_H is not a singleton.

For the second claim, suppose C⁰ = {C_H⁰ 0 : H⁰ ∈ P⁰} is an antisymmetric P⁰- scheme that is homogeneous onG⁰∩gHg⁻¹for someg ∈G. By Theorem 6.4, the inducedP-schemeC = {C_H⁰ : H⁰ ∈ P}is antisymmetric. AsG⁰ acts transitively onH\G, the double coset spaceH\G/G⁰has only one double cosetHg⁻¹G⁰, which implies thatφH,g :G⁰∩gHg⁻¹\G→H\Gis surjective. AsC⁰is homogeneous on G⁰∩gHg⁻¹, we knowCis homogeneous onH.

The proof for strong antisymmetry is the same.

Now letS be a finite G-set and let G be a subgroup of G. Fix m ∈ N⁺ and let P ={G_T : 1≤ |T| ≤m}be the system of stabilizers of depthmwith respect to the action ofGonS. Note that forT ⊆S, we haveG⁰ ∩gG_Tg⁻¹ =G⁰∩G^g_T =G⁰gT. So P⁰ = {G⁰∩gHg⁻¹ : H ∈ P}is exactly the system of stabilizers of depth m with respect to the action ofG⁰ onSrestricted from that ofG. Therefore we have:

Corollary 6.2. LetGbe a finite group acting on a finite setS,G⁰a subgroup ofG, andm ∈N⁺. LetP (resp. P⁰) be the system of stabilizers of depthmoverG(resp.

G⁰) with respect to the action ofG(resp. G⁰) onS.

1. Suppose all antisymmetricP-schemes are discrete onG_xfor allx∈S. Then all antisymmetricP⁰-schemes are discrete onG⁰_x for allx∈S.

2. Suppose all antisymmetricP-schemes are inhomogeneous on G_x0 for some x₀ ∈S, andG⁰acts transitively onS. Then all antisymmetricP⁰-schemes are inhomogeneous onG⁰_x for allx∈S.

The same claims hold if antisymmetry is replaced with strong antisymmetry.

In particular, we seed(G)andd⁰(G)(cf. Definition 2.8) are monotone with respect to inclusion of permutation groups:

Corollary 6.3. LetGbe a finite permutation group on a finite setS, and letG⁰ be a subgroup ofGonS. Thend(G⁰)≤d(G)andd⁰(G⁰)≤d⁰(G).

We also mention the following variant of Corollary 6.2, which allowsG⁰ ⊆Gto act on a proper subset ofS.

Corollary 6.4. LetGbe a finite group acting on a finite setS,G⁰a subgroup ofG, andm∈ N⁺. LetT a subset ofS such that the action ofG⁰ onSfixesT setwisely andS−T pointwisely. LetP (resp. P⁰) be the system of stabilizers of depthmover G(resp. G⁰) with respect to the action ofG(resp. G⁰) onS (resp. T). Suppose all antisymmetricP-schemes are discrete onG_xfor allx∈S. Then all antisymmetric P⁰-schemes are discrete onG⁰_xfor allx∈T. The same claims hold if antisymmetry is replaced with strong antisymmetry.

Proof. If S = T, the claim holds by Corollary 6.2. So assume S 6= T. LetP⁰⁰ be the system of stabilizers of depth m over G⁰ with respect to the action of G⁰ on S. Then P⁰⁰ = P⁰ ∪ {G}. A P⁰-scheme C always extends to a P⁰⁰-scheme

C :=C ∪ {C_G}, whereC_Gis the only partition of the singletonG\G, and such an extension clearly preserves antisymmetry and strong antisymmetry. The claim then follows from Corollary 6.2.