GROUPS AND DYNAMICS ARISTOTELIS PANAGIOTOPOULOS

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1 GROUPS AND DYNAMICS ARISTOTELIS PANAGIOTOPOULOS Introduction This is a topics course in topological groups and dynamics. We will be focusing on techniques and phenomena that go beyond the realm of locally compact groups. In the absence of classical techniques which rely on local compactness, we will develop various frameworks for analyzing the structure of large topological groups, such as: Fraïssé theoretic methods; Baire category methods; structured completions and compactifications. As we develop these techniques we will provide a wide range of applications and examples from topology, analysis, and logic. Here are three centerpieces that, among others, will be covered: Extreme amenability. Consider the automorphism group Aut(Q, <), of all order preserving bijections of the rationals. Endowed with the pointwise convergence topology it has the astonishing property that whenever it acts continuously on a compact space K, K has a fixed point under the action. As we will see, this is intimately tied to a combinatorial principle known as the Ramsey theorem. We will develop the theory of a more general phenomenon known as the KPT-correspondence (for Kechris, Pestov, Todorcevic). Anti-classification results. Dynamics play a very important role in logic because they allow us to prove negative results about classification projects in mathematics. We will see for example that there is no constructive way of classifying countable graphs up to isomorphism by assigning to them enough meaningful invariants. By developing more intricate tools (such as Hjorth s turbulence theory) we will see that similar classification projects from analysis and topology are even more hopeless (e.g. classifying unitary operators up to unitary equivalence). Exotic groups. Consider the group Homeo + ([0, 1]), of all orientation preserving homeomorphisms of the compact space [0, 1]. Endowed with the compact-open topology it has no non-trivial continuous representations on a Hilbert space. Even more, by a recent result of Megrelishvili, it has no non-trivial representation on any reflexive Banach space. As we will see, this is intimately tied to the the fact that the group Homeo + ([0, 1]) has no compactification on which the multiplication of Homeo + ([0, 1]) extends to a seperately continuous semigroup operation. 1

2 2 ARISTOTELIS PANAGIOTOPOULOS 1. Week 1 A Polish space is a topological space X that is separable and completely metrizable, i.e., it admits a complete metric d that induces the topology on X. Here are some facts that we are going to use regarding Polish spaces. Let X be a Polish space. A set A X is nowhere dense if for every non-empty open U in X, the set U \ A is non-empty. A subset M X is meager if M is a countable union of nowhere dense sets. A subset C X is co-meager if C c is meager. Fact 1. Let X be a Polish space. (1) If Z is a subspace of X (with the subspace topology) then Z is Polish if and only if Z is G δ in X. (2) Baire category theorem. If (C n ) is a sequence of dense and open sets in X then n C n is dense. (3) every comeager set C contains a dense G δ. A topological group is a group G endowed with a topology which makes the operations (g, h) g h, g g 1 continuous. Check. One can equivalently define topological groups as those groups endowed with a topology which makes (g, h) g h 1 continuous. Definition 2. A Polish group is a topological group whose topology is Polish. Countable discrete groups Γ such as (Z, +), (F 2, ), etc., or more generally, locally compact metrizable groups such as (R, +), SL 2 (R), etc., are Polish. However, our treatment will include Polish groups which are far from being locally compact. Here are some families of Polish groups which we will consider. (1) Permutation groups. Consider S to be the group consisting of all bijections from N to N where the group operation is composition (gh)(n) = g(h(n)). The topology on S is the pointwise convergence topology, i.e., the smallest topology with the property that (g k ) converges to g, if and only if, for all nn we have that lim k g k (n) = g(k), or equivalently, k 0 k > k 0 g k (n) = g(n). In other words, the collection of all sets of the form: V A,f : = {g S f A = g A}, where A varies over all finite subsets of N and f varies over elements of S, constitutes a basis for the topology on S. To see that S is a Polish group notice that the subgroup S fin of all permutations with finite support is countable and dense in S and that S is a G δ subset of the Baire space N N of all functions from N to N endowed also with the pointwise convergence topology. Alternatively one can directly define a complete metric for S as follows. First consider the metric d(g, h) = 2 n iff g h and n is the smallest natural number so

3 GROUPS AND DYNAMICS 3 that g(n) h(n). Let also D(g, h) = d(g, h) + d(g 1, h 1 ). Then D is a complete metric on S compatible with the topology. Check. The metric d is compatible with the topology (and it is left invariant) but it is not complete! (d keeps track of where n maps to, but not what is mapped on n) We will be interested in closed subgroups of S. As an example consider the group Aut(Q, ), of all order-preserving bijections from Q to Q. By fixing an enumeration of Q we can view Aut(Q, ) as a subgroup of S, which is moreover closed, since not being in Aut(Q, ) can be detected by the fact that a finite A = {a b} Q is mapped to f(a) f(b). In fact, we will see that all closed subgroups of S are of the form Aut(M), for some appropriate L-structure M on domain N. (2) Isometry groups. Let (X, ρ) be a Polish metric space and consider the group Iso(X, ρ) of all isometries of X with multiplication being the composition. An isometry of X is any bijection g from X to X with ρ(x, x ) = ρ(g(x), g(x )). As in (1), we view Iso(X, ρ) with the pointwise convergence topology, i.e., the collection of all sets of the form: V A,f,ε : = {g Iso(X, ρ) for all x A ρ(g(x), f(x)) < ε}, where A varies over all finite subsets of Iso(X, ρ), f varies over elements of Iso(X, ρ), and ε is any positive real, constitutes a basis for the topology on Iso(X, ρ). It is easy to see that Iso(X, ρ) is a G δ subset of the Polish space Emb(X, ρ) of all isometric embeddings from X to X endowed with the pointwise convergence topology. It is therefore Polish. We can build a left invariant metric d on Iso(X, ρ) by fixing some dense sequence (x n ) in X and setting d(g, f) = n 1 2 n ρ(f(x n ), g(x n )) 1 + ρ(f(x n ), g(x n )) As in (1), one can check that while the metric d is not complete in general, the metric D given by D(f, g) = d(f, g) + d(f 1, g 1 ) is a complete and compatible metric on Iso(X, ρ). In fact, a corollary of Theorem 8 implies that D is complete, given that d is left invariant. There are many Polish groups of interest, which are closed subgroups of Iso(X, ρ), for various (X, ρ). One example is the unitary group U(H) of any separable Hilbert space H. Moreover, notice that the examples under (1) are in fact examples of the form (2), since S is of the form Iso(X, ρ), where X is N and ρ is the discrete {0, 1}-valued metric. (3) Homeomorphism groups. Let X be any compact metrizable topological space and let Homeo(X) be the group of all continuous bijections from X to X. We view Homeo(X) as a topological group endowed with the compact-open topology. Since Homeo(X) is a G δ subset of the Polish space C(X, X) it is Polish. To find an explicit complete metric we can again invoke Theorem 8 after we provide a left invariant metric for Homeo(X). Let ρ be any compatible metric on X and let d u be

4 4 ARISTOTELIS PANAGIOTOPOULOS the uniform metric on Homeo(X) with respect to ρ, i.e., d u (g, f) = sup{ρ(f(x), g(x)) x X}. It is easy to see that d u (g, f) is a compatible right invariant metric. Hence d(f, g) = d u (f 1, g 1 ) is a compatible left invariant metric. The usual, by now, recipe gives as a complete metric D that is compatible with the topology. Exercise. Show that ρ is not complete, in general, by constructing a sequence (g n ) of elements of Homeo(X) which converge to the function whose graph is (1, 1) ( 1 4, 1 2 ) ( 3 4, 1 2 ) (0, 0) (4) Banach spaces. Any separable Banach space is an abelian Polish group with multiplication being vector addition. Important examples will be the Banach spaces (l p, +), p [1, ), of all p-summable sequences. When it comes to Polish groups the extra uniform structure gives a strengthening of Fact 1 (1). Lemma 3. Let G be a Polish group and let H be a Polish subgroup of G. Then H is closed. Proof. By 1(1) H is G δ and dense in the Polish group H. If g H \ H then gh would also be dense G δ in H but since it would be a different coset of H we would have H gh =, contradicting Fact1(2). The rigidity resulting from the combination of the Baire category theorem and the uniform structure of a Polish group is best exemplified by Banach-Pettis lemma and its corollaries. To motivate this result consider the following question: Question 4. Let End((R, +)) be the collection of all homomorphisms π : (R, +) (R, +), of the additive group (R, +). (1) Classify End((R, +)); (2) Classify the definable (e.g., Borel) members of End((R, +)); (3) Classify the continuous members of End((R, +)); For the first problem, we can view R as a vector space over Q and by axiom of choice fix a Hamel basis Hamel Q (R). End((R, +)) can be now seen to be in bijective correspondence with the collection of all functions from Hamel Q (R) to R. The

5 GROUPS AND DYNAMICS 5 last problem has also a simple solution. Let π be a continuous homomorphism. Continuity says that π is determined by π Q and the fact that π is a homomorphism says that π Q is completely determined by π {1}. The continuous members of End((R, +)) are in bijective correspondence with the collection of all functions from {1} to R,i.e., with R. While the second problem looks difficult, at a first glance, Banach-Pettis lemma reduces it to the third problem which we have already solved. Recall the collection of Borel sets in a topological space X is the smallest σ-algebra B(X) containing the open sets of X and the collection of Baire measurable sets is the smallest σ-algebra BP(X) containing the open sets and the meager sets of X. Alternatively, BP(X) consists of all sets of the form M U = (M/U) (U/M), where U is open and M is meager. Definition 5. Let X be a Polish space and let A be any subset of X. We will denote by U(A) the largest open set in which A is comeager. Before we proceed we give a brief justification that this definition makes sense, i.e., that such U(A) always exists. To see this, let O(A) be the collection of all open sets in which A is comeager and set U(A) := {U U O(A)}. It follows that U(A) is open and it overlays every open set in which A is comeager. So we are left to show that A is comeager in U(A). This follows by the Baire category theorem since by separability of X we can find a countable collection (U n ) of elements from O(A) so that n U n = {U U O(A)}. Lemma 6 (Banach-Pettis). Let G be a Polish group and let A, B BP(G). Then U(A)U(B) AB. Proof. If x is in U(A)U(B), then the set xu(b) 1 U(A) is open and non-empty, since x = a b = x b 1 = a. But xu(b) 1 U(A) = U(xB 1 )U(A), since inversion and translation are homeomorphisms of G. Hence both xb 1 and A are comeager in the non-empty open set U(xB 1 )U(A) and therefore xb 1 A, i.e., x AB. Theorem 7. If π : G H is a Baire-measurable homomorphism between Polish groups then π is continuous. Proof. It suffice to show that π is continuous at 1 G. Let therefore V be an open neighborhood of 1 H = π(1 G ) in H and notice that by continuity of (x, y) x y and x x 1 we can find an open W V in H, with 1 H W and W W 1 V. But then, π 1 (W ) is Baire-measurable and non-meager, since countably many translates of π 1 (W ) cover G (because the same is true for W and H). By Banach- Pettis lemma we have: 1 G U(π 1 (W ))U(π 1 (W )) 1 π 1 (W )π 1 (W ) 1 π 1 (V ). In the sequel we will see Polish groups which satisfy the much stronger property that arbitrary homomorphisms mapping out of them are continuous. In other words,

6 6 ARISTOTELIS PANAGIOTOPOULOS their algebra dictates their topology. A Polish group like this is S. However not every Polish group has this property. Exercise. Show that the abstract group (R, +) admits many different Polish topologies, i.e., there are two Polish groups which are not continuously isomorphic but if you forget the topology, both are isomorphic to the group (R, +). Show actually that (R, +) admits infinitely many different Polish topologies. Exercise. (Solecki) (1) Let (H n ) be a collection of subgroups of the Polish group G. Assume that H n has the Baire-property, for all n, and that G \ n H n is meager. Show that there is n so that G/H n is countable. (2) the group n (Z 2 n)n does not carry a Polish group topology. Project 1. There are many topological groups which do not admit any Polish topology. Moreover the arguments involved in showing that such a group is not Polish can be quite intricate. Collect examples of such groups and simplify/conceptualize existing proofs (see Shelah, Mann, Solecki, Rosendal,...) As we pointed out in the examples, for a Polish group G, in order to find a compatible complete metric it suffice to find a left invariant metric. Theorem 8. Let G be a topological group endowed with a compatible metric d that is left invariant. Let also D(g, h) := d(g, h)+d(g 1, h 1 ). Then D is also a compatible metric for G, and if (Ĝ, D) is the completion of (G, D) then the multiplication of G extends uniquely to Ĝ, so that Ĝ becomes a topological group, in the topology generated by D. Before we prove Theorem 8 we observe that together with Lemma 3 it implies the following corollary.. Corollary 9. If G is a Polish group and d is a compatible left invariant metric on G then D(g, h) := d(g, h) + d(g 1, h 1 ) is a compatible complete metric on G. Proof of Theorem 8. Uniqueness of multiplication follows from the fact that G is dense in Ĝ. So it suffice to show existence. For that we first record the following claim and we show how existence follows from this claim. Then we prove the claim Claim. If ˆx, ŷ Ĝ and ε > 0 then there exists a δ > 0 so that if x, x X are δ-close to ˆx (with respect to D) and y, y are δ-close to ŷ (with respect to D) then D(xy, x y ) < ε. Given this claim the rest follows easily. Since if x n ˆx and y n ŷ then the claim shows that (x n y n ) is Cauchy and therefore it converges to some ẑ Ĝ. Moreover by the claim, ẑ will not depend on the choice of (x n ), (y n ). This defines a continuous (by the claim) multiplication on Ĝ which extends the one on G, it is associative, and 1 is the identity. For inversion, notice that x x 1

7 GROUPS AND DYNAMICS 7 is a D-isometry and therefore it uniquely extends to a D-isometry ˆx ˆx. Using the fact that multiplication is continuous on Ĝ it follows that ˆx ˆx = ˆx ˆx = 1. So, we are left with proving the claim. Proof of Claim. First observation that by triangle inequality and left invariance of d we have that d(gh, 1) d(g, 1) + d(h, 1) for all g, h G. If now x, x, x 0 and y, y, y 0 are in G, by left invariance of d and the above observation we have that: D(xy, x y ) = d(xy, x y ) + d(y 1 x 1, (y ) 1 (x ) 1 ) = = d(1, y x x 1 y 1 ) + d(x y y 1 x 1, 1) ( d(1, y y 0 1 ) + d(1, y 0 x x 1 y 0 1 ) + d(1, y 0 y 1 ) ) + + ( d(x x 0 1, 1) + d(x 0 y y 1 x 0 1, 1) + d(x 0 x 1, 1) ) Fix now some x 0 and some y 0 which are ε/10=close to ˆx and ŷ respectively. Notice now that if we chose δ to be any number less than ε/10 and x, x, y, y are chosen to be δ-close to ˆx, ŷ respectively then each of the terms d(1, y y 1 0 ), d(1, y 0 y 1 ), d(x x 1 0, 1) and d(x 0 x 1, 1) is less than δ + ε/10 < ε/5. So, by restricting δ to be perhaps even smaller, we would like for these fixed x 0, y 0 to make each of the remaining terms d(1, y 0 x x 1 y 1 0 ), d(x 0 y y 1 x 1 0, 1) to be less than ε/10, as well. But notice that for these fixed x 0, y 0 the maps z x 0 zx 1 0 and z y 0 zy 1 0 are continuous and send 1 to 1. Let therefore δ be small enough so that additionally x 0 B(1, 2δ)x 1 0 B(1ε/10) and y 0 B(1, 2δ)y 1 0 B(1, ε/10), and notice that this choice finishes the proof. 2. Week 2 Theorem 10 (Birkhoff-Kakutani). Let G be a topological group. Then G is Hausdorff and admits a countable neighborhood basis at 1 if and only if it admits a compatible metric d which can be always taken to be left invariant. Proof. We show the only non-trivial direction. Let (U n ) be any countable neighborhood basis at 1 so that U 0 = G. We construct a sequence of open neighborhoods (V n ) of 1 so that (1) V 0 = U 0 = G and V n+1 V n ; (2) V n = V n 1 ; (3) V n V n V n V n+1 ; (4) V n U n. This is an easy task given the continuity of x x 1 and x x x x. Given this (V n ), for every x, y G we define δ(x, y) = inf{2 n y 1 x V n }.

8 8 ARISTOTELIS PANAGIOTOPOULOS Notice that δ is well defined due to (1), left invariant, δ(x, x) = 0, x y = δ(x, y), and δ(x, y) = δ(y, x) due to (2). The only thing δ fails, in general, to satisfy is the triangle inequality. We remedy this by defining n 1 d(x, y) = inf{ δ(x i, x i+1 ) x 0 = x, x n = y, x i G, n 1}. i=0 In the rest we show that d is a left invariant metric that is compatible with the topology. It is clear that d satisfies triangle inequality and that d inherits from δ left invariance, symmetry and the property d(x, x) = 0. So in order to show that d is a metric we are left with showing that d(x, y) = 0 = x = y. Claim. For every x, y G we have that d(x, y) 1 δ(x, y). 2 Proof of Claim. We will prove by induction on n 1 that for all x 0,..., x n, ( ) n 1 δ(x i, x i+1 ) 1 2 δ(x 0, x n ). i=0 For n = 1 is is clear. Assume now inductively that ( ) holds for all m with m < n. We will need to use the following property of δ which follows by (3) above: if δ(p, q), δ(q, r), δ(r, s) m then δ(p, s) 2m. Let S be the sum in the left hand side of ( ). If δ(x 0, x 1 ) is too large, i.e, 1S then n 1 2 i=1 δ(x i, x i+1 ) is too small, i.e., 1 S and therefore, by inductive 2 hypothesis δ(x 1, x n ) 2 1S = S. So, by the above property of δ we have that 2 δ(x 0, x n ) δ(x 0, x 1 ) + δ(x 1, x n ) 2S. The same argument works if δ(x n 1, x n ) is too large. We can therefore assume that n 3 and that both δ(x 0, x 1 ) and δ(x n 1, x n ) are < 1 S. This implies that the largest r with the property that 2 r 1 i=0 δ(x i, x i+i ) is too small, i.e., 1 S, satisfies 1 r n 2. By induction 2 hypothesis we have that r 1 δ(x 0, x r ) 2 δ(x i, x i+i ) S. Again by induction, and since r i=0 δ(x i, x i+i ) is too large, we have that δ(x r+1, x n ) 2 i=0 n 1 i=r+1 δ(x i, x i+i ) S. Trivially we also have that δ(x r, x r+1 ) S and therefore by the above property of δ we have that δ(x 0, x n ) 1S. 2 To finish with the proof of the Theorem we need to show that the metric d is compatible with the topology on G. This follows easily from the fact that (V n ) is a basis at 1, the definition of δ and the inequalities δ d δ/2.

9 GROUPS AND DYNAMICS 9 The argument provided in the proof of 10 is very useful. Variations and refinements of this argument can often be used in order to characterize a fixed class of Polish groups by whether all its elements admit a left invariant metric which satisfies certain properties. We will see examples in Theorem??, Theorem??. Another example is the subject of Project 2: Project 2. Simplify/conceptualize the proof of the theorem: let G be a Polish group. The family of continuous representation of G on reflexive Banach spaces generates the topology on G if and only if G admits a left invariant metric d that is stable, i.e., for all d-bounded sequences (g n ), (h k ) we have that whenever the two limits exist. lim n lim k d(g n, h k ) = lim k lim n d(g n, h k ), Lemma 11. Let G be a Polish group and let d, ρ both be compatible left invariant metrics on G. Then, a sequence (g n ) is d-cauchy if and only if it is ρ-cauchy. Proof. Assume that (g n ) is d-cauchy and let ε > 0. We will find n 0 so that for all m, l n 0 we have that ρ(g m, g l ) < ε. Since both d, ρ are compatible with the topology, we can find a δ > 0 so that B d (1, δ) B ρ (1, ε). Let now n 0 so that for all m, l n 0 we have that d(g m, g l ) < δ and notice that by left invariance we have d(g m, g l ) < δ = d(g 1 l g m, 1) < δ = ρ(g 1 l g m, 1) < ε = ρ(g m, g l ) < ε. The following definition is essentially a corollary of the previous lemma. The proof that ĜL is a a Polish monoid is a sub-argument of the proof of Theorem 8 and will be left to the reader. Definition 12. Let G be a Polish group. The left-completion ĜL of G is the completion of G with respect to any compatible left invariant metric. Then multiplication on G extends uniquely on ĜL turning ĜL into a Polish monoid. The group G is said to be CLI if it admits a complete left-invariant metric, i.e. if ĜL = G. One can easily compute now ĜL when G = S. In particular, consider the space N N of all functions from N to N endowed with the pointwise convergence topology. The metric usual metric defined by d(f, f ) = 2 n iff f f and n is the smallest natural number so that f(n) f (n) is compatible with the topology on N N and it is easy to check that it is complete. This metric restricts to a left invariant metric on S N N. Therefore, in order to identify Ŝ L it suffice to compute the closure of S inside N N this easily checks out to be the monoid of all injections from N to N (which are not surjective in general). The left completion of the unitary group U(H) consists of all linear embeddings from H to H which preserve the inner product. Similarly, the right completion Homeo + ([0, 1]) R is the collection of all continuous and (non-strictly) increasing surjective maps from [0, 1] to [0, 1].

10 10 ARISTOTELIS PANAGIOTOPOULOS Theorem 10 implies that every Polish group admits a left invariant metric d l. As a consequence the metric d r defined by d r (g, h) = d l (g 1, h 1 ) is a compatible right invariant metric. The next example shows that not every Polish group, in fact not every locally compact Polish group, admits a two sided invariant metric. A metric d on G is two sided invariant or simply invariant if it is both left and right invariant. Notice that if G admits a compatible invariant metric and (g n ), (h n ) are sequences in G with lim g n h n = 1 then lim h n g n is also 1, since d(h n g n, 1) = d(h 1 n h n g n h n, h 1 n h n ) = d(g n h n ) 1 Example 13. Consider the group topological group SL(2, R) of all real 2x2 matrices with determinant 1. The topology it inherits from R 4 makes it a locally compact Polish group. Notice that if A n = [ 1/n 1/n 0 n ], B n = [ ] n 1/n, C = 0 1/n [ ] 1 2, 0 1 then A n B n 1, while B n A n C. Hence by the discussion above SL(2, R) does not admit an invariant metric. The following example characterizes all topological groups which admit a compatible invariant metric. Theorem 14 (Klee). A topological group G admits an invariant metric compatible with its topology if and only if is Hausdorff and 1 admits a countable neighborhood basis (U n ) at 1 consisting of conjugation invariant sets, i.e., gu n g 1 = U n, for all g G. Proof. Notice that if d is an invariant metric then d(g, 1) = d(hg, h) = d(hgh 1, hh 1 ) = d(hgh 1, 1). Hence, if G admits an invariant metric d then U n = B d (1, 1/n) is conjugation invariant. Conversely, notice that if U is conjugation invariant set then U 1 = {g 1 g G} is also conjugation invariant since ( hu 1 h 1) 1 = huh 1 = U, for all h G. Therefore one can arrange in the proof of Theorem 10 so that the sequence (V n ) is additionally conjugation invariant. As a consequence, in the definition of δ there, if y 1 x V n then hy 1 h 1 hxh 1 V n. Hence δ(x, y) = δ(hxh 1, hyh 1 ) and the same property is inherited by d. We are left to show that d is right invariant, but: d(xh, yh) = d(hxhh 1, hyhh 1 ) = d(hx, hy). [ ] 1 ε In contrast with Theorem 14 notice that in the case of SL(2, R), if A ε = 0 1 is any fixed element arbitrary close to the identity [ then ] we can send it arbitrary far r 0 from 1 simply by conjugating it with B r =, for large enough r. 0 1/r

11 GROUPS AND DYNAMICS 11 Definition 15. A Polish group G is TSI is it admits a (two-sided) invariant metric. We have now the following picture among the various classes of groups we have seen so far: CLI locally compact T SI Solvable Compact Discrete Abelian Figure 1. The arrows denote strict inclusions in the class of Polish groups Proposition 16. All the inclusions in the diagram hold. In particular: (1) all compact Polish groups are TSI; (2) all discrete Polish groups are TSI; (3) all Abelian Polish groups are TSI; (4) all locally compact Polish groups are CLI; (5) all TSI Polish groups are CLI; (6) all solvable Polish groups are CLI (Hjorth-Solecki); Proof. (1) By Theorem 10 fix a left invariant metric d on G and define d : G G R by setting d (x, y) = sup h G d (xh, yh). By compactness, for every x and y the supremum is realized by some actual h G. It is immediate that d is two sided invariant. So we are left to check that it is compatible with the topology. Since (G, τ d ) is compact, it suffice to show that the identity map from (G, τ d ) to (G, τ d ) is continuous. But notice that if U = B d (x, ε) is the ε-ball around x in (G, τ d ) then the complement U c of U is a projection of a compact subset of (G, τ d ) (G, τ d ) to (G, τ d ): U c = proj G {(y, h) G G d(xh, yh) ε}. Therefore U is open in (G, τ d ) as well. (2) take d(x, y) = 1 x y. (3) Immediate by Theorem 14. (4) By Theorem 10 fix a left invariant metric d on G. Notice that if (g n ) is a d-cauchy sequence then there is some compact set K G so that for all (g n ) eventually lies in K and therefore it converges to some g K. (5) Let d is an invariant metric on G. By corollary 9 we have that D is a complete metric on G. But D is left-invariant in this case since by invariance of d we have that d(g 1, h 1 ) = d(g, h)

12 12 ARISTOTELIS PANAGIOTOPOULOS (6) If G is solvable then it built by a finite sequence of abelian extensions starting from the trivial group 1. Hence this follows by induction using (3),(5) and the next theorem. Theorem 17 (Gao). Let G be a Polish group and let H be a closed normal subgroup of G. The following are equivalent: (1) G is CLI; (2) Both H and G/H are CLI. We will not prove this theorem but it will be important to recall some facts about quotients of Polish groups. Let G be a Polish group and let H be any subgroup of it. We denote by G/H the collection {Hg g G} of all right cosets of H. If π : G G/H is the natural projection g Hg then a set U G/H is open in the quotient topology if π 1 (U) is open in G. This makes π both continuous and open. The topology is bad in general when H is not closed. However, when H is closed G/H is Polish. This follows from the following classical theorem: Theorem 18 (Sierpinski). If X is a Polish space, Y a metrizable space and π : X Y a continuous and open surjection. Then Y is Polish. It is not difficult to see that G/H is metrizable when H is closed but one can construct explicitly a metric d on G/H as follows: Proposition 19. If d is a left invariant metric on the Polish group G and H is a closed subgroup of H then d is a compatible metric on G/H where d(hx, Hy) = inf{d(g x, g y ) g x Hx, g y Hy}. It is a good exercise working out the details of the proof of the above proposition and trace the reasons as to why we use the left invariant metric d to define d on space of right cosets.

13 GROUPS AND DYNAMICS Week 3 Recall that we view the Polish group S as a G δ subset of N N endowed with the pointwise convergence topology. This topology is compatible with the metric given by d(g, h) = 2 n iff g h and n is the smallest natural number so that g(n) h(n), which is a left invariant metric on S. This metric is an ultrametric, i.e., it satisfies the following strengthening of the triangle inequality. A metric d is an ultrametric if and only if for all x, y, z we have that d(x, z) max{d(x, y), d(y, z)}. We say that a Polish group is non-archimedean if it admits a compatible left invariant ultrametric. There are many equivalent reformulations of this property. Theorem 20. Let G be a Polish group. The following are equivalent: (1) G is isomorphic to a closed subgroup of S ; (2) G admits a compatible left invariant ultrametric; (3) G admits a countable neighborhood basis at 1 consisting of open subgroups; (4) G admits a countable basis for the topology B that is invariant under multiplication from the left, i.e., gb B for all B B and g G. Proof. (1) (2) is immediate since we can always restrict the aforementioned left invariant ultrametric of S on G. For (2) (3), notice that if d(g, 1), d(h, 1) < ε then d(gh, 1) = d(h, g 1 ) max{d(h, 1), d(g 1, 1)} = max{d(h, 1), d(1, g)} < ε. Therefore, any countable collection of vanishing open d-balls around the identity will do. For (3) (4) let B 1 be a countable neighborhood basis at 1 consisting of open subgroups and notice that each B B 1 has countably many left cosets. Hence the required basis is the collection B = {gb g G, B B 1 }. For (4) (1) we enumerate B = {B 0, B 1,...} and we consider the representation π : G S given by π(g)(k) = l gb k = B l. It is immediate that π is a homomorphism. It is injective since B separates points and continuous since gb k = B l is witnessed by a whole open set, namely g 1 B k. So we are left to show that π is an open map from G to π(g). This will show that G is isomorphic to π(g) and that π(g) is Polish. By Lemma 3 the latter it implies that π(g) is closed. Let g B l B. It suffice to find an open set V S with π(g) V π(g) π(b l ). Let B k = g 1 B l B and set V = {f S f(k) = l}. Then V is clearly open with π(g) V. Moreover, if h G with π(h) V then hb k = B l = h B l. Hence V π(g) π(b l ).

14 14 ARISTOTELIS PANAGIOTOPOULOS Corollary 21. If (G n ) is a countable collection of closed subgroups of S then the group Π n G n is isomorphic to a closed subgroup of S. In particular, Z ω is a closed subgroup of S. Theorem 20 contained many equivalent topological reformulations of non-archimedeanity. The next result allows us to associate to each non-archimedean group G a countable structure M so that G = Aut(M). First recall some definitions. A language L = {R, S,..., f, g,...} is a collections of relational symbols R, S,... and function symbols f, g,..., together with the arity function α: L N which assigns to each symbol s L the size of the tuple x with respect to which s( x) is going to be treat as syntactically correct. An L-structure M = (M; R M, S M,... ; f M, g M,...) consists of a set M together with interpretations for every every symbol in L. In particular if R is a relation in L with arity α(r) = n then the interpretation R M of R in M is just a subset of M n. Similarly the interpretation f M of an n-ary function f in M is just a function from M n to M. The function symbols c L whose arity is 0 we call them constants and c M is by definition some point in M. Since the 0-ary Cartesian product M 0 of any set M is the set { } there are only 2 possible interpretations of a 0-ary relation symbol which we usually denote by and. Often when the context is set we will simplify our notation and use the letters R, f,... instead of R M, f M,... Examples. We can view the set of rationals endowed with the usual linear order as an L-structure (Q, ) where L consists of one relation symbol with arity α( ) = 2. Any countable discrete group Γ can be viewed as a a countable L-structure (Γ, 1, () 1, ) where L consists of three function symbols with α1 = 0, α(() 1 ) = 1, and α( ) = 2. Given an L-structure M we denote by Aut(M) the group of all automorphisms of M under composition. Recall that that ϕ Aut(M) is f is a bijection f : M M so that for every relation symbol R L and any tuple ā = (a 1,..., a n ) in M we have that (a 1,..., a n ) R M (ϕ(a 1 ),..., ϕ(a n )) R M, and for every function symbol f L, any tuple ā = (a 1,..., a n ) in M and b M we have that f M (a 1,..., a n ) = b f M (ϕ(a 1 ),..., ϕ(a n )) = ϕ(b) We view Aut(M) as a topological group with the pointwise convergence topology. One of the implications of the next theorem is that if M is a countable structure then Aut(M) is Polish. Theorem 22. Let G be a Polish group. The following are equivalent: (1) G is non-archimedean;

15 GROUPS AND DYNAMICS 15 (2) There is a countable structure M on a relational language and G Aut(M). Proof. If M is a countable L-structure then by enumerating M we can assume that Aut(M) is a subgroup of S. To see that it is closed notice that if f S \Aut(M) then there is some relation symbol R L (or function symbol, but this is left as an exercise) and a tuple ā in M so that ā R M but not ϕ(ā) R M (or, not ā R M but ϕ(ā) R M, but this is left to the reader). But then any g S in the open subset defined by the data ā ϕ(ā) will fail to be an automorphism. Assume now that G is non-archimedean, i.e. by Theorem 20, a closed subgroup of S. For every n 1 let {Ok n, k I} with On k N n, be the collection of all orbits in the diagonal action of G on N n. For each such orbit introduce an n-ary relation Rk n and define a structure M by setting the interpretation (Rn k )M of Rk n in M to be Ok n. It is immediate that G A (M) for the converse inclusion use the fact that G is closed. The structures M provided by Theorem 22 enjoy a very useful property know as ultrahomogeneity. In short, any isomorphism f : A B between finite substructures of M extends to a global automorphism f Aut(M). Many natural examples fail to be ultrahomogeneous. For example the substructures ({0, 1}, ) and ({0, 2}, ) of (Z, ) are isomorphic but this isomorphism cannot be extended to an element of Aut(Z, ). To put things into context we need some definitions. Rather than studying the various mathematical structures directly, i.e., up to isomorphism, logic is often interested in comparing them by means of the formal statements they satisfy. The usual first order logic cares about finitary statements, i.e., L-formulas. However when one wants to study countable structures up to isomorphism it sometimes is convenient to rely on infinitary statemetns known as L ω1,ω-formulas. Both logics are instances of the following definition where for historic and convenience purposes one writes L for L ℵ0,ω and L ω1,ω for instead of L ℵ1,ω. Definition 23. Let L be a language and let κ be an infinite cardinal. Let also X be a collection of variables.the set Φ κ,ω (L) of L κ,ω -formulas is the smallest collection of statements so that: (1) R(x 1,..., x n ) and f(x 1,..., x n = x n+1 ) are in Φ κ,ω (L), for all R, f L; (2) if ϕ( x) is in Φ κ,ω (L) then ϕ( x) is in Φ κ,ω (L); (3) if if ϕ( x, y) is in Φ κ,ω (L) then yϕ( x, y) and yϕ( x, y) are in Φ κ,ω (L); (4) If Ψ is a collection of L κ,ω -formulas with Ψ < κ and so that there is a finite tuple x = (x 1,..., x n ) of free variables so that the free variables of each φ Ψ are among x, then the following are also L κ,ω -formulas: φ( x), φ Ψ φ( x). φ Ψ

16 16 ARISTOTELIS PANAGIOTOPOULOS A quantifier free L κ,ω -formula is any L κ,ω -formula that is constructed without any use of the the operation (3) above. A L κ,ω -sentence is any L κ,ω -formula σ containing no free variables. If ϕ( x) is some L κ,ω -formulas and ā is a tuple in an L-structure M the we will write M = ϕ(ā) to denote that ā satisfies the statement ϕ( x) in M. This can be defined formally by induction on the complexity of the formula in the obvious way. The L κ,ω -theory of M is the collection Th κ,ω (M) of all L κ,ω -sentences satisfied by M. Let ā = (a 1,..., a n ) be a tuple in the L-structure M and let b = (b 1,..., b n ) be a tuples in N. We write (M, ā) qf (N, b) if the tuples ā and b have the same quantifier free type, i.e., if every quantifier free formula is satisfied by ā if and only if it is satisfied by b. Notice that we haven t specified a κ since this doesn t depend on which L κ,ω logic we chose. In fact (M, ā) qf (N, ā) holds if and only if the substructure ā M generated by ā in M is isomorphic to the substructure b N generated by b in N, by an isomorphism that sends a i to b i. Similarly we write (M, ā) L (N, b) if (M, ā) and (N, b) are elementarily equivalent, i.e., if ā and b satisfy the the same L-formulas (Laleph0,ω-formulas in the definition). Finally we write (M, ā) ω1,ω (N, b) if (M, ā) and (N, b) are L ω1,ω-elementarily equivalent, i.e., if ā and b satisfy the the same L ω1,ω-formulas (L ℵ1,ω-formulas in the definition). Definition 24. Let M be a countable L-structure. We say that M is ultrahomogeneous if for every two finite tuples ā, b in M, is (M, ā) qf (M, b) then there is a f Aut(M) with fā = b. We say that M is homogeneous if for every two finite tuples ā, b in M, is (M, ā) L (M, b) then there is a f Aut(M) with fā = b. One could define in an analogous homogeneity property for tuples which satify the stronger assumption that (M, ā) ω1,ω (M, b). However in Theorem 25 we will see that the assumption (M, ā) ω1,ω (M, b) is too strong. Examples. (1) The structure (Q, ) is ultrahomogeneous. One can see this by means of a standard back and forth argument: let ā, b be two tuples in Q which satisfy the same quantifier free type and fix some enumeration of Q. Step by step we will extend the assignment ā b to a bijection from Q to Q which preserves. Let ā, c 1,..., c n 1 b, d 1,..., d n 1 be the assignment before the step n. If n is odd, let c n be the first element (with respect to the fixed enumeration) in Q \ {ā, c 1,..., c n 1 }. Since (Q, ) is dense and without endpoints we can easily find d n so that (ā, c 1,..., c n 1, c n ) qf ( b, d 1,..., d n 1, d n ). If n is even, let d n be the first element Q \ { b, d 1,..., d n 1 } and again extend by finding appropriate c n. Odd steps guarantee that the resulting map will be total and even steps guarantee that it will be surjective. (2) The structure (Z, ) is not ultrahomogeneous however it is homogeneous. One can easily see this since the distance n k of two integers is first

17 GROUPS AND DYNAMICS 17 order definable in (Z, ) and automorphisms of (Z, ) are precisely the order preserving bijections which additionally preserve distance. (3) The structure M = (Z, ) (Z, ) is not homogeneous, where (Z, ) (Z, ) is has domain the set Z {0} Z {1}, for each i = 0, 1 the elements within Z {i} are ordered as in (Z, ) and each element in Z {0} is smaller that every element of Z {1}. In order to see this one has to use the quantifier elimination machinery in order to prove that (M, (0, 0)) L (M, (0, 1)). Given this it is easy to see why there is no automorphism sending the 0 of the first copy to the 0 of the second copy. Theorem 25 (Karp). Let M be a countable L-structure and let ā, b be tuples in M. Then, (M, ā) ω1 ω (M, b) if and only if there is an automorphism f of M sending ā to b. Proof. If f is an automorphism of M with fā = b then it immediate to see that (M, ā) ω1 ω (M, b). Formally one can use straight forward induction on the complexity on each L ω1 ω-formula. Conversely, notice that if (M, ā) ω1 ω (M, b) then for every c in M there is a d in M so that (M, ā, c) ω1 ω (M, b, d) because if not then for every d M we can find some L ω1 ω-formula φ d ( x, y) so that M = φ d (ā, c) but M = φ d ( b, d), and this would contradict (M, ā) ω1 ω (M, b) since then we have M = y φ d (ā, y) but M = y φ d ( b, y), d M Hence inductively we can produce a back and forth system resulting to an automorphism of M sending ā to b. Exercise. Check that the structures we constructed in the proof of Theorem 22 are actually ultrahomogeneous. Given any L-structure M, explain how one can define a language L with L L together with appropriate interpretations for the new symbols that will extend M to an ultrahomogenous L structure M (on the same domain with M). We conclude with the following characterization of CLI non-archimedean groups due to Su Gao. Theorem 26 (Gao). Let M be a countable structure. The following are equivalent: (1) Aut(M) is CLI; (2) there are no uncountable structures satisfying the L ω1 ω-theory of M. Before we proceed to the proof notice the immense difference of L ω1 ω-logic with the first order logic where compactness theorem guarantees that if M is infinite then the L-theory of M admits models in every cardinal. For the proof we first need a lemma. d M

18 18 ARISTOTELIS PANAGIOTOPOULOS Lemma 27. Let M be a countable structure. A map γ : M M is in left completion Aut(M) L if and only if γ is an L ω1 ω- elementary embedding from M to M, i.e., γ is an embedding and (M, ā) ω1 ω (M, γā) for every finite ā in M. Proof. Assume that there exists left-cauchy sequence (g n ) in Aut(M) converging to γ and let ā be any finite tuple in M. Then we can find n large enough so that g n ā = γā. Hence (M, γā) ω1 ω (M, g n ā) ω1 ω (M, ā), where the last equivalence holds because g n is an automorphism. Conversely, let γ be an L ω1 ω- elementary embedding from M to M. To build a left-cauchy sequence converging to γ it suffice to find for every finite tuple ā in M an automorphism g so that gā = γā. But since (M, ā) ω1 ω (M, γā) the same argument as in the proof of Theorem 25 produces a back and forth system from M to M starting from the assignment ā γā. The result of this back and forth system is the required automorphism. Sketch of proof of Theorem 26. We briefly sketch the proof relying on some facts regarding L ω1 ω-logic which we will use without proving them. The first fact is Scott s isomorphism theorem: If M is a countable L structure then there is a single L ω1 ω-sentence σ so that for every other countable L structure N we have that N = σ if and only if M and N are isomorphic. The second fact is the following analogue of what is model theory is known downward Löwenhein-Skolem theorem: If L is a countable language. If N is any infinite structure and let T be a countable L ω1 ω-theory, then for any countable subset A of N there exists a countable substructure M of N with A M so that the inclusion i: M N is elementary with respect to L ω1 ω-subformulas of sentences of T, i.e., if ϕ( x) is a subformula appearing within some σ T then for all ā in M we have that M = ϕ(ā) if and only if N = ϕ(ā). See [Kue, Theorem 2.3 and Theorem 1.1] for proofs in the more general L κλ setting of these statements. Let σ be the Scott sentence of M and assume that N is an uncountable structure satisfying σ. By downward Löwenhein-Skolem theorem applied on N for T = {σ} and A = we get a substructure M 0 of N that is countable so that so that the inclusion i 0 : M 0 N is elementary with respect to L ω1 ω-subformulas of sentences of T. Since T is a complete L ω1 ω-theory it follows that i 0 is an elementary L ω1 ω-embedding. By a second application of downward Löwenhein-Skolem theorem applied on N for T = {σ} and A strictly including M 0 we get a countable substructure M 1 of N so that, as above, the inclusion i 1 : M 1 N is elementary L ω1 ω-embedding. But then both M 0, M 1 satisfy the same Scott sentence and therefore they are isomorphic to M. Moreover the map γ : M 0 M 1 with γ = i 0 is an elementary L ω1 ω-embedding

19 since GROUPS AND DYNAMICS 19 (M 0, γā) ω1 ω (N, i 0 ā) ω1 ω (M 1, γā), for every ā in M 0. By Lemma 27 γ Aut(M) L, and since M 0 is strictly contained in M 1 we have that γ Aut(M). Conversely, if γ Aut(M) L \Aut(M) then by Lemma 27 γ is an elementary L ω1 ω- embedding that is not surjective. By transfinite induction we build an ω 1 -sequence of elementary L ω1 ω-embeddings γλ κm κ M λ, κ λ < ω 1 so that M κ is isomorphic to M and each embedding is strict. Given this it is easy to check that the colimit of all these structures is uncountable and that is satisfies the same L ω1 ω-theory as M. For the construction, if M κ has been defined and it is isomorphic to M, we define M κ+1 and γκ+1 κ so that γκ+1 κ : M κ M κ+1 is isomorphic to γ : M M. If λ is a countable limit ordinal we define M λ to be the colimit of what we have so far and γλ κ : M κ M λ the induced colimit maps. It is easy to check that each γλ κ is elementary L ω1 ω-embedding since for every a in M λ there should be κ < λ with a M λ. In particular M λ is countable and satisfies the same Scott sentence as each previous M κ. By Scott isomorphism theorem it is isomorphic to M. 4. Week 4 (the week of shame) One of the most effective tools for analyzing closed subgroups of S is the theory of Fraïssé limits. We will see that Fraïssé limit is just another term for countable ultrahomogeneous structure. Although it is not necessary, it is definitely convenient to restrict our attention to purely relational languages L. For purely relational languages there is bijective correspondence between subsets A of an L-structure and substructures A = A M. As a consequence a relational structure M is ultrahomogeneous if whenever g : A B is an isomorphism between finite substructures of M then g extends to an automorphism of M. Since the proof of Theorem 22 associates to each closed subgroup of S a countable purely relational ultrahomogenous structure, this restriction will not make us lose in generality. The big advantage in working with an ultrahomogeneous structure M boils down to the fact that most of the information regarding the symmetries of M is directly encoded on the isomorphism type of its finite substructures (local data) and not on the way that these substructures sit inside M (global data): if A is a finite structure, it doesn t matter how you embed it within M because all these embeddings can be exchanged by an automorphism. A reflection of this high symmetry is that ultrahomogeneous structures can be pieced together via a very canonical limiting process known as taking the Fraïssé limit. Definition 28. Let M be an L-structure (in a relational language). The age of M, denoted by Age(M), is the collection of all finite L-structures which embed in M. Notice that the age of any structure is a class (not a set) which we view as a category where arrows are embeddings. For example, Age((Q, )) is the class of all finite linear orderings.

20 20 ARISTOTELIS PANAGIOTOPOULOS We write A B when A embeds in B and A f B when we want to keep track of a particular map f : A B that is witnessing this embedding. Lemma 29. Let M be a countable L-structure (in a relational language). Then: (1) Age(M) is countable up to isomorphism; (2) Age(M) satisfies the Hereditary Property (HP), i.e., if B Age(M) and A B then A Age(M); (3) Age(M) satisfies the Joint Embedding Property (JEP), i.e., if B, C Age(M) then there is D Age(M) so that B, C D. If moreover M is ultrahomogeneous, then additionally (4) Age(M) satisfies the Amalgamation Property (AP), i.e., if A, B, C Age(M) and A f B, A g C, then there is D Age(M) and B f D, C g D so that f f = g g. Proof. For (4), let B, C be realized as substructures of M and let A B and A C be the copies of A within these two fixed substructures on M (realizing A f B and A g C). By ultrahomogeneity we can find ϕ Aut(M) so that ϕ(a B ) = A C. Let D be the structure generated by ϕ(b) C and f, g be the obvious embeddings. Definition 30. A Fraïssé class is any collection K of finite L-structures satisfying the properties (1) (4) from the above lemma. The following theorem is an inverse to Lemma 29. Theorem 31. Let K be a Fraïssé class of finite L-structures. Then there exists a unique up to isomorphism countable structure M which satisfies any of the following equivalent properties: (1) Age(M) = K and M is ultrahomogeneous; (2) Age(M) = K and M satisfies the extension property, i.e., if A, B K and f A : A M, f A B : A B are embeddings, then there is f B : B M so that f A = f B f A B ; (3) Age(M) = K and M satisfies the 1-point extension property which is defined as in (2) with the additional restriction B \ A. Proof. First we show that (1),(2) and (3) are equivalent. Property (2) clearly implies (3). Property (3) implies property (1) since it allows us to build a back and forth system from M to M starting from any isomorphism A A between finite substructures of M. We leave (1) implies (2) as an exercise. Before we show that such an M satisfying either of (1) (3) exists notice that uniqueness follows easily since if M and M satisfy property (3) then it easy to build a back and forth system between them. We show now that if K is as in the statement of the theorem then there exists M satisfying property (2) above. We will build M as the colimit (union) of a generic sequence in K: A 1 A 2 A 3 A m A n M

21 GROUPS AND DYNAMICS 21 That is, a coherent sequence ((A n ), gn m ) of embeddings gn m : A m A n, m n with A n K so that (i) every A from K embeds eventually in A n for large n; (ii) for every m > 0 and every embedding f : A m B to some B in K there is n > m and an embedding f : B A n so that f f = gn m. Condition (i) together with the fact that A n K imply that the age of M = n A n is indeed K. Property (ii) together with the fact that K satisfies the amalgamation property implies that M has the extension property from (2). In order to build a generic sequence in K we enumerate (B n ) all elements (up to isomorphism) of K. Since K contains finite structures there are at most finitely many embeddings between any two elements of K. Hence we can also enumerate (f n : C n D n ) all possible embeddings (up to isomorphism) that appear in K so that each embedding appears infinitely many often in the list. The construction of ((A n ), gn m ) is now by induction. Let A 1 be B 1. Assume now that we have built A n 1. To build A n we proceed as follows. First by JEP we can extend A n 1 to a structure C so that B n embeddds in C as well. By finitely many applications of AP enlarge C further to a structure D so that every embedding from C n to C extends to an embedding from D n to D. Set A n to be D. Definition 32. The structure M associated to a Fraïssé class K as in the statement of Theorem 31 is called the Fraïssé limit of K. Since Theorem 22 produces ultrahomogeneous structures we have, at least abstractly, many examples of Fraïssé structures. Here is a collection of more natural examples which will be important later on. Example 33. Let K LO be the collection of all finite linear orders. It is easy to see that K LO is a Fraïssé class. A way to see this is to recall that (Q, ) is ultrahomogeneous and its age is K LO. In other words, the Fraïssé limit of K LO is (Q, ). Let K Graphs be the collection of all finite graphs. It is easy to see that K Graphs is a Fraïssé class. It s limit is known as the random graph or Rado graph. While the most useful characterization of the Random graph is via the extension property (see Theorem 31) we point out that the Random graph can be defined as the isomorphism type one attains with probability 1 if you put an edge between any pair of a countable set of vertexes with probability 1/2. Similarly, for every n 3 the collectionkhenson n of all finite graphs which do not have a clique of size n forms a Fraïssé class. The Fraïssé limit of this class is known as the K n -free random graph or as the n-henson graph. Consider the collection K QU of all finite metric spaces with rational distances. We can view these metric spaces as L-structures where L = {d q q Q + } is the language that contains a binary predicate for every positive rational. Again, K QU is a Fraïssé class whose limit is known as the rational Urysohn space QU. The completion of QU is a very important metric spaces known as the universal Urysohn space U.