aa r X i v : . [ m a t h . L O ] M a r EQUIVALENCE RELATIONS AND DETERMINACY
LOGAN CRONE, LIOR FISHMAN, AND STEPHEN JACKSON
Abstract.
We introduce the notion of ( Γ , E )-determinacy for Γ a pointclass and E an equivalence relationon a Polish space X . A case of particular interest is the case when E = E G is the (left) shift-action of G on S G where S = 2 = { , } or S = ω . We show that for all shift actions by countable groups G ,and any “reasonable” pointclass Γ , that ( Γ , E G )-determinacy implies Γ -determinacy. We also prove acorresponding result when E is a subshift of finite type of the shift map on 2 Z . Introduction
For X = 2 ω or X = ω ω , E any equivalence relation on X , and Γ any pointclass (a collection ofsubsets of Polish spaces closed under continuous preimages, the reader can consult [8] and [3] for back-ground on the basic notions of descriptive set theory which we use throughout), there is a natural notionof ( Γ , E )-determinacy. Namely, this asserts that any A ⊆ X in Γ which is E -invariant is determined.Similarly, if G is a countable group, and we fix an enumeration of G , then there is a natural notion of( Γ , E )-determinacy for sets A ⊆ G or A ⊆ ω G in Γ (under the natural identification of 2 G with 2 ω viathe enumeration of G ). We will give a more general definition of ( Γ , E )-determinacy for arbitrary Polishspaces X and equivalence relations E on X in §
5. However, even the special cases just mentioned haverisen in various contexts. For example, when E is the Turing equivalence relation on 2 ω , then the questionof when Γ Turing-determinacy implies full Γ -determinacy has been an important question in modern logic.Harrington [2] showed that Σ Turing-determinacy is equivalent to Σ -determinacy. Woodin showed thatin L ( R ) Turing determinacy implies full determinacy. It is open in general for which pointclasses Γ wehave that Γ Turing-determinacy implies Γ -determinacy.In another direction, in recent years arguments involving Borel determinacy have had fruitful applicationsto the theory of Borel equivalence relations. The determinacy of Borel games is a fundamental result ofMartin [6], [7]. Despite the central significance of this result in modern logic, this result has until recentlyfound relatively few applications as a proof technique. Recently, however, Marks [5] uses Borel determinacyarguments to get lower-bounds on the Borel chromatic number χ B for free actions of free products of groups,in particular, the lower-bound that χ B (2 F n ) ≥ n + 1 for the chromatic number for the free part of theshift action (defined below) of the free group F n on the space 2 F n . See also [4] for a detailed account ofrecent advances in the theory of descriptive graph combinatorics including the use of Borel determinacy.There are currently no other proofs of this result, and so the introduction of determinacy methods into thesubject represents an important connection.In this paper we begin to investigate the general question of when ( Γ , E )-determinacy implies Γ -determinacy. Again, we formulate the notion of ( Γ , E )-determinacy more generally for arbitrary Polishspaces and equivalence relations in §
5. First, however, we investigate the special case where X = 2 G (or X = ω G ) and E is the equivalence relation induced by the shift action of G on X .One of our main results is that for any countable group G and any pointclass Γ satisfying some reasonableclosure properties (a “reasonable pointclass”) that ( Γ , E G )-determinacy implies Γ -determinacy. The proofpasses through a property of G which we call weak amenability which may be of interest elsewhere. In § § § § Γ , E )-determinacy and then in § E induced by subshifts of 2 Z of finite type and reasonable pointclasses Γ that ( Γ , E )-determinacy implies Γ -determinacy. The proof follows the outline of the first main theorem,but has extra complications involving the combinatorics of the subshift.2. Shift Actions
Let G be a countable group. The (left) shift action of G on S G is the action defined by g · x ( h ) = x ( g − h ).The cases of primary interest are when S = 2 = { , } and S = ω . In either of these cases, we refer to S G as the shift space, and note that the action of G on S G is continuous (with the usual product of thediscrete topologies on S G ). Let E G denote the equivalence relation on S G induced by the shift action.Let π : ω → G be a bijection, which we view as an enumeration of the group G , G = { π (0) , π (1) , . . . } .We also write g i for π ( i ) to denote the i th group element. The enumeration π induces a homeomorphism,which we also call π , between S ω and S G , namely π ( x ) = y where y ( g n ) = x ( n ). Definition 1.
Let Γ be a pointclass. Let G be a countable group and S = 2 or S = ω , and let E G be theshift equivalence relation on S G . We say ( Γ , E G )-determinacy holds if for all A ⊆ S G which are in Γ and E G -invariant we have that π − ( A ) ⊆ S ω is determined, for all enumerations π of G .Our main theorem will require a mild closure hypothesis on the pointclass Γ which we now state. Definition 2.
We say a pointclass Γ is reasonable if(1) Γ is closed under unions and intersections with ∆ sets.(2) Γ is closed under substitutions by ∆ -measurable functions.We note that all levels of the Borel hierarchy past the finite levels are reasonable, as are all levels of theprojective hierarchy.The next result is our main result connecting ( Γ , E G )-determinacy with full Γ -determinacy. Theorem 3.
For any countable group G and any reasonable pointclass Γ , ( Γ , E G ) -determinacy implies Γ -determinacy. The proof of Theorem 3 will involve a weak form of amenabilty of groups which we simply call weakamenability . We give this definition next.
Definition 4.
Let G be a countable group. We say G is weakly amenable if either G is finite, or if thereis an equivalence relation ∼ on G such that(1) G/ ∼ is infinite(2) ∀ g ∈ G ∃ b ( g ) ∈ N ∀ C ∈ G/ ∼ |{ C ′ ∈ G/ ∼ : gC ∩ C ′ = ∅}| ≤ b ( g )and an increasing sequence of finite sets A n ⊆ G/ ∼ , such that G/ ∼ = S n ∈ ω A n such that for any g ∈ G (1) lim n →∞ |{ C ∈ A n : gC ⊆ ∪ A n }|| A n | = 1We call an equivalence relation ∼ on a group G which satisfies both conditions (1) and (2) of Definition 4 appropriate . We note that the equivalence classes C in an appropriate equivalence relation need not befinite themselves, but in the definition of weak amenability, the sets A n are finite (i.e., they are finite setsof equivalence classes).We note that every amenable group G is weakly amenable. This follows taking ∼ to be the equalityequivalence relation on G . Note that the equality equivalence relation on G is an appropriate equivalencerelation (with b ( g ) = 1 for every g ∈ G ).We will prove Theorem 3 by showing two separate results, one of which is a purely algebraic resultconcerning weak amenability, and the other a pure game argument. The algebraic result is the following: Theorem 5.
Every infinite group has an infinite weakly amenable subgroup.
The game argument is given in the following theorem.
Theorem 6. If G is a countable group which has an infinite weakly amenable subgroup, then for everyreasonable pointclass Γ we have that ( Γ , E G ) -determinacy implies Γ -determinacy. QUIVALENCE RELATIONS AND DETERMINACY 3 Weak Amenability
In this section we establish that certain classes of groups are weakly amenable, including all the amenablegroups and free groups, and prove Theorem 5.The following two lemmas are easy and well-known.
Lemma 7. If G is a non-torsion group, then G has an infinite weakly amenable subgroup. Proof.
This is immediate as an element of infinite order generates an infinite cyclic subgroup, which isamenable and so weakly amenable. (cid:3)
Lemma 8. If G is locally finite then G is amenable, and so weakly amenable. Proof.
We may write G = S G n , an increasing union of finite subgroups. The G n can be used as Folnersets to witness the amenability of G . (cid:3) Proof of Theorem 5.
We may assume without loss of generality that G is an infinite countable group. ByLemma 7 we may assume that G is a torsion group. By Lemma 8 we may assume that G is not locallyfinite. Then G contains an infinite subgroup H = h F, g i generated by a finite subgroup F ≤ G and anelement g ∈ G of finite order. If suffices to show that any such group H is weakly amenable.Every element h ∈ H can be written (not uniquely) in the form h = f g a f g a · · · f n g a n where f i ∈ F and a i are positive integers less than the order of g . We call n the length of this representation of h . Welet | h | denote the minimum length of a representation of h . Note that | h − | ≤ | h | + 1 for any h ∈ H . Weeasily have that | h h | ≤ | h | + | h | and also | h h | ≥ || h | − | h || − ∼ be the equivalence relation on H given by h ∼ h iff | h | = | h | . Each equivalence class isfinite as F is finite and g has finite order. So, H/ ∼ is infinite. Let h ∈ H . By the above observations wehave that for any k ∈ H that | k | − | h | − ≤ | hk | ≤ | k | + | h | and we may take b ( h ) = 2 | h | + 2 to satisfy (2)of Definition 4. Thus ∼ is an appropriate equivalence relation on H .Let A n = { [ h ] ∼ : | h | ≤ n } , so | A n | = n + 1. For h ∈ H we have that { C ∈ A n : hC ⊆ ∪ A n } ⊇ { [ k ] ∈ A n : | k | ≤ n − | h |} , and so |{ C ∈ A n : hC ⊆ ∪ A n }| ≥ n − | h | + 1, and Equation 1 follows. (cid:3) The argument above for the proof of Theorem 5 in fact shows the following.
Theorem 9.
Every finitely generated group is weakly amenable.sketch of proof.
Assume G is infinite and finitely generated. Let S = { g , . . . , g n } be a finite generatingset for G . For g ∈ G , let | g | be the minimal length of a word representing g using the symbols g i , g − i , for g i ∈ S (we use only g i and g − i here, not other powers of the g i ). We define ∼ in the same way as before(i.e., g ∼ h iff | g | = | h | ). Note that G/ ∼ is still infinite with this modification. The rest of the argumentproceeds as before. (cid:3) In fact, as pointed out to us by Simon Thomas, this argument shows the following even more generalfact.
Theorem 10. If G is a countable group which has a Cayley graph with an infinite diameter, then G isweakly amenable. (cid:3) In particular, the free group F ω on infinitely many generators is weakly amenable. Question 11.
Is every countable group weakly amenable?4.
Proof of Theorem 3
In this section we use a game argument to prove Theorem 6, which in view of Theorem 5 impliesTheorem 3. For convenience we restate Theorem 6:
Theorem. If G is a countable group which has an infinite weakly amenable subgroup, then for everyreasonable pointclass Γ we have that ( Γ , E G ) -determinacy implies Γ -determinacy LOGAN CRONE, LIOR FISHMAN, AND STEPHEN JACKSON
Proof.
Let H ≤ G be an infinite weakly amenable subgroup. Let the equivalence relation ∼ on H and thethe sequence of sets { A n } n ∈ ω witness this. We recall here that the sets A n are finite subsets of the quotient H/ ∼ . Without loss of generality, assume that the sequence { A n } is such that lim n →∞ | A n +1 \ A n || A n +1 | = 1.Define H and H subsets of H by H = ∪ A ∪ [ n ∈ ω A n +2 \ A n +1 ! H = ∪ [ n ∈ ω A n +1 \ A n ! Let { g k H : k ∈ ω } enumerate the cosets of H in G . Let G I = S k ∈ ω g k H and G II = S k ∈ ω g k H . Clearly G I ∩ G II = ∅ and G = G I ∪ G II .Let A ⊆ X ω be in Γ . We’ll define an alternate payoff set ˜ A ⊆ X G which is E G -invariant and simulatethe game A by the game ˜ A in which player I makes moves corresponding to g ∈ G I and player II makesmoves corresponding to g ∈ G II (we assume the enumeration π satisfies π ( g ) is even for g ∈ G I , and π ( g )is odd for g ∈ G II ).We will define sets of rules, which if followed by both players will enforce that each player in the game ˜ A eventually specifies moves in A to play. First, we partition the positive even integers into infinitely manydisjoint subsequences n { c n,j } j ∈ ω o n ∈ ω , and the odd integers also into n { d n,j } j ∈ ω o n ∈ ω . Let B I n,j denote A c n,j \ A c n,j − and B II n,j denote A d n,j \ A d n,j − . Note that lim j →∞ | B I n,j || A cn,j | = 1, and likewise for B II n,j .We will have the players specify in ˜ A their n th move in A by playing more of those moves (by proportion)on the rounds corresponding to B I n,j (or B II n,j respectively). In order to successfully specify a move, theymust have that the limit as j → ∞ of the proportion of classes C for which all the moves in C are thesame goes to 1.Now we give the formal definition of the rules which will enforce the correct encoding of moves from A into ˜ A . R I n = ( x ∈ X G : ∃ m ∀ k lim j →∞ (cid:12)(cid:12)(cid:8) C ∈ B I n,j : ∀ h ∈ C x ( g k h ) = m (cid:9)(cid:12)(cid:12)(cid:12)(cid:12) B I n,j (cid:12)(cid:12) = 1 ) R II n = ( x ∈ X G : ∃ m ∀ k lim j →∞ (cid:12)(cid:12)(cid:8) C ∈ B II n,j : ∀ h ∈ C x ( g k h ) = m (cid:9)(cid:12)(cid:12)(cid:12)(cid:12) B II n,j (cid:12)(cid:12) = 1 ) We claim these rules are invariant. Suppose x ∈ R I n and g ∈ G . Let m witness the fact that x ∈ R I n ,and fix k ∈ ω . We want to show that m also witnesses g · x ∈ R I n , or in other words thatlim j →∞ (cid:12)(cid:12)(cid:8) C ∈ B I n,j : ∀ h ∈ C g · x ( g k h ) = m (cid:9)(cid:12)(cid:12)(cid:12)(cid:12) B I n,j (cid:12)(cid:12) = 1Let ℓ ∈ ω and h ′ ∈ H be so that g − g k = g ℓ h ′ and notice that, by definition of the shift, we areattempting to show that the following set is large. (cid:8) C ∈ B I n,j : ∀ h ∈ C g · x ( g k h ) = m (cid:9) = (cid:8) C ∈ B I n,j : ∀ h ∈ C x ( g − g k h ) = m (cid:9) = (cid:8) C ∈ B I n,j : ∀ h ∈ C x ( g ℓ h ′ h ) = m (cid:9) Now define S j by the formula S j = (cid:8) C ∈ B I n,j : ∀ h ∈ C x ( g ℓ h ′ h ) = m (cid:9) and T j by T j = B I n,j \ S j . It will suffice to show that T j is small compared to B I n,j as j → ∞ . To see this, recall that B I n,j = A c n,j \ A c n,j − and observe that each class C in T j is of one of the following three types: QUIVALENCE RELATIONS AND DETERMINACY 5 (1) C is moved by h ′ to intersect some class in A c n,j − ,(2) C is moved by h ′ to intersect some class outside A c n,j ,(3) or C is moved by h ′ to hit some other class C ′ which fails to specify m properly.Thus, T j ⊆ (cid:8) C ∈ B I n,j : ∃ C ′ ∈ A c n,j − h ′ C ∩ C ′ = ∅ (cid:9) ∪ (cid:8) C ∈ B I n,j : h ′ C * ∪ A c n,j (cid:9) ∪ n C ∈ B I n,j : h ′ C ⊆ ∪ B I n,j ∧ ∃ C ′ ( h ′ C ∩ C ′ = ∅ ∧ ∃ ˆ h ∈ C ′ x ( g ℓ ˆ h ) = m ) o . We want to show that lim j →∞ | T j | (cid:12)(cid:12) B I n,j (cid:12)(cid:12) = 0and we will do so by showing that the limit is 0 for each of the three sets above whose union contains T j .The first set has size at most b ( h ′ ) (cid:12)(cid:12) A c n,j − (cid:12)(cid:12) , which is small compared to (cid:12)(cid:12) B I n,j (cid:12)(cid:12) as j → ∞ . The secondset is small compared to | A c n,j | by the weak amenability hypothesis, and so also small compared to | B I n,j | .The third set has size at most b ( h ′− ) (cid:12)(cid:12) T ′ j (cid:12)(cid:12) , where T ′ j = n C ′ ∈ B I n,j : ∃ ˆ h ∈ C ′ x ( g ℓ ˆ h ) = m o , which is smallcompared to (cid:12)(cid:12) B I n,j (cid:12)(cid:12) as j → ∞ since m witnesses x ∈ R I n .Thus the rule sets R I n and R II n are all invariant.Now we define the payoff for the auxiliary game. Via the bijection π : ω → G we have a natural bijectionbetween X ω and X G . The auxiliary game is officially a subset of X ω , but we view it as a subset of X G with this bijection. Thus, for a position n in the game, the move y ( n ) is viewed as giving the value ˜ x ( π ( n )),where ˜ x ∈ X G is the function the players are jointly building. The payoff ˜ A ⊆ X G for player I in theauxiliary game is given by:˜ A = [ n ∈ ω \ i ≤ n R I i \ R II n ∪ \ n ∈ ω (cid:0) R I n ∩ R II n (cid:1) ∩ f − ( A ) ! where f is the following decoding function with domain T n ∈ ω (cid:0) R I n ∩ R II n (cid:1) . For ˜ x ∈ T n ∈ ω (cid:0) R I n ∩ R II n (cid:1) ,define f (˜ x )(2 n ) = m ⇔ ∀ k lim j →∞ (cid:12)(cid:12)(cid:8) C ∈ B I n,j : ∀ h ∈ C ˜ x ( g k h ) = m (cid:9)(cid:12)(cid:12)(cid:12)(cid:12) B I n,j (cid:12)(cid:12) = 1 f (˜ x )(2 n + 1) = m ⇔ ∀ k lim j →∞ (cid:12)(cid:12)(cid:8) C ∈ B II n,j : ∀ h ∈ C ˜ x ( g k h ) = m (cid:9)(cid:12)(cid:12)(cid:12)(cid:12) B II n,j (cid:12)(cid:12) = 1Since all the rule sets R I n and R II n are invariant, and the function f is invariant, ˜ A is invariant. We wantto show that whichever player has a winning strategy in the game ˜ A has a winning strategy for A .The rule sets R I n , R II n are easily Π if X is finite, and Σ if X = ω . This easily gives that ˜ A is theunion of a Σ set with the intersection of a Π set and f − ( A ). A simple computation gives that f is ∆ -measurable. Since Γ is reasonable, ˜ A ∈ Γ .Suppose now ˜ τ is a winning strategy for player II in ˜ A (the case for player I is similar but slightlyeasier). We will define a winning strategy τ for player II in A .We call a class C declared (at position p ) if C ∩ dom( p ) = ∅ . At any position p in the game ˜ A onlyfinitely many digits of the resulting real ˜ x ∈ X G have been determined, and thus only finitely many classes C have been declared. For each position p and declared class C relative to p , we have exactly one of thefollowing:(1) for some m ∈ X , for all moves p ( g ) so far played with g ∈ C we have p ( g ) = m , (we say in thiscase that C is a m -class )(2) or there moves p ( g ), p ( h ) played so far with g, h ∈ C for which p ( g ) = p ( h ). (we say in this casethat C is an invalid class ). LOGAN CRONE, LIOR FISHMAN, AND STEPHEN JACKSON
For every position p of the game, any class C is either undeclared, an m -class for some unique m , or aninvalid class. Over the course of the game, a class can change from an undeclared class to an m -class, andwill then either remain an m -class or become an invalid class at some point. Note that invalid classes cannever change. Thus the players’ progress towards following the rules can actually be measured at a finiteposition p .We say a ring B I n,j or B II n,j is declared relative to a position p if all of the classes C in this ring aredeclared relative to p . Consider one of the sets B I n,j (or B II n,j ). Suppose p is a position of the auxiliarygame and B I n,j is a declared ring. We say (relative to the position p ) that B I n,j is an invalid ring if at least1 /
10 of the classes C ∈ B I n,j are invalid at position p . We say B I n,j is an m -ring if at least 1 / C ∈ B I n,j are m -classes. If B I n,j is invalid at some position p , and q is a position which extends p , then B I n,j is also invalid with respect to q . If B I n,j (or B II n − j ) is declared but not an m -ring with respect to p ,then it is not an m -ring with resprct to any extention q of p .Consider the first round of the game A . Suppose I makes first move m in this game, and we define τ ( m ). Consider the set P m of positions p of even length in the auxiliary game ˜ A satisfying:(1) p is consistent with ˜ τ .(2) For every j , every class C ∈ B I ,j , and every g ∈ C ∩ dom( p ), we have that p ( g ) = m .First note that we cannot have a sequence p , p , . . . in P m with p n extending p n − for all n , and foreach n there is a j such that B II ,j is invalid with respect to p n but either not invalid or not declared withrespect to p n − . For otherwise the limit of the p n would give a run by ˜ τ for which I has satisfied the rule R I but II has not satisfied R II , contradicting that ˜ τ is winning for II . Let q ∈ P m be such that thereis no extension of q in P m which a new ring B II ,j becomes invalid. So, for all sufficiently large j and any q extending q , the ring B II ,j is not invalid. Likewise we cannot have a sequence q ⊆ q ⊆ · · · of postionsextending q such that for each n there is a j so that B II ,j is declared and not an m -ring (for any m ) at q n , but is not declared at q n − . For in this case each of these rings B II ,j would remain not m -rings in thelimiting run, which again violates R II . By extending q we may assume that for all sufficiently large j andall q extending q , B II ,j , if it is declared at q , is an m -ring for some m at q . Note that this ring will remainan m -ring for all further extension r of q , since it cannot change to become an invalid ring or an m ′ ring forany m ′ = m . Finally, a similar argument shows that we cannot have a sequence q ⊆ q ⊆ · · · extending q such that for each n there are two rings B II ,j and B II ,j ′ declared at q n but not declared at q n − with B II ,j an m -ring and B II ,j ′ an m ′ -ring with m = m ′ . By extending q further we may asssume that we havea declared m -ring B II ,j with respect to q , and such that for all extexions q of q and all j ′ > j , if B II ,j ′ isdeclared at q then it is also an m -ring We define τ ( m ) = m .In general, suppose I has played m , m , . . . , m k in A . Suppose inductively we have defined positions q ⊆ q ⊆ · · · ⊆ q k − . Let P m ,...,m i , for i ≤ k , be the set of positions p in ˜ A extending q i − such that forall g ∈ (dom( p ) \ dom( q i − )) ∩ B I i ′ ,j , for some i ′ ≤ i , we have p ( g ) = m i ′ . We inductively assume that forall sufficiently large j and any q extending q k − in P m ,...,m k − , and for any i < k , if B II i,j is declared at q then it is an m i +1 -ring, where m i +1 = τ ( m , . . . , m i ). We now consider extensions of q k − in P m ,...,m k .By the same arguments as above, there is a q k ∈ P m ,...,m k extending q k − such that for all large enough q and all extensions q of q k , if B II k,j is declared at q , then it is an m k +1 -ring for some fixed integer m k +1 .We let τ ( m , . . . , m k ) = m k +1 . This complete the definition of the strategy τ .To see that τ is winning, note that for any run x according to τ , we have a sequence of positions q , q , . . . which give us a run ˜ x consistent with ˜ τ which follows all the rules. From the definition of the q i we havethat f (˜ x ) = x . Thus since ˜ τ is winning in ˜ A , we know that ˜ x / ∈ f − ( A ), and so x / ∈ A , resulting in a winfor player II in the game A . (cid:3)
5. ( Γ , E ) -determinacy We now present a notion of ( Γ , E )-determinacy for more general equivalence relations on Polish spaces. QUIVALENCE RELATIONS AND DETERMINACY 7
Definition 12.
Let Γ be a pointclass and E an equivalence relation on a Polish space X . We say( Γ , E )-determinacy holds if for all continuous, onto π : ω ω → X and all A ⊆ X which are E -invariant Γ sets, we have that π − ( A ) is determined.First we note that the restriction that the coding maps π be onto is necessary to avoid trivialities. Forby the Silver dichotomy, for every Borel equivalence relation E on X with uncountably many classes, thereis a continuous map π : ω ω → X such that x = y implies ¬ π ( x ) E π ( y ). Given A ⊆ ω ω in some pointclass Γ ,let B ⊆ X be the E -saturation of π ( A ). Then B is an E -invariant subset of X which also lies in Γ for mostpointclasses (in particular for all pointclasses closed under ∃ ω ω or ∀ ω ω ). Since π − ( B ) = A we see that ifwe allow non-surjective maps π in Definition 12 then ( Γ , E )-determinacy trivially implies Γ -determinacy(even restricting the maps to be continuous).Another possible variation of Definition 12 would be to allow Borel onto maps π : ω ω → X . Althoughwe do not see that this version trivializes the notion, it seems more natural to require the coding maps tobe as effective as possible.A common situation is that we wish to impose a set of “rules” on the players in the game π − ( A ). Wenext show that a more general version of ( Γ , E )-determinacy which allows for rules imposed on the gameis in fact equivalent to ( Γ , E )-determinacy as in Definition 12. Definition 13.
Let T ⊆ ω <ω be a pruned tree (i.e., T has no terminal nodes). We say ( Γ , E, T )-determinacyholds if for every continuous onto map π : [ T ] → X and every E -invariant Γ set A ⊆ X , we have that game G ( π − ( A ) , T ) with payoff set π − ( A ) and rule set T is determined. Theorem 14.
For every pointclass Γ , for every equivalence relation E , we have that ( Γ , E ) -determinacyimplies ( Γ , E, T ) -determinacy for every T .Proof. Assume ( Γ , E )-determinacy and let T ⊆ ω <ω be a pruned tree, and let π : [ T ] → X be a continuous,onto map. Fix an E -invariant Γ set A ⊆ X . We define a continuous onto map π ′ : ω ω → X which extends π . Let x ∈ ω ω \ [ T ], and we define π ′ ( x ). Let s be the least initial segment of x with s / ∈ T . Let s ′ = s ↾ ( | s | − N s ′ ∩ π − ( A ) = ∅ and N s ′ ∩ π − ( X \ A ) = ∅ . Fix x s , y s in [ T ] with π ( x s ) ∈ A and π ( y s ) / ∈ A . If | s | − I was responsible for first violatingthe rules) then we set π ′ ( x ) = π ( y s ). Likewise, if | s | − π ′ ( s ) = π ( x s ). Next suppose that N s ′ ∩ [ T ] ⊆ π − ( A ) or N s ′ ∩ [ T ] ⊆ π − ( X \ A ). Note that the game is essentially decided at this point, soour intention is to ignore further violations of the rules. In this case let π ′ ( x ) = π ( ℓ ( x )) where ℓ : ω ω → [ T ]is a fixed Lipschitz continuous retraction of ω ω to [ T ]. We clearly have that π ′ is continuous and extends π . By the assumption of ( Γ , E )-determinacy, the game π ′− ( A ) ⊆ ω ω is determined. Say without loss ofgenerality that σ ′ is a winning strategy for I in π ′− ( A ). Let σ = ℓ ◦ σ ′ (as ℓ is Lipschitz, we may view σ ′ as defined on sequences s ∈ ω <ω ). We show that σ is winning for I in G ( π − ( A ) , T ). Since σ = ℓ ◦ σ ′ , I following σ will never first move off of the tree T . So we assume therefore II always moves in the tree T .Consider a run x of G ( π − ( A ) , T ) where I follows σ and both players move in T . If for every even n wehave that σ ′ ( x ↾ n ) = σ ( x ↾ n ), then x is also a run of σ ′ and so x ∈ π − ( A ). Suppose that there is a least(even) n so that σ ′ ( x ↾ n ) = σ ( x ↾ n ), that is ( x ↾ n ) a σ ′ ( x ↾ n ) / ∈ T . Let s ′ = x ↾ n and s = s ′ a σ ′ ( s ′ ). We cannotbe in the first case above (that is, both N s ′ ∩ π − ( A ) and N s ′ ∩ π − ( X \ A ) are non-empty), as otherwise π ′ ( x ) = y s ′ / ∈ A , and would be a loss of I in π ′− ( A ), a contradiction. In the second case we have either N s ′ ∩ [ T ] ⊆ π − ( A ) or N s ′ ∩ [ T ] ⊆ π − ( X \ A ). We cannot have that N s ′ ∩ [ T ] ⊆ π − ( X \ A ) since then N s ′ ⊆ π ′− ( X \ A ), which contradicts σ ′ being winning for I . So we have N s ′ ∩ [ T ] ⊆ π − ( A ) and so since σ = ℓ ◦ σ ′ , x ∈ N s ′ ∩ [ T ] ⊆ π − ( A ), and so I has won the run following σ . (cid:3) Subshifts of finite type
In this section we consider ( Γ , E )-determinacy where E is the equivalence relation corresponding to asubshift X ⊆ Z of finite type. Recall this means that there is a finite set of “forbidden words” w , . . . , w e ∈ <ω and X = { x ∈ Z : ∀ k ∈ Z ∀ ℓ ≤ e x ↾ [ k, k + | w ℓ | ] = w ℓ } , where | w | denote the length of the word w . LOGAN CRONE, LIOR FISHMAN, AND STEPHEN JACKSON C C Figure 1. a double loop
Theorem 15.
Let E be the shift equivalence relation on a subshift X ⊆ Z of finite type, and assume E has uncountably many classes. Then for all reasonable pointclasses Γ , ( Γ , E ) -determinacy implies Γ -determinacy.Proof. Let w , . . . , w e be the forbidden words of the subshift X . Fix N ≥ max {| w i | : 1 ≤ i ≤ e } . Let G bethe finite directed graph, the de Bruijn graph, corresponding to the forbidden words and N . That is, thevertices of G are elements of 2 N which don’t contain any forbidden words, and ( u, v ) is an edge in G iff v ↾ [0 , N −
1] = u ↾ [1 , N ].Throughout the rest of this section, G will denote this fixed de Bruijn graph (and not a countable group). Definition 16.
Let u ∈ G . We say u is good to the right there are uncountably many directed paths p = ( u, u , u , . . . ) in G starting from u . Likewise we say u is good to the left if there are uncountablymany ( . . . , u , u , u ) paths starting from u and moving in the reverse direction in G (i.e., ( u n +1 , u n ) is anedge in G ).If v is good to the right and there is a path u = u , u , . . . , u n = v from u to v in the graph G , then u is good to the right as well. Likewise in this case, if u is good to the left, then v is also good to the left.This simple observation will be used throughout.Note that an element of X can be identified with a bi-infinite path through G . Definition 17. A double loop in a directed graph G is a directed subgraph consisting of the union of twocycles with vertex sets C and C such that C ∩ C = ∅ and C = C . See Figure 1. Lemma 18. If G is a finite directed graph with uncountably many paths, then G contains a double loop. Proof.
Let u be a vertex in G for which there are uncountably many directed paths in G starting from u .Inductively define ( u , u , . . . , u n ), a directed path in G , such that there are uncountably many directedpaths in G starting from u n . We can clearly continue this construction until we reach a least n so that u n ∈ { u , . . . , u n − } (since G is finite). Let j ≤ n − u n = u j , Let p denote this directedpath ( u , u , . . . , u n ). Let ℓ = ( u j , u j +1 , . . . , u n − ) be the “loop” portion of p . There must be a vertex v = u k of ℓ such that there is an edge ( v , v ) in G where v = u k +1 and such that there are uncountablymany directed paths in G starting from v . For if not, then the only directed paths in G starting from v would be, except for a countable set, those which continually follow the loop ℓ . This is a contradiction tothe definition of v as there are only countably many such paths. If v ∈ p then we have a double loopin the graph. If not, then we repeat the process starting at v forming a path p = ( v , v , . . . , v m − , v m )where v m ∈ p ∪ ( v , . . . , v m − ). If v m ∈ p , then we have a double loop in G . Otherwise v m = v i for some i ≤ m −
2, and ℓ = ( v i , . . . , v m − ) gives another loop in G . Since G is finite, we must eventually producea double loop in G . (cid:3) Note that every vertex in a double loop is good to the right and left. Returning to the proof of thetheorem, by our assumptions, the de Bruijn graph G for the subshift has a double loop, and thus G has avertex which is good to the right and left.We now define the continuous onto map π : [ T ] → X , where T will be be a pruned tree on ω which wewill be implicitly defining as we describe π . Let x ∈ ω ω , and we describe the conditions on x which give QUIVALENCE RELATIONS AND DETERMINACY 9 p ℓ p ℓ · · · p n ℓ n Figure 2. x ∈ [ T ], and in this case describe π ( x ) ∈ X ⊆ Z . First, view every digit i as coding a binary sequence u i of length N (recall N was maximum size of the forbidden words, and was used to construct the de Bruijngraph G ). Fix a fast growing sequence 0 = b < b < · · · , with say lim i P j
0, except (for notational convenience) we let the B I n,j correspond to odd i , and the B II n,j to even i .In order to define the rule sets for the game we will make use of the following notion. Definition 19.
Given y ∈ X and a double loop ( C , C ) in G , and given integers a < b , we say y ↾ [ a, b ) traces the double loop with pattern s ∈ <ω if v = y ↾ [ a, a + N ) ∈ C ∩ C , and if v i = y ↾ [ a + i, a + i + N ),then the sequence v , v , . . . , v b − a − N +1 of nodes in G is a path in G of the form C ′ s (0) a C ′ s (1) a · · · C ′ s ( | s |− where C ′ , C ′ are the same cycles as C , C except we start at the vertex v . If | s − ( i ) | > | s − (1 − i ) | , wecall C i the majority loop and C − i the minority loop .We define the conditions R I n , R II n and the decoding function f : X → ω as follows. We fix an orderingon the cycles of G (the de Bruijn graph) as so write each double loop in G as ( C , C ) where C < C . This makes the representation of each double loop unique. In the definition of R I n (and likewise for R II n )we require that the following hold: y ∈ R I n ↔∃ i ∈ { , } ∀ ǫ > ∃ j ∀ j ≥ j ∃ double loop ( C , C ) such that the following hold:(2) ∃ [ a, b ) ⊆ B I n,j ( b − a ) > (1 − ǫ ) | B I n,j | and y ↾ [ a, b ) traces ( C , C ) with pattern s ∈ <ω | s − ( i ) | ≥ (1 − ǫ ) 23 | s || s − (1 − i ) | ≥ (1 − ǫ ) 110 | s | Likewise we define R II n using the B II n,j sets. If there is some direction so that every subword of y is goodin that direction and y ∈ T n R I n ∩ T n R II n , then we define the decoding map f at y by f ( y )(2 n ) is thewitness i ∈ { , } to y ∈ R I n , and likewise f ( y )(2 n + 1) is the witness to y ∈ R II n .To ensure f is well-defined, we note that if i witnesses y ∈ R I n , then 1 − i does not. This is becausefor small enough ǫ , for all large enough j if y traces ( C , C ) with pattern s over a subinterval I of length(1 − ǫ ) | B | of some fixed B = B I n,j then the double loop ( C , C ) is unique (for this j ). Say C i ( i ∈ { , } )is the majority loop. The majority loop C i is traced say M times, where M ≥ (1 − ǫ ) | s | many times.The minority loop is traced say m times, where m ≤ ( + ǫ ) | s | + ǫ | B | . The condition m ≤ M becomes ǫ | B | ≤ ( − ǫ ) | s | . Since | s | ≥ (1 − ǫ ) | B || G | it suffices to have ǫ < ( − ǫ ) (1 − ǫ ) | G | . Clearly this is satisfied for ǫ small enough. On the other hand, m ≥ (1 − ǫ ) | s | ≥ (1 − ǫ ) | B || G | . Any other loop can be traced only in B \ I , and so can be traced at most ǫ | B | many times. But (1 − ǫ ) | B || G | ≥ ǫ | B | holds for all small enough ǫ . So, there is a fixed ǫ , independent of j , so that if y ↾ B I n,j satisfies Equation 2 for this ǫ then the doubleloop ( C , C ) is well-defined as is the integer i ∈ { , } with C i being the majority loop.On the invariant set of y such that there is at least one direction so that all subwords are good in thatdirection we define ˜ A by: ˜ A = [ n ∈ ω \ i ≤ n R I i \ R II n ∪ \ n ∈ ω (cid:0) R I n ∩ R II n (cid:1) ∩ f − ( A ) ! This completes the definitions of π , and the auxiliary game ˜ A .We next observe that the auxiliary game ˜ A ⊆ Z is shift invariant. Given y ∈ Z , the case split as towhether there is a direction so that all all subwords of y are good in that direction is clearly shift invariant(and the set of such directions is also invariant). In the case where there is at least one such good direction,whether y ∈ ˜ A is decided by putting down the sets A ′ i for i ∈ Z \ { } , using these to define the sets B I n,j , B II n,j , then defining the sets R I n , R II n as in Equation 2 which gives the decoding function f and finally askingwhether f ( y ) ∈ A . It suffices to show that the sets R In , R II n are invariant (in that y ∈ R In iff m · y ∈ R In for all m ∈ Z ), as this implies that the decoding function f is also invariant. The intervals B I n,j ( m · y )as defined for the shift m · y are just the shifts m · B I n,j ( y ) of the corresponding sets B I n,j ( y ) for y . Inparticular, | B I n,j ( y ) ∩ B I n,j ( m · y ) | / | B I n,j ( y ) | tends to 1 as j goes to infinity. Thus the asymptotic conditionof Equation 2 holds for y iff it holds for m · y . The value of i in Equation 2 is therefore the same for both y and m · y . This shows that f ( y ) = f ( m · y ) and so y ∈ ˜ A iff m · y ∈ ˜ A .By the assumption of ( Γ , E )-determinacy the game π − ( ˜ A ) on ω is determined. First consider the casewhere I has a winning strategy σ in π − ( ˜ A ). Claim 20. If x (0) , x (1) , . . . , x (2 n ) is a position of the game π − ( ˜ A ) consistent with σ in which all of II ’smoves u x (2 k +1) , k < n , are good in both directions, then u x (2 n ) is good in both directions. Proof.
We first note that u x (2 n ) is not bad in both directions. If n > u x (0) is good inboth directions and the last move is legal, and so good in (at least) the direction pointing back to u x (0) .If n = 0, and u x (0) is bad in both directions, then there is no direction for which every subword of the QUIVALENCE RELATIONS AND DETERMINACY 11 resulting y is good, which is a loss for I , a contradiction. Suppose now that u x (2 n ) were bad in exactlyone direction, say bad to the right. If the definition of π we gave full control of all future moves to II (weignored I ’s moves after this point). But II can now play moves to violate R I and thus produce a loss for I , a contradiction. For example, II can move (for each of the two directions) to a cycle of G , and simplytrace this cycle forever. (cid:3) We will construct a position p of odd length which is consistent with σ , so that σ is committed to aparticular witness i to its following of R I n , which will be our first move. Fix ǫ small enough so that for alllarge enough j , if a double loop is traced in B I n,j meeting the conditions of Equation 2 for this ǫ , then thatdouble loop is unique.Now consider the tree of positions of odd length which(1) are consistent with σ ,(2) in which player II has made only moves which are good in both directions,(3) there is some j such that for the partial sequence y ↾ [ c, d ) constructed so far, we have [ c, d ) ∩ B I ,j = B I ,j and [ c, d ) ∩ B I ,j +1 = ∅ ,(4) and that y ↾ [ c, d ) doesn’t satisfy the existence of a double loop as in Equation 2 on B I ,j with ǫ = ǫ .This tree must be wellfounded, as a branch could be used to produce a loss for I which is consistent with σ , a contradiction. Let p ′ be a terminal position in this tree. Next, we consider the tree of positions ofodd length extending p ′ which(1) are consistent with σ ,(2) in which player II has made only moves which are good in both directions,(3) there is some j so that for the partial sequence y ↾ [ c, d ) constructed so far, we have [ c, d ) ∩ B I ,j +1 = B I ,j +1 and [ c, d ) ∩ B I ,j +2 = ∅ ,(4) if y ↾ [ c, d ) traces ( C , C ) in B I ,j and ( C ′ , C ′ ) in B I ,j +1 satisfying the conditions of Equation 2with ǫ = ǫ and the majority loop which y ↾ [ c, d ) traces in B I ,j is C i , then the majority loop y ↾ [ c, d )traces in B I ,j +1 is C ′ − i .Again, this tree must be wellfounded, since a branch would result in a y which fails to satisfy R I . Let p be any terminal node of this tree and let j be such that if y ↾ [ c , d ) is the portion of y constructed at p , then [ c , d ) ∩ B I ,j = B I ,j and [ c , d ) ∩ B I ,j +1 = ∅ . Notice that for any extension q of p whichis consistent with σ and in which II ’s moves are good in both directions, with y ↾ [ c, d ) the portion of y constructed at q , we will have that y ↾ [ c, d ) traces a double loop in each B I ,j for all j ≥ j , satisfying theconditions in Equation 2 with ǫ = ǫ . Furthermore, there will be some i ∈ { , } so that for any j ≥ j , if y ↾ [ c, d ) traces ( C , C ) in B I ,j , then C i will be the majority loop and C − i will be the minority loop.Our first move in the game A will be to play this i . Next suppose our opponent plays i . We will againconsider a tree of positions consistent with σ . We consider only positions of odd length extending p which(1) are consistent with σ ,(2) in which player II has made only moves which are good in both directions,(3) in each new B II ,j player II has moved as quickly as possible to the nearest double loop ( C , C )in G , and in B II ,j declares C i the majority loop and C − i the minority loop (say by tracing theminority loop sufficiently many times first, then the majority loop for the rest of the time) to satisfythe conditions in Equation 2,(4) for the partial sequence y ↾ [ c, d ) constructed so far, we have for some j that [ c, d ) ∩ B I ,j = B I ,j and[ c, d ) ∩ B I ,j +1 = ∅ ,(5) and that y ↾ [ c, d ) doesn’t satisfy the conditions of Equation 2 on B I ,j with ǫ = ǫ .This tree must be wellfounded, since a along a branch, II would be satisfying R II (with i ) but player I isn’t satisfying R I , which would be a loss consistent with σ , a contradiction. Let p ′ be a terminal positionin this tree. Next, we consider the tree of positions of odd length extending p ′ which(1) are consistent with σ ,(2) in which player II has made only moves which are good in both directions, (3) in each new B II ,j player II has moved as quickly as possible to the nearest double loop ( C , C ) in G , and in B II ,j declares C i the majority loop and C − i the minority loop to satisfy the conditionsin Equation 2,(4) there is some j so that for the partial sequence y ↾ [ c, d ) constructed so far, we have [ c, d ) ∩ B I ,j +1 = B I ,j +1 and [ c, d ) ∩ B I ,j +2 = ∅ ,(5) if y ↾ [ c, d ) traces ( C , C ) in B I ,j and ( C ′ , C ′ ) in B I ,j +1 satisfying the conditions of Equation 2with ǫ = ǫ and the majority loop which y ↾ [ c, d ) traces in B I ,j is C i , then the majority loop y ↾ [ c, d )traces in B I ,j +1 is C ′ − i .This tree also must be wellfounded, since I must commit to a particular i with which to satisfy R I , anda branch through this tree would have I changing its answer infinitely often, resulting in a loss consistentwith σ . Let p be any terminal position in this tree, and let i be the digit which I will declare in eachnew B I ,j from p onwards.We play the move i in A and continue playing in this manner, using each new move i n +1 by ouropponent to satisfy in the auxilliary game π − ( ˜ A ) an additional rule R II n , and then by moving to terminalpositions p n in wellfounded trees, fix digits i n which we will play in A . By the construction of thesepositions p n , we will have that if y is the resulting element of X corresponding to the sequence of moves i , i , · · · in A , then we will have f ( y )( n ) = i n , and that y ∈ T n ( R I n ∩ R II n ), so that our strategy is winningfor I in A .Consider now the case where II has a winning strategy τ in the game π − ( ˜ A ). The argument is similarto the case above where I had the winning strategy, so we just sketch the differences. By Lemma 18 thereis double loop in the graph G , and therefore there is a word of length N which is good in both directions.Have I play in ˜ A such a word as their first move u x (0) . Suppose I plays i in the game A . We proceedas in the argument above having I move as quickly as possible (only making moves which are good inboth directions) to a double loop within each B I ,j , and moving to encode i within B I ,j . An analogousclaim to Claim 20 shows that as long as I plays in this manner, τ ’s moves are also good in both direction.since I is satisfying R I (with the digit i ), a wellfoundedness argument as before will produce a position p consistent with τ and a digit i so that for all runs y consistent with τ in which I plays as just described wehave that y satisfies R II with i . Continuing in this manner, as previously, this defines a winning strategyfor II in A . 7. Conclusion
We have introduced the general notion of ( Γ , E )-determinacy for arbitrary equivalence relations E onPolish spaces and pointclasses Γ . Since the definition involves the use of continuous coding maps from ω ω onto X (which seems necessary to have a reasonable definition), it is not immediately clear to what extentthe structure of the particular Polish space X plays a role. Since all uncountable Polish spaces are Borelisomorphic, it is reasonable to ask the following: Question 21.
Suppose X , Y are uncountable Polish spaces, ϕ : X → Y is a Borel isomorphism, and E X is a Borel equivalence relation on X . Let E Y be the corresponding Borel equivalence relation on Y , thatis y E Y y iff ϕ − ( y ) E X ϕ − ( y ). Then for any pointclass Γ closed under substitution by Borel functions,countable unions and countable intersections, is it the case that ( Γ , E X )-determinacy is equivalent to( Γ , E Y )-determinacy?If E, F are equivalence relation on X with E ⊆ F , then we immediately have that ( Γ , E )-determinacyimplies ( Γ , F )-determinacy (because F -invariant sets are also E -invariant). Since the shift equivalence rela-tion E Z on 2 Z is a subset of the Turing equivalence relation on 2 Z , we have that ( Γ , E Z )-determinacy implies Γ Turing-determinacy for any Γ . We have shown that ( Γ , E Z )-determinacy implies full Γ -determinacy forany reasonable Γ . We recall that Harrington showed that Σ Turing-determinacy implies Σ -determinacyand Woodin showed that in L ( R ), Turing-determinacy is equivalent to full determinacy. Thus, extendingour results to more general equivalence relations is expected to be a difficult problem. Nevertheless, inTheorem 15 we extended the result to include subshifts of 2 Z of finite type. QUIVALENCE RELATIONS AND DETERMINACY 13
We recall that the Feldman-Moore theorem states that every countable Borel equivalence relation E ona Polish space is generated by the Borel action of a countable group G (one can also choose the Polishtopology to make the action continuous). We also recall the result [1] that every equivalence relation E generated by the action of a countable group G Borel (equivariantly) embeds into the shift action of G × Z on 2 G × Z . Theorem 6 applies to shift actions of arbitrary countable groups, so the problem of passingto general (not necessarily closed) subshifts embodies the general question of whether ( Γ , E )-determinacyimplies Γ -determinacy. In particular, we can ask: Question 22.
For which subshifts (closed, or more generally Borel, invariant subsets X of 2 Z with the shiftmap) of 2 Z do we have that ( Γ , E )-determinacy implies Γ -determinacy, where E is the shift equivalencerelation restricted to X .If E is generated by the continuous action of a countable group G on a compact 0-dimensional space X ,then [1] shows that ( X, E ) equivariantly and continuously embeds into a subshift of 2 Z × G . Thus, given apositive answer to Question 21, the question of whether ( Γ , E )-determinacy implies Γ -determinacy reducesto considering the question for subshifts of 2 G , for countable groups G .Aside from the observation above on subequivalence relations, it is not clear how the notion of ( Γ , E )-determinacy interacts with other aspects of the theory of Borel equivalence relations. So we ask: Question 23.
How does the notion of ( Γ , E )-determinacy interact with the notions of Borel reducibilityof equivalence relations, products of equivalence relations, increasing unions of equivalence relations, etc.? (cid:3) References
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Logan Crone, University of North Texas, Department of Mathematics, 1155 Union Circle
E-mail address : [email protected] Lior Fishman, University of North Texas, Department of Mathematics, 1155 Union Circle
E-mail address : [email protected] Stephen Jackson, University of North Texas, Department of Mathematics, 1155 Union Circle
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