Topics in relational Hrushovski constructions - Fall

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1 Topics in relational Hrushovski constructions - Fall Omer Mermelstein UW-Madison Fall Notation Denote by P(X) the set of subsets of X and by Fin(X) the set of finite subsets of X. Write Y fin X for Y Fin(X). For X, Y sets and c, d elements, when there is no confusion we may abuse notation and omit the union operation: Xc = X {c}, cy d = Y {c, d}, XY = X Y, cd = {c, d}, and so on. Similarly, for tuples ā = (a 1,..., a n ), b = (b 1,..., b k ) abuse notation and write ā b for (a 1,..., a n, b 1,..., b k ). 2 Combinatorial Pregeometries Definition 2.1. A matroid is a pair (M, cl), where M is a set and cl : P(M) P(M) satisfies, for all X, Y M and a, b M, C1. X cl(x) C2. X Y = cl(x) cl(y ) C3. cl(cl(x)) = cl(x) C4. a cl(xb) \ cl(x) = b cl(xa) (Exchange) Definition 2.2. A combinatorial pregeometry is a finitary matroid, i.e. a matroid satisfying the additional requirement C5. cl(x) = X 0 fin X cl(x 0) (Finite character) Exercise 2.3. Show the following naturally occurring closure operators give rise to combinatorial pregeometries i. A set M with the trivial closure operator cl(x) = X. ii. A vector space V over a field F with cl(x) = Span(X). 1

2 iii. An algebraically closed field with the algebraic closure operator cl(k) = K alg. Many of the phenomena we are familiar with from the examples above, e.g. linear independence and linear dimension, can be generalized to any combinatorial pregeometry. Independence: A point a is said to depend on X if a cl(x). A set X is said to be independent if whenever a X, then a does not depend on X \ {a}. For Y X, Y is said to be a base for X if Y is independent and X cl(y ). Dimension: Define r(x), the rank of a set X, to be the cardinality of any base for X. This does not depend on the choice of the base. Proposition 2.4. If Y 1, Y 2 are bases for X, then Y 1 = Y 2. Proof. We prove by induction. Enumerate Y 1 = {a i : i < Y 1 }. For some n < Y 1, denoting A n = {a i : i < n}, assume that there is an injection f <n : A n Y 2 such that B := A n (Y 2 \ f <n [A n ]) is a base for X. Since a n cl(b), there is some B 0 fin B, minimal under inclusion, such that a n cl(b 0 ). As a n / cl(a n ), there is some y B 0 \ A n. By exchange, y B 0 := (B 0 \{y}) {a n }. Moreover, cl(b 0 ) = cl(b 0). In particular, cl(ab 0 ) = cl(ab 0) for any set A. Thus, letting B = (B \ {y}) {a n }, we have that cl(b ) = X. So B is a base for X. Proceed inductively by extending f <n to f n = f <n {(a n, y)}. Taking the ascending union of the functions we have constructed, we get an injection f : Y 1 Y 2 such that Y 1 (Y \ f[y 1 ]) is a base for X (note that this assertion uses finite character). By independence, it must be that (Y \f[y 1 ]) = and f is bijective. Corollary 2.5. The following are equivalent: 1. B is a base for X 2. B is a maximal independent subset of X 3. B is a minimal subset of X such that X cl(b). Observation 2.6. A combinatorial pregeometry (M, cl) is uniquely determined by either of i. Its collection of finite independent subsets ii. The restriction of its rank function to finite sets For the first item, note that a cl(x) if and only if there is some independent X 0 fin X such that X 0 a is not independent. For the second, note a finite set X is independent if and only if r(x) = X. One can define the notion of a combinatorial pregeometry by imposing conditions on its declared collection of independent sets. For I P(M) consider the following axioms: 2

3 I1. I I2. If Y X and X I, then Y I I3. Whenever X, Y I with Y < X, then there is some a X \ Y such that Y a I. I4. For any X M, X is an element of I if and only if every X 0 fin X is an element of I. Similarly, combinatorial pregeometries can be defined via rank functions. For r : P(M) N { } consider the following axioms: R1. For all X M, r(x) X R2. For all X M and a M r(x) r(xa) r(x) + 1 R3. For all X, Y M, (Submodular) r(x Y ) + r(x Y ) r(x) + r(y ) R4. If X M is such that r(x 0 ) = X 0 for any X 0 fin X, then r(x) = X. Let us show the equivalence of all three definitions. Lemma 2.7. If cl : P(M) P(M) satisfies C1-C5, then I, the collection of independent sets according to cl, satisfies I1-I4. Proof. Axioms I1-I2 clearly hold by definition of I. Axiom I4 immediately follows from C5. We show that I3 holds. Let X, Y I be such that Y < X, and assume for a contradiction that there is no a X \ Y such that Y a I. Fix some a X \ Y. For any b Y, if b cl(y a \ {b}), then by Y I and exchange we have a cl(y ). Since Y a / I, by what we have shown it must be that a cl(y ). As a was arbitrary, X cl(y ), so cl(x) = cl(y ). So X, Y are bases for the same set and as we have shown previously Y = X, in contradiction. Lemma 2.8. Let I P(M) be a collection satisfying I1-I4. For every X let r(x) = sup{ X : X P(X) I}. Then r satisfies axioms R1-R4. Proof. Axioms R1-R2 clearly hold by definition of r. If X is a set as in R4, then I4 guarantees that X I, and so r(x) = X. We show that R3 holds. First, observe that I3 guarantees that there is always some X P(X) I such that r(x) = X. Taking a sequence (X i ) i<r(x) of sets X i P(X) I with sup{ X i : i < r(x)}, by iterating I3 we may assume that the sequence is ascending, and so i<r(x) X i P(X) I. 3

4 Choose I 0 P(X Y ) I such that I 0 = r(x). By I3, choose I 1 P(X) I extending I 0 with I 1 = r(x). Similarly, choose I 2 P(X Y ) I extending I 1 with I 2 = r(x Y ). So r(x Y ) + r(x Y ) = I 2 + I 0 = I 1 + (I 2 \ I 1 ) I 0 I 1 + I 2 = r(x) + r(y ) Lemma 2.9. Let r : P(M) N { } satisfy axioms R1-R4. For every X M let cl(x) = {a M : X 0 fin X such that r(x 0 a) = r(x 0 )}. Then cl : P(M) P(M) satisfies axioms C1-C5. Proof. Axiom C5 is by the definition of cl. Axioms C1-C2 are easy to prove. Observe that for any X Y M, iterating R2 gives r(x) r(y ). We show C3. Let a cl(cl(x)), and choose some B := {b 1,..., b n } fin cl(x) such that r(ab) = r(b). We may assume a / B. For each i n choose some Y i fin X such that r(y i b i ) = r(y i ). Let Y = n i=1 Y i. Without loss of generality, by reordering, assume B Y = {b 1,..., b j }, for j = B Y. Then by submodularity, for any j < i n, r(y b i ) + r(y i ) r(y ) + r(y i b i ). So r(y b i ) r(y ) and by R2 equality holds. Again by submodularity r(y b j... b i ) + r(y ) r(y b j... b i 1 ) + r(y b i ) hence r(y b j... b i ) r(y b j... b i 1 ) and by R2 equality holds. Since this holds for all j < i n, we get r(y B) = r(y ). Another application of submodularity yields r(y Ba) + r(b) r(y B) + r(ba) so r(y Ba) r(y ). But also r(y ) r(y a) r(y Ba), so r(y a) = r(y ) and a cl(x). We show C4. Let a cl(xb) \ cl(x). It cannot be that b cl(x), since then by C3 we d have a cl(x). Let X 0 fin X be such that r(x 0 ba) = r(x 0 b). Since a, b / cl(x 0 ), r(x 0 a) = r(x 0 b) = r(x 0 ) + 1. So r(x 0 ab) = r(x 0 a) and b cl(xa). Remark The axioms C5, I4, R4 are only relevant in the case of a matroid of infinite rank, and they guarantee its finite character. The axiom sets C1-C4, I1-I3, R1-R3 are equivalent (and sufficient) for matroids of finite rank. We may think of a combinatorial pregeometry (M, cl) as a first order structure in the language L PG = {I n : n ω} by interpreting I i as the set of independent n-tuples of elements of M. An isormorphism of combinatorial pregeometries is an isomorphism between their first order representations. By what we have shown, it suffices to show that a bijection respects either the independent subsets, the rank function, or the closure operator. 4

5 Definition Say that a combinatorial pregeometry (M, cl) is a combinatorial geometry if every point in M is closed. i.e. cl(x) = {x} for any x M. Remark For convenience, from now on omit the combinatorial from combinatorial pregeometry and combinatorial geometry. To each pregeometry (M, cl) there is a canonically associated geometry. Denote M = M \cl( ) and define the equivalence relation x y cl(x) = cl(y) on the elements of M. For any X M let cl ({[x] : x X}) = {[y] : y cl(x) M }. Then (M, cl ) is the geometry gotten from M by taking every closed set of rank 1 to be a point, and removing all points of rank 0. This process is analogous to the construction of the projective space P n (F ) from the vector space F n+1 over F. Definition Say that a pregeometry (M, cl) is n-trivial 1, if n is maximal such that every subset of cardinality n or less is closed. If no such n exists because every set is closed in (M, cl) is closed, say that (M, cl) is trivial. With the definition of n-triviality, we can redefine a geometry as a pregeometry that is n-trivial for some n > 0. The above association of a geometry to a pregeometry can be seen as a two-staged process. First we associate canonically to an arbitrary pregeometry a 0-trivial pregeometry by removing all the dependent points. Then, to the resulting 0-trivial pregeometry we associate canonically a 1-trivial pregeometry (geometry) by replacing each closed set of rank 1 with a single point. Observe that the first stage cannot be skipped, since then we are left with a dependent point (the closure of the empty set). Observation If two pregeometries (M 1, cl 1 ), (M 2, cl 2 ) are such that cl i (X) is countably infinite for any X M i, and their associated geometries are isomorphic, then (M 1, cl 1 ) = (M 2, cl 2 ). Question Is there a canonical procedure that associates to each 1-trivial geometry a 2-trivial geometry? Namely, a canonical association such that if (M 1, cl 1 ) and (M 2, cl 2 ) are geometries with isomorphic associated 2-trivial geometries according to this procedure, and every closed set of rank 2 in either pregeometry is countably infinite, then (M 1, cl 1 ) = (M 2, cl 2 ). Definition For (M, cl) a pregeometry and A M, define the localization of (M, cl) in A to be to be the pregeometry (M, cl A ) where cl A (X) = cl(a X). Example Let A be some affine space over a field F. The affine span operator (the affine span of a set of vectors S is the collection of linear combinations s S α s s such that s S α s = 1) is a closure operator defining a geometry on A. Localizing this geometry at any one point produces the pregeometry of a vector space over F of dimension n 1, where n is the dimension of A. Localizing by a point essentially makes that point into the origin. 1 If you are familiar with the notion of a trivial pregeometry, we will call these disintegrated or degenerate 5

6 Definition Say that pregeometries (M 1, cl 1 ), (M 2, cl 2 ) are locally isomorphic if there exist finite sets A i M i such that (M 1, cl 1 A 1 ) = (M 2, cl 2 A 2 ). 3 The algebraic closure operator in minimal structures We wish to see how combinatorial pregeometries arise naturally in model theory. Definition 3.1. Fix some first order L-structure M with universe M. For each ϕ(x) L(M) (L-formulas with parameters from M) say that ϕ(x) is an algebraic formula if there are only finitely many a M such that M = ϕ(a). Say that a is algebraic over A M if there is some algebraic formula ϕ(x) L(A) such that M = ϕ(a). For every X M define acl M (X), the algebraic closure of X in M to be the set of elements algebraic over X. Exercise 3.2. In any structure M, the operator acl M : P(M) P(M) satisfies axioms C1-C3 and C5. Hint: ϕ(x) has exactly k solutions is expressible in first order logic. Example 3.3. Let us see that C4 need not necessarily hold. Let G be the countable ω-regular connected tree. Then acl G (xy) is the set of vertices forming the unique path between x and y. Let (x 1, x 2, x 3 ) be a path in G, then x 2 acl G (x 1 x 3 ) \ acl G (x 1 ), but x 3 / acl G (x 1 x 2 ). Definition 3.4. Say that a model M is minimal if for any ϕ(x) L(M), either ϕ(x) is algebraic or ϕ(x) is algebraic. Equivalently, every definable set is either finite or cofinite in M. Say that a model (a theory) is strongly minimal if every model of T h(m) is minimal. Lemma 3.5. If M is minimal, then acl := acl M satisfies C4. Proof. Let a acl(ab) \ acl(a), note that this implies a / acl(a). Then there is some formula ϕ(x, y) L(A) such that a ϕ(m, b) and ϕ(m, b) = k for some k N. Let ψ(y) be the L(A)-formula stating that ϕ(m, y) = k. As b / acl(a), it must be that ψ(m) is cofinite. For each c M consider the definable set D c = {y M : ϕ(c, y) ψ(y)}. Then b D a, and so if D a is finite, then b acl(aa). So we may assume that D a is cofinite, hence M \ D a l for some natural fixed l. Consider now the A-definable set E = {x M : M \ D x l}, that is the set of x M such that there are at most l solutions to (ϕ(x, y) ψ(y)). Since a / acl(x), the set E must be cofinite. In particular, E > k, so choose some distinct a 1,..., a k+1 E. For each i k+1, the set D ai is cofinite, hence D = k+1 i=1 D a i is cofinite, and in particular non-empty. Let d D, then k+1 i=1 ϕ(a i, d). But this is a contradiction to ψ(d), which is asserted by d D ai. 6

7 Remark 3.6. In strongly minimal structures, the rank function associated to the acl operator is exactly the Morley Rank. Remark 3.7. In the setting of a minimal structure, localizing the pregeometry in a set A is synonymous with adding the parameter set A as constants to the language. The examples of Exercise 2.3 are all strongly minimal sets. The pregeometries of these examples increase in complexity, in a quantifiable way. Definition 3.8. i. Say that a pregeometry is disintegrated or degenerate if its associated geometry is trivial. Equivalently, cl(x) = x X cl(x) for every X. ii. Say that a pregeometry is modular if for any closed X, Y r(x Y ) = r(x) + r(y ) r(x Y ) iii. Say that a pregeometry is locally modular if its localization by any one point is modular An example of a strongly minimal theory with a disintegrated pregeometry is the theory of Z with the successor function s, or similarly the theory of an infinite path in the language of graphs. The equality in the definition of modularity is precisely the dimension theorem of linear algebra. The theory of a vector space over a fixed field has a modular pregeometry. As in Example 2.17, an affine space has a locally modular pregeometry. Fact 3.9. The pregeometry of an algebraically closed field of infinite transcendence degree is not locally modular. Maybe-an-example: Consider C[x, y, z] with x, y, z transcendental over each other. Look at the sets {x, y}, {z, x(y + z)}. Zilber s trichotomy conjecture Zilber famously conjectured that pregeometries of strongly minimal structures come in one three distinct flavours of increasing complexity 2 : 1. Disintegrated (essentially binary, or less) 2. Locally modular (essentially a linear space) 3. Field-like (essentially an algebraically closed field) The guiding principle is that geometrical complexity comes from algebraic complexity. This is where Hrushovski comes in: 2 Disclaimer: This is a very simplistic, inaccurate statement of the conjecture. It is purposefully not made precise, as this formulation suits our needs. 7

8 Theorem 3.10 (Hrushovski). There exists a strongly minimal structure not interpreting an infinite group, whose geometry is not locally modular (in particular not disintegrated). In the next few sections we will cover the construction of such a structure. 4 Hrushovski s construction method The hypergraph δ Notation. A hypergraph (V, E) is a set of verices V and an arbitrary set of finite edges E Fin V. For a hypergraph A = (V, E), we often (always) abuse notation and write A to mean the set of vertices of A. Write R A for the set of edges of A and write r(a) := R A. If B A, write r(a/b) := R A \ R B. Definition 4.1. For a finite hypergraph A define δ(a) := A r(a) For A B such that B \ A is finite, write δ(b/a) := B \ A r(b/a) where if B is finite this equals δ(b) δ(a). If A, B D, write δ(b/a) := δ(b A/A). Observation 4.2. If A, B D then δ(b/a) δ(b/a B), i.e. δ is submodular. Proof. Observe that R B R A = R A B, so R B \ R A B R A B \ R A. Also, (A B) \ A = B \ A = B \ (A B). Therefore, δ(b/a) = B A \ A r(b A/A) B \ A B r(b/a B) = δ(b/a B) Note that in the above observation equality holds if and only if R A B = R A R B. Remark 4.3. The next few portions we will do on arbitrary predimension functions, but there is no real harm in always assuming δ is the function defined above. 8

9 Predimension Definition 4.4. For a class of finite structures C closed under isomorphism and substructures, say that δ : C Z is a predimension function for C if 1. δ( ) = 0 2. δ(x) X 3. δ is preserved under isomorphism 4. δ is submodular Lemma 4.5. Let A be some structure and let δ be an integer valued submodular function defined on the age of A. Define for each B fin A: d(b) = inf{δ(x) : B X fin A} Then d is submodular. If δ is a predimension function, then R1 R3 hold for d. If additionally δ only takes non-negative values, then d is the rank function of a pregeometry. Proof. Let X 1, X 2 fin A, we may assume d(x 1 X 2 ) >. Choose X i fin A realizing d(x i ). Then X 1 X 2 X 1 X 2 and X 1 X 2 X 1 X 2. Hence, d(x 1 X 2 ) δ( X 1 X 2 ) and d(x 1 X 2 ) δ( X 1 X 2 ). So d(x 1 X 2 ) + d(x 1 X 2 ) δ( X 1 X 2 ) + δ( X 1 X 2 ) δ( X 1 ) + δ( X 2 ) = d(x 1 ) + d(x 2 ) For the additional part, assume that δ is a predimension function on age(a). Then d(x) δ(x) X, so by submodularity d(xa) d(x) + 1, and by definition d(x) d(xa). Lastly, if δ only takes non-negative values, then the same is true of d. Self-sufficiency Fix some class of finite structures C closed under isomorphism and substructures and fix δ a predimension function for C. Denote by C the class of all countable structures M with age(m) C. Assume that whenever X M C then d M (X) >. Definition 4.6. For N M C, say that N M if for every X fin M, δ(x/x N) 0. Observation 4.7. For N fin M, the following are equivalent: 1. N M 2. δ(n) = d M (N) 3. For every N X fin M, δ(x/n) 0. 9

10 Proof. 1 2: If δ(n) d M (N) then δ(n) > d M (N). So there is N X fin M with δ(x) < δ(n), hence δ(x/x N) = δ(x/n) < : If δ(x/n) < 0 then d M (N) δ(x) < δ(n). 3 1: By submodularity δ(x N/N) δ(x/x N), so if the right is negative, the same is true for the left. Lemma 4.8. N M if and only if d N = d M P(N). Proof. Assume N M, then δ(x/x N) < 0. Choose some X N Y N realizing δ(y ) = d N (X N), so Y N. By submodularity 0 > δ(x/x N) δ(x Y/Y ), so d M (Y ) < d N (Y ). Assume d N (X) d M (X) for some finite X, then for Y a realization of d N (X), δ(y/y N) < 0. Corollary 4.9. The relation is transitive. Observation N M if and only if whenever Y M, then Y N Y. Corollary If X, Y M then X Y M. Proof. If X M, then X Y Y. If additionally Y M, by transitivity X Y M. Now we see that whenever X M C, there is a unique minimal (under inclusion) X M containing X. Call X the self-sufficient closure of X (not to be confused with the geometric closure). Note that if X is finite, then also X is finite. Definition A strong-embedding is an embedding f : N M such that f[n] M. Since δ is preserved under isomorphism, every isomorphism is a strong embedding. By transitivity of, the strong embeddings are closed under composition. Flatness Recall that our goal is to find a strongly minimal structure whose geometry does not fall into Zilber s classification. In order to do that, we define the property of flatness. We will see that a strongly minimal structure whose acl-pregeometry is flat, cannot interpret an infinite group, i.e. not (essentially) a vector space nor ACF. Notation. Whenever we list E 1,..., E k sets in some ambient pregeometry (M, cl M ): For any s [k] which is not empty, E s = i s E i. E = k i=1 E i Est M ({E 1,..., E k }) = =s [k] ( 1) s +1 d M (E s ) 10

11 Diff M ({E 1,..., E k }) = s [k] ( 1) s +1 d M (E s ) We omit M if it is clear from context. Definition Say that a pregeometry (M, cl) is flat if whenever E 1,... E k are closed of finite dimension, then Diff M ({E 1,..., E k }) 0. Proposition If M is a saturated strongly minimal structure whose aclpregeometry is flat, then any group interpretable in M is finite. Proof. Assume for a contradiction that G is an infinite group interpreted in M. Denote the dimension of G (Morley rank) by g. Let a 1, a 2, a 3 be generic elements of G. For i = 1, 2, 3 let E i = cl({a 1, a 2, a 3 } \ {a i }) and let E 4 = cl(a 1 1 a 2, a 1 1 a 3). Then the dimension of the union is 3, the dimension of each E i is 2, the dimension of each intersection E i E j is 1, and intersection of three of the E i is empty. By flatness, 0 Diff({E 1,..., E 4 }) = (4 2g 6 g) 3g = g So g = 0 in contradiction to G being infinite. We will show that for the case of hypergraphs and δ of before, the pregeometries that arise are flat. Exercise An intersection of closed sets is closed. Lemma A closed set in M is self-sufficient in M. Proof. If N M then there is some X N such that d N (X) > d M (X). Letting Y realize d M (X), there is some a Y \ N. So d M (Xa) = d(x) but a / X. Definition Define C to be the class of finite hypergraphs A with A. By Lemma 4.5, for every M C the d M defined using delta is the rank of a predimension. Proposition For any M C, the pregeometry given by d M is flat. Proof. Let E 1,..., E k be closed subsets. For each = s [k] let B s be a base for E s, and let B be the self-sufficient closure of s [k] B s in M. Let F s = B E s and note δ(f s ) = d(e s ), and finally denote F = k i=1 F i. Est({E 1,..., E k }) = ( 1) s +1 δ(f s ) =s [k] = =s [k] ( 1) s +1 F s F R F = δ(f) =s [k] Since F contains a base for E, in particular δ(f) d(e ). ( 1) s +1 R Fs 11

12 Remark In the above proof we get equality if and only if R E k = i=1 REi and E M. So in order to refute Zilber s conjecture, it will suffice to come up with a hypergraph that is strongly minimal such that its Morley rank coincides with the dimension d. Fraïssé s Theorem For some language L, fix some class of finite L-structures C, and a class of distinguished embeddings of L-structures, closed under isomorphism and composition, such that if A C then there is some A C with A A. Definition For (C, ), define the following properties: Hereditary Property If B C and A B, then A C. Joint Embedding Property If A, B C, then there is some C C such that A, B C. Amalgamation Property If A B 1, B 2 are in C, then there is some D C and strong embeddings f i : B i D such that f[b i ] D and f 1 A = f 2 A. If (C, ) has HP, JEP and AP, then we say it is an amalgamation class. Remark AP implies JEP if there is a prime structure in C, embedding into any element of C. This is not always the case, for example consider the class of countable ACF, which has AP but not JEP (take fields of different characteristics). Definition Say that M is a generic structure for (C, ) if age (M) := {A M} = C, whenever A fin M there is some A A M finite, and Whenever A M is finite and A B C, there exists some strong embedding f : B M such that f A = Id. Proposition If M 1, M 2 are countable generic structures for (C, ), then any finite partial isomorphism f : A 1 A 2 with A i M i extends to an isomorphism between M 1 and M 2. In particular, if C then there is at most one countable (up to isomorphism) generic structure for (C, ). Proof. We show that f can be extended by any one point b 1 M 1 and proceed by back-and-forth. Let B M 1 be finite such that A 1 b 1 B 1, In particular B 1 C. Letting B 2 be the structure B 1 after renaming the elements of A 1 using f, we have A 2 B 2. By genericness of M 2 there is some strong embedding g : B 2 M 2 over A 2. Extending f by g (B 2 \ A 2 ), we are done. Theorem 4.24 (Fraïssé). If (C, ) is a countable (up to isomorphism) amalgamation class with HP,JEP and AP, then there exists a generic model for (C, ). 12

13 Proof. Let M be a countable infinite set this will end up being our universe. Let A be a countable subset of C, closed under substructures and containing a set of representatives of the countably many isomorphism classes of C. Let F = {f : A M : A A} and note that it is countable. Consider T, the family of tuples (f, A, B) where A B A and f F with dom(f) = A. As T F A A, T it is also countable. Let {(f i, A i, B i ) : i ω} be an enumeration of T such that each tuple of T is repeated infinitely many times. We inductively build an ascending chain of structures M 0 M 1 M 2... such thatm i A for every i N. Given M i, if f i is not a strong embedding of A i into M i, by JEP choose some M i+1 C to which both M i and B i strongly embed. Otherwise, by AP choose some D C with strong embeddings g 1 : M i D and g 2 : B i D such that g 1 f i = g 2 A. Without loss of generality, we may assume that g 1 is the identity, hence M i D. Since M \ M i is infinite, we may assume that D M \ M i. Define M i+1 = D. Choose arbitrarily some M 0 C and produce M = i<ω M i. We need to check that 1. M has the extension property 2. whenever A fin M there exists some finite A A M 3. age (M) = C. Assume A M and A B C, then there is some large enough i < ω such that A M i and (f i, A i, B i ) is such that A i = A, B i = B. Then by construction B strongly embeds into M i+1, and so into M (1). Whenever A fin M, there is some i < ω such that A M i M (2). In particular, if A M then A M i C so by HP A C. For the other direction of inclusion, for any B C there is some i < ω such that B i = B, so by construction B embeds strongly into M i+1 M (3). Remark If you are careful, the theorem will also works when replacing finite by finitely generated and all the structures are countable. Exercise For M the countable generic structure for C, if A M is finite and A N C, then there is a strong embedding of N into M over A. Exercise Given a class C of relational structures and a notion of strong embedding, decide whether (C, ) is an amalgamation class and, if it is, describe its generic structure 1. The class of all finite graphs with =. 2. The class of all finite trees (forests) with A B if for any b B \ A there is at most one a A such that (a, b) is an edge in B. 3. The class of all finite linear orders with =. 4. The class of all finite linear orders with A B if A is an interval in B. 13

14 Non-collapsed construction We now return to C, the class of hypergraphs with A and the δ we defined for that class. Definition For hypergraphs B 1,..., B k with B i B j = A for all i j, define their free join over A to be the hypergraph with vertex set k i=1 B i and edges k i=1 RBi. Observation If D is the free join of B 1,..., B k over A, then for any X D k δ(x/x A) = δ(x B i /X A) i=1 Corollary The class (C, ) has AP. Proof. Let A C strongly embed into B 1, B 2 C. By renaming elements we may assume A B 1, B 2, B 1 B 2 = A. Then the last observation, letting D be the free join of B 1, B 2 over A we have that B 1, B 2 D. In particular, since B 1 D and B 1 C, also D C. By Fraïssé s Theorem, there is a unique (up to isomorphism) countable generic structures M C such that whenever A M, A B C, then B embeds into M over A. We call this construction Hrushovski s non-collapsed construction. It is not strongly minimal, but it is very similar to the strongly minimal (collapsed) version we will soon see. Lemma Every countable hypergraph N for which M1. age(n) = C M2. Whenever A fin N and A B C, then there is an embedding of B into N over A. M3. There is an infinite independent set under d N then N is a generic structure for C. Proof. We only need to show self-sufficient extension. Let A N and A B C. Let X be a basis of B such that X A is a basis of A. Let n := X \ A and choose arbitrarily b 1,..., b n N independent over A in N. By M2 we may embed B into N over Ab 1... b n, and this must be a strong embedding because d N (B) X and δ(b) = X. Corollary Every countable elementary extension of M is isomorphic to M. In particular, M is saturated. Proof. The properties M1-M3 hold for M, and they are all preserved under elementary extensions. M1-M2 are given by T h(m), whereas M3 is a realized 14

15 ω-type. Thus, an elementary extension is also a generic for C, so isomorphic to M. To see that M is saturated let p(x) be some type over a finite set A. We may assume that A M by taking its self-sufficient closure. Then p(x) is realized by b in some countable elementary extension M M. The identity on A is a finite partial isomorphism between strong substructures of M and M, so extends to an isomorphism f : M M. Then f(b) M realizes the type p(x). Proposition The theory of M has quantifier elimination to the level of boolean combinations of existential formulas Proof. Let A 1, A 2 M be such that for any existential formula ϕ( x), M = ϕ(a 1 ) ϕ(a 2 ). The isomorphism type of the self-sufficient closure of A i is the ismorphism type of the largest hypergraph A i B i fin M such that δ(b i /A i ) < 0 and for any A i X B i, δ(x/a i ) > δ(b i /A i ). So the self-sufficient closures of A 1 and A 2, say Ā1 and Ā2, are isomorphic strong substructures of M, so there is some f automorphism of M with f(ā1) = Ā2. In particular, f(a 1 ) = A 2, so tp M (A 1 ) = tp M (A 2 ). As M is saturated, this is enough to prove the statement. The collapse What we d like to do now, is construct a structure similar to M, that is strongly minimal, with Morley Rank being d. So in other words, dependent means algebraic. So for each minimal isomorphism type of d-dependency, we will assign a uniform bound. Definition For A B finite, say that B \ A is simply algebraic over A if: δ(b/a) = 0 For any A X B, δ(x/a) > 0. Say that B \ A is minimally simply algebraic over A if there is no A A such that B \ A is simply algebraic over A. Example Here s a non-trivial example of A B such that B \ A is minimally simply algebraic over A. Let A = {a 1, a 2, a 3 }, R A = {{a 1, a 2, a 3 }}. Let B = A {b 1,..., b 4 }, R B = R A {{a i, b i, b i+1 } : i 3} {{b 1, a 2, b 4 }}. It is easy to extend this example so that B is of any finite size. Lemma If B 1,..., B n are simply algebraic over A and distinct such that A A n i=1 B i, then the B i are pairwise disjoint. Moreover, A n i=1 B i is a free join over A. 15

16 Proof. Denote B 0 := B 1 B 2. As δ(b i /AB 0 ) = δ(b 0 /A), we have δ(b 1 B 2 /AB 0 ) = δ(b 1 /AB 2 ) + δ(b 2 /AB 0 ) δ(b 1 /AB 0 ) + δ(b 2 /AB 0 ) = 2δ(B 0 /A). So δ(b 1 B 2 /A) δ(b 0 /A). If B 0, then by simple algebraicity of B 1 over A, δ(b 0 /A) > 0. This will contradict A B 1 B 2. Exercise Show that if B is simply algebraic over A, then B is either contained in the self-sufficient closure of A or disjoint from it. Definition Let A M C and A B C. The multiplicity of B over A in M is the maximal number of disjoint realizations of atp(b/a) in D. Call a function µ : C C N { } a µ-function or a multiplicity function if it is preserved under isomorphism, i.e., if A i B i, B 1 = B2 such that the restriction to A 1, A 2 is also an isomorphism, then µ(a 1, B 1 ) = µ(a 2, B 2 ). Definition Let C µ be the class of finite hypergraphs A C such that whenever F C A with C \ F minimally simply algebraic over F, then the multiplicity of C over F in A is at most µ(f, C). Clearly C µ has HP and JEP for any µ. It is not clear exactly for which µ amalgamation works. We will see that if µ is large enough, things work out. The setup: Fix some A 1, A 2 C, denote A 0 := A 1 A 2, and let D be the free join of A 1, A 2 over A 0. We want to see how multiplicity behaves in a a free join. Keep in mind that later we will expect A 0 A 1, A 2, since we are doing self-sufficient amalgamation. Later, we will see it is sufficient to assume that either A 1 \ A 2 or A 2 \ A 0 is simply algebraic over A 0. Fix some F D and C D minimally simply algebraic over F, and let C 1,..., C k be all the realizations of atp(c/f ) in D. For every X D, denote X i = X A i. Note that X is a free join of X 1 and X 2 over X 0. We will categorize the copies of C according to which parts of the Venndiagram of D they intersect. Label the parts of the diagram A 0, A 1 \ A 0, A 2 \A 0 by 0,1,2 respectively. Say that C i is a type-ijk copy if it intersects the parts of the diagram listed in ijk. e.g., if C i is type-01, then C i 0, C i 1\C i 0, and C i 2 \ C i 0 =. We will make a series of observation regarding different possible configurations of F and the C i. Claim 1. It cannot be that C is type-12. Proof. As D is a free join, we d have δ(c 1 /F C 2 ) = δ(c 1 /F ). But C is simply algebraic over F, so δ(c 1 /F C 2 ) = δ(c 2 /F ) < 0 and δ(c 1 /F ) > 0. Claim 2. If C is type-1, then F A 1. The statement is symmetric in 1 and 2. Proof. Since D is a free join and C A 2 =, we have δ(c/f 1 ) = δ(c/f ). But C is minimally simply algebraic over F, so F = F 1. 16

17 Claim 3. Let b 1 the number of type-02/012 copies of C, and assume these are exactly C 1,..., C b. Then δ(c 1... C b /A 1 ) + b 1 δ(f 2 /A 0 ). In particular, if A 0 A 2, then b 1 δ(f 2 /A 0 ). Proof. For every such C i, by simple algebraicity over F, δ(c i /F C i 1) < 0. By submodularity, this gives δ(c i /F A 1 ) δ(c i /F C i 1) < 0. Applying submodularity for all bad C i at once, we have δ(c 1... C b /F A 1 ) b δ(c i /F A 1 ) b 1. i=1 As D is a free join, δ(f/a 1 ) = δ(f 2 /A 0 ). Adding δ(f/a 1 ) + b 1 to both sides of the inequality above, we get the inequality of the statement. For the additional part, note that for any X, δ(x/a 1 ) = δ(x A 2 /A 0 ), because D is a free join. So A 0 A 2 implies δ(c 1... C b /A 1 ) 0. Claim 4. If F A 1, denoting by b 2 the number of i such that C i A 1, we have b 2 δ(f 2 /F 0 ) δ(f 2 /A 0 ) Proof. For every C i A 1, since F 1 F and C i is minimally simply algebraic over F, δ(c i /F 1 ) > δ(c i /F ). This means r(c i /F 1 ) < r(c i /F ), so there exists some relation r i involving both an element from C i and an element from F 2 \ F 0. Since D is a free join, all the elements of C i appearing in r i must come from C i 0. By disjointedness of the C i, if i j then r i r j. Hence, As a free join, δ(f 2 /F 1 ) = δ(f 2 /F 0 ). δ(f 2 /A 0 ) δ(f 2 /F 1 ) b 2. Proposition If A 0 A 2, one of the following conditions holds: 1. Everything happens inside A i for some i = 1, 2. i.e., F, C 1,..., C k A i. 2. F A 0 and there is some type-2 copy of C. 3. There are no type-2 copies and k δ(f ). 4. There is some type-2 copy of C, and there is some type-01/012 C i such that δ(c i 1/C i 0F 0 ) < 0. In particular, A 0 A 1. Proof. Assume condition 1 does not hold. Consider the case F A 1. As condition 1 does not hold, choose some C i A 1. By claims 1,3 and the fact δ(f/a 1 ) = 0, it must be that C i is type-2. By Claim 2, F A 2 so in particular F A 0, and condition 2 holds. So assume from now that F A 1. Assume there exists some type-2 copy of C. Then F A 2, by Claim 2. Condition 1 does not hold, so choose C i such that C i A 2. As F A 1, by Claim 2 C i is not type-1, so it is either type-01 or type-012. In particular, by simple algebraicity, δ(c1/c i 2F i ) < 0. As D is a free join, δ(c1/c i 2F i ) = δ(c1/c i 0F i 0 ). Condition 4 holds. So assume there are no type-2 copies of C. 17

18 Then we are left only with type-02/012 copies, of which there are at most b 1 δ(f 2 /A 0 ), and type-0/01 copies, of which there are at most b 2 δ(f 2 /F 0 ) δ(f 2 /A 0 ). Over all, k δ(f 2 /F 0 ) δ(f ). Corollary Let A 0 A 1 C µ and A 2 \ A 0 simply algebraic over A 0, where µ(f, C) δ(f ) for any F, C. Then the free join of A 1 and A 2 over A 0 is in C µ, unless A 2 \ A 0 is minimally simply algebraic over some F A 0 and the multiplicity of A 2 \ A 0 over F in A 1 is µ(f, A 2 \ A 0 ). Proof. Let D be the free join of A 1 and A 2 over A 0. Let F, C 1,..., C k D be such that C 1 is minimally simply algebraic over F and C i = F C j for all i, j k. We know that one of the conditions of the above proposition must hold. By A 0 A 1, we know that condition 4 fails. If condition 1 holds, then k µ(f, C 1 ) since A i C µ. If condition 3 holds, then by choice of µ we are okay. Then assume condition 2 holds F A 0 and for some i k C := C k A 2 \ A 0. Since C is simply minimally algebraic over F A 0, A 0 A 0 C, A 0 C =, C is simply algebraic over A 0. So it must be that A 2 \ A 0 = C. The only way for k > µ(f, C) is to have k 1 = µ(f, C) copies of C in A 1. Theorem If µ is a multiplicity function such that µ(a, B) δ(a), then C µ has AP with respect to. Proof. We prove by induction on A 2 \ A 0. Case 0: A 2 \ A 0 = {x} and δ(a 2 ) > δ(a 0 ). In the free join, the point x is not involved in any instance of minimal simple algebraicity. Case 1: There exists some A 0 X A 2 such that X A 2. By induction, amalgamate A 1 with X over A 0 to receive D. As X strongly embeds into D and into A 2, again by induction, amalgamate D with A 2 over X to receive D. Case 2: Cases 0, 1 fail. Then A 2 \ A 0 is simply algebraic over A 0. By Corollary 4.41, if the free join of A 1 and A 2 is not in C µ, then for F A over which A 2 \A 0 is minimally simply algebraic, in A 1 there are µ(f, A 2 \A 0 ) copies of A 2 \A 0. By Exercise 4.37, since A 0 A 1, each of these copies is either contained in A 0 or in A 1 \ A 0. It cannot be that all copies are in A 0, for then A 2 will have µ(f, A 2 \ A 0 ) + 1 copies, in contradiction to A 2 C µ. So there is some copy C A 1 \ A 0. Then CA 0 A 1, and by identifying A 2 \ A 0 with C, we can see A 1 C µ as an amalgam of A 1 and A 2 over A 0. For a multiplicity function µ, if C µ has AP, we denote by M µ its generic countable structure. From now on assume that µ only takes integer values (no ), and that µ(a, B) δ(a). We show that the theory of M µ is model complete Lemma Every model of T := T h(m µ ) is existentially closed. i.e., T is model complete and has QE up to existential statements. 18

19 Proof. Fix A, B with B minimally simply algebraic over A. Fix some (F, Q) with Q minimally simply algebraic over F such that F Q AB. Note that there are only finitely many options for F,Q. For every M, some finite hypergraph containing A such that in the free amalgam of B and M over A there are Q 1,..., Q µ(a,b)+1 disjoint copies of Q over F, let T M ( v, A) be the atomic type of Q i M over A. Enumerating over all M, note that there exist only finitely many distinct T M, since these are atomic types on (µ(a, B)+1) Q + A points. Let θ (F,Q) ( x) be the disjunction over all formulas of the form ȳ T M ( x, ȳ) for some M. One should read θ (F,Q) ( x) as saying the reason A cannot be extended freely by a copy of B is that this will introduce too many copies of Q over F. Now let Φ (A,B) ( x) be the disjunction of all θ (F,Q) ( x) for F Q AB. We have seen in Proposition 4.40 that failure of the free amalgam to be in C µ is a result of some Q B having too large a multiplicity over some F AB. So Φ (A,B) ( x) enumerates all the reasons that A cannot be extended freely by a copy of B. So T = x ( A( x) Φ (A,B) ( x) ) where A( x) means x realizes the atomic type of A. Now let P N both be models of T. We wish to show P is existentially closed in N. Let ɛ be quantifier-free with set of parameters A fin P, satisfied by some B N. We may assume ɛ is an atomic type over A pass to DNF and reduce to the conjunctive clause that B satisfies. We may also assume B N \ P by replacing A with A (B P ). Let B B be minimal non-empty such that δ(b /A) 0, if such a B exists. Then B is minimally simply algebraic over some A A. But A, by virtue of P = Φ (A,B )(A ), has exhausted the amount of copies of B it is allowed over it by some reason found in P. This reason is also in N \ B, and N is bound by C µ. So there is no such B. In fact, we have shown that P is closed in N. Then δ(b /A) > 0 for any non-empty B B. By examining Proposition 4.40 we see that the free amalgam of any extension of A and B over A is in C µ. So the existence of a copy of B over Ā in P is guaranteed by T in Mµ one can (strongly) embed a copy of B over the self-sufficient closure of any copy of A, so in particular strongly embed B over any copy of A. Corollary M µ is saturated. Proof. As before (Lemma 4.32), any countable elementary extension of M µ will be generic for C µ. For fun, let s faff about this time around: Let M µ N. Then N C µ, so embeds into M µ. Call the embedded copy N, then by model completeness N M µ, in particular N M µ. Let A N, A B C µ, then B embeds strongly into M µ over A. By existential closedness of N, there is a copy of B over A inside N. Since N contains an infinite independent set, we can make sure the embedding of B into N is strong, with the trick of

20 Lemma In M µ, the algebraic closure operator and the closure operator given by d are equivalent. Proof. Let A M µ be finite and let x cl(a), we will show x acl(a). Let Ax D fin cl(a) with δ(d) = d(a). Choose A B D minimal such that B D. Then B is simply algebraic over A, and therefore algebraic, since every element of B solves an existential equation with at most µ(a, B) solutions in M µ. If x B, we are done. Otherwise, Repeat the process, replacing A with B, until D is exhausted. If A fin M µ (not necessarily strong) and x cl(a), then there is some A A independent such that x cl(a ). Being independent, A M µ so x acl(a ) acl(a). For the converse assume that A M µ and x 1, x 2 / cl(a). Letting Ā be the self-sufficient closure of A, we have Āx i M µ and Āx 1 = Āx 2. So there is an isomorphism of M µ over A taking x 1 to x 2. Therefore, there is a unique type independent over A with infinite orbit, so any point outside cl(a) is not algebraic over A. Corollary M µ is strongly minimal. Proof. There is a unique independent type over A, whose orbit is infinite. Any other type is algebraic over A, so has a finite orbit. Since M µ is saturated, that implies every formula not in the independent type has finitely many realizations. If not, say ϕ is not in the independent type and defines an infinite set, take the type containing ϕ and diagonalizing over all algebraic formulas over A. This type is realized in M µ, not the unique independent type, but also not algebraic, hence not dependent on A. Exercise Going back to M, the non-collapsed construction: Show that if B M is simply algebraic over A M then MR(B/A) = 1. Show that if x depends on A then its Morley Rank over A is finite. In particular, MR(x/A) is the minimal length of a chain of simply algebraic extensions leading up from A to x. Show that the Morley Rank of the generic (independent) type over A is exactly ω. Conclude that M is ω-stable. Remark Note that µ(a, B) > 0 does not guarantee there are any realizations of B over A, even if A is self sufficient. It is possible that B itself already violates µ(f, C) for some F B. Continuum many non-isomorphic (pre)geometries Observation It is enough that µ(f, C) δ(f ) whenever C 2 for C µ to be an amalgamation class. 20

21 Proof. Recalling the conditions of Proposition 4.40, note that if C = 1 then C is type-0/1/2. We need to check that k δ(f ) can be omitted from condition 3 without anything breaking. If there are no type-2 copies, all copies of C are in A 1. If F A 1 then condition 1 holds, so assume F A 1. By Claim 2, all copies of C are type- 0. By simple algebraicity δ(c/f 1 ) > δ(c/f ), so F 2 and C are related, hence δ(c/f 2 ) C 1 = 0 so C is simply algebraic over F 2. By minimality of F, F A 2 and again condition 1 holds. If there is a type-2 copy of C, then by Claim 2 F A 2. If condition 1 doesn t hold, there is a type-1 copy of C, implying F A 0, so condition 2 holds. Exercise In the proof above we used explicitly our definition of δ. Assuming all copies of C are type-0, continue the proof using only that δ is a predimension and C = 1. (this is straightforward, but can be a bit tricky). Definition Fix n. Let C n be the class of finite n-regular hypergraphs A, such that whenever X [A] n, then X A. Observation For A C n, a set X [A] n is dependent if and only if it is an edge. Otherwise, δ(x) = X but d(x) < X, hence X A. C n is the class C νn for the multiplicity function ν such that - ν(f, C) = 0 whenever F n and F C is not an edge - ν(e \ {a}, a) = 1 when E is an edge - ν(f, C) = whenever F > n. The pregeometry of any A C µ is (n 2)-trivial. In particular, if n 3 (and it is), then the pregeometry of any A C µ is a geometry. Remark It may seem from the above observation that setting things to zero does not disturb amalgamation. This is not true. Consider A 1 = {c 1, c 2, c 3, f 1, f 2 }, A 2 = {c 1, c 2, c 4, f 3, f 4 } R 1 = {(c 1, f 1, c 3 ), (c 2, f 2, c 3 )}, R 2 = {(c 1, f 3, c 4 ), (c 2, f 4, c 4 )} Then C = {c 1,..., c 4 } is minimally simply algebraic over F = {f 1,..., f 4 }. Setting µ(f, C) = 0 forbids this configuration, but it forms whenever µ allows edges. Proposition There are continuum many amalgamation classes C µ whose associated (pre)geometries are not isomorphic. Proof. Choose some integer-valued µ such that µ(f, C) = ν(f, C) whenever F n, and µ(f, C) δ(c) otherwise. Then C µ is an amalgamation class with generic structure M µ. 21

22 We know that a set X [M µ ] n is an edge if and only if it is dependent, so the structure M µ can be read off of its pregeometry. Hence, multiplicity functions giving rise to non-isomorphic structures also give rise to non-isomorphic pregeometries. Finding pairs (F i, C i ) i<ω such that µ 1 (F i, C i ) µ 2 (F i, C i ) = µ 1 µ 2 is left as an exercise to the reader. Remark We can vary C, our notion of an acceptable hypergraph, in several ways such that all the proofs we ve done go through: Symmetry: Consider a directed hypergraph where edges are tuples instead of sets. One can also choose an intermediate level of symmetry for every arity for arity n quotient out the n-tuples by a subgroup of S n. Irreflexivity: For the undirected case, one can allow or disallow repetition of elements in the same tuple. Arity: Restrict the arity of the edges of the hypergraph. n-triviality: Only consider hypergraphs whose pregeometry is n-trivial. Colored edges: Have several sets of edges, each representing a different color. (Need to be careful if introducing infinitely many, some things may break) 5 Isomorphism of pregeometries For this section fix (C 1, 1 ) and (C 2, 2 ), two strong amalgamation classes which we will use for general statements. Assume both have finite closures and A i for any A i C i. Omit the subscript from i when it is clear from context. Assume that for any A C i there is a canonically associated pregeometry, invariant under isomorphism. Denote that pregeometry by PG i (A), or PG(A) when i is clear from context. Assume further that if A B C i, then PG i (A) PG i (B). Write PG(C) for the pregeometry of the generic structure for C. Definition 5.1. Say that C 1 C2 if, given structures A i C i such that PG 1 (A 1 ) = PG 2 (A 2 ), Whenever A 1 B 1 C 1, there exist D 1 C 1 and A 2 D 2 C 2 such that PG 1 (D 1 ) = PG 2 (D 2 ) (extending the isomorphism PG(A 1 ) = PG(A 2 )). Write C 1 C 2 if an addition one may always choose D 1 = B 1. Exercise 5.2. Assume that is in all classes and strongly embeds into everything (under all notions of ). Then: 1. Is transitive? 2. Is transitive? 3. C 1 C 2 C3? = C 1 C3 4. C 1 C2 C 3? = C 1 C3 22

23 5. Show that the relation is an equivalence relation such that is well defined on the quotient induces. Proposition 5.3. Assume C 1 C2 and C 2 C1. Let f 0 : PG(A 1 ) PG(A 2 ) be a finite partial isomorphism of pregeometries for A 1 M 1 and A 2 M 2. Then f 0 extends to an isomorphism of pregeometries f : PG(M 1 ) PG(M 2 ). In particular, PG(M 1 ) = PG(M 2 ) Proof. By back and forth. We show the forth direction, with the back being symmetric. Choose arbitrarily some x M 1 and let B 1 be the self-sufficient closure of A 1 x in M 1. Then by C 1 C2 there exist some B 1 D 1 C 1 and A 2 D 2 C 2 such that PG(D 1 ) = PG(D 2 ), extending f 0. As a generic structure, embed D 1 strongly into M 1 over B 1. Similarly, embed D 2 strongly into M 2 over A 2. Since D i M i, we have PG(D i ) PG(M i ). Then we have extended the isomorphism of pregeometries so that x is in the domain. Taking the union of our nested isomorphisms, we have an isomorphism of pregeometries with domain M 1 and image M 2. Since M 1, M 2, we may start with f 0 = to receive PG(M 1 ) = PG(M 2 ). We now go over some variants of the (non-collapsed) construction, and show that the changes do not affect the pregeometry. Fix some natural n. Definition 5.4. C the class of all n-regular undirected (edges are sets of size n) hypergraphs A with A. C the class of all n-regular irreflexive directed (edges are tuples of length n without repeating elements) hypergraphs A with A C the class of all n-regular directed (edges are arbitrary tuples of length n) hypergraphs A with A In all of the above δ is adjusted accordingly: #vertices - #edges. Notation. If A is a structure and X is a subset of the universe of A, write A[X] for the substructure of A induced on X. Lemma 5.5. i. C C r ii. C C Proof. i. Let A 1 B 1 C, A 2 C r such that PG(A 1) = PG(A 2 ). assume the universes of A 1 and A 2 are identical, say the set A. We may Let B 2 be the structure obtained from B 1 by replacing A 1 with A 2. i.e., the universe of B 2 is the universe of B 1, call it B, and R B2 = (R B1 \R A1 ) R A2. Then B 2 is a directed hypergraph, and for any X B δ r (B 2 [X]/A 2 [X A]) = δ (B 1 [X]/A 1 [X A]). 23

24 So A 2 B 2, and PG(B 1 ) = PG(B 2 ) via the identity. ii. Let A 1 B 1 C, A 2 C r such that PG(A 1) = PG(A 2 ). Again assume the universe of A 1, A 2 is A. For each set r R B1 \R A1 choose some ordered r tuple whose elements are exactly those of r. Let B 2 be the directed hypergraph with the same set of vertices as B 1 and set of (directed) edges R A2 {r : r R B1 \ R A1 }. As in the previous item, A 2 B 2 and PG(B 1 ) = PG(B 2 ). Exercise 5.6. Let C Rainbow be the class of all colorful n-regular undirected (an edge carries with it one of countably many colors) hypergraphs A with A. Show that C Rainbow C and C C Rainbow. Showing the converse directions to those of Lemma 5.5 is not so difficult, but requires a bit of technical detail. Also, a weakening to is necessary. Observation 5.7. If C 1 C 2, then {PG(A 1 ) : A 1 C 1 } {PG(A 2 ) : A 2 C 2 }. Just think of A 1 C 1 as extending the empty set, then there is some A 2 C 2 with PG(A 1 ) = PG(A 2 ). Corollary 5.8. C C, C C. Proof. Consider A C with vertex set {a 1,..., a n } and set of edges R A = {(a 1,..., a n ), (a n,..., a 1 )}. Then A = n and d A (A) = n 2. No such structure exists in C, because on n points one can have at most one relation in C. So PG(A) / {PG(A ) : A C }. For the second statement apply the same trick, this time noting that, unless n = 1 (which it is not), for A = a with R A = A n we have A = 1, d A (A) = 0 which cannot happen in C because every set of size < n has no relations on it. Lemma 5.9. Assume (C i, i ) and its associated scheme of pregeometries come from a predimension function δ i (for i = 1, 2). Let B i C i have universe B and let A B be such that B i [A] := A i B i. Assume that PG(A 1 ) = PG(A 2 ). For any X B, denote X i := B i [X], δ i (X) := δ i (X i ) and d i (X) := d i (X i ). Then: 1. If Y B is d 1 -closed in B 1, then δ 1 (Y A) = δ 2 (Y A) 2. If Y B is d 1 -closed in B 1 and δ 1 (Y/Y A) δ 2 (Y/Y A), then for any X B such that cl 1 (X) = Y, it holds that d 1 (X) d 2 (X). 3. If δ 1 (Y/Y A) = δ 2 (Y/Y A) for any d 1 -closed or d 2 -closed Y B, then PG(B 1 ) = PG(B 2 ). Proof. 1. Then Y A is d 1 -closed in A 1, and since PG(A 1 ) = PG(A 2 ) also Y A is d 2 -closed in A 2, with d 1 (Y A) = d 2 (Y A). A closed set is self-sufficient, so A i [Y A] A i and d i (Y A) = δ i (Y A). 24