Borel Equivalence Relations

§1 MATH 569 CLASS NOTES MATH 569 CLASS NOTES SIMON THOMAS' LECTURES ON SET THEORY James Holland

20191209

CAUTION: these notes were typed up during the lectures, and so are probably full of typos along with misheard or misread parts (and perhaps genuine misun derstandings).

Section 1. Polish Spaces, standard Borel spaces, and Borel equivalence relations

1 • 1. Deﬁnition A topological space X; T is Polish iff it admits a complete separable metric d. h i Notation: If X; T is a topological space, then B.T / is the algebra of Borel subsets of X. Normally when you talk about Borel equivalenceh i relations, you don’t care about the topology, but whether something is Borel. 1 • 2. Deﬁnition Let B be a algebra on a set X. Then X; B is a standard Borel space iff there is a Polish topology T on X such that B B.T /. h i D To what extent is the topology determined by the Borel subsets? The answer is not at all, by the following theorem, which has a proof to be found in Kechris’ book. 1 • 3. Theorem

Let X; T be a Polish space and let Y B.T /. Then there exists a Polish topology TY T on X such that h i 2 Ã B.TY / B.T / and Y is clopen in X; TY . D h i And this has a nice corollary. 1 • 4. Corollary If X; B is a standard Borel space, and Y B, then Y; BY is a standard Borel space. h i 2 h i Here, BY just means Z B Z Y . This follows just because we may make Y into a clopen set, and Y X being closed implies that¹ the2 correspondingW Â º metric is still complete. Â

Theorem 1 • 3 suggests that two standard Borel spaces should look alike. In fact, there is a unique uncountable one. 1 • 5. Theorem There is a unique uncountable, standard Borel space up to isomorphism.

1 • 6. Deﬁnition If X is a standard Borel space, then an equivalence relation E on X is Borel iff E is a Borel subset of X X. So what we care about is comparing the relative complexity of Borel equivalence relations. Three examples we can define straight away are the following.

1. If X is a standard Borel space, then the identity relation id X idX is Borel. D

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N 2. E0 is the Borel equivalence relation of 2 defined by

x E0 y x.n/ y.n/ for all by finitely many n N. $ D N 2 3. The Turing equivalence relation T on P .N/ 2 is defined by A T B iff A 6T B and B 6T A. And T is a Borel equivalence relation. Á D Á Á 1 • 7. Deﬁnition A Borel equivalence relation E is countable iff every Eclass is countable.

We want to compare the complexity of Borel equivalence relations by way of reduction. Now we don’t allow arbitrary maps in the sense of reduction, but instead focus on Borel maps. 1 • 8. Deﬁnition If X, Y are standard Borel spaces, a map f X Y is Borel iff graph.f / is a Borel subset of X Y . W ! There is an equivalent definition of Borel maps by the following theorem. 1 • 9. Theorem If X, Y are standard Borel spaces, and f X Y , then the following are equivalent: 1. f is Borel. W ! 2. f 1"Z is Borel in X for each Borel subset Z Y . Â 1 • 10. Deﬁnition Suppose E, F are Borel equivalence relations on the standard Borel spaces X, Y . 1. E is Borel reducible to F , written E 6B F , iff there is a Borel map f X Y such that W ! x E x0 f .x/ F f .x0/. $ 2. E and F are Borel bireducible, written E B F iff E 6B F and F 6B E. Á 3. E

For example, if X and Y are uncountable, standard Borel spaces, then idX B idY . As another example, we can define N N f _ _ Á _ _ _ a Borel reduction f 2 2 from id N to E0 by x x1 x2 x3 xn , because any difference W ! 2 7! between x and y occurs infinitely often in f .x/ and f .y/. As a challenge, find an explicit Borel reduction from E0 to T. Á The following is a theme of the course: Supposing that E and F are Borel equivalence relations, what techniques are available to prove that E 6B F ? Often a proof that E 6B F shows that if f is a Borel homomorphism from E to F , then there exists a “large6 set” which is mapped to a6 single F class. So the idea is that if the reason is E is too complicated, then you’d expect the “kernel” to be large, which is similar to this statement. So what we need is suitable notions of largeness, and in this course, we will use three.

• Category 1 • 11. Deﬁnition

If X is a topological space, then Z X is comeager iff there exist dense, open subsets Dn n ! such Â ¹ W 2 º that Tn ! Dn Z. 2 Â

So if Zn n ! are comeager sets, then Tn ! Zn is comeager. And this is supposed to motivate that this is a notion¹ ofW largeness.2 º 2

Now if we want to separate id2N and E0, i.e. that id2N

• Measure

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In particular, we look at probability measures. 1 • 12. Deﬁnition Suppose X; B is a standard Borel space. Then a Borel probability measure on X is a function B Œ0; 1 suchh that i W ! – . / 0, .X/ 1; ; D D – If An B are disjoint, then .S An/ P .An/ 2 n D n

Remark: if Zn n ! satisfy .Zn/ 1, then .Tn ! Zn/ 1. Note that this depends on the choice of measure. ¹ W 2 º D 2 D

• Martin’s measure 1 • 13. Deﬁnition For each A P .N/ 2N , the corresponding cone is 2 D DA B P .N/ A 6T B . D ¹ 2 W º

As you might expect by now, if Dn n ! are cones, then there exists a cone D Tn ! Dn. To see this, ¹ W 2 º Â 2 choose An P .N/ such that Dn DA . Then there exists an A P .N/ such that An 6T A for all n !. 2 D n 2 2 Clearly DA Tn ! DAn Tn ! Dn. Â 2 D 2 We get the following picture illustrating the limitations of these various notions of largeness. In particular, category is only useful in distinguishing E0 from id2N . Measure and ergodic theory isn’t able to deal with turing equivalence. It’s also not yet known whether E B T. 1 Á Á E Martin's measure 1

determinacy arguments T other notions of largeness ˚ Á

measure ergodic theory « etc. id2N category

˚ E0 1 • 14. Figure: Countable Borel Equivalence Relations

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Section 2. Countable Borel Equivalence Relations: The Feldman–Moore Theorems

So far we’ve learned the language, and it’s time we did more. As a matter of notation, write G Õ X to indicate a group action. 2 • 1. Deﬁnition Õ X If G X, then EG is the corresponding orbit equivalence relation: x EX y g G.g x y/. G $ 9 2 D Suppose is a countable group and X is a standard Borel space. An action Õ X is Borel iff for every , 2 the map x x is Borel. In this case, we say X is a standard Borel space. 7! X For example, if E has all of its classes as countable, then we can write E EZ for a suitable group action from Z. D X Of course, to do this naïvely, we will use choice. But Feldman–Moore tells us that E E for some and Borel group action. It turns out in the absence of choice that not every countable Borel equivalenceD relation can be realized by a group action from Z.

The theorem itself is the source of a great many of connections between other branches ofmath. 2 • 2. Theorem (Feldman–Moore Theorem) If E is a countable Borel equivalence relation on a standard Borel space X, then there exists a countable group and a Borel action Õ X such that E EX . D The proof of Feldman–Moore Theorem (2 • 2) will use a suitable uniformization theorem. Suppose X and Y are any sets, and P X Y . As a mater of notation, Â • x X has Px y Y x; y P ; 2 D ¹ 2 W h i 2 º • y Y has P y x X x; y P ; and 2 D ¹ 2 W h i 2 º • proj .P / x X y Y. x; y P/ . X D ¹ 2 W 9 2 h i 2 º 2 • 3. Deﬁnition

For P X Y , P P is a uniformization iff for all x projX .P /, there exists a unique y Y with x; y ÂP . Â 2 2 h i 2

So in essence, P is the graph of a function f where f projX .P / Y such that f .x/ Px holds. So AC implies every P X Y can be uniformized. W ! 2 Â Now suppose that X, and Y are standard Borel spaces, and P X Y is Borel. Does P necessarily have a Borel uniformization? The answer is that it need not have one.Â For example, consider the following, due toLuzin. 2 • 4. Result With P X Y and X, Y standard Borel spaces, if P has a Borel uniformization P , then proj .P / is Borel. Â X Proof Sketch .:.

Suppose that P is a Borel uniformization. Then projX P X is injective, and so the image is Borel (by a famous theorem of Luzin). But proj .P / proj .P /.W ! X D X a 2 • 5. Corollary There exists a Borel P X Y with X;Y standard Borel spaces, and with no Borel uniformization. Â So in general, Borel sets in the plane don’t have Borel uniformizations. But there are several cases where there are, as can be found (along with others) in Kechris’ book. This is due to Luzin–Novikov.

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2 • 6. Theorem

Suppose X, Y are standard Borel spaces and P X Y is a Borel subset such that Px is countable (perhaps empty) for all x X. Therefore, Â 2 1. projX .P / is Borel, and P has a Borel uniformization; and

2. Moreover, we can express P Sn ! Pn, where each Pn is the Borel graph of a partial function; i.e. if D 2 Pn.x; y/ and Pn.x; z/, then y z. D As an application of this, we get the following. 2 • 7. Corollary Suppose that X, Y are standard Borel spaces and f X Y is a countabletoone Borel map. Then im f is Borel, and there exists a Borel map g im f X suchW that!f .g.y// y for all y im f . W ! D 2 Proof .:. Apply Theorem 2 • 6 to P y; x x; y f . D ¹h i W h i 2 º a We will continue to use Corollary 2 • 7 with smoothness.

Section 3. Smooth Countable Borel Equivalence Relations

3 • 1. Theorem 6 If E is a Borel equivalence relation with uncountable many classes, then id2N B E.

3 • 2. Deﬁnition

A Borel equivalence relation is smooth iff E 6B idZ for some (equivalently every) uncountable standard Borel space Z.

We immediately get the following observation from this definition. 3 • 3. Result

If E is a smooth, countable, Borel equivalence relation on an uncountable Borel space, then E B id N . Á 2 We also get a nice, fairly easy theorem. 3 • 4. Theorem If E is a countable Borel equivalence relation on a standard Borel space X, then the following are equivalent. 1. E is smooth. 2. There exists a Borel set T X which intersects every Eclass in a unique point. (We say T is a Borel transversal for E.) Â 3. There exists a Borel map s X X such that s.x/ E x, and if x E y then s.x/ s.y/. (We say s is a Borel selector for E.) W ! D

Proof .:.

For (3) (1), clearly s is a Borel reduction from E to idX . !

For (1) (2), suppose f X Y is a Borel reduction from E to idY for some uncountable, standard Borel space Y!. So f is a countabletooneW ! map. Hence A im f is a Borel subset of Y , and there is an “inverse” g A X such that f .g.a// a for all a A. ThenDg is injective and so T im g is Borel and satisfies the desiredW ! property. D 2 D

For (2) (3), we can define a Borel selector by s.x/ y iff x E y and y T . ! D 2 a

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Note that this applies only to countable Borel equivalence relations. For example, there exists a smooth Borel equiv alence relation with no Borel transversal. To see this, let X, Y , Z be such that X and Y are standard Borel spaces while Z X Y is Borel with proj .Z/ nonBorel. Then Z is also a standard Borel space, and we can define a Borel Â X equivalence relation E on Z by x; u E y; v iff x y. The map x; u x is a Borel reduction from E to idX , and so E is smooth. But any proposedh i T h ZiwhichD is a Borel transversalh i 7! for E has x; u x as injective on T . Hence proj .T / proj .E/ proj .Z/Âis Borel, contradicting construction. h i 7! X D X D X Furthermore, there is a countable Borel equivalence relation which isn’t smooth. At this point, this is easiest to show using measure, although an argument using category can be given. 3 • 5. Theorem

E0 is not smooth

Proof .:. N N N Let be the usual uniform product probability measure on 2 . For each n N, let n 2 2 be the Borel n 2 W ! bijection x0; ; xn; x0; ; 1 xn; , flipping just the nth entry. h i 7!h i Õ N Let Ln ! Cn, where Cn n . Then 2 ; as a group of measurepreserving transformations. D 2 2N D h i h N i Also clearly E0 E . Furthermore, acts freely on 2 ; i.e. if 1 , then x x for all x X. D ¤ 2 ¤ 2 N N Suppose that E0 is smooth. Then there exists a Borel transversal T 2 . Since acts freely, 2 F "T . 1 Â D 2 This is because 1t1 2t2 implies 2 1t1 t2. So as it’s a transversal, t1 t2. But as it act’s freely, we get a contradiction. D D D

Since T is Borel, T is measurable. Since preserves , . "T/ .T / for all . Hence D 2 1 .2N / X . "T/ X .T /, D D D 2 2 which is a contradiction a Let’s return to Feldman–Moore Theorem (2 • 2). So more than just this, we have the following. 3 • 6. Theorem Feldman–Moore Theorem (2 • 2) holds, and moreover and Õ X can be chosen such that x E y x y or there exists 1 g with g2 1 such that g x y $ D ¤ 2 D D Proof .:. Let E be a countable Borel equivalence relationon a standard Borel space X. Clearly we can suppose that X is uncountable. (Otherwise, the theorem is trivial.) Applying Theorem 2 • 6, since E X X has countable Â sections, we can express E Sn ! Fn where each Fn is the Borel graph of a partial function. Out of this, we want to get a group action. D 2

1 1 For each n; m N, let Fn;m Fn F where F y; x x; y F . Then each Fn;m is the Borel 2 D \ m D ¹h i W h i 2 º graph of an injective partial function; and E S Fn;m. Let x x; x x X . D n;m D ¹h i W 2 º Claim 1 2 We can express X x Sp N Ap Bp where each Ap, Bp is a pair of disjoint Borel subsets. n D 2 Proof .:. Since X is Borel isomorphic to R, it suffices to work with R. But this is easy for R, since we just take sufficiently small discs with rational centers and radiuses. a

For each n; m; p, let Fn;m;p Fn;m .Ap Bp/. So we’re getting an injective function from a subset of Ap to a D \ subset of Bp, and we can get the inverse just by going back. Explicitly, Fn;m;p is the graph fn;m;p for some Borel

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bijection between disjoint Borel sets Dn;m;p and Rn;m;p. Hence we can define a corresponding Borel bijection gn;m;p by

8fn;m;p.x/ if x Dn;m;p ˆ 1 2 gn;m;p.x/

As a nice story, Thomas found this Popper’s superrigidity result on accident when searching something onGoogle, noticing that this was precisely what was needed for this idea about free groups. As he investigated who Popper was, he found that Popper was going to give a talk on the topic at UCLA, where Kechris works. And as it happened, Thomas was able to get to the result first.

Section 4. Hyperfinite, and Universal Borel Equivalence Relations

Why do we focus on countable Borel equivalence relations? One big reason is Theorem 3 • 6, suggesting connections with group theory, which yields lots of applications. In particular, the isomorphism relation on a class of (countable) structures. As it turns out, this is a countable Borel equivalence relation iff the structures are all “finitely generated” in a certain sensei. And then from here, it’s a matter of thinking about how complicated these relations can get. To do this, it’s useful to develop milestones. 4 • 1. Deﬁnition

A countable Borel equivalence relation E is universal iff F 6B E for every Borel equivalence relation F .

So by definition, for any two universal, countable Borel equivalence relation E, F , E B F . But this raises the question, does such a relation exist? Á

The following result by Friedman–Stanley shows that there is no universal Borel equivalence relation. 4 • 2. Theorem

If E is any Borel equivalence relation, then there exists a Borel equivalence relationEC such that E

So instead, consider the following definition. 4 • 3. Deﬁnition Õ Let F2 be the free group on 2 generators. Then E is the orbit equivalence relation of the Borel action F2 P .F2/ g 1 via S g"S g s s S . 7! D ¹ W 2 º F2 1 Equivalently, consider the shift action of F2 on 2 defined by . f /.x/ f . x/ for f F2 2. D W ! 1 Note that we have this to preserve that ı . f /.x/ .ı / f .x/, as otherwise ı . f /.x/ . f /.ı x/ f . ıx/ which might not be f .ı x/. D D D 1 These two actions are the same by the following: consider f S . Then . S /.x/ 1 iff S . x/ 1 iff 1 D D D x S iff x S. Hence S "S . 2 2 D iin the sense that fixing just finitely many elements gives only one transformation

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We have the following target theorem. 4 • 4. Theorem (Target Theorem) E is a countable Borel equivalence relation. 1 Now we look at a more general version of the shift action. 4 • 5. Deﬁnition Let X be a standard Borel space and G a countable group. Then define X G p p G X with the product Õ G D ¹ 1W W ! º Borel structure. The shift action G X is the Borel action g p.h/ p.g h/. The corresponding orbit equivalence relation is denoted E.G; X/. D ı

For example, E E.F2; 2/. 1 D 4 • 6. Proposition X 6 N Let G be a countable group and let X be a standard Borel Gspace. Then EG B E.G; 2 /.

Proof .:.

Let Un n ! be a sequence of Borel subsets of X which separates points, meaning for any two points ¹ W 2 º N G x y, there are n; m ! with x Un Um and y Um Un. Define the Borel map f X .2 / by ¤ 21 2 n 2 n W ! Œf .x/.g/.n/ 1 iff g .x/ Un. D 2 X N We claim that f is a Borel reduction from EG to E.G; 2 /. Mostly this consists in going through the definitions. Firstly, suppose that y h x for some h G. Then D 2 Œf .y/.h/.n/ 1 iff Œf .hx/.g/.n/ 1 D D 1 iff g hx Un 2 1 1 iff .h g/ x Un 2 1 iff Œf .x/.h g/.n/ 1. D Thus f .y/.g/ f .x/.h 1g/ and so f .y/ h f .x/. D D Conversely, suppose that f .y/ h f .x/ for some h G. Then for all n N, D 2 2 y Un iff Œf .y/.1/.n/ 1 2 D iff Œh f .x/.1/.n/ 1 D 1 iff Œf .x/.h /.n/ 1 D iff hx Un. 2 So as Un n ! separates points, y h x. ¹ W 2 º D a 4 • 7. Proposition N Let F! be the free group on countably many generators. Then E.F! ; 2 / is a universal countable Borel equivalence relation.

Proof .:. Let E be a countable Borel equivalence relation on a standard Borel space X. By Theorem 3 • 6, there exists a Õ X countable group and a Borel action X such that E is the corresponding orbit equivalence relation E . Õ g Let F! be a surjective homomorphism. Then we can define a Borel action F! X by x .g/ x. Clearly,W E !EX . So by Proposition 4 • 6, 7! D F! X N E E 6B E.F! ; 2 / D F! a

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4 • 8. Proposition Suppose that G and H are countable groups and H G is a surjective homomorphism. If X is any standard Borel space, then W ! E.G; X/ 6B E.H; X/ via the Borel reduction p p where p .h/ p..h//. 7! D The proof of this is left to the reader that wants to be complete. Intuitively, the result holdssince H is bigger than G. Similarly, although we require a bit of a trick this time, we get the following proposition. 4 • 9. Proposition Suppose G, H are countable groups and G 6 H. If X is a standard Borel space, then

E.G; X/ 6B E.H; X/.

Proof .:. G H Fix some x0 X and consider the Borel map f X X by p p where 2 W ! 7! ´p.h/ if h G p.h/ 2 D x0 if h G. … We claim that f is a Borel reduction from E.G; X/ to E.H; X/. Clearly if g p q for some g G, then g p q . What we want is the converse. D 2 D

Conversely, suppose that h p q for some h H. If h G, then clearly h p q. So suppose h H G. D 2 21 1 D 2 n In this case, we must have that for all g G, q.g/ p.h g/ x0 since h g G. This means q.h/ x0 for all h H, and so p q , and p 2q. D D … D 2 D D a Applying the (unproved) Target Theorem (4 • 4) and Proposition 4 • 9, we see that if G is a countable group with a free P .G/ nonabelian subgroup, then EG B E . Á 1 4 • 10. Proposition Z 0 If G is a countable group, then E.G; 2 / 6B E.G Z; 3/. n¹ º Proof .:. G Consider the Borel map f 2Z 0 3G Z by p p by W n¹ º ! 7! ´p.g/.n/ if n 0 p.g; n/ ¤ D 2 if n 2. D Z 0 Z 0 G We claim that f is a Borel reduction from E.G; 2 n¹ º/ to E.G Z; 3/. First, suppose that p; q 2 n¹ º and that g p q for some g G. Then it is easily checked that g; 0 p q . 2 D 2 h i D

Conversely, suppose that q g; n p for some g; n G Z. If n 0, then clearly q g p. Suppose D h i h i 2 1 D D n 0. Then for all h; m G Z, q.h; m/ p.g h; m n/. In particular, q.h/.n/ q.h; n/ p ¤.g 1h; 0/ 2, a contradiction.h i 2 D D D D a

N One of the issues with E.F! ; 2 / is that it’s basically impossible to visualize. We want to show E E.F2; 2/ is universal countable Borel instead by some reductions. 1 D 4 • 11. Proposition

Let G be a countable group and let C2 0; 1 be the cyclic group of order 2. Then D ¹ º E.G; 3/ 6B E.G C2; 2/

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Proof .:. Consider the Borel map f 3G 2G C2 where p p defined by W ! 7! 80 if p.g/ 0 ˆ D ˆ0 if p.g/ 1 and i 0 p.g; i/ < D D D 1 if p.g/ 1 and i 1 ˆ D D ˆ1 if p.g/ 2. : D G We claim that f is a Borel reduction from E.G; 3/ to E.G C2; 2/. First suppose p; q 3 and there exists 2 g G such that q g p. Then clearly q .g; 0/p. Next suppose that q .g; i/p. If i 0, then clearly q 2 gp, as you don’tD disturb the coding. D D D D So suppose i 1. First consider the case where there exists an h G such that q.h/ 1. Then q .h; 0/ 0 D 2 D D and q.h; 1/ 1. Note that for all a; j G C2, D h i 2 1 q.a; j / .g; 1/p.a; j / p.g a; j 1/. 1 D 1 D C 1 In particular, p.g h; 0/ q.h; 1/ 1 and so p.g h/ 2 while p.g h; 1/ q.h; 0/ 0, which is a contradiction. Thus q 0;D 2 G , and itD follows that q g pD. D D 2 ¹ º D a Now we can go through to prove Target Theorem (4 .• 4)

Proof of Target Theorem (4 • 4) .:. Let E be a countable Borel equivalence relation. Then N Z 0 E 6B E.F! ; 2 / B E.F! ; 2 n¹ º 6B E.F! Z; 3/ 6B E.F! Z C2; 2/ 6B E.F! ; 2/ Á since there is a surjection from F! to F! Z C2. And this Borel reduces to E.F2; 2/ since F! , F2. !

Recall that by HKL (Harris–Kechris–Louveau?), E0 is the

F0 F1 Fn Â Â Â Â of finite Borel equivalence relations on X such that E Sn ! Fn. D 2 Remark: If F is a finite Borel equivalence relation on X, then F is smooth. This is just because X is Borel isomorphic to R: there is a Borel linear order < of X, and so we can define a Borel selector s by taking s.x/ min.ŒxE /. D N Remark: E0 is hyperfinite. To see this, define Fn on 2 by x Fn y iff x.`/ y.`/ for all ` n. Then F0 F1 D Â Â Fn , and E0 Sn ! Fn. Â Â D 2 4 • 13. Open Problem

Suppose that E1 E2 En for n ! are hyperfinite Borel equivalence relations. Is E Sn ! En hyperfinite? Â Â Â Â 2 D 2

Despite looking trivial, this isn’t so easy. And Simon himself conjectures that it’s probably false. Consider the following characterization of hyperfinite Borel equivalence relations due to Dougherty–Jackson–Kechris. 4 • 14. Theorem

If E is a nonsmooth hyperfinite Borel equivalence relation, then E B E0. Á The new target theorem is the following:

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4 • 15. Theorem (Second Target Theorem) If E is a countable Borel equivalence relation on a standard Borel space X, then the following are equivalent: (a) E is hyperfinite. (b) There exists a Borel action Z Õ X such that EX E. Z D Proof of (a) implies (b) .:.

First we show that (a) implies (b). Express E Sn ! Fn as an increasing chain F0 F1 Fn of D 2 Â Â Â Â finite Borel equivalence relations such that F0 idX . Let < be a Borel linear ordering of X. Then we can define D an increasing sequence of Borel partial orderings

• <0 . D; • x

2. If C D are Fnclasses such that ŒC Fn 1 ŒDFn 1 , then either C

(iii) !, the reverse of !; or (iv) ! !, the order type of Z. C We obtain a Zaction by defining a Borel bijection T X X as follows. W !

Case i. Suppose ŒxE has order type 1 n < ! under

Case iv. Finally, if ŒxE has order type ! ! under

Hence we can restrict our attention to aperiodic equivalence relations, i.e. those equivalence relations such that every class is infinite. 4 • 16. Deﬁnition Let E be an aperiodic countable Borel equivalence relation. Then a vanishing sequence of markers is a decreasing sequence A0 A1 An such that Ã Ã Ã Ã 1. Each An is a complete Borel section for E, i.e. An ŒxE for all x X; and \ ¤ ; 2 2. Tn ! An . 2 D; 4 • 17. Lemma (The Marker Lemma) Every aperiodic countable Borel equivalence relation admits a vanishing sequence of markers.

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Proof .:. N N Without loss of generality, E is a relation on 2 . For each x 2 , and n !, let Sn.x/ be the lexiographicleast n n2 2 s 2 such that ŒxE Us is infinite, where Us f 2 s f . Then we define x An iff xn Sn.x/. 2 j \ j D ¹ 2 W D º 2 D Then An n ! is a decreasing sequence of Borel subsets which intersects each Eclass in infinitely many h W 2 i elements. Note that A Tn ! An intersects each Eclass in at most one element. Thus An A n ! is a vanishing sequence of markers.D 2 ¹ n W 2 º a Now we prove that (b) implies (a) from above.

Proof of (b) implies (a) from Second Target Theorem (4 • 15) .:. We want to show that if Z Õ X is a Borel action on a Standard Borel space, then E EX is hyperfinite. D Z So without loss of generality, E is aperiodic. Let T X X be a Borel bijection which generates the Zaction and let < be the Borel partial order on X defined byW ! x < y iff n > 0.T n.x/ y/. 9 D Then < gives a Zordering of every Eclass. By The Marker Lemma (4 • 17), let An n ! be a vanishing sequence of markers for E. Define ¹ W 2 º

Y x X there is an n ! such that An ŒxE has a < least or greatest element . D ¹ 2 W 2 \ º Then Y is an Einvarant Borel subset such that EY is smooth, because we have a Borel selector f Y Y W ! for E. Now for each n !, we can define a finite Borel equivalence relation Fn by 2 ` x Fn y iff x y x; y T .f .x// n ` n . D _ 2 ¹ W Ä Ä º Then F0 F1 Fn , and E Sn ! Fn. Thus E Y is hyperfinite. Ä Ä Ä Ä D 2

So without loss of generality, Y . Thus for each x X and each n !, An ŒxE is unbounded in both directions. Hence we can defineD; a finite Borel equivalence2 relation by 2 \

x Fn y iff x y An Œx; y . D _ \ D; Then F0 F1 Fn and E Sn ! Fn. Thus E is hyperfinite. Â Â Â Â D 2 a Now we get some closure operations on the class of hyperfinite Borel equivalence relations. 4 • 18. Theorem Let E, F be countable Borel equivalence relations on standard Borel spaces X, Y . (a) If X Y , and E F with F hyperfinite, then E is hyperfinite (trivial). D Â (b) If F is hyperfinite and E 6B F , then E is hyperfinite (nontrivial). (c) If E is hyperfinite and A X is Borel, then EA is hyperfinite (trivial). Â (d) If A is a complete Borel section for E and EA is hyperfinite, then E is hyperfinite (nontrivial). (e) If E and F are hyperfinite, then E F on X Y is hyperfinite (trivial) Proof .:.

(d) By Feldman–Moore Theorem (2 • 2), there exists a countable group G gn n ! and a Borel action Õ X D ¹ W 2 º G X such that E EG . For each n !, let n.x/ be the least n ! such that gn x A. Let D 2 2 2 E A Sn ! Fn where Fn n ! is an increasing sequence of finite Borel equivalence relations. D 2 ¹ W 2 º

Let En be the finite Borel equivalence relation such that x En y iff either x y or n.x/; n.y/ < n and D gn.x/ x Fn gn.y/ y. Clearly, E Sn ! En is hyperfinite. D 2 (b) Let f X Y be a Borel reduction from E to F . Since f is countabletoone, A f "X is a Borel subsetW of Y!, and there exists a Borel inverse g A X. By (c), F A is hyperfinite.D Also, since g is injective, it follows that B g"A is Borel, andW it! is clearly a complete section, since each x X is mapped to something in A, whichD is taken to another representative by g. Since EB F A, it follows2 that EB is hyperfinite. By (d), E is also hyperfinite. Š a

12 §4 MATH 569 CLASS NOTES

Recall Open Problem 4 • 13: if E Sn ! En is the union of hyperfinite Borel equivalence relations, is E hyperfinite? Consider a related idea using the dominationD 2 order. 4 • 19. Deﬁnition N 6 If f; g N , then f g iff f .n/ g.n/ for all but finitely many n N. Write f g iff f .n/ g.n/ for all but finitely2 many n, so that f Äg iff f 6 g and g 6 f . 2 D D DC

Note that is clearly a countable Borel equivalence relation, and it is easily checked that B E0. Observe that D N N D Á if fn n N N , then there exists a g N such that fn 6 g for all n !. To see this, just define ¹ W 2 º Â 2 2 g.`/ 1 maxn ` fn.`/. D C Ä Observe also that if E is a countable Borel equivalence relation on a standard Borel space X, and ' X NN is a Borel map, then there exists a map X NN such that W ! W ! • if x E y, then .x/ .y/; D • '.x/ .x/ for all x X. Ä 2 Note that we can’t ensure that is Borel unless E is smooth. 4 • 20. Theorem There exists a Borel map ' 2N NN such that there doesn’t exist a Borel map 2N NN satisfying W ! W ! • x E0 y implies .x/ .y/; D • '.x/ 6 .x/ for all x 2N . 2 The proof of this will be delayed until the next section. Suppose we have thismap '. When we look at , mapping to the Baire space, we are sending E0 things to identical things. Since E0 isn’t smooth, this shouldn’t be a Borel reduction. So this map must have a huge kernel in the sense that we get a counter example to the second property. To show this, we will use category. Why not use measure? Because it cannot work here.

We will make use of one of the only two theorems Simon knows from probability theory: the Borel–Cantelli Lemma below. 4 • 21. Lemma (Borel–Cantelli)

Suppose X; is a standard Borel probability space, and En X is a Borel subset for each n N. If h i Â 2 Pn N .En/ < , then 2 1 . x X x En for infinitely many n N / 0. ¹ 2 W 2 2 º D 4 • 22. Theorem Let X; be a standard Borel probability space. Let ' X NN be any Borel map. Then there exists a fixed h hNN suchi that W ! 2 x X '.x/ 6 h 1. ¹ 2 W º D Proof .:. .n 1/ For each n N, there exists an h.n/ N such that . x X '.x/.n/ > h.n/ / < 2 C . By Borel–Cantelli (4 • 21), 2 2 ¹ 2 W º . x X '.x/ 6 h / 1 ¹ 2 W º D a 4 • 23. Deﬁnition Let E be a countable Borel equivalence relation on a standard Borel space. Then E is BorelBounded iff for every N N 6 Borel map ' X N , there exists a Borel homomorphism X N from E to such that '.x/ .x/ for all x XW. ! W ! D 2 4 • 24. Theorem If E is hyperfinite, then E is BorelBounded.

13 MATH 569 CLASS NOTES §5

Proof .:. N Express E Sn ! Fn as the union of a chain of finite Borel equivalence relations. Define X N by D 2 W ! .x/.n/ max '.y/.n/ y Fn x . D ¹ W º Then is a homomorphism from E to , and '.x/.n/ .x/.n/ for all n ! D Ä 2 a 4 • 25. Open Problem Is every BorelBounded countable Borel equivalence relation hyperfinite? Does there exist a nonBorelBounded, countable Borel equivalence relation?

ii N N Remark: assuming Martin’s Conjecture, T isn’t BorelBounded . In fact, it’s enough to show that if 2 2 is Á W ! a Borel homomorphism from T to E0, then sends a cone to a single E0 class. Á Exercise 1

Suppose E and F are countable Borel equivalence relations. If E 6B F and F is BorelBounded, then E is also BorelBounded.

Section 5. Baire Category Methods

5 • 1. Deﬁnition Let X be a Polish space. • C X is comeager iff there are dense open subsets Dn for n < ! such that T Dn C . Â n ! Â • M X is meager iff X C is comeager. 2 Â n Recall the following theorem. 5 • 2. Theorem (Baire Category Theorem) If C is a comeager subset of a Polish space, then C is dense in X.

Usually, we only require C to be nonempty. 5 • 3. Deﬁnition Let X be a Polish space. Then A X has the Baire property (BP) iff there exists an open U X such that A U is meager. Â Â 4

Which sets have the Baire property? The following theorem tells us that we get the Borel sets together with themeager sets, in the sense that the σalgebra generated by these has the Baire property. 5 • 4. Theorem Let X be a Polish space and be the set of subsets of X with the Baire property. Then is the algebra generated by the open sets and the meager sets.

5 • 5. Corollary If A X is Borel, then A has the Baire property. Â

Using this and the following couple of results, it’s very easy to see that E0 isn’t smooth. 5 • 6. Theorem Suppose X, Y are Polish and f X Y is Borel. Then there exists a comeager C X such that f C is continuous. W ! Â

ii Martin's Conjecture is that the only homomorphisms from T to T are the jumps on cones. Á Á

14 §5 MATH 569 CLASS NOTES

Proof .:. 1 Let Un n < ! be an open basis for the toppology of Y . Since each f "Un is Borel, there exists an open ¹ W º 1 Vn such that Mn f "Un Vn is meager. Let Cn X Mn. Then C Tn ! Cn is comeager. Also 1 D 4 D n D 2 f "Un C Vn C . But this is saying precisely that f C is continuous. \ D \ a 5 • 7. Proposition Suppose that Õ X is a continuous action of a countable group on a Polish space X. If A X is comeager, then x X x A is comeager. Â ¹ 2 W Â º Proof .:. Õ 1 1 Since X is continuous, for each g , g A is also comeager. Hence B T g A is also comeager. 2 D g If x B, then x g 1A for all g . And so g x A for all g . 2 2 2 2 2 2 a

We next give a category proof of the fact that id2N

Case 2. Otherwise let an n ! be the increasing enumeration of A. Then we define '.x/.n/ an. ¹ W 2 º D Claim 1 N N For each h N , Dh x 2 '.x/ 6 h is comeager. 2 D ¹ 2 W 6 º

15 MATH 569 CLASS NOTES §5

Proof .:. For each m !, 2 h N Dm x 2 n m.'.x/.n/ > h.n// D ¹ 2 h W 9 º is clearly dense open. Hence Dh Tm ! DM is comeager. Ã 2 a

N N Suppose such a map 2 N exists, i.e. a Borel homomorphism from E0 to idNN . Then there exists a N W ! N comeager C 2 and a fixed h N such that .x/ h for all x C . Also, Dh C is comeager; and so Â 2 D 2 \ there exists an x Dh C . But then '.x/ 6 h .x/, a contradiction. 2 \ 6 D a The next target theorem is that category isn’t useful for anything else. 5 • 10. Theorem (Third Target Theorem) If E is a countable Borel equivalence relation on a Polish space X, then there exists a comeager Einvariant Borel subset C X such that EC is hyperfinite. Â Fortunately, this is false using measure instead of category. To prove this, we will use the Kuratowski–Ulam theorem. 5 • 11. Deﬁnition

ith each A X, we associate the property A.x/ iff x A, and we write xA.x/ iff A is comeager. Here ‘ ’ is sometimesÂ known as a category quantifier. 2 8 8

In essence, the Kuratowski–Ulam theorem says that category quantifiers commute: x X y Y.A.x; y// iff y Y x X.A.x; y//. 8 2 8 2 8 2 8 2 5 • 12. Theorem (Kuratowski–Ulam) Suppose that X, Y are Polish and A X Y has the Baire property. Therefore Â y A is comeager iff x X.Ax is comeager/ iff y Y.A is comeager/. 8 2 8 2 Despite looking rather unimportant, the usefullness of this theorem is immense. 5 • 13. Deﬁnition Let E be a countable Borel equivalence relation on a Polish space X. Then a cascade is a sequence

X S0 S1 S2 Sn D Ã Ã Ã Ã Ã of Borel complete sections for E together with Borel retractions fn Sn Sn 1. (Here retraction means W ! C fn Sn 1 idSn 1 and fn.x/ E x for all x Sn.) C D C 2

Given a cascade Sn; fn n ! , we can define a sequence of Borel equivalence relations En by x En y iff ¹ W 2 º fn fn 1 ::: f0.x/ fn fn 1 ::: f0.y/. Then each En is smooth; and E0 E1 En E. ı ı ı D ı ı ı Â Â Â Â Â 5 • 14. Lemma

With the above hypothesis, if each fn is finitetoone, then E! Sn ! En is hyperfinite. D 2 So the plan will be to construct a suitable cascade such that for most elements of the space, the union of these equivalence relations is the whole space E.

Proof of Third Target Theorem (5 • 10) .:. 2 Applying Theorem 3 • 6, let gn n ! be a sequence of Borel bijections. gn X X with g 1 such that ¹ W 2 º W ! n D g0 idX and x E y iff n.gn.x/ y/. Also, fix some Borel linear ordering < of X. For each Borel subset D 9 S D S X, and each n !, let Fn be the Borel equivalence relation on S defined by Â 2 S x Fn y iff x y gn.x/ y. S D _ D As involutions, this is symmetric. Here each Fn class has at most 2 elements. Hence the set ˆn.S/ of <minimal S elements of each Fn class is also Borel. Also, if S is a complete section, so is ˆn.S/.

16 §5 MATH 569 CLASS NOTES

S Let f S ˆ .S/ be the Borel map which sends each x S to the <minimal element of Œx S . The n n Fn strategy forW the! rest of the proof is to define for each element2 of Baire space, there is a a corresponding cascade that uses these ingredients, and then use Kuratowski–Ulam (5 • 12).

N ˛ ˛ For each ˛ N , we define a cascade Sn ; fn n ! by ˛ 2 ¹ W 2 º • S0 X; ˛ D ˛ • Sn 1 ˆ˛.n/.Sn /; and C D ˛ • f ˛ f Sn . n D ˛.n/ ˛ ˛ ˛ Let En n < ! be the corresponding sequence of finite Borel equivalence relations and let E! Sn ! En E be¹ theW correspondingº hyperfinite Borel equivalence relation. The theorem follows fromD the 2followingÂ seriesof claims. Claim 1 There exists an ˛ NN and a comeager Einvariant Borel C X such that EC E˛C . 2 Â D ! In fact, we prove the stronger claim below. Claim 2 N ˛ N x X .ŒxE Œx ˛ /. 8 2 8 2 D E! By Kuratowski–Ulam (5 • 12), it is enough to show the following claim instead. Claim 3 N x X ˛ N .ŒxE Œx ˛ /. 8 2 8 2 D E! Proof .:.

Fix some x X. It is enough to show that for each y ŒxE , 2 2 N ˛ ˛ A ˛ N y ŒxE! [ŒxEn D ¹ 2 W 2 D n ! º 2 is dense open, since then the countable intersection is comeager. Claim 4 A is open.

Proof .:. Suppose that ˛ A. Then there exists an n ! such that y Œx ˛ . It follows that 2 2 2 En N N˛n 1 ˇ N ˇn 1 ˛n 1 A. C D ¹ 2 W C D C º Â is a neighborhood of ˛, and so A is indeed open. a Claim 5 A is dense.

Proof .:. n 1 Fix some basic open Ns where s N . Consider the “finite cascade” defined by 2 C S0; f0;S1; f1; ;Sn; fn;Sn 1. C Where these are the functions determined by the initial values. Let x0 fn f0.x/ and y0 D ı ı N D fn f0.y/. Since x0 E y0, there exists a k ! such that gk.x0/ y0. Let ˛ N be such that s ı˛ and ı ˛.n 1/ k. Then ˛ A. And so A2 is also dense. D 2 Â C D 2 a

17 MATH 569 CLASS NOTES §6

Section 6. Measure theoretic methods

Question: suppose that G is a countable group and G Õ X is a Borel action on a standard Borel space. Does the X complexity of EG reflect the complexity of G? To some extent, there is some truth here. For example, if G Z, we can only get hyperfinite things. Unfortunately, there is an easy counterexample. So let G be any countableD group, and let G Õ G Œ0; 1 be the Borel action G Œ0;1 g h; r gh; r . Then the Borel map g; r 1G ; r shows that E is smooth. h i D h i h i 7! h i G So what is wrong with this, and how can we eliminate this kind of triviality? One issue with the above idea is that the action is free, but also there is no invariant probability measure. 6 • 1. Proposition Suppose G is a countable group and G Õ X is a free Borel action on a standard Borel space X. If there exists a X Ginvariant, Borel, probability measure on X, then EG isn’t smooth.

Proof .:. X Õ Suppose that EG is smooth. Then there exists a Borel transversal T X. Since G X is free, X Fg G g"T . Â D 2 So 1 Pg G .g"T/ Pg G .T /, which is a contradiction. D 2 D 2 a

6 • 2. Deﬁnition Let G be countable and let X be a standard Borel Gspace. A Ginvariant probability measure is ergodic iff whenever A X is a Ginvariant Borel subset, then .A/ 1 or .A/ 0. Â D D We have another characterization due to the following theorem. Note that f X Y is Ginvariant iff f .g x/ f .x/ for all x X and g G. W ! D 2 2 6 • 3. Theorem If is a Ginvariant, Bore, probability measure, then the following are equivalent: 1. is ergodic; 2. If Y is a standard Borel space, and f X Y is Ginvariant, then there is a Ginvariant Borel M X with .M / 1 and such that f M isW constant.! Â D Proof .:. (2) (1). Suppose A X is a Ginvariant, Borel subset. Define f X 2 by f .x/ 1 iff x A. Since ! A is Ginvariant,Â f is Ginvariant. Hence there exists a GinvariantW ! Borel M D X with2.M / 1 such that f M is constant. So .A/ 1 or .A/ 0. Â D D D (1) (2). Without loss of generality, Y Œ0; 1, since we can expand to an uncountable space if necessary, and ! D it will be isomorphic to Œ0; 1. Let Z0 Œ0; 1. Suppose inductively that we have defined an interval D Ä an an 1 Zn n 1 ; nC1 D 2 2 n 1 1 for some 0 an < 2 such that .f "Zn/ 1. Let Ä D Ä2an 2an 1 Â2an 1 2an 2 In 1 ; C Jn 1 C ; C . C D 2n 2n C D 2n 2n 1 1 1 Then f "In 1 and f "Jn 1 are Ginvariant, Borel subsets. And so either .f "In 1/ 1 or 1 C C C D .f "Jn 1/ 1. In the former case, let Zn 1 In 1; in the latter, set Zn 1 Jn 1, the closure C D C D C C D C 18 §6 MATH 569 CLASS NOTES

of Jn 1. C 1 Clearly, M Tn ! f "Zn is a Ginvariant, Borel subset with .M / 1, and such that f M is constant. D 2 D a Now we want something stronger than merely being ergodic. 6 • 4. Deﬁnition Let G be a countable group, and let X be a standard Borel Gspace. The action G Õ X is uniquely ergodic iff there is a unique Ginvariant probability measure on X.

Clearly, we should have the following. 6 • 5. Theorem If G Õ X; is uniquely ergodic, then G Õ X; is ergodic. h i h i Proof .:. Suppose that G Õ X; isn’t ergodic. We need to find two different probability measures. Then there existsa Borel A X such thath 0i < .A/ < 1, and 0 < .X A/ < 1 with both A and X AGinvariant. Then we can Â n n define Ginvariant probability measures 1 2 by ¤ .Y A/ 1.Y / \ D .A/ .Y .X A// 2.Y / \ n D .X A/ a n 6 • 6. Example (Abstract Example) Let K be a separable, compact group. Then there exists a unique probability measure on K such that is K invariant under the lefttranslation action of K Õ K; namely, the Haar measure. If < K is a countable, dense subgroup, then Õ K; is uniquely ergodic. h i 6 • 7. Example (Concrete Example)

Let K Qn ! Cn where each Cn 0; 1 of order 2. Then K is compact and the Haar measure is the usual D 2 D ¹ º K uniform product probability measure. Let Ln ! Cn, which is a dense subgroup. Then E E0 and Õ K; is uniquely ergodic. D 2 D h i

This then gives a theorem about E0, which is a third proof that E0 isn’t smooth. In some sense, this states that the kernel is large in terms of measure, whereas last time we proved that the kernel is large in terms of category. 6 • 8. Theorem N N N Let be the usual product measure on 2 . If f 2 2 is a Borel homomorphism from E0 to id2N , then there N W ! exists an E0invariant Borel subset M 2 with .M / 1 such that f M is constant. Â D We often want every infinite subgroup H < G to act ergodically. Here’s a nonexample of this happening. 6 • 9. Example (Non-example) 1 Let K Qn ! Cn be as above. Let Ln 1 Cn and H Qn 1 Cn. Then H is invariant and .H / 2 , since it’sD an index2 2subgroup. D D D

19 MATH 569 CLASS NOTES §6

6 • 10. Deﬁnition Let G be a countable group and let X be a standard Borel Gspace with invariant probability measure . Then Õ G X; is strongly mixing iff for any Borel subsets A; B X, and any sequence gn n ! of distinct elementsh of iG, either .B/ 0 or Â h W 2 i D .gn"A B/ lim \ .A/. n .B/ D !1

Note that if .B/ 0 we still have limn .g"A B/ 0 .B/ .A/. Note further that the literature often has “weakly mixing”D in addition to “strongly!1 mixing”,\ so oneD mustD be careful with which version is referred to if just “mixing” is used.

With the above hypotheses, if H 6 G is an infinite subgroup, then H Õ X; is also strongly mixing. It also follows that if you’re strongly mixing, then you’re ergodic, just by taking A Bh for aiGinvariant A. D 6 • 11. Theorem If G Õ X; is strongly mixing, then G Õ X; is ergodic. h i h i Proof .:.

Suppose A X is Borel and Ginvariant. Let G gn n ! . Thus Â 2 D ¹ W 2 º .A/ lim .gn"A A/ lim .A/ .A/. D n \ D n D !1 !1 This implies .a/ is either 0 or 1. . a

As a convention, if G is countably infinite, then the uniform product probability measure on 2G is denoted by . Clearly G Õ 2G ; is measurepreserving. This is usually called the Bernoulli action. h i 6 • 12. Theorem G Õ 2G ; is strongly mixing. h i Proof .:. First suppose that there exist finite S;T G and subsets F 2S , H 2T such that Â Â Â A f 2G f S F B f 2G f T H . D ¹ 2 W 2 º D ¹ 2 WÕ 2 º Suppose gn n ! is a sequence of distinct elements of G. Since G G is free, for all but finitely many h W 2 i n !, gn.S/ T ; and so the events gn.A/ and B are independent: 2 \ D; .gn"A B/ \ .gn"A/ .A/. .B/ D D G Thus the limit limn .gn"A B/ .A/.B/. In general, if C 2 is Borel, then for every " > 0, there exists a finite S !1G and an F \ 2S suchD that Â Â Â C ®f 2G f S F ¯ < ". 4 2 W 2 The result follows easily. a 6 • 13. Deﬁnition If G is countably infinite, then G G .2/ x 2 g G 1G .g x x/ D ¹ 2 GW 8 2 n ¹ º ¤ º Õ G .2/ is the free part of G 2 . We define EG E . D G Fortunately, we keep the probability measure around. 6 • 14. Proposition ..2/G / 1, where is the uniform product probability measure on 2G . D

20 §6 MATH 569 CLASS NOTES

Proof .:. It is enough to show that for each 1 g G, ¤ 2 . x 2G g x x / 0. ¹ 2 W D º D Let H g be the subgroup generated by g. Then g x x iff x is constant on each coset Ht. The result followsD easily. h i D a

Note that if H 6 G, then EH 6B EG . To see this, we can define a Borel reduction x x by 7! ´x.g/ if g H x.g/ 2 D 0 otherwise. Here’s a question: to what extent does the converse hold? The answer is that it doesn’t hold in general, since it’s obviously false for hyperfinite things, but we will move to a setting where the it is almost true. In particular, wewill move to the world where we can apply Popa superrigidity, and in particular, groups with a normal, Kazhdan group. 6 • 15. Deﬁnition A countable Borel equivalence relation E on a standard Borel space X is free iff there is a free Borel action G Õ X of a countable group such that E EX . D G

For example, EG is clearly free. 6 • 16. Deﬁnition

E is essentially free if there exists a free countably Borel equivalence relation F such that E 6B F .

Z For example, take .2/ x0 , since we can just add a few elements to make sure the action on x0 is free. A question is now brought up: is everythingt ¹ º essentially free? In particular, is E essentially free? This was answered by Simon using an easy consequence of Popa Superrigidity. First some definitions.1 6 • 17. Deﬁnition Let E be a countable Borel equivalence relation on a standard Borel space X. A probability measure on X is Õ X Einvariant if for some (equivalently every) Borel action G X of a countable group such that E EG , is Ginvariant. D

6 • 18. Deﬁnition Suppose E, F are countable Borel equivalence relation on X, Y , and is an Einvariant probability measure. Then a Borel homomorphism f X Y from E to F is tivial iff there exists a Borel Z X with .Z/ 1 such that f sends Z to a single FW class.! Â D

6 • 19. Deﬁnition

If G, H are countable groups, then a homomorphism G H is a virtual embedding iff ker < ℵ0. W ! j j The following is an easy consequence of Popa Superrigidity. 6 • 20. Theorem (Black Box)

Let S be any countable group and let G SL3.Z/ S. Let H be any countable group and let HDÕ Y be a free Borel action on a standard Borel space Y . Y Therefore, if there exists a nontrivial Borel homomorphism from EG to EH , then there exists a virtual embedding from G to H .

Now we give a theorem that shows how we use this, as we are not yet prepared to prove it yet. 6 • 21. Theorem

If E is an essentially free, countable Borel equivalence relation, then there exists a countable G such that EG 6B E. 6 6 • 22. Corollary E isn’t essentially free. 1

21 MATH 569 CLASS NOTES §6

In fact, this says that of the essentially free Borel equivalence relations, there isn’t a universal one. To prove Theorem 6 • 21, we will make use of two grouptheoretic results. 6 • 23. Theorem (B.H. Neumann) There exist uncountably many finitely generated groups up to isomorphism.

6 • 24. Proposition If L is any group, then the free product L Z has no nontrivial, finite, normal subgroups. Proof of Theorem 6 • 21 .:. Y Õ Without loss of generality, we can suppose that E EH for some free Borel action H Y . Since there are uncountably many finitely generated groups. DUp to isomorphism, there exists a finitely generated L such that L doesn’t embed in H, since H has only countably many finitely generated subgroups. Let S L Z. Then S D has no nontrivial finite normal subgroups. Finally, let G SL3.Z/ S. D G Y Y Suppose f .2/ EH is a Borel reudction from EG to EH E. Then f is a nontrivial Borel homomor phism. HenceW by Black! Box (6 • 20), there exists a virtual embeddingD G H. Since ker 1, it follows that S embeds in H ; and hence L embeds in G, a contradiction. W ! D a 6 • 25. Theorem There exist uncountably many (continuum) free, countably Borel equivalence relations up to Borel bireducibility.

In fact, we have a rich supply of nonessentially free ones. 6 • 26. Theorem There exist uncountably many nonessentiallyfree countable Borel equivalence relations up to Borel bireducibility.

First we prove Theorem 6 • 25. What examples of essentially free things do we know? So far, EG . One issue we encounter is to make sure our coding S from Black Box (6 • 20) doesn’t embed in SL3.Z/. To proceed, we will make use of the following theorem 6 • 27. Theorem

SL3.Z/ contains a torsionfree subgroup of finite index.

Proof .:. To quote a famous theorem, by Selberg’s theorem, every finitely generated, linear group over a field of charac teristic 0 contains a torsionfree subgroup of finite index. a One last thing we require (which it just so happens is something we can prove) is the following. 6 • 28. Lemma If H;K 6 G are any groups, then ŒK K H 6 ŒG H. W \ W Proof .:.

Let ti i I be coset representatives for K H in K. It is enough to show that if i j , then Hti Htj . ¹ W 2 º \ 1 ¤1 ¤ Suppose Hti Htj . Then there exist a; b H such that ati btj and so ti t a b H K. So that D 2 D j D 2 \ .H K/ti .H K/tj , a contradiction. \ D \ a

Proof of Theorem 6 • 25 .:.

Let P be the set of primes. For each p P, let Ap Ln ! Cp be the direct sum of countably many copies of 2 D 2 the cyclic group Cp of order p. 6 For each C P, let GC SL3.Z/ Lp C Ap. We will show that EGC B EGD iff C D, which is certainly enough to getÂ uncountablyD many of˚ these.2 Â

22 §6 MATH 569 CLASS NOTES

6 6 6 If C D, then GC GD, and so EGC B EGD . Conversely, suppose EGC B EGD . Then there exists a virtualÂ embedding (by Black Box (6 • 20))

SL3.Z/ M Ap SL3.Z/ M Aq. W ! p C q D 2 2 To make sure the coding actually works, let N E SL3.Z/ be a torsionfree subgroup of finite index, and let F SL3.Z/=N , a certain finite group. Let D ' SL3.Z/ M Aq F M Aq W ! q D q D 2 2 be the canonical surjection. The kernel of this will be N , and so it will be torsion free. Fix some p C , and 6 E 2 let Bp "Ap SL3.Z/ Lq D Aq. Then there exists a possibly trivial finite subgroup Np Ap such that D 2 Bp Ap=Np Ap. Also note that Bp ker ' is trivial, and hence Ep '"Bp Ap. Finally, note that Š Š \ D Š hEp Ep M Aqi hF M Aq M Aqi F < ℵ0 W \ Ä W D j j q D q D q D 2 2 2 Thus Ep Lq D Aq Ap. In particular, there is an element of order p, meaning p D and thus C D. \ 2 Š 2 Â a

Now we want to show Theorem 6 • 26. Doing this isn’t so simple as with Theorem 6 • 25. The proof of it makes use of the following group theoretic concept and fact. 6 • 29. Deﬁnition The groups G, H are isomorphic up to finite kernels iff there exist finite normal subgroups N E G, M E H such that G=N H=M . Š 6 • 30. Theorem (Group Theoretic Fact) N There exists a Borel family Sx x 2 of finitely generated groups such that if Gx SL3.Z/ Sx, then for all x y, ¹ W 2 º D ¤ (i) Gx, Gy are not isomorphic up to finite kernels; and

(ii) Gx doesn’t virtually embed in Gy .

(i) is only used in showing the nonessentially free part, and (ii) is only used in showing the Borel bireducibility part.

Proof of Theorem 6 • 26 .:. N Using the Gxs from Group Theoretic Fact (6 •, 30) for each Borel subset A 2 , let EA Fx A EGx be the Â D 2 smooth disjoint union of EGx x A , i.e. EA is the countable Borel equivalence relation on the standard ¹ W 2 º Gx Borel space XA x; f x A; f .2/ defined by x; f EA x ; f iff x x f EG f . D ¹h i W 2 2 º h i h 0 0i D 0 ^ x 0 Claim 1 N If A 2 is an uncountable Borel subset, then EA isn’t essentially free. Â Proof .:. 6 Y Õ Suppose EA B EH , where G Y is a free Borel action of a countable group H. Then for each x A, 6 Y 2 EGx B EH , just by restricting to each piece. This allows us to use Black Box (6 • 20), and so there exists a virtual embedding x Gx H. Since A is uncountable, and each Gx is finitely generatedH ( W ! has only countably many finitely generated subgroups), there exist x y such that x.Gx/ y .Gy /, ¤ D contradicting that Gx, Gy aren’t isomorphic up to finite kernels. a Claim 2

EA 6B EB iff A B. Â

23 MATH 569 CLASS NOTES §6 A

Proof .:.

Clearly if A B then EA 6B EB . So suppose EA 6B EB and there exists an x A B. Then there exists Â Gx Gy Õ 2Gx n a Borel reduction f .2/ Fy B .2/ from EGx to EB . Since Gx ..2/ ; x/ is ergodic, there Gx W ! 2 Gy exists Z .2/ with x.Z/ 1 such that f "Z .2/ for some fixed y B. But then f Z yields a Â D Â 2 xnontrivial Borel homomorphism from EGx to EGy . (Everything outside of Z, we can send to a single element.) And so by Black Box (6 • 20), Gx virtually embeds in Gy , which is a contradiction a

a Finally we sketch the proof of Group Theoretic Fact (6 • 30).

Proof sketch of Group Theoretic Fact (6 • 30) .:.

6 • 31. Deﬁnition An infinite group G is quasifinite iff every proper subgroup of G is finite.

6 • 32. Theorem (Ol'shanskii)

Let P be the standard Borel space of strictly increasing sequences of primes x pn n ! such 75 D h W 2 i that p0 > 10 . Then there exixts a Borel family Tx x of 2generator groups such that for each ¹ W 2º x pn n ! D h W 2 i • Tx contains a cyclic subgroup of order pn for each n !. 2 • Every proper subgroup of Tx is cyclic of order pn for some n !. 2 • Tx is simple.

Provint the above theorem takes a huge amount of work, as Ol’shanskii’s book shows. Using this result, though, Group Theoretic Fact (6 • 30) follows from the following. 6 • 33. Proposition

For each x P , let Gx SL3.Z/ Tx. Then the Borel family Gx x P satisfies that if x y, then 2 D ¹ W 2 º ¤ • Gx, Gy are not isomorphic up to finite kernels; and • Gx doesn’t virtually embed in Gy .

Proof .:.

Since each Tx is simple and SL3.Z/ has no nontrivial, finite, normal subgroups; it follows that Gx has no nontrivial, finite, normal subgroups. Thus it is enough to show thatif x y, then Gx doesn’t embed in Gy . ¤ Suppose that Gx Gy is an embedding. Certainly isn’t an isomorphism, since we have different W ! primes. In particular, "Gx is an infinite, proper subgroup of Gy , which contradicts Ol’shanskii (6 • 32), as they are all quasifinite. a

§ 6 A. The Black Box

Now we should attempt to understand why Black Box (6 • 20) is true. To do this, we will need to introduce cocycles. Until further notice, let G Õ X; be a Borel action of a countable group G on a standard Borel space X with Ginvariant probability measureh . i 6 A • 1. Deﬁnition If H is a countable group, then a Borel map ˛ G X H is a cocycle iff for all g; h G, ˛.hg; x/ ˛.h; g x/˛.g; x/ for almost every x X. W ! 2 D 2 These are somewhat disgusting, but the theme will be to make these less disgusting, and Popa saysyoucan.

To explain where these come from, cocycles arise naturally as follows. Suppose H Õ Y is a free Borel action of a 24 §6 A MATH 569 CLASS NOTES

X Y countable group H and that f X Y is a Borel homomorphism from EG to EH . Then we can define a Borel cocycle ˛ G X H by W ! W ! ˛.g; x/ the unique h H such that h f .x/ f .g x/. D 2 D Since H Õ Y is free, this is how we get uniqueness, as per the following diagram: f .x/ must be in the same orbit as f .gx/ via ˛.g; x/. f .gx/ must be in the same orbit as f .hgx/ iva ˛.h; gx/. And so f .x/ must be in the same orbit as f .hgx/ via ˛.hg; x/.

x f .x/

g ˛.g; x/

hg g x ˛.hg; x/ f .g x/ h ˛.h; g x/

hg x f .hg x/ X Y

We are lucky that the following holds. If ˛.g; x/ ˛.g/ depends only on the g variable, then the cocycle identity reduces to ˛.hg/ ˛.h/˛.g/, meaning ˛ G DH is a group homomorphism. Also, in this case using the same setup as the aboveD example, if we look at .G;W X/! .H; Y / using ˛ and f , we get that ˛.g/f .x/ f .gx/ is a homomorphism of permutation groups. So if we can! take a function and perturb it to get a function ofD one variable, then we are in good shape: not only is there a group homomorphism in play, but there’s one that respects the group actions. So we’ll try to eliminate variables.

Quesetion: can we “adjust” f X Y so that ˛ becomes a group homomorphism? (And what does this “adjusting” mean?) W !

Suppose that b X H is a Borel map. Then we can define f 0 X Y by f 0.x/ b.x/f .x/ Hf .x/. Then f 0 W ! X Y W ! D 2 is also a Borel homomorphism from EG to EH (we haven’t changed the orbit). If f is a Borel reduction, then so is f . Similarly, if f is nontrivial, then so is f . Let ˇ G X H be the Borel cocycle corresponding to f . 0 0 W ! 0

f .x/ b.x/ f 0.x/ x g ˛.g; x/ ˇ.g; x/ b.g x/ g x f .g x/ f .g x/ 0 X Y

Thus ˇ.g; x/ b.gx/˛.g; x/b.x/ 1. This motivates the following definition. D 6 A • 2. Deﬁnition The cocycles ˛; ˇ G X H are equivalent, written ˛ ˇ, iff there exists a borel b X H such that for all g G, W ! W ! 2 1 ˇ.g; x/ b.gx/˛.g; x/b.x/ D for almost every x X. 2

25 MATH 569 CLASS NOTES §6 A

6 A • 3. Theorem (Popa Superrigidity) Let be a countable infinite Kazhdan group (defined later) andlet G;K be countable groups such that E G 6 K. Therefore, If H is any countable group, then every Borel cocycle ˛ G .2/K H is equivalent to a group homomorphism from G into H. W !

In most applications, G K, and so we have a cocycle ˛ G .2/G H and the corresponding orbit equivalence D W ! relation is EG . In many (not all, e.g. Black Box (6 • 20)) applications, G as well. D 6 A • 4. Deﬁnition Let be a countable group. Then is a Kazhdan group iff there exists a finite subset F and " > 0 such that the following holds. Â (*) If U.H/ is any unitary representation such that there is a unit vector v H with . /v v < " for allW !F , 2 k k 2 then there exists a invariant unit vector v H. 2 Note that every countable Kazhdan group is finitely generated, and we can let F be any finitely generating set. Fur thermore, any homomorphic image of a Kazhdan group is Kazhdan. 6 A • 5. Example

SLn.Z/ is Kazhdan for all n 3. Recall Black Box (6 • 20), restated below. We now are in a position to prove it. 6 A • 6. Theorem (The Black Box) Õ Let S be any countable group and let G SL3.Z/ S. Let H be any countable group and let H Y be a free D Y Borel action. If there exists a nontrivial Borem homomorphism f X Y from EG to EH , then there exists a virtual embedding G H. W ! W ! Proof .:. Y Suppose that f X Y is a nontrivial Borel homomorphism from EG to EH . Then we can define a Borel cocycle ˛ G W .2/!G H by taking ˛.g; x/ to be the unique h H such that hf.x/ f .hx/. By Popa SuperrigidityW (6 A • 3), there! exists a Borel map b .2/G H, a2 group homomorphism D' G H and a subset X .2/G with .X/ 1 such that for all gW G, '.g/! b.gx/˛.g; x/b.x/ 1 for all xW X!. Â D 2 D 2

So let f 0 X Y be the borel map f 0.x/ b.x/f .x/. Then f 0 is also a nontrivial Borel homomorphism W Y! D from EG to E . Also for all g G and x X, H 2 2 1 f 0.gx/ b.gx/f .gx/ b.gx/˛.g; x/f .x/ b.gx/˛.g; x/b.x/ f 0.x/ '.g/f 0.x/.(**) D D D D To see that ' is a virtual embedding, suppose that N ker ' is infinite. By (**), for all g N and x X, f .gx/ f .x/. Thus f X Y is N invariant.D Since G Õ X; is strongly mixing,2 N Õ X;2 is 0 D 0 0 W ! h i h i ergodic; and so there exists a Z X with .Z/ 1 such that f 0 sends Z to a single point y0 Y . But then f Y Â D 2 sends Z into the single EH class containing y0 (we’ve only adjusted within the same class), which contradicts the fact that f is nontrivial. a The next application uses the following definition. 6 A • 7. Deﬁnition Let E be a Borel equivalence relation on a standard Borel space X and let 1 n !. Then nE E E (n times) is the Borel equivalence relation on X n defined by Ä Ä D ˚ ˚ i; x nE j; y iff i j x E y h i h i D ^

Clearly nE 6B mE for n m. But it’s difficult to show that E 1E

26 §6 A MATH 569 CLASS NOTES

6 A • 8. Theorem There exists a countable Borel equivalence relation E such that

Fact 1a. SL3.Z/ has no nontrivial, finite, normal subgroups.

Fact 1b. If SL3.Z/ SL3.Z/ is an injective homomorphism, then is surjective; i.e. SL3.Z/ is coHopfian. W !

As a corollary, if SL3.Z/ SL3.Z/ is a virtual embedding, then is an automorphism. W ! Fact 2. Suppose G is a countable group and X; is a standard Borel Gspace with invariant probability h i measure . If is ergodic, then there exists a Ginvariant, Borel subset X0 X with .X0/ 1 such Õ Â D that G X0 is uniquely ergodic (with unique invariant probability measure X0).

G Õ So let G SL3.Z/ and let X .2/ be a Ginvariant Borel subset with .X/ 1 such that G X; D ÂX 6 6 D h i is uniquely ergodic. Let E EG . Then clearly E B E E B . So we want to show that there are no reductions holding in the oppositeD direction. ˚ Claim 1 If f X X is a nontrivial Borel homomorphism from E to E, then .G f "X/ 1. W ! D Assuming the claim, suppose ' X .n 1/ X n is a Borel reduction from .n 1/E to nE. For each W C ! ÕC 0 i n, let Xi X i ; and for each 0 j n, let 'j 'Xj . Since G X; ergodically, and Ä Ä D ¹ º Ä Ä D h i each is isomorphic to X; , for each 0 j n, there exists 0 kj n 1 and Zj Xj ; with measure h i Ä Ä Ä Ä Â .Zj / 1 such that 'n"Zj Xk . By the claim, .G f "Zj / 1. D Â j D

There exist by the pigeonhole principle i j with ki kj . But then ¤ D .G f "Zi G f "Zj / 1, \ D and so G f "Zi G f "Zj , which is a contradiction: elements which are nonequivalent going into the same class. \ ¤ ; a So all that remains to show is the claim. To do this, we make use of some basic observations in measure theory. 6 A • 9. Observation If X; is a standard Borel probability space and f X Y is a Borel map, then we can define a probability measureh i f defined by .A/ .f 1"A/. W ! D D 6 A • 10. Observation With the above hypotheses, suppose G Õ X; is a measurepreserving Borel action, H Õ Y is a Borel action and ' G H is a group homomorphismh suchi that '.g/f .x/ f .gx/. Then f is '.G/invariant. W ! D D Proof .:. 1 Let A Y be Borel and g G. Let B f "A. Then f .gB/ '.g/f .B/ '.g/A. Hence .'.g/A/ .gB/Â. Since is Ginvariant,2 .gB/ D.B/ .A/. D D D D D a Using these observations with Popa Superrigidity (6 A • 3) gives this result easily. 6 A • 11. Result With X, G, and E as in the proof of Theorem 6 A • 8, if f X X is a nontrivial Borel homomorphism from E to E, then .G f "X/ 1. W ! D

27 MATH 569 CLASS NOTES §6 B

Proof .:. Suppose f X X is a nontrivial Borel homomorphism from E to E. Then we can define a Borel cocycle ˛ G X W G!by taking ˛.g; x/ to be the unique h G such that hf.x/ f .gx/. Now we want to adjust thisW to become! a homomorphism. Since G is Kazhdan,2 by Popa SuperrigidityD (6 A • 3), there exists a Borel map b X G, a group homomorphism ' G G, and a Borel subset Z Z with .Z/ 1 such that—letting W ! W ! Â D f 0.x/ b.x/f .x/—for all g G, and x Z, f 0.gx/ '.g/f 0.x/ (so we are unravelling the cocyle to get the homomorphism).D Furthermore,2 we can suppose2 that ZDis Ginvariant (otherwise we take the intersection of all of the translates). Note that G Õ Z; is (still) strongly mixing. h i Claim 1 ' is a virtual embedding.

Proof .:. Suppose N ker ' is infinite. Since G Õ Z; is strongly mixing, N Õ Z; is ergodic. Since D h i h i f 0 Z X is N invariant, it follows that f 0 is almost everywhere constant. But this means almost everyW x! Z is sent by f to a single Eclass, a contradiction with nontriviality. 2 a

Thus ' G G is an automorphism. Let f 0 be the probability measure on X defined by .A/ 1 W ! D D .f 0 "A/.

Since f 0.gx/ '.g/f 0.x/, it follows that is '.G/invariant. But '.G/ G, so is Ginvariant. Since Õ D D G X; is uniquely ergodic, it follows that . Thus .f 0".Z// .f 0"Z/ which, by definition of the pushforward,h i is just .Z/ 1. Since f "Z GDf "X, we have that .GD f .x// 1. D 0 Â D a

§ 6 B. Weak Borel reductions

6 B • 1. Deﬁnition Suppose that E;F are countable Borel equivalence relations on X;Y . Then E is weakly Borel reducible to F , written E 6w F , iff there exists a countabletoone Borel homomorphism f X Y from E to F . B W ! We remark the following about this definition. 1. If f is a Borel reduction, then f is a weak Borel reduction. 6w 6w 6w 2. If E B E0 and E0 B E00, then E B E00. 3. If E F are countable Borel equivalence relations on X, then idX is a weak Borel reduction from E to F . Â In essence, this is all that weakly Borel reductions are: Borel maps with inclusions. 6 B • 2. Theorem If E, F are countable Borel equivalence relations on X, Y , then the following are equivalent: 6w (i) E B F ; (ii) There exists a countable Borel equivalence relationR E on X such that R 6B F . Ã Proof .:. 6w 6 6w For (ii) implies (i), we have that E B R and R B F . Hence E B F . For the other direction, suppose 1 f X Y is a weak Borel reduction from E to F . Let R f .F /. Then R E is a countable Borel equivalenceW ! relation and f is a Borel reduction from R to F . D Ã a 6w 6 So we explore the connection between B and B. 6 B • 3. Deﬁnition A countable Borel equivalence relation E is weakly smooth iff there is a smooth, countable, Borel F such that 6w E B F .

We clearly have that all smooth Borel equivalence relations are weakly smooth, but we also have the reverse. 28 §6 B MATH 569 CLASS NOTES

6 B • 4. Theorem A countable Borel equivalence relation E is smooth iff it is weakly smooth.

Proof .:. 6w Let E be weakly smooth on X, and let F be a smooth countable Borel equivalence relation such that E B F . Then there exists a countable Borel R E on X such that R 6B F . Thus R is smooth. Ã Hence there exists a Borel transversal T for R. By Feldman–Moore Theorem (2 • 2), there exists a countable Õ X group G gn n ! and a Borel actionG X such that R E . Hence we can define a Borel selector D ¹ W 2 º D G s X X for E by s.x/ gn t where T ŒxR t and n is minimal such that gn t E x. W ! D \ D ¹ º Thus E has a selector and so is also smooth. a 6 B • 5. Deﬁnition A countable Borel equivalence relation E is weakly hyperfinite iff there is a hyperfinite, countable, Borel F such 6w that E B F .

Again, we have the same characterization as before. 6 B • 6. Theorem If E is a weakly hyperfinite Borel equivalence relation on X, then E is hyperfinite.

Proof .:. Let F be a hyperfinite, Borel equivalence relation such that E 6w F . Then there exists a countable Borel R E B Ã on X such that R 6B F . Hence R is hyperfinite, and thus E is too. a 6 6w So for idX , there’s no difference between B and B. Similarly, there’s no difference for E0. 6 B • 7. Theorem A countable Borel equivalence relation E is weakly universal iff for every countable Borel equivalence relation F , 6w 6w F B E; equivalently E B E. 1 This yields to the following conjecture from 2001. 6 B • 8. Open Problem (Hjorth's Conjecture) Every weakly universal countable Borel equivalence relation is universal.

After this, we arise at Thomas’ question in the early 2000s: do there exist countable Borel equivalence relations E F Â such that E 6B F ? It turns out that there are. 6 6 B • 9. Theorem

There exist countable Borel equivalence relations E, F on a standard Borel X such that E F and E 6B F . Â 6 6 6w Hence B and B are in fact distinct notions. In proving this, as usual, the point is to find/Google suitable “big guns”. 6 B • 10. Theorem

Suppose that G is a proper subgroup of finite index in SL3.Z/. Therefore, 1. G is a Kazhdan group. 2. G has no nontrivial, finite normal subgroups.

3. SL3.Z/ does not embed in G.

6 B • 11. Theorem Suppose H Õ X; is a strongly mixing, Borel action on a standard Borel probability space. Then there exists h i an H invariant Borel subset X0 X with .X0/ 1 such that the action of every infinite, finitely generated subgroup of H is uniquely ergodic.Â D

29 MATH 569 CLASS NOTES §6 B

Proof of Theorem 6 B • 9 .:.

Let S SL3.Z/ and let T ker ' where ' SL3.Z/ SL3.F7/ acts surjectively and where F7 is the field D D W ! with 7 elements (the ‘7’ is unimportant, we just want a finite index). Then 1 < ŒS T < ℵ0. It follows that T is also finitely generated. Note that T is also a Kazhdan group. W

Let X .2/S be an Sinvariant Borel subset with .X/ 1 such that the action of every infinite, finitely generatedÂ subgroup of S on X is uniquely ergodic. Let E ¨ DF be the orbit equivalence relations of T Õ X and Õ S X. We will show that E 6B F . 6 Suppose that f X X is a Borel reduction from E to F . Then we can define a Borel cocycle ˛ T X S by taking ˛.t; x/W to be! the unique s S such that sf .x/ f .tx/. By Popa Superrigidity (6 A • 3),W since T ! S is Kazhdan, after deleting null subset2 and adjusting f Dif necessary, we can suppose that ˛ T S is a groupÂ homomorphism. W ! Claim 1 ˛ T S is an embedding. W ! Proof .:.

Suppose not. Since ŒSL3.Z/ T < ℵ0, T has non nontrivial, finite, normal subgroups. Thus N ker ˛ is infinite. Since S Õ .2/S ; W is strongly mixing, N Õ X; is ergodic. But then the N invariantD Borel map f X X is h almosti everywhere constant, contradictingh i that we have a Borel reduction. W ! a Also, since T S, it follows that ˛"T is a proper subgroup of S. Since the actions of S, T on X; are free and ˛.t/f .x/ f6Š .tx/ for t T , x X, it follows that f X X is an injection. Thus we hhavei an embedding of D 2 ˛;f 2 W ! permutation groups: T;X S;X . We more or less want f to be surjective, and to do this, we use unique ergodicity. h i !h i

1 Hence we can define an ˛"T invariant probability measure f on X by .A/ .f "A/. Since ˛"T is finitely generated and infinite, ˛"T Õ X; is uniquelyD ergodic. Thus andD hence .f "X/ .f "X/ .f 1".f "X// .X/ 1. So theh mapi is onto. D D D D D As a proper subgroup of S, let s S ˛"T . Then .f "X s.f "X// 1. Hence there exist x; y X such that f .x/ sf .y/ f "X s.f2"X/n. Thus f .x/ F f .y/ and\ so x E yD. Hence there exists a t T2such that D 2 \ 1 2 x ty. It follows that ˛.t/f .y/ f .ty/ f .x/ sf .y/. But then s ˛.t/f .y/ f .y/, which contradicts theD fact that S Õ X is free. D D D D a So that was the last consequence of Popa Superrigidity (6 A • 3) we will look at. 6 B • 12. Theorem (Miller) If E is a countable, Borel equivalence relation on a standard Borel space X, then the following are equivalent. (i) There is a universal countable Borel F on X such that F E. Â (ii) E is weakly universal.

Proof .:. Showing (i) implying (ii) is easy, but (ii) implying (i) is very technical and never used. a 6 B • 13. Corollary

T is weakly universal. Á Proof .:. Let F a; b be the free group on two generators. Then E is the orbit equivalence relation of the shift action Õ DF h2 i Õ1 Õ F2 2 . We can identify the left translation action F2 F2 with the action N of a suitable group of recursive permutations. With this identification, E T. 1 ÂÁ a

30 §7 MATH 569 CLASS NOTES

For later use, we record the following. 6 B • 14. Theorem If E is a weakly universal, countable Borel equivalence relation, then E isn’t essentially free.

Proof .:. Y Õ Otherwise, we can suppose that E EH for some free Borel action H Y . But then there exists a countable D Y w G such that there’s no nontrivial Borel homomorphism from EG to E E . And in this case, EG 6 E. D H 6 B a 6 B • 15. Open Problem

Is T countable universal? Á For a more vague open problem, we have the following. 6 B • 16. Open Problem Õ N 2N Is there a “natural” action 2 of a countable group such that E is T? Á

We next try to develop an analog of ergodicity for T. Á

Section 7. Martin's Measure

Recall the following definition. 7 • 1. Deﬁnition N N For each r 2 , the corresponding cone is C s 2 r 6T s . 2 D ¹ 2 W º

Note that if Cn n ! is a countable set of cones, then T Cn contains a cone. The following can be regarded ¹ W 2 º n

As a remark, X is Tinvariant iff whenever y T x X, then y X. Before we prove this (from Borel determinacy), we state a few corollaries.Á Á 2 2 7 • 3. Corollary N N N If ' 2 2 is a Tinvariant Borel map, then there exists a cone C 2 such that 'C is a constant map. W ! Á Â Proof .:. Z For each n !, there exists an "n 0; 1 such that Xn x 2 '.x/n "n contains a cone Cn. Then 2 2 ¹ º D ¹ 2 W D º C Tn ! Cn contains a cone and ' C is a constant function. D 2 a 7 • 4. Corollary N If X 2 is a Tinvariant Borel subset, then the following are equivalent. 1.ÂX containsÁ a cone.

2. X is a 6Tcofinal.

Proof .:. (1) implies (2) obviously. For (2) implies (1), clearly 2N X cannot contain a cone. n a

31 MATH 569 CLASS NOTES §7

Proof of Martin’s Theorem (7 • 2) .:. N Suppose X is a Tinvariant Borel subset of 2 . Consider the 2 player game GX where I wins iff sn n ! Á h W 2 i 2 X (here I plays s2n 0; 1 , and II plays s2n 0; 1 for n < !). Suppose, for example, that I has a winning

Let t tn n < ! C . Consider the play of GX , where D h W i 2 • II plays s1; s3; s5; , where t s2n 1 n < ! ; D h C W i • I uses and plays s0; s2; . Then s sn n ! X as is a winning strategy. Clearly t 6T s. Also, since 6T t, s 6T t. Thus D h W 2 i 2 t 6T s X, and so t X. 2 2 a We will explore some consequences of Martin’s Conjecture (MC). 7 • 5. Open Problem (Martin's Conjecture) N Z If f 2 2 is a Borel homomorphism from T to T, then exactly one of the following holds: W ! N Á Á 1. There exists a cone C 2 such that f maps C into a single Tclass. Â N Á 2. There exists a cone C 2 such that x 6T f .x/ for all x C . Â 2 There is a stronger version (namely, the actual version), where (2) is replacedby N .˛/ .˛/ 2’. There exists a cone C 2 and a countable ˛ < !1 such that f .x/ T x for all x C , where x is the ˛th Turing jumnp. Â Á 2 Since the 1980s, there’s been a single instance of progress on MC. Namely, the following theorem, proven some time in the 80s. 7 • 6. Theorem (Slaman–Steel) N N N Suppose that f 2 2 is a Borel homomorphism from T to T. If there exists a C 2 such that W ! Á Á Â f .x/

(MC) T

Suppose that (i) fails. By MC, there exists a cone C such that x 6B f .x/ for all x C . Clearly f C is countable to one (since there are only countably many predecessors to any turing degree).2 Hence f C is a weak Borel reduction from TC to T. Á Á

Now we let D be any cone, and let D0 C D. Since f C is countable to one, it follows that f "D0 is Borel. D \ 6 N Hence Œf "D0 T is also Borel. Since Œf "D0 T is a Tinvariant, Tconfinal, Borel subset of 2 , Martin’s Á Á Á Theorem (7 • 2) implies that Œf "D0 T contains a cone. Á a

As a matter of notation, for x; y 2N , then x y is the usual recursive join: x is placed on the evens while y is placed on the odds. 2 ˚

32 §7 MATH 569 CLASS NOTES

7 • 9. Corollary N (MC) If A 2 is a Tinvariant, Borel subset, then the following are equivalent. Â Á 1. TA is weakly universal. Á 2. A contains a cone.

Proof .:. N For (ii) (i), suppose that A contians the cone C r 2 z 6T r . We want to show that this is weakly ! D ¹ 2 W º universal. We can define an injective, weak Borel reduction from T to TA by x x z. Á Á 7! ˚ N For the other direction, suppose that TA is weakly universal. Let f 2 A be a weak Borel reduction Á N W ! from T to T A. By Theorem 7 • 7, Œf "2 T A contains a cone. Á Á Á Â a

Remark: there are currently no naturally occurring Borel sets of Turing degrees D for which it is known that TD isn’t weakly universal. In particular, it is not known when D is the set of minimal degrees. Á 7 • 10. Deﬁnition

Let E be a countable Borel equivalence relation on a standard Borel space X. Then T is Emergodic iff for every N Á Borel homomorphism f 2 X from T to E, there exists a cone C such that f maps C to a single Eclass. W ! Á

So by Martin’s Theorem (7 • 2), T is id N mergodic. One can also see that if E 6B F and T is F mergodic, then Á 2 Á T is Emergodic. Á 7 • 11. Open Problem

Is T E0mergodic? Á But assuming MC, we understand everything. 7 • 12. Theorem (MC) If E is a countable Borel equivalence relation, then exactly one of the following holds: a. E is weakly universal.

b. T is Emergodic. Á Proof .:.

If E is weakly universal, then there exists a weak reduction from T to E; and so T isn’t Emergodic. So (a) and (b) are mutually exclusive. Á Á

So it suffices to show that if T isn’t Emergodic, then E is weakly universal. So suppose the Borel map N Á f 2 X witnesses the failure of (b): a nontrivial, Borel homomorphism from T to E. Since T is weakly W ! N Á Á universal, there exists a weak Borel reduction g X 2 from E to T. Let h g f . Then h is a Borel W ! Á D ı homomorphism from T to T. Á Á N Suppose there exists a cone C 2 such that h sends C to a single T class, say Œx T . Then f sends a cone Â Á 1 Á to the countable preimage of a single class: f maps C into the countable set Y g "Œx T . We’d like to have the inverse image of one of the points is large, but this follows from the fact thatD there mustÁ be some y Y with 1 6 1 2 f .y/ as Tcofinal. By Martin’s Theorem (7 • 2), it follows that Œf .y/ T contains a cone D. But then f Á maps D into ŒyE , a contradiction.

It follows that there exists a cone C such that hC is countable to one. Thus f C is countable to one, and f C is a weak Borel reduction from TC to E. Since TC is weakly universal, E is weakly universal. Á Á a

Finally, we give two striking applications of MC, which don’t mention T. Recall Definition 4 • 23, that a Borel equivalence relationE on X is BorelBounded iff every Borel ' X NNÁhas a Borel homomorphism X NZ from E to (eventual equality) which dominates ' for all x: '.x/W !6 .x/ for all x X. W ! D 2 Also recall Open Problem 4 • 25, where it’s an open problem whether there exist any countable Borel equivalence

33 MATH 569 CLASS NOTES §7 relations that are not BorelBounded. MC answers the open problem by saying that any weakly universal one isn’t BorelBounded. 7 • 13. Theorem (MC) If E is weakly universal, then E is not BorelBounded.

Note that even assuming MC, nothing is known when E0

(MC) T isn’t BorelBounded. Á 7 • 15. Lemma isn’t weakly universal. D Proof .:.

It is easily seen that is Borel bireducible with E0. Since E0 is free, E0 isn’t weakly niversal. D a Proof of Theorem 7 • 14 .:. Identifying each r 2N with the corresponding subset of N, let ' 2N 2N be the Borel map such that 2 W ! • if r 2N is infinite, then '.r/ is the strictly increasing enumeration of r 2N; \ \ • otherwise '.r/ is the identically zero function. Now we claim the following. Claim 1 N N N For each h N , the Tinvariant Borel set Sh r 2 s 2 .s T r h < '.s// contains a cone. 2 Á D ¹ 2 W 9 2 Á ^ º

Proof .:. Z Z First fix a strictly increasing e N such that h < e. Now suppose r 2 satisfies e 6T r. Consider s N defined by 2 2 Â s 2e.n/ n N 2` 1 ` r . D ¹ W 2 º [ ¹ C W 2 º Then clearly s T r, and h < e < '.s/. Á a N N 6 Finally, suppose that 2 N is a Borel homomorphism from T to such that '.s/ .s/ for all N W ! Á D s 2 . Since isn’t weakly universal, T is mergodic, and so there exists a cone C such that maps C to2 a single class,D say, Œh . But this isÁ a problem, because then '.s/ 6 h for all s C and so C S , h which is aD contradiction. D 2 \ D; a To see that every weakly universal E isn’t BorelBounded from MC, we we prove the following lemma. 7 • 16. Lemma Suppose E, F are countable Borel equivalence relations, and that E is BorelBounded. Therefore, (i) If F 6B E, then F is BorelBounded. (ii) If F E, then F is BorelBounded. Â Proof .:. For (ii), suppose that F E are equivalence relations on X. Let ' X NN be any Borel map. Therefore Â N W ! 6 there exists a Borel homomorphism X N from E to such that '.x/ .x/ for all x X. So is also a Borel homomorphism from F toW !. D 2 D For (i), suppose f X Y is a Borel reduction from F to E. Then Z im f is a Borel subset of Y and there W ! D

34 §7 A MATH 569 CLASS NOTES

exists a Borel map g Z X such that f g id Z. W ! ı D Suppose that ' X NN is Borel. Ideally, we’l like to consider the function g ' (and 0 elsewhere on Y ), but W ! ı Õ X this won’t work. Let n n ! be a countable group with a Borel action X such that E F by D ¹ W 2 º D Feldman–Moore Theorem (2 • 2). Let ' Y NN be the Borel map defined by O W ! ´max '. i g.z// i n if z Z '.z/.n/ ¹ W Ä º 2 O D 0 otherwise.

N And now we just check that this works. Let O Y N be a Borel homomorphism from E to such that 6 W ! Z D '.z/ .z/O for all z Y (since E is BorelBounded). Define X N by .x/ .O f /.x/. Clearly isO a Borel homomorphism2 from F to . Now we just check domination.W ! D ı D

Fix some x X and let z f .x/ Z. Then there exists an n ! such that x ng.z/. So if m satisfies 2 D 2 2 D • m n; and • m max ` '.z/.`/ > .z/.`/ ; ¹ WO O º then '.x/.m/ '. ng.z//.m/ '.z/.m/ .z/.m/ .f .x//.m/ .x/.m/. D ÄO Ä O D O D Thus '.x/ 6 .x/ for all x X. 2 a So the following theorem tells us that measure is useless near the top of the countable Borel equivalence relations. 7 • 17. Theorem (MC) Let E be a countable Borel equivalence relation on a standard Borel space X and let be a (not necessarily Einvariant) probability measure on X. Then there exists a Borel Y X with .Y / 1 such that EY isn’t weakly universal. Â D

We will make use of the following consequence of Borel–Cantelli (4 • 21). 7 • 18. Lemma If X; is a standard Borel probability space and X NN is Borel, then there exists an h NN such that .h x iX .x/ 6 h / 1. W ! 2 ¹ 2 W º D Proof .:. Let ' 2N NN be the Borel map such that W ! • if r 2N is infinite, then '.r/ is the strictly increasing enumeration of r 2N; and \ \ • otherwise, '.r/ is identically 0. Õ N By Feldman–Moore Theorem (2 • 2), there exists a countable group n n ! and a Borel action 2 2N N N D ¹ W 2 º such that E is T. Let 2 N be the Borel map defined by Á W ! .x/.n/ max '. nx/ m n . N D ¹ W Ä º Then for all r; s 2 with s T r, '.s/ 6 .r/. 2 Á N N Let f X 2 be a weak Borel reduction from E to T. Define X N by .x/ .f .x//. Then there existsW ! an h NN such that Y x X .x/ 6 Áh satisfies thatW .Y! / 1 by Borel–CantelliD (4 • 21). 2 D ¹ 2 W º D

Let Z Œf .x/ T . Therefore Z is Borel as f is countabletoone, and saturation is Borel. Moreover, for each D 6 Á N r Z, '.s/ h for all s T r. By an earlier claim, there exists a cone C 2 Z. Hence Z doesn’t contain 2 Á w Â n a cone. By MC, TZ isn’t weakly universal. Since E 6 TZ, it follows that E isn’t weakly universal. Á B Á a

Now return to consequences ZFC, where the remainder of the course takes place.

§ 7 A. Actual Theorems of ZFC

35 MATH 569 CLASS NOTES §7 A

7 A • 1. Deﬁnition

f E, F are Borel equivalence relations on Polish spaces X, Y ; write E 6c F iff there exists a continuous (in the sense of X and Y ) reduction from E to F .

A problem of Kanovei is to find nontivial instances of countable Borel equivalence relations E, F such that E 6B F and E 6c F . What we mean by trivial is given by the following example. 6 7 A • 2. Example (Trivial Example)

Consider idŒ0;1 and id N . Then idŒ0;1 6B id N , but idŒ0;1 6c id N , just because of topological reasons. 2 2 6 2 Now in descriptive set theory, almost every equivalence relation is on a totally disconnected space. 7 A • 3. Theorem (Target Theorem 4)

T 6c E . Á 6 1 7 A • 4. Deﬁnition N Z et 1 be the relation of recursive isomorphism on 2 (regarded as P .N/; i.e. if x; y 2 , then Á 2 x 1 y iff there is a recursive permutation of N such that '"x y. Á D 7 A • 5. Theorem (Folklore Theorem)

The map x x is a Borel reduction from T to 1. 7! 0 Á Á Note that the following theorem implies that the above map is not continuous (as we will see later). 7 A • 6. Theorem (Target Theorem 5)

T 6c 1. Á 6 Á

It is an open problem whether 1 is universal. However, let Rec.N/ be the group of recursive permutations of N. 3N Á 2N Marks has shown that ERec.N/ is universal (the question is then why we can’t with ERec.N/). Both Target Theorem 4 (7 A • 3) and Target Theorem 5 (7 A • 6) are immediate consequences of the following. 7 A • 7. Theorem (Theorem AA)

2N N N Suppose G is a group of recursive permutations of N and E EG . Then whenever 2 2 is a continuous D N W ! homomorphism from T to E, then there exists a cone C 2 such that maps C into a single Eclass. Á Â This is more generally true when G 6 Sym.N/ is any countable subgroup. The proof just gets more technical, and less transparent. 7 A • 8. Theorem If f 2N 2N , then the following are equivalent: i.W f is! continuous. ii. There exist e N and z 2N such that f .x/ 'z x for all x 2N . 2 2 D e ˚ 2 N r To introduce some notation, 'e is the eth oracle Turing machine. If r 2 , then ' =e is the (partial) function computed 2 by 'e with oracle r.

Before proving it, we have the following corollary. 7 A • 9. Corollary N N N If f 2 2 is continuous, then there exists a cone C 2 such that f .x/ 6T x for all x C . W ! Â 2

36 §7 A MATH 569 CLASS NOTES

Proof of Theorem 7 A • 8 .:. To show that (ii) implies (i), suppose that f .x/ y. We need to show that if you’re close to x, then you’re close D to y. In particular, for n N, there exists an m N such that x0m xm, then (because we’re only using finitely many values 2from the oracle) 2 D z x0 z x f .x0/n ' ˚ n ' ˚ n f .x/n. D e D e D Then f is continuous.

Now suppose f 2N 2N is continuous. Let W ! `. Then y.`/ .`/. 2 h i 2 Â j j D a The following definition is deceptively disgusting, but it is the proper notion toprove Theorem AA (7 A • 7). 7 A • 10. Deﬁnition If T 2

Proof .:. If not, then by Borel determinacy, I has a winning strategy 2

37 MATH 569 CLASS NOTES §7 A

For u 2

• Next, let u.00/ and u.01/ be the lexicographically least extensions of u.0/ whose even parts agree with O such that u.00/, u.01/ are incompatible. • And continue in this fashion to create a binary tree. Define

Let T0 be the subtree such that at each 2nth branching point, we always go left if z.n/ 0, and always go right D if z.n/ 1. Then clearly T0 6T z. Let y ŒT0 be the leftmost branch. Then y 6T T0; and since y ŒT , it D 2 2 follows that T 6T y and therefore T 6T T0. And it follows that z 6T T0 just by looking at what happens at the even levels compared to T . Therefore T0 T z. Finally, suppose that x ŒT0 ŒT . Therefore T 6T x, and so Á 2 Â by the same idea (considering the even branching points compared to T ), z 6T x and therefore T0 6T x. a So just by refining the pointed tree given by Theorem 7 A • 12, we can get as complicated a pointed tree as we’d like. 7 A • 14. Deﬁnition If E F are Borel equivalence relations on X, then F is smooth over E iff there exists a Borel homomorphism XÂ X from F to E such that .x/ F x for all x X. (This implies is a Borel reduction from F to E). W ! 2

In the proof of Theorem AA (7 A • 7), we will use the following. (For example, 1 T.) Á ÂÁ 7 A • 15. Theorem (Theorem BB) 6 N 2N If H Sym.N/ is any countable subgroup and D 2 is a cone such that EH D T D. Therefore T D N Â ÂÁ Á 2 isn’t smooth over EH D.

Proving this is where pointed trees will come into play. But assuming this theorem, we can prove Theorem AA (7 A • 7).

Proof of Theorem AA (7 A • 7) .:. Suppose G 6 Sym.N/ is a group of recursive permutations and 2Z 2N is a continuous homomorphism 2N 2N W ! from T to E . Since E T, we can also regard as a homomorphism from T to T. Á G G ÂÁ Á Á

38 §7 A MATH 569 CLASS NOTES

Z Since is continuous, is computable on a cone. Explicitly, there exists a cone C 2 such that .x/ 6T x for all x 2N . Applying Martin’s Theorem (7 • 2), there exists a cone D C such thatÂ either 2 Â (i) .x/ T x for all x D; or Á 2 (ii) .x/

There are two major pieces of content: this nonsmoothness resultof Theorem BB (7 A • 15) and of course Slaman–Steel (7 • 6).

Proof of Theorem BB (7 A • 15) .:. 6 Z 2N Let H Sym.N/ be a countable subgroup and D 2 be a cone such that EH D T D. Suppose Â 2N ÂÁ D D is a Borelhomomorphism from TD to E D such that .x/ T x for all x D. W ! Á H Á 2 Since is countable to one (since it’s going inside it’s own equivalence class) it follows that "D is a Borel subset N

Let x ŒT be the leftmost branch. Then clearly x T T as T 6T x as a pointed tree, and x 6T T as just the 2 Á leftmost branch. Now define an increasing sequence of nodes yn T as follows. 2 • y0 . D; • For yn defined, ynC is such that yn ynC T is the next branching node. Let ynC `n. If hn.`n/ x, _ Â 2 _ j j D … then let yn 1 ynC 1. Otherwise, let yn 1 ynC 0. C D C D 6 6 Taking y Sn

There are two results that suggest this might be interesting. The first, also due to Marks, is reallyquiteweird. 7 A • 18. Theorem (i) Recursive isomorphism on 3N is countable universal. (ii) Recursive isomorphism on 2N is measure universal.

Using MC, Thomas has shown the following.

39 MATH 569 CLASS NOTES §8

7 A • 19. Theorem (MC) If F is any countable Borel equivalence relation on X and is any Borel probability measure, there exists a Y X with .Y / 1 such that F Y isn’t weakly universal. Â D Eventually, we will prove Theorem 7 A • 18, but it will require some Borel combinatorics.

Another open problem, practically an open area, is due to Marks. 7 A • 20. Observation

There are very few countable Borel equivalence relations E for which it is known that E 6B T. Á 7 A • 21. Deﬁnition

Õ F2 Let ET be the orbit equivalence relation for F2 .2/ . 1 The following conjecture due to Marks is open. 7 A • 22. Open Problem

ET 6B T. 1 6 Á A slightly stronger conjecture of Thomas is the following. 7 A • 23. Open Problem If E is a nonhyperfinite countable Borel equivalence relation, then there exists a weakly universal countable Borel F such that E 6B F . 6

Section 8. Borel Combinatorics

8 • 1. Deﬁnition A Borel graph X;R consists of a standard Borel space X and a symmetric, irreflexive, Borel relation R X X. h i Â The first thing we will look at with these is chromatic number. 8 • 2. Deﬁnition

Let X;R be a Borel graph. Then the Borel chromatic number B./ is the least cardinality of a standard Borel spaceD h Y suchi that there exists a Borel map c X Y such that if x R y then c.x/ c.y/. Such a map c is called a Borel coloring. W ! ¤

We need the Y to be contained in X so that we have a notion of Borel. Note that clearly ./, the actual chromatic number, is no more than B./. 8 • 3. Example Z 1 Let X .2/ and let 0 X;E where x E y iff T .x/ y or T .x/ y where T generates the action Z Õ XD. D h i D D

We can actually calculate the Borel chromatic number of 0. 8 • 4. Theorem

For 0 as in Example 8 • 3, 2 .0/ < B.0/ 3. D D Proof .:.

Since every connected component of 0 is a copy of Z, it’s clear that .0/ 2. D

To see that B.0/ > 2, suppose that c X 2 is a Borel 2coloring. For i 0; 1, let Xi x X c.x/ i . W ! D D ¹ 2 W D º Therefore the restriction of c to each connected component gives alternating colors on Z. Thus X X0 X1 is D t 40 §8 MATH 569 CLASS NOTES

a partition of X into two Borel subsets, each of which is invariant under 2Z Õ X. Let be the usual uniform product probability measure on X. Since Z Õ X; is strongly mixing, 2Z acts ergodically on X. Therefore h i either X0 or X1 has measure 1. But then .X0/ .T .X0// .X1/ implies both have the same measure of 1, a contradiction. D D

The fact that B.0/ 3 follows from the next theorem. D a

ℵ0 As far as we know thus far, B.0/ Œ3; 2 . To prove that it is exactly 3, we need some notation for the next theorem. 2 8 • 5. Deﬁnition If X;E is a graph and x X, then h • E.x/i y X y2 E x . D ¹ 2 W º • deg.x/ E.x/ . D j j The next theorem is fairly easy if we leave out the “Borel”. 8 • 6. Theorem

If X;R is a Borel graph such that deg.x/ k for all x X. Then B./ k 1. D h i Ä 2 Ä C First we need some basic results in Borel combinatorics. 8 • 7. Deﬁnition A graph V;E is locally finite iff deg.v/ < for all v V . D h i 1 2 8 • 8. Result

If X;R is a locally finite Borel graph, then B./ !. D h i Ä Proof .:.

Let X; T be a Polish space realizing the standard Borel structure of X. Let Un n ! be a basis for h i ¹ W 2 º the topology T . Then we can define a Borel !coloring by taking c.x/ to be the least n such that x Un and 2 R.x/ Un . \ D; a This, of course, is not the most efficient way of proceeding, since thenullgraph X; would use up countably many colors, for example. h ;i 8 • 9. Deﬁnition Let V;E be a graph. (i) AD subset h iD V is discrete iff no two elements of D are joined by an edge. Â (ii) A maximal discrete subset D V is called kernel Â Of course, there is always a kernel, but the issue is whether there is a kernel that is Borel. 8 • 10. Result If X;R is a locally finite Borel graph, then there exists a Borel kernel D V . D h i Â Proof .:. For each Borel subset Y X, let R.Y / x Y y Y .y R x/ . Â D ¹ 2 W 9 2 º Claim 1 f Y X is Borel, then R.Y / is Borel. Â Proof .:.

Let P R .X Y/. Then P is a Borel subset of X X, and each section Px countable (and actually finite Das is\ locally finite). By Theorem 2 • 6, R.Y / proj .P / is Borel D X a

41 MATH 569 CLASS NOTES §8

Next, let c X ! be a Borel !coloring; and for each n !, let Xn x X c.x/ n . Then W ! 2 D ¹ 2 W D º X Fn ! Xn is a partition of into discrete Borel subsets. We use this to inductively define our Borel kernel. D 2

Define inductively Borel subsets Yn X by Â • Y0 X0; D • Yn 1 Yn .Xn 1 R.Yn//. C D t C n Then Y Sn ! Yn is a discrete Borel subset. To see that Y is a kernel, let x X Y . Then x Xn 1 for some D 2 2 n 2 C n 0. Since x Yn 1, it follows that x R.Yn/. … C 2 a And now we can prove Theorem 8 • 6.

Proof of Theorem 8 • 6 .:. We argue by induction on k 0. The result is clear when k 0. Suppose the result holds for some k > 0. Let X;R be a Borel graph such that deg.x/ k 1 forD all x X. Let Y X be a Borel kernel, and let Z DX h Yi. Then deg .v/ k for all v Z Ä(sinceC everyone needs2 to be connectedÂ to someone in Y as D n Z Ä 2 otherwise Y wouldn’t be maximal). Hence there exists a Borel .k 1/coloring c0 Z 0; ; k . Extend C W ! ¹ º c0 to a .k 2/coloring of X by c.y/ k 1 for all y Y . C D C 2 a

Õ Z Let 0 X;R be the graph associated with Z .2/ as before. Let Y X be a kernal. Then each connected D h i Â component of X Y is either an isolated point, or else a pair. So it is very easy to define a Borel 3coloring of 0. n Next we start working towards the following theorem of Marks, which uses Borel determinacy. 8 • 11. Theorem

For each n 1, there exists an nregular acyclic Borel graph with B./ n 1 D C 8 • 12. Deﬁnition

A marked group is a group with a specified set S 1 of generators. Â n 8 • 13. Deﬁnition For a marked group and X a standard Borel space, G.; X/ is the Borel graph with vertex set Free.X / y X y y for all 1 y D ¹ 2 W ¤ ¤ 2 º and edge set E defined by x E y iff S . x y y x/. 9 2 D _ D 8 • 14. Deﬁnition

If and are marked groups, then is the free product with generating set S S. [ 8 • 15. Theorem If , are finitely generated marked groups, then

B.G. ; N// > B.G.; N// B.G.; N// 1. C

To try to understand this, consider C2. Then each connected component of G.; N/ is just a pair. Thus D D B.G.; N// .G.; N// 2. Each connected component of G. ; N/ looks like Z: it’s just a line (but generated in aD different way).D By Theorem 8 • 15,

B.G. ; N// 2 2 1 3. C D Since each vertex has degree 2, B.G. ; N// 3. D The only thing mysterious or “yucky” abou thte theorem is the appearance of N. Fortunately, Seward–TuckerDrob have shown that every finitely generated group and all n 2 have B.G.; n// B.G.; N//. We, however, rely on N for the proof of Theorem 8 • 15. D

42 §8 MATH 569 CLASS NOTES

8 • 16. Corollary

For each n 1, let n C2 C2 (n times). Therefore G.n; N/ is a cyclic nregular graph with D B.G.n; N// n 1. D C Proof .:.

Clearly the Cayley graph of n is the nregular tree. It is also clear that B.G.1; N// B.G.C2; N// 2. D D Suppose inductively that B.G.n; N// n 1. Therefore by Theorem 8 • 15, D C B.G.n 1; N// B.G.n C2; N// C D B.G.n; N// B.G.C2; N// 1 C .n 1/ 2 1 n 2. C C D C Since the graph of n 1 is .n 1/regular, it follows that B.G.n 1; N// n 2. Hence B.G.n 1; N// n 2. C C C Ä C C D C a Theorem 8 • 15 is a consequence of the following main theorem of Marks. 8 • 17. Theorem (Marks' Main Theorem) If , are nontrivial countable groups and A Free.N / is Borel, then at least one of the following holds: (i) There exists a continuous, injective equivariantÂ ( x y implies f .x/ f .y/) map f Free.N / Free.N / such that im f A. D D W ! Â (ii) There exists a continuous injective equivariant (ıx y implies ıf.x/ f .y/) map f Free.N/ Free.N / such that im f N A. D D W ! Â n The proof of Marks’ Main Theorem (8 • 17) uses Borel determinacy. In fact, Marks’ Main Theorem (8 • 17) is equivalent to Borel determinacy modulo Z †1-Replacement DC. C C Before proving Marks’ Main Theorem (8 • 17), we derive some consequences.

Proof of Theorem 8 • 15 .:. Suppose that

B.G.; N// n 1 D C B.G.; N// m 1. D C Suppose that c G. ; N/ .n m/ is a Borel .n m/coloring. Let W ! C C A x G. ; N/ 0 c.x/ n 1 . D ¹ 2 W Ä Ä º Case 1. Suppose there exists a continuous, injective, equivariant f G.; N/ G. ; N/ such that im f A. Suppose that x; y Free.N / G.; N/ are adjacent.W Then! without loss of generality, Â 2 D there exists a S such that y x. Since f is equivariant, f .y/ f . x/ f .x/ and so f .x/, f .y/ are2 adjacent and soDc.f .x// c.f .y//. But this means c Df is a BorelDncoloring of ¤ ı G.; N/, a contradiction with the fact that B.G.; N// n 1. D C Case 2. By Marks’ Main Theorem (8 • 17), there then exists an injective, continuous, equivariant map f G.; N/ G. ; N/ such that W ! im f G. ; N/ A x G. Z/ n c.x/ .n m/ 1 . Â n D ¹ 2 W Ä Ä C º It follows as before that c f is a Borel mcoloring of G.; N/, a contradiction. ı a 8 • 18. Deﬁnition A graph V;E is bipartite iff there exists a partition V A B such that every edge e E joins a vertex v A to a vertexh w i B. D t 2 2 2

43 MATH 569 CLASS NOTES §8

8 • 19. Deﬁnition A perfect matching of a graph V;E is a collection M E such that every vertex v V lines on a unique edge e M . h i Â 2 2 8 • 20. Theorem (König's Theorem) An nregular, bipartite graph has a perfect matching.

8 • 21. Observation There exists a 2regular, bipartite, Borel graph with no Borel perfect matching.

Proof .:. Let V;E be Free.Z; 2Z/. Then every connected component has a graph that looks like Z. So each element’s edgeh is determinedi by the first choice of our edge in M , yielding an ability to choose elements. Formally, let M E be a perfect matching and let < be a Borel linear ordering of V . Then Â X0 v V v is the < least element of v e M D ¹ 2 W 2 2 º is 2Zinvariant and we reach a contradiction as before. a Question: What about Borelbipartite (meaning the pieces of the partition are Borel) graphs? Answer: the answer still is that there may not be a perfect matching. Consider the following theorem due to Marks. 8 • 22. Theorem For every n 1, there exists an nregular, acyclic, Borelbipartite graph with no Borel perfect matching. To prove this, we make use of the following theorem. 8 • 23. Theorem Õ Let , be countable groups and let E , E be the orbit equivalence relations for the Borel actions Õ Free.N / and free.N /. Then E , E do not have disjoint, Borel complete sections.

Proof .:.

Suppose that S, T are disjoint, Borel complete sections for E , E. We will apply Marks’ Main Theorem (8 • 17) with A T . Thus D S Free.N / A Free.N / T . Â n D n Case 1. Suppose there exists a continuous, injective, equivariant map f Free.N / Free.N / with im f A T . Then im f is a invariant subset such thatWS im f ! . Thus S is not a Â D ¤ ; \ D; complete section for E . Case 2. Otherwise, there exists a continuous, injective, equivariant map f Free.N / Free.N / with W ! im f Free.N / T S. This implies as before that T is not a complete section for E, a contradiction.Â n D a Proof of Theorem 8 • 22 .:. n Let , be cyclic of n 1. Let Y Free.N / be the standard Borel space consisting of the E classes Â and Eclasses.

Let be the intersection graph on i.e. s; t Y are adjacent iff s t . Note that each edge joins an E class 2 \ ¤ ; to an Eclass and so is Borel bipartite. Also is clearly nregular, since the action is free. Since is bipartite, it contains no odd cycles. Suppose contains an even cycle; say

s1; t1; s2; t2; ; s`; t` Without loss of generality, we can suppose the si are E classes.

Let x s1 t1. Then there exists 1 ı1 such that ı1x t1 s2. Similarly, there is a 1 1 ¹ º D \ ¤ 2 ¹ º D \ ¤ 2

44 §8 MATH 569 CLASS NOTES

such that 1ı1x s2 t2. Continuing in this fashion, there exist 1 ıi and 1 i for 1 i `¹ 1 suchº D that \ ¤ 2 ¤ 2 Ä Ä ` 1ı` 1 1ı1x s` t`. 2 \ But then as a cycle, there is a 1 ı` such that ¤ 2 ı` ` 1ı` 1 1ı1x t` s1. D \ But this is in the same orbit as x, and thus there is a such that 2 ı` ` 1ı` 1 1ı1x x, D contradicting the fact that acts freely. Now suppose that M is a Borel perfect matching of , and let A x Free.N / there exists an s; t M such that s t x . D ¹ 2 W ¹ º 2 \ D ¹ ºº Then A, Free.N / A are disjoint, Borel complete sections for E , E, contradiction. n a The following condition allows you to have disjoint complete sections. 8 • 24. Deﬁnition

Suppose that E0, E1 are countable Borel equivalence relations on a standard Borel space X. • E0 E1 is the smallest equivalence relation which contains E0, E1. _ • E0, E1 are everywhere nonindependent iff whenever C is an .E0 E1/class, then there exists a sequence _ of distinct elements x0; ; xn C with n 1 and a sequence i0; i1; ; in 0; 1 with ij ij 1 for 2 2 ¹ º ¤ C j < n and in i0 such that ¤ x0 Ei x1 Ei x2 xn Ei x0. 0 1 n Note that E F is a countable Borel equivalent relation. To see this, by Feldman–Moore Theorem (2 ,• 2) there exist _ Õ X X countable groups G0, G1, and Borel actions Gi X such that E E and F E . Let G G0 G1, and let D G0 D G1 D G Õ X be the corresponding action. Then E F EX . _ D G 8 • 25. Open Problem

Let fg be the space of finitely generated groups, and let R be the Borel relation defined by R iff there exist isomorphic (unlabelled) Caley graphs of and .

R isn’t an equivalence relation. Is the transitive closure of R Borel?

The proof of Marks’ Main Theorem (8 • 17) will make use of the following theorem. 8 • 26. Theorem

If E0, E1 are everywhere nonindependent countable Borel equivalence relations on the stadard Borel space X, then there exists a partition X X0 X1 such that each Xi is a Borel complete section for Ei . D t The above theorem makes use of yet another result. 8 • 27. Deﬁnition Let E be a countable Borel equivalence relation on X. < < • ŒE 1 is the Borel subset of the standard Borel space ŒX 1 consisting of the nonempty finite S X such that S is contained in a single Eclass. Â

B.E / !. Ä

45 MATH 569 CLASS NOTES §8

Proof .:. 2 First let gn n ! be a sequence of Borel permutations gn X X with g 1 such that if x; y X, then h W 2 i W ! n D 2 x E y x y or there exists n ! such that gn.x/ y. $ D <2 D Also fix some Borel linear order < of X. Given S ŒE , let S x0; ; xn where x0 < < xn. 2 1 D ¹ º Then c.S/, the color of the set, is the lexicographically least sequence ki;j i j such that for all i < j n, ¹ º ¤ Ä gk xi xj . i;j D < We will show that if S T ŒE 1 with S T , then c.S/ c.T /. So suppose S T are a counterexample. Then ¤clearly S2 T . Let \ ¤ ; ¤ ¤ j j D j j S x0; ; xn D ¹ º T y0; ; yn D ¹ º be the <enumerations. Let i; j n be such that xi yj . Ä D

Case 1. Suppose that i j . Then without loss of generality, i < j . Hence i < j implies xi < xj gk .xi /. ¤ D i;j But this says that yj < gk .yj / yi , which requires j < i, a contradiction. i;j D Case 2. So we must have that i j . But then for each ` i, we have x` gki;` .xi / gki;` .yi / y`. And so S T , a contradiction.D ¤ D D D D a Proof of Theorem 8 • 26 .:.

For each " 0; 1, let A";0 consist of those x X such that there exist S x0; ; xn B and j n such D 2 D ¹ º 2 Ä that x xj and ij ". D D

For example: if we start with x0 E0 x1 E1 x2 E0 x3 E1 x0 then x0; x2 A0;0 and x1; x3 A1;0. 2 2

Clearly A0;0 A1;0 . Also, note that for each " 0; 1, and x A";0, there exists a y ŒxE such that \ D; D 2 2 " y A1 ";0. We now inductively construct disjoint Borel sets A0;n, A1;n satisfying: 2 (a) A";n A";n 1, Â C (b) for each x A";n, there exists a y ŒxE" such that y A1 ";n. 2 2 2 Case 1. Suppose n is even. Then we define

A0;n 1 A0;n ŒA0;nE0 A1;n C D [ n A1;n 1 A1;n ŒA0;nE1 A0;n 1 . C D [ n C

Clearly A0;n 1, and A1;n 1 are disjoint and (a) holds. To see that (b) holds, first suppose that x A0;n 1 C C 2 C n A0;n. Then there exists (by saturation by E0) an x A0;n such that ŒxE Œx E , and by induction there 0 2 0 D 0 0 exists a y A1;n A1;n 1 such that y Œx0E0 ŒxE0 . Next, suppose that x A1;n 1 A1;n. As it’s in 2 Â C 2 D 2 C n the saturation of the previous one, there exists a y A0;n A0;n 1 such that ŒxE1 ŒyE1 . Thus (b) holds. 2 Â C D Case 2. Suppose n is odd. Then we define

A0;n 1 A0;n ŒA1;nE0 A1;n C D [ n A1;n A1;n ŒA1;nE1 A0;n 1 . C D [ n C 46 §8 MATH 569 CLASS NOTES

Arguing as above, the inductive hypotheses still hold.

Let X0 Sn ! A0;n and X1 Sn ! A1;n. Then it suffices to prove the following claims. D 2 D 2 Claim 1 X X0 X1. D t Claim 2

If " 0; 1, then X" is a complete Borel section for E1 ". D Proof .:.

Assuming Claim 1, consider the case where " 0. Let x X. If x X0, then clearly ŒxE X0 . D 2 2 1 \ ¤ ; Otherwise, by Claim 1, x X1 and hence there exists a y ŒxE such that y X0. 2 2 1 2 a Proof of Claim 1 .:.

Suppose z X. Then there exists x0 A0;0 ŒzE0 E1 . It follows that there exists a sequence 2 2 \ _ x0 Ei x1 Ei x2 xm Ei z, 0 1 m for i` 0; 1 . We can suppose inductively that xm X0 X1. Suppose, for example, that xm X0. Then 2 ¹ º 2 [ 2 there exists an even n such that xm A0;n. If z A0;n A1;n, we are done. If not, and z ŒA0;nE , then 2 2 [ 2 0 z A0;n 1. Otherwise, z ŒA0;nE1 and so z A1;n 1. 2 C 2 2 C a

a So we are almost ready to start proving Marks’ Main Theorem (8 • 17). Before finally beginning the proof, we need one more technical lemma. 8 • 29. Deﬁnition Let Y N be the set of all elements y N such that for all g , Â 2 2 gy gy for all 1 , and ¤ 2 n ¹ º ı gy ıy for all ı 1 . ¤ 2 n ¹ º Note that Y is invariant. Clearly, Free.N / Y . So what concerns us is the difference. Â Let E be the orbit equivalence relation on N and E be the orbit equivalence relation. 8 • 30. Lemma E .Y Free.N // and E.Y Free.N // are everywhere nonindependent. n n Proof .:. Let C Y free.N / be a . /orbit. Then there exists an x C and g . / . / such that Â n 2 2 n [ g x x. Suppose, for example, that g 1ð1 nın where i 1 , ıi 1 for 1 i n and D . We also suppose n is minimalD within C . 2 n ¹ º 2 n ¹ º Ä Ä 2

Case 1. Suppose 1. Then x, ınx, nınx, ..., ı1 nınx, 1ı1 ınx x witnesses nonidependence. D D Case 2. Suppose 1. Suppose 1 1. Then replacing 1 by 1 and x by x x, we obtain ¤ ¤ 10 D 0 D ı1 nınx x . And so we are in (Case 1). 10 0 D 0 Case 3. Suppose 1 1. Then replacing x by x0 x, we obtain ı1 2ı2 ın 1 nın x0 x0. By minimality of n, we haveD a contradiction. D D a Now we will prove Marks’ Main Theorem (8 • 17).

47 MATH 569 CLASS NOTES §8

Proof of Marks’ Main Theorem (8 • 17) .:. Each nonidentity element of can be uniquely written as a finite product of the form (i) i ıi i ; or 0 1 2 (ii) ıi i ıi ; 0 1 2 where ij 1 , and ıij 1 . Words of the form (1) are called words and words of the form (ii) are 2 n ¹ º 2 n ¹ º called words. We will make use of games for building an element y N , where I decides y on words, and II* decides y on words. 2

First we fix injective (possibly finite) listings 0, 1, ...; and ı0, ı1, ... of 1 and 1 respectively. n ¹ º n ¹ º Next we define the turn function t . / 1 N as follows. Suppose ˛ . / 1 has the form (i) W n ¹ º ! 2 n or (ii) with associated sequence i0, i1, ..., im (where the indices are from these fixed enumerations). Then t.˛/ is the least n such that ij j n for all j m. C Ä Ä

For example, the elements with t.˛/ 0 are 0, ı0. The elements with t.˛/ 1 are 1, 0ı0, 1ı0, and similarly D D ı1, ı0 0, and ı1 0. And so on.

B Write e for 1 . For each Borel subset B Y and k N, the following game G produces an element N k Â 2 B y N with y.e/ k. First we set y.e/ k. On the nth turn of Gk , first I defines y.˛/ on the words with2 t.˛/ n; and II definesD y.˛/ on the wordsD with t.˛/ n. D D B The winning conditions for Gk : • If y Y , then II wins iff y B. 2 2 1 1 Suppose that y Y . Then there exists an ˛ such that either 1 . ˛ y ˛ y/ or … 1 1 2 9 2 n ¹ º D ı 1 .ı˛ y ˛ y/. In the former case, we say that .˛; / witnesses that y Y , and in the latter case,9 2 .˛;n / ¹ witnessesº D that y Y . In both cases, we say that ˛ witnesses that y Y . … … … • If y Y and .e; / witnesses that y Y then I loses. Otherwise,… if .e; / witnesses that y … Y , then II loses. If neither of the above cases hold, then… I wins iff there is a word ˛ witnessing that y Y such that for all words ˇ with t.ˇ/ t.˛/, ˇ doesn’t witness that y Y . … Ä … Finally, recall that E .Y Free.N // and E .Y Free.N // are everywhere nonindependent. Hence n n there exists a Borel subsets C Y Free.N / such that C meets every Eclass on Y Free.N / and the c Â n n complement C meets every E class.

Now let A Free.N / be Borel. Then we define BA A C Y . So by Borel Determinacy, for each Â DBA [ Â k N, either I or II has a winning strategy in the game Gk . Hence one of the players has a winning strategy for2 infinitely many k N. 2 Suppose, for example, that S k N II has a winning strategy in GBA is infinite. Clearly there exists an D ¹ 2 W k º injective, continuous, equivariant map from Free.N / to Free.S /. Hence it’s enough to show that there is an injective, continuous, equivariant map f Free.S / Free.N / such that im f A. W ! Â We will define f such that for all x Free.S /, the following hold: 2 (i) f .x/. / x. / for all ; D 2 BA (ii) f .x/ will be a winning outcome for II’s winning strategy in Gx.e/. Clearly (i) ensures that f is injective. Suppose, for the moment, that f is equivariant and that f .x/ Y for all x Free.S /. I will help to ensure that these are both true. 2 2 Claim 1 With the suppositions above, f .x/ A for all x Free.S /. 2 2

48 §8 A MATH 569 CLASS NOTES

Proof .:.

BA Since f .x/ Y , and and f .x/ is an outcome in II’s winning strategy in Gx.e/, it follows that f .x/ BA 2 2 D A C . Suppose that f .x/ C Y Free.N /. Since Free.N / is invariant, it follows that if [, then f .x/ f . x/Y2 andÂ in factn is in C . And so f .x/ C . But this contradicts the fact that every2 orbit on Y DFree.N / intersects C c. Â n a The basic idea: in order to deal with the assumptions for asingle x, while II uses the winning strategy in the BA games Gx. 1/, I plays to ensure that f is equivariant and that f .x/. / x. /. D

BA Finally fix some x Free.S /. Then for each , we will play an instance of G 1 to produce an element x. / 2 2 1 y N which will be equal to f .x/. We begin by setting f .x/.e/ x. / for each (we’re playing infinitely2 many games simultaneously). D 2

Now suppose that f .x/.˛/ is defined for all and ˛ with t.˛/ < n. Then the nth move of I in 2 2 each game is as follows. Suppose ˇ is a word with t.ˇ/ n. Then ˇ i ˛ where i n and t.˛/ < n. 1 D D Ä For every , we define f .x/. i ˛/ to be . f .x//.˛/, which has been defined. Then II uses the winning 2 i strategy in all of these games to define . f .x//.ıi ˛/ for every word ıi ˛ with t.ıi ˛/ n. Clearly f is equivariant by the way I has played, and f .x/. / x. /, as a special case of equivariance.D Also, it is clear that f is continuous, since if the words agree on largeD initial segments, then the players play along those long initial segments. Thus it only remains to check that f .x/ Y for all x Free.S /. 2 2 First, since x Free.S / and f .x/ x, we have that f .x/ f .x/ for all 1 . Thus .e; / doesn’t witness2 that f .x/ Y . SinceDf .x/ is an outcome for II’s¤ winning strategy, 2.e; /n ¹doesn’tº witness f .x/ Y . Now we prove inductively… that ˛ doesn’t witness f .x/ Y for all x Free.S / and all ˛ with t.˛/… n. This is certainly true for the identity, so now… we proceed2 by inductionon n. First, suppose2˛ ˇ is a wordD with t.˛/ n and t.ˇ/ < n. Since D D 1 1 1 1 1 ˛ f .x/ ˇ f .x/ ˇ f . x/, 1 D D and ˇ doesn’t witness that f . x/ Y , it follows that ˛ doesn’t witness f .x/ Y . Let ˛ be a word with t.˛/ n. Hence we can suppose that… if ˇ is a word with t.ˇ/ n, then ˇ doesn’t… witness that f .x/ Y . SinceDf .x/ is an outcome of II’s winning strategy, it cannot be thatÄ˛ is the first time that a witness …appears (as that means that II would lose). It follows that ˛ cannot witness that f .x/ Y . Hence f .x/ Y , as desired. This … 2 B completes the proof, as the case is symmetric where S is the k N where I has a winning strategy for G A . 2 k a

§ 8 A. Recursive Isomorphism

We now state another theorem of Marks (recall F2 is the free group on two generators). 8 A • 1. Theorem

Let G 6 Sym.F2 N/ be a countable group of permutations such that for each g F2, there exists g G such 2 2 that g .h; n/ gh; n for all h; n F2 N. Then Õ FD2 hN i h i 2 (i) G 2 is a measure universal countable Borel equivalence relation. Õ F2 N (ii) G 3 is a universal Borel equivalence relation.

It’s open whether the action in (1) is universal. As stated, of course, this is hard to really understand. So instead we have the following corollary. 8 A • 2. Corollary (i) Recursive isomorphism on 2N is measure universal. (ii) Recursive isomorphism on 3N is countable universal.

49 MATH 569 CLASS NOTES §8 A

Proof .:.

N F2 N Let Y 2; 3 . Via computable bijection between N and F2 N, we can identify 2 and 2 . Then G 2 ¹ º D Rec.F2 N/ (recursive permutations of F2 N) satisfies the hypotheses of Theorem 8 A • 1. a Proof of Theorem 8 A • 1 .:. Throughout, Y 2; 3 , as much will be the same. Let E be the universal countable Borel equivalence relation 2Õ ¹ º F2 N1 arising from F2 X where X 2 . Let f X Y be any Borel map. We modify this to get a map from F2 N D W ! X to Y .

F2 N 1 Consider the associated fO X Y defined by fO .x/.h; n/ f .h x/.n/. Suppose x; y X and g F2 satisfies gx y. Then W ! D 2 2 D 1 1 .g f .x//.h; n/ f .x/.g h; n/ f .h gx/.n/ f .gx/.h; n/ f .y/.h; n/. O D O D D O D O Õ Õ F2 N Thus g f .x/ f .gx/ f .y/, and f is a Borel homomorphism from F2 X to F2 Y . O D O D O O We will define an injection f X Y N so that f witnesses either (i) or (ii) of Theorem 8 A • 1. W ! O Basic Idea: we will construct f such that for all G and x X, either 2 2 • f .x/ im f ; or O … O • f .x/ f .y/ for some y E x. O D O 1 We now introduce a definition to help with this. 8 A • 3. Deﬁnition 1 • G has type I iff for every k > !, there exist m; n > k with n m such that .1; n/ F2 m . 2 ¤ 2 ¹ º • G has type II if it is not type I, and there exists an m such that for infinitely many n N, 21 2 .1; n/ F2 m . 2 ¹ º • G has type III iff it is not type I nor type II. 2 1 Note that if g F2, then g has type III. If G has type III, then for all but finitely many n N, .1; n/ 2 2 2 2 F2 n . ¹ º 1 For each G of type II, fix some m such that there exist infinitely many n N with .1; n/ F2 m . We say that2 each such n witnesses that has type II. 2 2 ¹ º

To begin, we inductively construct a partition N S0 S1 S2 S3 such that S2 and S3 are infinite, and for every G, D t t t 2 1 • if has type I, then there exist an n S1 and m S0 such that .1; n/ F2 m ; 2 2 2 ¹ º • if has type II, then there are infinitely many n S2 and infinitely many n S3 such that n witnesses that has type II. 2 2 Now we make another definition, and then we actually prove something. 8 A • 4. Deﬁnition A set S N is good if whenever G has type II, then infinitely many n S witness this. Â 2 2

Hence S2 and S3 are good by definition. Furthermore, any good subset can be partitioned into two good subsets. During the proof, we will successively add more constraints to the map f X Y N . W !

´0 if m S0 Constraint 1. For every x X, define f .x/.m/ 2 2 D 1 if n S1. 2 Claim 1 If has type I, then f .x/ im f for all x X. O … O 2

50 §8 A MATH 569 CLASS NOTES

Proof .:. 1 There exists an n S1 and m S0 such that .1; n/ h; m for some h F2. Hence 2 2 D h i 2 1 . f .x//.1; n/ f .x/.h; m/ f .h x/.m/ 0, O D O D D since m S0. On the other hand, for all y X, 2 2 f .y/.1; n/ f .y/.m/ 1. O D D Thus f .x/ ran f . O … O a Claim 2 If 1 has type I, then f .x/ ran f for all x X. O … O 2 Proof .:. 1 Otherwise, there exist x; y X such that fO .x/ fO .y/; and so fO .y/ fO .x/, which contradicts Claim 1. 2 D D a

So we only need to worry about the elements where it and its inverse have type II or III. Let 0; 1; G 1 2 enumerate the elements of G such that both and have type II or III (not necessary both with the same type). Since S2 is good and every good set can be partitioned into two good sets, we can inductively define infinite, disjoint subsets S2;0, S2;1, of S2 such that: • If i has type II, then every n S2;i witnesses this; 21 • if i has type III, then every .1; n/ F2 n for every n S2;i . 2 ¹ º 2 S Constraint 2. For each i, let hi X 2 2;i be a Borel bijection. Then for every x X, define f .x/.n/ W ! 2 D hi .x/.n/ iff n S2;i . 2

So we’ve defined f on S0 and S1, and we’re dealing with S2 and S3. Next, let S3;00 , S3;0, S3;10 , S3;1, ..., be (finite and possibly empty) disjoint subsets of S3 such that: 1 • If i or have type II, then S 0 contains mi and/or m 1 provided they’re not already included in i 3;i i

S0 S1 S2 S .S 0 S3;j /, where mi is as in Definition 8 A • 3 for type II. [ [ [ j

Constraint 3. We define f .x/.n/ 0 if n S2 S S2;i or n S3 S S3;i . D 2 n i 2 n i

Finally, we will define f .x/.n/ for n S3;i by induction on i !. Suppose we have done this for j < i. In essence, there is only one candidate. 2 2 Claim 3 1 Suppose that i ; i . Then there exists a fixed Borel map g X X such that for any Borel map N 2 ¹ º W ! f X Y satisfying all our previous constraints before step i, if f .x/ f .y/, then y g.x/. W ! O D O D

51 MATH 569 CLASS NOTES §8 A

Proof .:.

First suppose that has type II. Let j (possibly j i). Then f .x/.m/ has already been defined for D each x X (being either 0 or else defined at an earlier stage). Thus for every n S2;j , 2 2 1 1 f .x/.1; n/ f .x/. .1; n// f .x/. n; m/ f . x/.m/, O D O D O D n for some n F2, has already been defined. Suppose that x; y X and that f .x/ f .y/. Then for each 2 2 O D O n S2;j , by constraint 2, 2 hj .y/.n/ f .y/.n/ fO .y/.1; n/ fO .x/.1; n/. S D D D 1 Since hj X 2 2;j is a bijection, there exists at most one such y; namely g.x/ h .n W ! D j 7! fO .x/.1; n//, which has already been defined.

Next, suppose that has type III. Then for every n S2;j , 2 1 1 f .x/.1; n/ f .x/. .1; n// f .x/. n; n/ f . x/.n/, O D O D O D n for some n F2, has been defined. As above, there exists at most one y such that f .x/ f .y/; namely 2 O D O h 1 of this function: y h 1.n f .x/.1; n//. j D j 7! O a

For each i, let gi X X be the Borel partial function defined by W ! gi .x/ y iff gi .x/ y and g 1 .y/ x. D D i D Therefore,

• If i f .x/ f .y/, then gi .x/ y. O D O D • gi is an injection. To finish the proof, it is enough to complete the construction of f such that for all x X and i, 2 either gi .x/ E x or i f .x/ f .gi .x//. (?) 1 O ¤ O We now continue with step i of the construction. First, suppose that i has type II. Choose some ni S3;i such 1 2 that ni witnesses that i has type II. Then there exists a i F2 such that i .1; ni / i ; mi . Recall that 1 2 D h i i f .x/.1; ni / f . x/.m / has been defined. Hence to ensure that i f .x/.1; ni / f .gi .x//.1; ni /, where O D i i O ¤ O then f .gi .x//.1; ni / f .gi .x/.n/, it is enough for each y X to define D 2 ´ 1 1 1 1 f . i gi .y//.mi / if gi .y/ is defined f .y/.ni / D 0 otherwise. And thus we’ve “killed off” all of type II. More precisely, we have the following. Claim 4 N If i has type II and f X Y satisfies our current constrains, then for all x X, i f .x/ im f . W ! 2 O … O

To sum up, for each i, there exists a Borel partial map gi X X such that W ! • i f .x/ f .y/, then gi .x/ y; and O D O D • gi is an injection. Moreover, we’ve dealt with types I and II.

Continuing with step i, suppose i has type III. Then S3;i 2 and for each n S3;i , there exists a n F2 such 1 j j D 2 2 that i .1; n/ n; n . For each n S3;i (all two of them), let gi;n X X be the Borel partial function D h i1 1 2 W ! defined by gi;n.y/ g .y/. Then if gi .x/ y and i f .x/ f .y/, for every n S3;i , D n i D O D O 2 f .y/.n/ f .y/.1; n/ i f .x/.1; n/ D O D O 1 f .x/. .1; n// f .x/. n; n/ D O i D O 1 f . x/.n/ f .gi;n.y//.n/ D n D

52 §8 A MATH 569 CLASS NOTES

In other words, f .y/.n/ f .gi;n.y//.n/. (†) D Until further notice, let Y 3. Let i;n be the Borel graph on X such that if x y, then x; y are adjacent iff D ¤ gi;n.x/ y or gi;n.y/ x. Then each vertex of i;n has degree at most two. Hence there exist Borel 3colorings D D ci;n X 3. For each n S3;i and y X, define W ! 2 2 f .y/.n/ ci;n.y/ (‡) D

Claim 5

If i has type III and i f .x/ f .y/, then x E y. O D O 1 Proof .:. 1 1 If i f .x/ f .y/, then y gi .x/, the unique candidate, and so g .y/ x is defined as well as g .y/ O D O D i D i;n for each n S3;i . Fix some n S3;i . 2 2 1 1 1 Case 1. Suppose gi;n.y/ y. Then y n gi .y/ n x and so they are in the same orbit and thus are E equivalent.D D D 1 Case 2. Next suppose gi;n.y/ y. Then the ci;n.y/ ci;n.gi;n.y//; and hence (†) and (‡) imply that ¤ ¤ i f .x/ f .y/. O ¤ O a

Õ F2 N This completes the proof of part (ii)of Theorem 8 A • 1. So now we wish to show part (i): that G 2 is measure universal. 8 A • 5. Observation Suppose that for every Borel probability measure on X, there exists a Borel subset A X with .A/ 1 Õ F2 N Â Õ F2 ND such that E A is Borel reducible to (the orbit equivalence relation of) G 2 . Then G 2 is measure universal.1

Proof .:. Let E be a countable Borel equivalence relation on a standard Borel space Z. Let be any Borel probability measure on Z. Let ' Z X be a Borel reduction from E to E . Let W '! , the pushforward. 1 D By the hypothesis, there then exists a Borel subset A X with .A/ 1 such that E A is Borel Õ F2 N 1 Â D 1 reducible to G 2 . Let Z0 ' .A/ by definition of the pushforward. Then .Z0/ 1 and Õ F2DN D EZ0 is Borel reducible to G 2 . a Let be any Borel probability measure on X. Now we use the following lemma to deduce the theorem. Then we prove the lemma. 8 A • 6. Lemma

For any standard Borel space X and Borel partial injections g0; g1 X X, there exists a Borel subset W ! A X with .A/ 1 and Borel maps c0; c1 A 2 such that for all x A either Â D W ! 2 1. there exists i 2 such that .gi A/.x/ is undefined; 2 2. there exists i 2 such that gi .x/ x; or 2 D 3. there exists i 2 such that ci .x/ ci .gi .x//. 2 ¤

Assuming Lemma 8 A • 6, we can complete the proof of part (i) as follows. Suppose i has type III so S3;i 2. j j D As before, for each n S3;i , let gi;n X X be the partial Borel injection defined by 2 W ! 1 gi;n.y/ n gi;n.y/, 1 D where .1; n/ n; n . Applying Lemma 8 A • 6, let Ai X be Borel with .Ai / 1 and let ci;n Ai 2 i D h i Â D W !

53 MATH 569 CLASS NOTES §8 A

be as in the lemma. For each n S3;i and y X, we define 2 2 ´ci;n.y/ if Ai f .y/.n/ 2 D 0 otherwise.

Claim 1

If i has type III and i f .x/ f .y/ for x; y Ai , then x E y. O D O 2 1 Proof .:. The proof is (practically) identical to Claim 5. a Let A T Ai i has type III . Then fO A is a Borel reduction from E A. This completes the proof of (i) of TheoremD ¹ 8 AW • 1 assuming Lemmaº 8 A • 6. So finally, we turn to the1 proof of the lemma.

Proof of Lemma 8 A • 6 .:.

For each i 2, let i be the Borel graph such that x y are adjacent iff gi .x/ y or gi .y/ x. Let Si be the Borel2 subset of those x X such that either ¤ D D 2 • x has no neighbors in i ; or • the connected component of x has an element of degree 1.

Then there exists a Borel 2coloring of Si . Hence, to simplify notation, we can suppose that S0 S1 , D D; since we can deal with these easily. Let Ei be the countable Borel equivalence relation on X defined by x Ei y iff x and y lie in the same connected component in i . Note that each Ei is aperiodic. 8 A • 7. Lemma

There exists a Borel partition X B C such that .ŒBE / .ŒC E / 1. D t 0 D 1 D

Assuming Lemma 8 A • 7, we can prove Lemma 8 A • 6 as follows. Let A ŒBE ŒC E . Then: D 0 \ 1 • for all a A, ŒaE B ; 2 0 \ ¤ ; • for all a A, ŒaE C . 2 1 \ ¤ ; Also .A/ 1. Let Ti be the Borel subset of a A such that ŒaE A. For each a Ti , either D 2 i 6Â 2 • a has no neighbors in i A; or

• the connected component of a in i A has an element of degree 1.

Thus there exist Borel 2colorings ci Ti 2. Hence, it is enough to consider the case when T0 T1 ; W ! D D; i.e. A is both E0invariant and E1invariant.

Let be the graph obtained from 0A by removing edges x; g0.x/ , where x B; and let be the 0 ¹ º 2 1 graph obtained from 1A by removing edges x; g1.x/ for x C . Then every connected component in ¹ º 2 is either a singleton, or else contains a vertex of degree 1. Hence there exist Borel 2colorings ci A 2 i W ! and of . Note that if x A, then either x B or x C . Hence either x; g0.x/ is an edge of or i 2 … … ¹ º 0 x; g1.x/ is an edge of . Thus either c0.x/ c0.g0.x//, or c1.x/ c1.g1.x//. ¹ º 1 ¤ ¤ a So all that remains is the (second) lemma.

54 §9 MATH 569 CLASS NOTES

Proof of Lemma 8 A • 7 .:. i i Since E0, E1 are aperiodic, by The Marker Lemma (4 ,• 17) there exist decreasing sequences C0 C1 i 0 1 Ã Ã of complete Borel Ei sections such that T Cn . Let Cn Cn Cn . Then C0 C1 and n D; D [ Ã Ã T Cn . Also, each Cn is a Borel complete section for both E0 and E1. n D;

Let A0 be any Borel complete E0section (e.g. the whole space). We will define inductively a sequence An n is even of Borel complete E0sections and a sequence Bn n is odd of Borel complete E1 ¹ W º ¹ W º sections, together with a strictly increasing sequence in of natural numbers.

Given An, we define Bn 1 as follows. Since An An T` C` S` An C`, it follows that X C D n D n D S`ŒAn C`E0 since An was a complete section. Hence there exists an in > in 1 such that .ŒAn Cin E0 / 1 n n c n 1 . =2/ . We define Bn 1 .An Cin / . Since Bn 1 Cin , Bn 1 is a Borel complete section for E1 c C D n 1 n C Ã C c which satisfies .ŒBn 1E0 / > 1 . =2/ . Similarly, given Bn, we define An 1 .Bn Cin / where C 1 n C D n in > in 1 satsifies .ŒBn Cin E1 / > 1 . =2/ . n c c c c Note that An and Bn 1 An Cin are disjoint; as are Bn and An 1. Also, c C D n c C c c An 2 .Bn 1 Cin 1 / .An Cin / Cin 1 .An Cin / Cin 1 An C C C C C D n Dc n c n D [ c n Ãc since Cin 1 Cin 1 An. Similarly, Bn 2 Bn. It follows that Sn even An, Sn odd Bn are disjoint Borel C Â Âc C Ã c sets such that S Bn meets almost every E0class and S An meets almost every E1class. n odd n even a

Section 9. Odds and Sods

[class missed, notes transcribed from lecturer’s notes]

Our next target is the following consequence of MC. 9 • 1. Theorem (MC) There exist uncountably many weakly countable universal Borel equivalence relations up to Borel reduction.

We first introduce yet another strong ergodicity notion. 9 • 2. Deﬁnition Suppose that E and F are countable Borel equivalence relations on the standard Borel spaces X and Y , and that is an Einvariant, Borel probability measure on X.

E is F ergodic iff for every Borel homomorphism f X Y from E to F , there exists a Borel subset Z X with .Z/ 1 such that f maps Z to a single F class.W ! Â D Note that if E is F ergodic and f X Y is a measurable homomorphism from E to F , then there exists a Borel subset Z X with .Z/ 1 suchW that! f maps Z to a single F class. Â D To see this, let g X Y be a Borel map where g.x/ f .x/ for almost every x X. Then W ! D 2 W x X g"ŒxE is not contained in a single F class D ¹ 2 W º is an Einvariant Borel subset of X with .W / 0. Hence, after adjusting g on W , we can suppose that g is a Borel homomorphism from E to F . The result follows.D

In Section 6, we proved the following theorem.

55 MATH 569 CLASS NOTES §9

9 • 3. Theorem N There exists a Borel family G˛ ˛ 2 of finitely generated groups, each with underlying set N, such that the following conditions hold.D ¹ W 2 º (a) G˛ has a normal subgroup N˛ SL3.Z/. Š (b) G˛ has no nontrivial, finite, normal subgroups.

(c) If ˛ ˇ, then Gˇ does not embed into G˛. ¤

N Õ G˛ N N For each ˛ 2 , consider the shift action G˛ 2 2 . Then the uniform product probability measure on 2 2 D is G˛invariant and N X˛ x 2 g x x for all 1 g G˛ D ¹ 2 W ¤ Õ ¤ 2 º satisfies .X˛/ 1. Let E˛ be the orbit equivalence relation of G˛ X˛. Applying Popa Superrigidity (6 A • 3), we obtain the followingD theorem. 9 • 4. Theorem

If ˛ ˇ, then Eˇ is E˛ergodic. ¤

In particular, E˛ is not weakly universal. On the other hand, T E˛ is clearly weakly universal. Á 9 • 5. Theorem

(MC) If ˛ ˇ, then . T Eˇ / 6B . T E˛/. ¤ Á 6 Á We will need to work in a forcing extension VP MA CH. So we first need to establish some absoluteness results. C: Recall the definition of MC from Martin’s Conjecture (7 • 5). We want to understand the complexity of this statement. For this, we need a parametrization of the Borel relations R 2N 2N . Â 9 • 6. Theorem (Classical Theorem) There exist subsets D 2N and P;S .2N /3 such that: 1 Â1 1 Â 1. D is …1, P is …1, and S is †1; 2. If d D, Pd Sd , where 2 D Pd x; y d; x; y P D ¹h i W h i 2 º Sd x; y d; x; y S . D ¹h i W h i 2 º For each d D, let Dd Pd Sd . 2 DN DN 3. Dd d D R 2 2 R is Borel . ¹ W 2 º D ¹ Â W º

Now consider F d D Dd is a function . D ¹ 2 W º 9 • 7. Observation 1 F is …1.

Proof .:. d F iff d D and 2 2 x y z .ŒS.d; x; y/ S.d; x; z/ y z/ 8 8 8 ^ ! D a

As a matter of notation, for each d F , let Fd be the corresponding Borel map. 2 1 It is now easily seen that MC is a …3 statement. We need to find a less complex formulation. 9 • 8. Deﬁnition N N (MC0) If f 2 2 is a Borel homomorphism from T to T, then either: W !N Á Á (a) for all x 2 , there exists x 6T y such that f .y/

56 §9 MATH 569 CLASS NOTES

9 • 9. Theorem MC MC . $ 0 Proof .:. N N Assume MC and let f 2 2 be a Borel homomorphism from T to T. Suppose there exists a cone N W ! ÁZ Á 6 C 2 such that f maps C to a single class Œr T . Then for each x 2 , there exists y C such that x T y Â Á 2 2 and f .y/ T r

N N Conversely, assume MC0 and let f 2 2 be a Borel homomorphism from T to T. If (a) holds from Definition 9 • 8, then W ! Á Á N A y 2 f .y/

9 • 11. Theorem Suppose that E, F are Borel equivalence relations on the standard Borel spaces X, Y and that f X Y is a Borel reduction from E to F . Therefore, if P is any notion of forcing then f P X P Y P is a BorelW ! reduction from EP to F P . W !

[end of class missed]

Let’s recall what we’ve done and what we’re working towards. 9 • 12. Theorem (MC) There exist uncountably many weakly countable universal Borel equivalence relations up to Borel reduction.

ℵ0 The idea is to have a whole bunch of countable Borel equivalence relations E˛, ˛ < 2 such that they are “mutually ergodic”, so that are really incompatible. They are not weakly universal, however.

The following are upwards absolute: • MC

• E 6B F for Borel equivalence relations E and F The thing we’re trying to prove is the following. 9 • 13. Theorem

(MC) If ˛ ˇ, then . T Eˇ / 6B . T E˛/. ¤ Á 6 Á We will use the following consequence of MA CH. C: 9 • 14. Theorem If is a Borel probability measure on a standard Borel space X and Z X is †1, then Z is measurable. Â 2 Proof .:. Z †1 A ˛ < ! Z A Z Since is 2, there exist Borel subsets ˛, 1 such that S˛

57 MATH 569 CLASS NOTES §9

Let B Z be a Borel subset such that Z B has inner measure 0 (meaning there’s no Borel subset of positive Â ˛ < ! .A nB/ 0 A B Z B measure). Then for each 1, ˛ . By MA CH, the union S˛

Suppose that y Eˇ x and r Cx. Then since r; y . T Eˇ / r; x , 2 h i Á h i y .r/ .r; y/ E˛ .x; y/ x.r/, D D and so y .r/ dx. Hence if y Eˇ x, then dy dx. 2 D

Consider the relation R Xˇ X˛ defined by Â R.x; z/ iff s r .s 6T r .r; x/ E˛ z/. 19 8 ! 1 Intuitively, this translates z dx. Then R is † . By Kondô’s theorem, there exists a † uniformization function 2 2 2 h Xˇ X˛ for R. Thus h.x/ dx for all x Xˇ . If U X˛ is open, W ! 2 1 2 2 h "U x Xˇ y .y U h.x/ y/ 1 1 D ¹ 2 W 9 1 2 ^ D º and so h "U is † . By MA CH, it follows that h "U is measurable. Thus h Xˇ X˛ is a 2 C: W ! measurable homomorphism from Eˇ to E˛. Since Eˇ is E˛ergodic, there exists a Borel Z Xˇ with Â .Z/ 1 such that h maps Z to a single E˛class; say, c: for every x Z, h.x/ dx c. D 2 D D N N For each x Z, let x 2 2 be the Borel map defined by x.r/ .r; x/. Then r is a Borel 2 W ! D homomorphism from T to T. If r; s Cx, then .r; x/; .s; x/ c; and hence r T s iff x.r/ T x.s/. Á Á 2 2 Á Á Thus x is a enduces a Borel reduction from T Cx to T. Hence by MC, there exists a cone Dx Œx"Cx T . Á Á Â Á

In particular, choosing x; y Z with ŒxE ŒyE , there exist r Cx and s Cy such that x.r/ T y .s/. 2 ˇ ¤ ˇ 2 2 Á But then f .r; x/ . T E˛/ f .s; y/, a contradiction. Á a Now we have a concept by Simon, with a name by Kechris. 9 • 15. Deﬁnition X A countable group G is (weakly) action universal iff there exists a standard Borel Gspace such that EG is (weakly) universal.

For notation, if G is countable and X is a standard Borel space, • E.G; X/ is the orbit equivalence relation of G Õ X G . • F.G; x/ is the free part of E.G; X/. We have the following easy theorems. 9 • 16. Theorem If the countable group G has a nonabelian free subgroup, then G is action universal.

Proof .:.

Since F2 embeds in G, it follows that E E.F2; 2/ 6B E.G; 2/. 1 D a

58 §9 MATH 569 CLASS NOTES

9 • 17. Theorem If G is a countable, amenable group, then G isn’t action universal.

The proof of Theorem 9 • 17 is delayed. First, we should define what “amenable” means. 9 • 18. Deﬁnition countable group G is amenable iff there exists a finitely additive probability measure P .G/ Œ0; 1 such that for all A G and g G, .Ag/ .A/. W ! Â 2 D Alternatively, under choice, G is amenable iff for every finite S G and " > 0, there is a nonempty finite A G such that for all s S, A As = A < ". This shows that amenabilityÂ is absolute. Â 2 j 4 j j j The two theorems above suggest the possibility of a “dynamical” version of the von Neumann conjecture: isittruethat if G is a countable group then the following are equivalent? i. G is action universal. ii. G contains a nonabelian free subgroup. If we replace (i) by “amenable”, this is von Neumann’s conjecture, and is false. Thomas’ conjecture is that the answer is no. Marks’ conjecture is that the answer is yes.

We begin working towards a proof of Theorem 9 • 17. First, the following theorem due to Day. 9 • 19. Theorem If G is a countable group, then the following are equivalent: 1. G is amenable. 0 2. There exists a sequence of functions fn G R such that fn `1.G/, fn 1 1 and such that for all g W g! 2 k k D g G, limn fn fn 1 0 where fn .h/ fn.hg/. 2 !1 k k D D

Remark: each fn can be regarded as a probability measure on G. Proof of Theorem 9 • 19 .:. The proof of (ii) from (i) involves functional analysis, and will be skipped. For (i) from (ii),foreach n !, 2 define n P .G/ Œ0; 1 by n.A/ Pa A fn.a/. W ! D 2

Let U be a nonprincipal ultrafilter on !. Then .A/ limU n.A/ satisfies our requirements. D

Here, for xn n ! a bounded sequence, limU xn is the unique ` R such that for each " > 0, n ! h W 2 i 2 ¹ 2 W rn ` < " U . It’s not difficult to show such an ` exists. j j º 2 a 9 • 20. Deﬁnition 0 et E be a countable Borel equivalence relation on X. Then E is 1amenable iff there exist Borel fn E R x W ! such that, letting fn .y/ fn.x; y/, x D x 1. fn `1.ŒxE / with fn 1 1. 2 k k x D y 2. If x E y, then limn fn fn 1 0. !1 k k D 9 • 21. Proposition X If G is a countable amenable group and X is a standard Borel Gspace, then EG is 1amenable.

Proof .:. X 0 Let fn n ! witness that G is amenable as in Theorem 9 • 19. For each n, define gn E R by h W 2 i W G ! gn.x; z/ X fn.g/. D g x z D x x X Clearly, if x X, then gn `1.ŒxE X / and gn 1 1. Suppose that x E y. Then there exists an h G such 2 2 G k k D G 2

59 MATH 569 CLASS NOTES §9

x that hx y. For each z ŒxE X , gn .z/ Pg x z fn.g/ and D 2 G D D y h 1 h 1 g .z/ X fn.g/ X f .gh/ X f .g/. n D D n D n gh x z gh x z g x z D D D h 1 x y Since fn f 1 tends to 0, it follows that g gn 1 also tends to 0. k n k k n k a In particular, hyperfinite Borel equivalence relations are 1amenable (because they can be realized by a Z action, and Z is amenable). It is not known whether the converse of this statement holds: it’s conceivable that hyperfiniteness and 1amenability are the same.

The converse of Proposition 9 • 21 doesn’t hold. And there is an interesting counterexample. Consider the action of GL2.Z/ (2 2matrices with determinant 1) on R given by ˙ [ ¹1º a b ar b Ä r C . c d D cr d C This action is hyperfinite. 9 • 22. Proposition Suppose G is a countable group and X is a standard Borel space with invariant probability measure . If G Õ X X is almost everywhere free, and EG is 1amenable, then G is amenable.

9 • 23. Corollary E isn’t 1amenable. 1 Proof .:.

Õ F2 F2 2 ; is almost everywhere free and F2 isn’t amenable. h i a Proof of Proposition 9 • 22 .:. X Let 'n n ! witness that E is 1amenable. Define h W 2 i G Z x fn.g/ ' .g x/ d.x/. D n Then clearly fn 0. Also Z x Z x Z x X fn.g/ X ' .g x/ d.x/ X ' .g x/ d.x/ X ' .y/ d.x/, D n D n D n g G g G g G y Œx 2 2 2 2 since the action is almost everywhere free. Note that by definition, this is just 1.

Finally, if h G, then by definition, 2 h fn f 1 X fn.g/ fn.gh/ k n k D j j g G 2 ˇZ x Z x ˇ X ˇ ' .g x/ d.x/ ' .gh x/ d.x/ˇ D ˇ n n ˇ g G ˇ ˇ 2 ˇZ x Z h 1x ˇ X ˇ ' .g x/ d.x/ ' .g x/ d.x/ˇ D ˇ n n ˇ g G ˇ ˇ 2 1 X Z ˇ'x.g x/ 'h x.g x/ˇ d.x/ Ä ˇ n ˇ g G ˇ ˇ 2 1 1 Z ˇ x h x ˇ Z x h x X 'n .g x/ ' .g x/ d.x/ 'n 'n d.x/, Ä ˇ ˇ D 1 g G ˇ ˇ 2 which goes to 0 as n goes to . 1

60 §9 MATH 569 CLASS NOTES

Thus to prove Theorem 9 • 17, it is enough to prove another proposition. Note that it suffices to prove (i), but doing so we need to prove (ii) and (iii). 9 • 24. Proposition Let E, F be countable Borel equivalence relations on X, Y . (i) If E is 1amenable and F 6B E, then F is 1amenable. (ii) If E is 1amenable and A X is Borel, then EA is 1amenable. Â (iii) If A X is a complete Borel Esection and EA is 1amenable, then E is 1amenable. Â Proof .:. Assuming (ii) and (iii), we can prove (i) as follows. Let f Y X be a Borel reduction from F to E. Then A f "Y is a Borel subset of X and there exists a Borel mapW g !A Y such that f .g.a// a for all a A. Also,D B g"A is a Borel complete F section. Note that gA isW a Borel! isomorphism betweenDEA and F2B. Also, byD (ii), EA is 1amenable and hence so is F B. Since B is a Borel complete F section, (iii) implies that F is 1amenable. 2 (ii) Let fn n ! witness the 1amenability of E. If A is Einvariant, then fnE A witnesses the 1amenabilityh W 2 ofiEA. Hence we can assume that A is a Borel complete Ehsection.\ Leti ' X A xW ! be a Borel map such that '.x/ E x for all x X. For x; y A with x E y, define gn .y/ to be x x 2 2 †z ' 1.y/fn .z/. Then gn n ! witnesses the 1amenability of E A. 2 h W 2 i (iii) Let ' X A be a Borel map such that '.x/ E x for all x X. For x; y X with x E y, define W ! 2 2 '.x/ ´fn .y/ if y A gx.y/ 2 n D 0 otherwise.

Then gn n ! witnesses the 1amenability of E. h W 2 i a 9 • 25. Deﬁnition G If G is a countable group, then Sg.G/ is the space of subgroups H 6 G (a compact subset of 2 ); and G is the conjugacy relation on Sg.G/: 1 K G L iff g G .gKg L/. 9 2 D 9 • 26. Theorem (MC) If G is a countable group, then the following are equivalent. (i) G is weakly universal. (ii) G is weakly action universal.

Proof .:. X That (i) implies (ii) is trivial. So suppose (ii) holds. Let X be a standard Borel Gspace such that EG is weakly universal. Suppose G isn’t weakly universal. Sonsider the Borel map ' X Sg.G/ given by '.x/ Gx X W ! D (the stabilizer). Then ' is a Borel homomorphism from E to G . G N X Next, let 2 X be a weak Borel reduction from T to EG ; and let ' . Then is a Borel W ! Á N D ı homomorphism from T to G . By MC, there exists a cone C 2 such that maps C to a single G class. Á Â By adjusting if necessary, we can suppose that there exists a fixed K 6 G such that G .r/ K for all r C . D 2 For later use, note that TC is weakly universal. Á 6w X X Let X0 x X Gx K . Then we have just seen that T C B EG X0 and so EG X0 is weakly D ¹ 2 W D º X Á universal. However, we will next show that EG X0 is essentially free, a contradiction.

X 1 1 Suppose x; y X0 and x E y. Then there exists a g G such that g x y. Since gKg gGxg 2 G 2 D D D Gy K, it follows that g NG .K/ (the normalizer). If h G also satisfies hx y, then hK gK. Hence X D 2 2 D D EG X0 is the orbit equivalence relation of the associated free Borel action of NG .K/=K. It follows that X D E X0 is essentially free, contradicting that it is weakly universal. G a

61 MATH 569 CLASS NOTES §9

But what can we prove from ZFC? Without MC, we can just prove the following much weaker version. 9 • 27. Theorem

If G is essentially free, then G is not weakly action universal. To give an idea of how little is known, consider the following question: is the converse true? Thomas’ conjecture (and belief) is “no”. Although he changes the conjecture in the last minute of class to be “yes”.

We will make use of the following theorem of Hjorth et al. 9 • 28. Theorem

If the countable group G has a nonabelian, free subgroup, then G is countable universal. Proof of Theorem 9 • 27 .:.

Suppose G is weakly action universal but that G is essentially free. Then there exists a countable H and a free 6 Z Z standard Borel Hspace Z such that G B EH . Let ' Sg.G/ Z be a Borel reduction from G to EH . Let L be a finitely generated group with no nontrivialW finite! normal subgroups suchthat L doesn’t embed into H and let SL3.Z/ L. D X Let X be a standard Borel Gspace such that EG is weakly universal and let 2 X be a weak Borel X W ! reduction from E.; 2/ to EG . Let X Sg.G/ be the Borel homomorphism defined by .x/ Gx (the stabilizer). So we have W ! D ' 2 X Sg.G/ Z. ! ! ! Z Let 2 Z be defined by ' . Then is a Borel homomorphism from E.; 2/ to EH . By Popa SuperrigidityW ! (6 A • 3), since L doesn’tD ı embedı into H , there exists a Borel subset Y 2 with .Y / 1 such Z Â D that maps Y to a single EH class.

Since ' is a Borel reduction, maps Y to a single G class. After adjusting if necessary, we can suppose ı that there exists a single subgroup K 6 G such that (the stabilizer) G .y/ K for all y Y . Let X0 X D 2 D x X Gx K . Then E X0 can be realized by the corresponding free action of NG .K/=K. Since ¹ 2 W D º G D Y is nontrivial, by Popa Superrigidity (6 A • 3), there exists an embedding , . Since SL3.Z/ 6 , ! it follows that contains a nonabelian, free subgroup. It follows that NG .K/ has a nonabelian, free subgroup. By Theorem 9 • 28, since G has a nonabelian, free subgroup, G is countable universal and hence not essentially free, a contradiction. a

Most questions concerning G are open. For example, note the following observation to motivate the first question. 9 • 29. Observation

If G, H are countable and there exists a surjective homomorphism G H , then H 6B G . W ! Proof .:. 1 Let f Sg.H/ Sg.G/ be defined by f .K/ "K. Then f is a Borel reduction from H to G . W ! D a 9 • 30. Open Problem

Suppose H 6 G. Does it follow that H 6B G ? 9 • 31. Deﬁnition subgroup H 6 G is malnormal if gHg 1 H 1 for all g G H . \ D 2 n

For example, let F2 a; b < F3 a; b; c . Then F2 is malnormal in F3. D h i D h i

Note that if H is a malnormal subgroup of G, then clearly H 6B G . Also, if H 6 G is a counterexample to Open Problem 9 • 30, then G has no nonabelian free subgroups. 

62 §9 MATH 569 CLASS NOTES

9 • 32. Open Problem

Let E be any countable Borel equivalence relation. Does there necessarily exist a countable G such that E B G ? Á 

Really, the question is what kinds of relations can be realized as G for some G? For now, we at least know there are uncountably many. 9 • 33. Theorem ℵ0 There exists an uncountable family G˛ ˛ < 2 of finitely generated, nonamenable groups such that if ˛ ˇ, ¹ W º ¤ then G and G are incompatible with respect to 6B. In particular, none are universal. And furthermore, each ˛ ˇ G is essentially free and hence not weakly universal. ˛ To prove this, we require a bit of background. 9 • 34. Deﬁnition

If H is any group, then the (restricted) wreath product C2 wr H is defined as follows. For each h H , let Ch ch Ì 2 1 D h i be cyclic of order 2. Then the base subgroup is B Lh H Ch, and C2 wr H B H where gchg cgh for g; h H. D 2 D D 2 9 • 35. Lemma

If H is a countable group and G C2 wr H, then E.H; 2/ 6B G . D Proof .:. For each A H, let Â KA M Ca 6 B 6 G. D a A Let g G be any element. Then there exist h H 2and b B such that g hb. Since B is abelian, 2 1 2 1 12 1 D gKAg hbKAb h hKAh KhA. D D D Thus A KA is a Borel reduction from E.H; Z/ to G . 7! a 9 • 36. Corollary

If H is an infinite sum of cyclic groups of order 2 and G C2 wr H , then G is nonsmooth and hyperfinite. D 

Each of our gropus will have the form G˛ C2 wr H˛ where H˛ is a finitely gnerated, simple, quasifinite group. In fact, every simple, quasifinite group is Dnecessarily finitely generated. So the “finitely generated” can be removed,as it’s redundant.

To see this, suppose that S is a counterexample. Then S is locally finite. Every infinite, locally finite group hasan infinite abelian subgroup (a nontrivial result). Since every proper subgroup is finite, it followsthat S must be abelian. But the only abelian, quasifinite groups are Z.p1/ (which isn’t simple) and Q (which isn’t simple). 9 • 37. Lemma

Suppose H is a simple, quasifinite group and X is a standard Borel H space. Let Y x X Hx 1 be the X X D ¹ 2 W ¤ º nonfree part of EH . Therefore EH Y is smooth.

Proof .:. X First let Z x X Hx H . Then E Z is a clearly smooth. So we can suppose that Z . Fix an D ¹ 2 W D º H D; element F of each of the countably many conjugacy classes of nontrivial finite subgroups of H . If x Y , 2 then Hx is a nontrivial finite subgroup of H. Let x be the corresponding cojugacy class containing Hx; and define .x/ y H x Hy F . D ¹ 2 W D x º We claim that .x/ is a nonempty, finite subset of Y . To see that .x/ is nonempty, choose g H such that 1 1 2 gHxg F and let y g x. Then Hy gHxg F and so y .x/. D x D D D x 2

63 MATH 569 CLASS NOTES §9

Next suppose that y; z .x/ and let h y z. Then 2 D1 1 hF h hHy h Hz F . x D D D x Thus h NH .Fx /. Since H is simple, NH .Fx / is a proper subgroup and hence is finite. Thus .x/ is finite. Clearly2 if H x H y, then .x/ .y/. Thus Y Y

Let H be a simple quasifinite group and let G C2 wr H . Then there exists a free standard Borel space Z such Z D that E 6B G . H Proof .:.

Let G H be the canonical surjection. Then Sg.G/ X0 X1 X1 X2, where W ! D t t t X0 K Sg.G/ "K H D ¹ 2 W D º X1 K Sg.G/ "K is a finite nontrivial subgroup of H D ¹ 2 W º X2 K Sg.G/ "K 1H D ¹ 2 W D ¹ ºº Let’s see how complicated each of these are. Let B be the base group for G as usual: B Lh H Ch. D 2 Claim 1

G X0 is smooth. Proof .:.

Suppose K X0 and g hb G be any element where h H and b B. Since "K H, there exists a c B such2 that k hcD K2. It follows that, as B is abelian,2 2 D 2 D 2 1 1 1 1 1 g.K B/g hg.K B/g h h.K B/h k.K B/k , \ D \ D \ D \ as B is normal and k normalizes K, this is just K B. Thus K B C G. Also since K=.K B/ H, \ \ \ Š it follows that K is finitely generated over K B and hence there are only countably many K0 X0 such that K B K B. \ 2 0 \ D \

Let be the equivalence relation on X0 defined by K K iff K B K B. Thus is a smooth Á Á 0 \ D 0 \ Á (witnessed by sending K K B) countable Borel equivalence relation. Since G X0 , it follows 7! \ ÂÁ that G X0 is smooth. a Claim 2

G X1 is smooth. Proof .:. Let F be a set of representatives of the counjugacy classes of countably many nontrivial finite subgroups 6 of H . For each F F , let XF K Sg.G/ "K F . Then G X1 B FF F G XF ; and so 2 D ¹ 2 W D º 2 it is enough to show that each G XF is smooth.

Fix some F F and let K XF . Since K=.K B/ F , there are only countably many K XF such 2 2 \ Š 0 2 that K0 B K B. Hence if is the equivalence relation on XF defined by \ D \ 1 K K0 iff h NH .F / .h.K B/h K0 B/, 9 2 \ D \ then is a countable Borel equivalence relation.

Since H is simple, NH .F / is a proper subgroup and hence is finite and so is smooth. Hence it is enough to show that G XF . Â 1 Suppose K;K0 XF and gKg K0. Let g hb, where h H and b B. Then clearly h NH .F /. Also, K B 2g.K B/g 1 Dh.K B/h D1 and so K K2 . 2 2 0 \ D \ D \ 0 a

64 §9 MATH 569 CLASS NOTES

By Claim 1 and Claim 2, G is Borel bireducible with G X2. Suppose K X2 and g hb G, where 1 1 2 D 2 h H and b B. Then gKg hKh . Thus G X2 can be realized by the corresponding Haction. Let 2 2 D Z Z X2 be the free part of this action. By the previous lemma, G X2 E as a simple, quasifinite group. Â H a We will make use of the following theorem of Ol’shanskii (and folklore). 9 • 39. Theorem ! If is a noncyclic, torsionfree, hyperbolic group, then has a family H˛ =N˛ ˛ < 2 of uncountably many nonisomorphic simple, quasifinite quotients. ¹ D W º

To apply Popa Superrigidity (6 A • 3), we want a Kazhdan group. So to use Theorem 9 • 39, we need to be careful about the group we’re taking quotients of. Luckily there is a Kazhdan group with these properties. Since the quotients are Kazhdan, we will be able to use Popa Superrigidity (6 A • 3). 9 • 40. Theorem There exists a noncyclic, torsionfree hyperbolic Kazhdan group.

Proof of Theorem 9 • 33 .:.

Let be a noncyclic, torsionfree, hyperbolic Kazhdan group as per Theorem 9 • 40; and let H˛ =N˛ ˛ < ! ¹ D W 2 be a family of nonisomorphic, simple, quasifinite quotients. Then each H˛ is also a Kazhdan group. º

Let G˛ C2 wr H˛. Suppose that there exist ˛ ˇ such that G 6B G . Since E.H˛; 2/ 6B G , D ¤ ˛ ˇ ˛ we have that E.H ; 2/ 6 . Let Z be a free standard Borel H space such that G 6 EZ . Then ˛ B Gˇ ˇ ˇ B Hˇ Z E.H˛; 2/ 6B E . So by Popa Superrigidity (6 A • 3), there exists a virtual embedding H˛ Hˇ . Since H˛ H W ! is simple, is an embedding. Since H˛ Hˇ , .H˛/ is an infinite, proper subgroup of Hˇ , which contradicts 6Š that Hˇ is quasifinite. a Now we discuss the conjecture of Marks. Really we will look at a slight weakening of his actual conjecture (self described as “ridiculously optimistic”). 9 • 41. Deﬁnition Let G be a countable group and let X be a standard Borel Gspace. X Then EG is uniformly universal iff whenever H is a countable group and Y is a standard Borel H space, then there Y X exists a Borel reduction f Y X from EH to EH such that there exists a map (just a function) u H G satisfying W ! W ! f .hy/ u.h/f .y/ (*) D for all y Y and h H . 2 2 In this case, we can suppose that • u.1/ 1; D 1 1 1 • if h h , then u.h / u.h/ (as we will check). ¤ 1 D To see this, suppose h h . By assumption, for all y Y , ¤ 2 1 f .hy/ u.h/f .y/ and so f .y/ u.h/ f .hy/. D D Let z Y be arbitrary and let y h 1z. Then f .h 1z/ u.h/ 1f .z/. 2 D D 9 • 42. Open Problem (Marks) If E EX is countable universal, then EX is uniformly universal (for each G realizing E EX ). D G G D G As a convention, from now on, .2! /G f f G 2! . D ¹ W W ! º 9 • 43. Example ! E.F! ; 2 / is uniformly universal. (Here F! is the free group on infinitely many generators.)

65 MATH 569 CLASS NOTES §9

Proof .:. Let H be a countable group and Y be a standard Borel Hspace. Recall that there exists an injection ' Y ! H W ! .2 / such that for all y Y and h H , '.hy/ h'.y/. Let F! H be a surjective homomorphism and define .2! /H .22! /F! by 2 D W ! W ! ! H .x/.g/ x..g// x .2 / ; g F! . ! D ! 2 2 Then is a Borel reduction from E.H; 2 / to E.F! ; 2 /.

1 1 ! H Let u H F! satisfy .u.h/ / h for h H. If x .2 / , h H , and g F! , then W ! D 1 2 2 1 2 12 .hx/.g/ .hx/..g// x.h .g// x..u.h/ g// .x/.u.h/ g/ u.h/ .x/.g/. D D D D D Thus .hx/ u.h/ .x/, as desired. D a We have the following theorem due to Marks. 9 • 44. Theorem X If G is a countable group, then there exists a standard Borel Gspace such that EG is uniformly universal iff G has a nonabelian, free subgroup.

Proof .:.

( ) Suppose G has a nonabelian, free subgroup. Then there exists an embedding F! G. Fix some ! ! F! ! G W ! p0 2 , and define f .2 / .2 / by 2 W ! ´x. 1.g// if g im f .x/.g/ 2 D p0 otherwise. ! ! Then f is a Borel reduction from E.F! ; 2 / to E.G; 2 /; and it is easily checked that f .hx/ .h/f .x/ ! F! D for all h F! and x .2 / . Taking compositions with the uniformly universal action from Example 9 • 43 yields2 that E.G;2 2! / is uniformly universal. X ( ) Suppose there exists a standard Borel Gspace X such that EG is uniformly universal. Let ¨i ! i ! ! ! XD 2 where each i F2. Then there exists a Borel reduction f .2 / X from F.; 2 / to EG and a map u G suchD that for all y .2! / and , W ! W ! 2 2 f . y/ u. /f .y/. D Applying Theorem 2 • 6 to R ® f .y/; y y .2! / ¯ , ! D h i W 2 there exists a partition .2 / Fi ! Ai (seen as the free part) into Borel pieces Ai such that f Ai is D 2 injective. By an extension of Marks’ Main Theorem (8 • 17), there exists an i ! and a i equivariant, ! ! F2 2 ! F2 injective, Borel map g going from .2 / i .2 / to Ai . Let ' f g. Then ' .2 / X is an ! F2 D D ı W ! injection; and if x .2 / and F2, then 2 2 '. x/ f . g.x// u. /'.x/. D D 1 1 1 Let F2 ˛; ˇ . Then, after adjusting u if necessary, we can suppose u.˛ / u.˛/ and u.ˇ / 1D h i ! F2 D D u.ˇ/ . Let a u.˛/ and b u.ˇ/. If x .2 / and w.˛; ˇ/ is a nontrivial reduced word in ˛; ˇ, then '.w.˛; ˇ/x/D w.a; b/'.x/D . Since w.˛;2 ˇ/x x and ' is an injection, it follows that w.a; b/ 1. Then a; b is a freeD subgroup of G. ¤ ¤ h i a Finally, we prove the following theorem. 9 • 45. Theorem

There exists a periodic group G of bounded exponent such that G isn’t essentially free. To prove this, we need some preparation in the form of two theorems of Ol’shanskii. 9 • 46. Theorem n If H is a noncyclic, torsionfree, hyperbolic group, then there exists an integer nH such that H=H is infinite for all odd n nH .

66 §9 MATH 569 CLASS NOTES

9 • 47. Theorem ! For every sufficiently large, odd n, there exists a family G˛ ˛ < 2 of nonisomorphic, 2generator, simple groups of exponent n. ¹ W º

So we have our big guns, and can move on to proving Theorem 9 • 45

Proof of Theorem 9 • 45 .:. Let H be a noncyclic, torsionfree, hyperbolic, Kazhdan group and let n be a sufficiently large, odd integer. Then n ! K H=H is an infinite Kazhdan group of exponent n. Also, let G˛ ˛ < 2 be a family of nonisomorphic, 2generator,D simple groups of exponent n. Let K be a dgenerator¹ group,W and letº B be the free Burnside group ! on d 2 generators of exponent n. Then for all ˛ < 2 , K G˛ is a homomorphic image of B and so C E.K G˛; 2/ 6B E.B; 2/. Claim 1 E.B; 2/ isn’t essentially free.

Proof .:. 6 Y Suppose there exists a countable H and a free standard Borel H space such that E.B; 2/ B EH . For ! Y each ˛ < 2 , we have that E.K G˛; 2/ 6B E.B; 2/ 6B E . Then Popa Superrigidity (6 A • 3) implies H that there exists an embedding ˛ G˛ H . Since we have uncountably many such ˛, there exist W ! uncountably many ˛ ˇ such that ˛"G˛ ˇ "Gˇ and hence G˛ Gˇ , a contradiction. ¤ D Š a

Let G C2 wr B. Then E.G; 2/ 6B G and so G isn’t essentially free. D a