The main results of this section are Theorems6.1 and6.2, where the later implies Theorem 1.2. In Remark 6.9 we observe that we do not have a zero-one law in Theorems6.2or1.2.
Theorem 6.1 Suppose that A ∈ S has no fixed point and let H be a subgroup of Aut(A) without any fixed point. Then S(A,H) has a zero-one law.
Before proving Theorem6.1we derive:
Theorem 6.2 (i) Let A1, . . . ,Am∈S be such that none of them has any fixed point. Suppose that for every i=1, . . . ,m andj=1, . . . ,li, Hi,j is a subgroup of Aut(Ai) without any fixed point. Then
m [ i=1 li [ j=1
S(Ai,Hi,j)has a limit law. Moreover, there is a finite setQof rational numbers that depends only on theAi andHi,j such that for every
sentenceϕ, the proportion ofn-element structures in the double union in whichϕis true converges to a number inQasn→ ∞.
(ii) The following sets have limit laws: for every finite group G, {M ∈ S : G ∼= Aut(M)} and {M ∈S:G≤Aut(M)}, and for every integerm≥2,S(spt∗ =m),
S(spt≥m) andS(spt∗≥m). Moreover, in each case there is a finite setQ⊆Qsuch
that for every sentenceϕ, the proportion ofn-element structures in whichϕ is true converges to a number inQasn→ ∞.
Proof Part (i) is immediate from Theorem6.1and Proposition5.9. For part (ii) letX
be any one of the sets of structures considered. By Lemmas4.2and4.3(and in one case the proof of Proposition5.15), there are structures A1, . . . ,Al ∈Swithout any fixed point and for everyi=1, . . . ,landj=1, . . . ,li, a subgroupHi,j of Aut(Ai) without any fixed point, such that ifXn =X∩Sn then
|Xn| ∼ l [ i=1 li [ j=1 Sn(Ai,Hi,j) . Now part (ii) follows from part (i).
6.1 Proof of Theorem6.1
Suppose thatA ∈Shas no fixed point and letHbe a subgroup of Aut(A) without any fixed point. We will define a theoryTA,H and show that it is consistent and complete and that for every finite subset∆⊆TA,H, the proportion of M ∈Sn(A,H) such that M |= ∆ approaches 1 as n → ∞. Then the compactness theorem implies that if
TA,H|=ϕthen the proportion of M ∈Sn(A,H) in which ϕis true approaches 1 as
n→ ∞; otherwise that proportion approaches 0. The argument follows a well known path, using so-called extension axioms. What makes things more complicated here, compared with Fagin’s original proof Fagin ([8] or Ebbinghaus and Flum [6]) that for everyk∈Nthe proportion ofM ∈Sn satisfying thek-extension axiom approaches 1 asn→ ∞, is that all members of Sn(A,H) have nonempty support (of cardinality |A|). We will defineTA,H to consist of a sentenceψimplying that the support is isomorphic toAand, for eachk∈N, ak-extension axiomϕk which takes the support in careful consideration.
To make the main ideas of the argument more transparent, to avoid heavy formulations and notation and to expose more clearly how our argument differs from the “standard argument” (in [6,8] for example), we will prove Theorem6.1in the special case when
the vocabulary consists of only one binary relation symbolR. It is straightforward to generalise the argument to any finite relational vocabulary with at least one relation symbol of arity at least 2, but it comes at the expense of longer definitions and heavier notation and formulations in order to keep track of all data and how it can be combined. Moreover, the arguments can be modified to the case when we assume that some (possibly all) relation symbols are always interpreted as irreflexive relations, or when we assume that some (possibly all) relation symbols are always interpreted as irreflexive
andsymmetric relations.
For any structureMand formulaϕ(x) we let
ϕ(M) = {a∈M:M |=ϕ(a)}.
Letp=|A|,A={a1, . . . ,ap} and let x1, . . . ,xp be distinct variables. For ai,aj∈A, letai ≈aj mean that ai and aj belong to the same orbit ofH. Let χA(x1, . . . ,xp) be a formula which describes the isomorphism type ofA. More precisely, χA(x1, . . . ,xp) is the conjunction of all formulas of the following form: xi6=xj fori=6 j;R(xi,xj) if A |=R(ai,aj); and¬R(xi,xj) ifA |=¬R(ai,aj).
We will define formulasθ(x),ξ(x,y) such that the proportion ofM ∈Sn(A,H) such that the following hold approaches 1 asn→ ∞:
(I) For everya∈M, M |=θ(a) if and only ifa∈Spt∗(M). (II) Msatisfies the following sentence, denoted by ψ:
∃x1, . . . ,xm χA(x1, . . . ,xm) ∧ ∀y θ(y) ←→ m _ i=1 y=xi ∧ ^ ai≈aj ξ(xi,xj) ∧ ^ ai6≈aj ¬ξ(xi,xj)∧ m ^ i,j=1 ∀y ¬θ(y) ∧ ξ(xi,xj) −→ ξ(y,xi)←→ξ(y,xj) ∧ ξ(xi,y)←→ξ(xj,y) . Then ψsays that there exists a copy of Awhose universe is defined byθ and whose orbit equivalence relation is defined by ξ. It is straightforward to define, for every
k∈N, a sentenceϕk such that for every, possibly infinite, structureM: (III) IfM |=ϕk then the following hold:
(a) |θ(M)|=mand the relation defined by ξ(x,y) restricted toθ(M) is an equivalence relation.
(b) For every choice of i∈ {0,1}, B⊆M\θ(M) with |B|=k, sets E,E0
ofξ-equivalence classes on θ(M) andY,Y0 ⊆B, there isc∈M\θ(M) such that
(i) M |=R(c,c) ⇐⇒ i=1, (ii) for everya∈θ(M),
M |=R(c,a) ⇐⇒ abelongs to some class in Eand M |=R(a,c) ⇐⇒ abelongs to some class in E0, and (iii) for everyb∈B,
M |=R(c,b) ⇐⇒ b∈Y and M |=R(b,c) ⇐⇒ b∈Y0.
We call ϕk the k -extension axiom. Assuming that we have θ(x), ξ(x,y), ψ and ϕk,
k∈N, as above, we let
TA,H = {ψ} ∪ {ϕk :k∈N}.
Note that every model ofTA,H is infinite. We postpone the proof thatTA,H has a model to the end of the argument. By Łos’ and Vaught’s categoricity theorem (see Rothmaler [19, Theorem 8.5.1], for instance),TA,H is complete if we can prove that it is countably categorical.
Lemma 6.3 IfM1 andM2 are countable models ofTA,H thenM1∼=M2.
Proof This is a standard back-and-forth argument, so we only sketch it. Suppose that M1 and M2 are countable models of TA,H. Since both M1 and M2 satisfy ψ it follows that θ(M1) and θ(M2) are finite and that there is an isomorphism
f0 :M1θ(M1)→ M2θ(M2) such that for alla,b∈θ(M1),M1|=ξ(a,b) if and only ifM2|=ξ(f0(a),f0(b)). Therefore it suffices to prove the following statement:
Claim Suppose thatB1⊆M1\θ(M1),B2 ⊆M2\θ(M2)and that
f :M1(θ(M1)∪B1)→ M2(θ(M2)∪B2)
is an isomorphism. Ifc1∈M1\(θ(M1)∪B1) (orc2∈M2\(θ(M2)∪B2)) then there isc2 ∈M2\(θ(M2)∪B2) (orc1 ∈M1\(θ(M1)∪B1)) such thatf can be extended to an isomorphism fromM1(θ(M1)∪B1∪ {c1}) toM2(θ(M2)∪B2∪ {c2}). Letk=|B1|=|B2|. The claim follows in a straightforward way sinceMandN are models of{ψ, ϕk}.
It remains to show that there areθ(x) andξ(x,y) such that, for every k, the proportion ofM ∈Sn(A,H) that satisfy (I) and the sentencesψand ϕk approaches 1 asn→ ∞. Recall that, with the notation from Section4,
Sn(A,H) = [ X [ AX SX(AX,H),
where the first union ranges over all subsets of [n] with cardinalitym=|A|, and for each such subsetX, the second union ranges over all structuresAX with universeXthat are isomorphic toA. As observed in that section, if X6=X0 thenSn(AX,H) is disjoint from Sn(AX0,H). Moreover, if AX and A0X are different structures with universe X thenSn(AX,H) is disjoint from Sn(A0X,H). Recall our assumption in this proof that there is only one relation symbolR and it has arity r =2. In Section4we also saw (recall (4–4)) that for eachSn(AX,H),
Sn(AX,H) = [
Π1
Sn(AX,Π1),
where the union ranges over all (AX,H)-partitions Π1 ofX; see Definition4.8. The number of (AX,H)-partitions of X is the same for every sufficiently largen, every
X⊆[n] with|X|=mand everyAX ∼=A. Therefore it suffices to prove that there are θ(x),ξ(x,y) and, for every k, 0< α <1 such that for every X ⊆[n] with|X|=m, every AX ∼= A with universe X and every (AX,H)-partition Π1, the proportion of M ∈Sn(AX,Π1) that satisfy (I) and the sentencesψandϕk is at least 1−αn.
For the rest of this section we fix X ⊆ [n] with |X| = |A| = m and AX ∼= A
with universeX. The results below refer to all large enough n with respect to other parameters that occur.
Definition 6.4 We say thatM ∈Sn(AX,Π1) has thek -extension propertyif (III) (b) holds when ‘θ(M)’ is replaced by ‘X’ and ‘ξ-equivalence classes’ with ‘parts of the partitionΠ1 (ofX)’.
Lemma 6.5 For everyk∈Nthere is0< αk <1, depending only onk andA, such that the proportion ofM ∈Sn(AX,Π1)which does not have the k-extension property is at mostαnk.
Proof Recall thatSn(AX,Π1)⊆Tn(AX,Π1), whereTn(AX,Π1) was defined in (4–3) in Section4. From Lemma4.9 we know that |Sn(AX,Π1)|
.
|Tn(AX,Π1)| → 1 as
n→ ∞, so it suffices to prove the statement of the lemma forTn(AX,Π1) in the place of Sn(AX,Π1). The reason for doing this is that Tn(AX,Π1) is easier to work with
because its members do not have the constraint that the support of the structure is exactly
X (but from the arguments in Section4 we know that for every M ∈ Tn(AX,Π1),
X⊆Spt∗(M)).
LetM ∈Tn(AX,Π1). A subsetB⊆[n]\X of cardinalitykcan be chosen in no more thannk ways. OnceB⊆[n]\X with|B|=k is fixed, the number of ways to choose
i∈ {0,1},E,E0 ⊆Π1 andY,Y0 ⊆B is bounded where the bound depends only onk andA. Therefore it suffices to show, for an arbitrary fixed B⊆[n]\X with|B|=k
and an arbitrary choice ofi∈ {0,1},E,E0 ⊆Π1 andY,Y0 ⊆B, that the proportion ofM ∈Tn(AX,Π1) such that there isno c∈Msuch that the conjunction of (i)–(iii) of (III) is satisfied is at mostαn
k for some constant 0< αk <1 that depends only on k andA.
For arbitraryc∈[n]\(X∪B) we estimate the probability that at least one of (i)–(iii) of (III) fails. We considerTn(AX,Π1) as a probability space by giving each member the same probability. From the definition ofTn(AX,Π1) we see that the probability that M ∈ Tn(AX,Π1) satisfies (i)–(iii) of (III) is 2−β for some β > 0 depending only on|X|and |B|, and independently of what the case is for other elements than c
in [n]\(X∪B). The probability that, for every c ∈ [n]\(X∪B), the conjunction of (i)–(iii) does not hold is therefore
1−2−βn−|X∪B|.
AsBcan be chosen in at mostnkways it follows that the probability that the conjunction of (i)–(iii) is not satisfied inM ∈Tn(AX,Π1) is at most αnk for some 0< αk <1 that depends only onk andA.
Remember thatm=|A|=|X|. Letθ(x) denote the following formula: ∃y1, . . . ,ym−1 m−1 ^ i=1 x6=yi ∧ ^ i6=j yi 6=yj ∧ ∀z z6=x ∧ m−1 ^ i=1 z6=yi −→ R(x,z) ←→ R(y1,z) .
Lemma 6.6 Suppose thatM ∈ Sn(AX,Π1)has the2-extension property. Then for alla∈M,a∈X=Spt∗(M) if and only ifM |=θ(a).
Proof Suppose thatM |=θ(a). Then there are distinct b1, . . . ,bm−1∈M different fromasuch that for all cdifferent fromb1, . . . ,bm−1 and froma,
As Mhas the 2-extension property this is only possible ifa,b1 ∈X=Spt∗(M) and
aandb1 belong to the same part of Π1.
Now suppose thata∈Spt∗(M)=X. Letb1, . . . ,bm−1 be such that
X={a,b1, . . . ,bm−1}
andaandb1 belong to the same part of Π1. By the definition ofSn(AX,Π1), there is an automorphism ofMwhich sends atob1 and fixes every element outside ofX and therefore we must have
M |=∀z z6=a ∧ m−1 ^ i=1 z6=bi −→ R(a,z) ←→ R(b1,z) .
Letξ(x1,x2) be the formula ∀z ¬θ(z) −→ R(z,x1) ←→ R(z,x2) .
Lemma 6.7 Suppose thatM ∈ Sn(AX,Π1)has the2-extension property. Then for all a1,a2 ∈ X = Spt∗(M), a1 and a2 belong to the same part of Π1 if and only if M |=ξ(a1,a2).
Proof Suppose thata1,a2∈X=Spt∗(M) and a1 anda2 belong to the same part of Π1. By the definition of Sn(AX,Π1), for every c∈M\X there is an automorphism which sendsa1 toa2 and fixes every element outside of X. From Lemma6.6it follows thatM |=ξ(a1,a2).
Now suppose that a1,a2 ∈X = Spt∗(M) andM |=ξ(a1,a2). From Lemma6.6it follows that for allc∈M\X
M |=R(c,a1)⇐⇒R(c,a2).
Since we assume thatMhas the 2-extension property this is only possible ifa1 and
a2 belong to the same part of Π1.
According to the arguments before Definition6.4and the compactness theorem, the following corollary concludes the proof of Theorem6.1.
Corollary 6.8 For every k ∈ N, there is 0 < α < 1, depending only on k and A,
such that the proportion ofM ∈Sn(AX,Π1)that satisfy (I) and the sentencesψand ϕl forl=0, . . . ,k is at least1−αn.
Proof Let k0 =max(2,k,m) (wherem=|A|). By Lemma6.5there is 0< α < 1, depending only on k0 and A such that the proportion of M ∈Sn(AX,Π1) with the
l-extension property for every l≤k0 is at least 1−αn. From Lemmas6.6and6.7
it follows that all M ∈ Sn(AX,Π1) with the l-extension property for every l≤ k0 satisfy (I) and the sentencesψandϕl for l=0, . . . ,k.
Remark 6.9 Let S0 be any one of the sets of structures in part (ii) of Theorem6.2
and letS0n=S0∩Sn. We assume that if a finite groupG is involved in the definition ofS0 thenG is nontrivial. We will show thatS0 does not satisfy a zero-one law. By Lemmas4.2and4.3(and in one case the proof of Proposition5.15), there are mutually nonisomorphic A1, . . . ,Al ∈ S without any fixed point and, for i = 1, . . . ,l and
j=1, . . . ,li, subgroups Hi,j⊆Aut(Ai) without any fixed point such that |S0n| ∼ l [ i=1 li [ j=1 Sn(Ai,Hi,j) .
If S0 is {M ∈S : G ≤Aut(M)} or {M ∈S : G ∼=Aut(M)}, then we may also assume thatG≤Hi,j orG=∼Hi,j, respectively, for alliandj.
Now observe the following: Suppose that A ∈S has no fixed point and that H is a subgroup of Aut(A) without any fixed point. LetA0 andA00 have the same universe
Aas Aand assume that for every relation symbolR, RA0 =∅ and RA00 =Ai if Ris
i-ary. Then H is a subgroup of Aut(A0) and of Aut(A00) and, from Proposition4.4, it follows that |Sn(A0,H)| |Sn(A,H)| and |Sn(A 00,H)| |Sn(A,H)|
converge to the same c∈ Qas n → ∞. From the assumption that S0 is one of the sets of structures in part (ii) of Theorem6.2(and G is assumed to be nontrivial) it follows that there must bei,i0,j,j0 such thatAi 6∼=Ai0 and both |Sn(Ai,Hi,j)||S0n|and |Sn(Ai0,Hi0,j0)||S0
n|converge to positive numbersc andc0 asn→ ∞. With the help of the formulaθfrom the proof of Theorem6.1one can easily construct a sentence ϕ which, in almost allM ∈S0, expresses that “MSpt∗(M)∼=Ai”. Then the proportion ofM ∈S0n in whichϕis true converges to some number 0<d<1.