Isomorphism of free G-subflows (preliminary draft)

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1 Isomorphism of free G-subflows (preliminary draft) John D. Clemens August 3, 21 Abstract We show that for G a countable group which is not locally finite, the isomorphism of free G-subflows is bi-reducible with with the universal countable Borel equivalence relation E. For a countable group G, the Bernoulli flow 2 G consists of all functions from G to 2 together with the shift action of G given by (g α)(h) = α(g 1 h). A G-subflow is a compact shift-invariant subset X 2 G. Two subflows are isomorphic, X = Y, if there is a homeomorphism ϕ : X Y which commutes with the shift, i.e., ϕ(g α) = g ϕ(α). Various classes of subflows are of particular interest. A subflow X is free if g α α for all g e in G and α X. A subflow is minimal if it contains no proper subflow, i.e., for all α X, the orbit closure of α, O α = G α, is equal to X. The authors of [?] asked to determine the exact complexity of the isomorphism relation for several classes of G-subflows for various countable groups G, namely the classes of all subflows, of free subflows, and of minimal subflows. This complexity is to be gauged using the notion of Borel reducibility of equivalence relations, explained below. The authors showed that for any countably infinite group G, the isomorphism relation for G-subflows reduces the equivalence relation E, and when G is a locally finite group (i.e., for any finite F G, the subgroup F generated by F is finite), then the isomorphism problem for minimal subflows is bi-reducible with E. In [?] it was shown that the isomorphism problem for all n-dimensional subshifts (i.e., Z n -subflows) is bi-reducible with E, the universal countable Borel equivalence relation, for each n 1. The construction there made use of periodic points in a fundamental way, producing subshifts which are not free let alone minimal. The purpose of this article is to improve that result in two ways: by showing that isomorphism of free subflows is already of 1

2 maximal complexity, and by extending the result from Z n to any countable group G which is not locally finite. In the next two sections we briefly review important notions from the theory of Borel equivalence relations and basic facts about subflows. In the following section prove the main result, and list some remaining open questions in the final section. 1 Countable Borel equivalence relations A Borel equivalence relation is an equivalence relation E on a Polish space X such that E is Borel as a subset of X 2. A standard method of comparing Borel equivalence relations is via the Borel reducibility relation. We say that an equivalence relation E on a space X is Borel reducible to an equivalence relation F on a space Y, E B F, if there is a Borel-measurable function f : X Y such that for all x 1, x 2 X we have x 1 E x 2 if and only if f(x 1 ) F f(x 2 ). We write E B F when we can find such an f which is injective. Note that having E B F amounts to having a definable injection from the quotient space X/E into Y/F, so that when E B F we may view the relation F as being at least as complicated as E. We say that an equivalence relation E is countable if every equivalence class of E is countable. We say E is a universal countable Borel equivalence relation if E is a countable Borel equivalence relation and for every countable Borel equivalence relation F we have F B E. Several such examples are known; one which we will use here is the (left) shift relation of the free group on two generators F 2 on the space 2 F 2, denoted E(F 2, 2). A good introduction to the theory of countable Borel equivalence relations is [?]. 2 G-subflows A basic observation is: Theorem (Curtis-Hedlund-Lyndon). If ϕ : X Y is a morphism of G- subflows, i.e., a map commuting with the shift action, then ϕ is given by a block code, i.e., there is a finite B G and a function π : 2 B 2 such that ϕ(x)(h) = π((h 1 x) B). The following lemma provides a necessary and sufficient condition for minimality: Lemma (Gao-Jackson-Seward). An element α 2 G is minimal (i.e., O α is a minimal subflow) iff for every finite A G there is a finite T G such 2

3 that i.e., α is uniformly recurrent. g G t T a A(α(gta) = α(a)), We will use the following terminology. By a pattern we mean u 2 <G, a finite partial function from G to 2. We say a pattern u occurs in α 2 G (or v 2 <G ), u α at h if α extends h u. G acts on patterns by shift. For a pattern u and α 2 G we define α u 2 G by { u(h) if h dom(u) α u(h) = α(h) otherwise and similarly for multiple patterns (we will always use patterns with disjoint domains). We now explain how to view the isomorphism relation on subflows as a countable Borel equivalence relation. Definition 1. For a countable group G, we let S G be the set of closed, S-invariant subsets of 2 G, where S is the left shift action. We view this as a subspace of the compact Polish space F(2 G ) = K(2 G ) in the Vietoris topology (see [?]). It is straightforward to note that S G is a closed subspace of K(2 G ). We can also code subflows via a set of forbidden patterns. We now consider the isomorphism relation on the spaces S G. Definition 2. For X, Y S G, we set X E Y if (X, S) = (Y, S), i.e., there is a homeomorphism of X and Y which commutes with S. It is again straightforward to verify that E is a countable Borel equivalence relation on each S G. 3 Group theoretic preliminaries Let G be a countable group which is not locally finite. We wish to develop a suitable sequence of group elements which will be used to encode information in the subflows we define below. Fix a finite F G such that H = F G infinite; we can assume that no proper subset of F generates an infinite 3

4 group. let Let F = F F 1 = {f,..., f N 1 } and define the following: F n = all products of n elements of F F <n = m<n F m F n = F n \ F <n Note that H = n= F n. Since H is infinite we must have F n for all n. Lemma 1. There is a sequence i(n) : n Z with i(n) < N such that if we let s = e and s n+1 = s n f i(n) for n and s n 1 = s n f 1 i(n 1) for n and S = {s n : n Z} then we have the following: 1. s n s m for n m. 2. For all n and m with m < n, s 1 m s n = f i(m) f i(n 1) / F <(n m). Proof: If some f i has infinite order we can simply let i(n) = i for all n, so suppose not. Define a tree T whose nodes are finite sequences of elements of F as follows. Let the root of T be T = { e }, and for n 1 let T n = { a n,... a, a 1,..., a n 1 : a i F a n a a 1 a n 1 / F <2n }, with T ordered by extension. T is a tree (i.e., it is closed under restrictions) since if a m a a 1 a m 1 inf <2m for some m < n then a n a a 1 a n 1 inf <2n. T is finitely branching since F is finite, and T is infinite since F 2n for all n. Hence T contains an infinite branch which defines a bi-infinite sequence β F Z. Take i(n) = β(n) and define s n accordingly. Now if m < n, let k = max( m, n ). If s 1 m s n F <(m n), then s 1 k s k F <2k contradicting our construction of T. Lemma 2. For any finite T G there is a k 1 such that the sets {s kn T : n Z} are pairwise disjoint. Proof: Choose k such that H T T 1 F <k. Then if s km T s kn T for m < n we have t 1, t 2 T such that s km t 1 = s kn t 2 so s 1 km s kn = t 1 t 1 2 H T T 1, contradicting that s 1 km s kn / F <(kn km) F <k. 4

5 4 Isomorphism of free G-subflows Theorem 3. Isomorphism of free G-subflows (and hence of all G-subflows) is a universal countable Borel equivalence relation for each countable group G which is not locally finite. Proof: As noted above, it suffices to show that isomorphism of free G- subflows reduces E(F 2, 2). Fix µ 2 G such that the orbit closure of µ, O µ is a free minimal G-subflow (the existence of free minimal subflows for all infinite G follow from [?] and [?], and uncountable many are produced in [?]). Let m 2 Z generate a minimal subshift, e.g., we can take m to be the bi-infinite Morse sequence. Claim 1. There is a finite T G with F {e} T such that neither of the patterns with constant value or 1 on T occurs in µ, and for any g G with g e we have g T T. Proof: We can find a T such that the constant patterns on T do not occur in µ by freeness of O µ ; we may assume that F {e} T. Now choose h such that h / T 2 cupt T 1 and let T = T {h}. Since e T, if g / T then g T T since g g T. We have that h T is disjoint from T which has size at least 2, so h T T. Finally, if g T with g e then h / g T (since h / T 2 ) and g h h so h / g T T. (claim 1) In particular, any translate of T in µ contains both a and a 1. Let T = T 2 T T 1 and define the pattern d with domain T with d T = 1 and d (T \ T ) =. Let d 1 = d T. Claim 2. Any two occurrences of the pattern d in some element α 2 G can not overlap on the T block of 1 s, and if we replace some T block in µ by a copy of d then the only occurrence of d is the one we added. Proof: Suppose first that α is the result of replacing a T block in µ by d, and suppose we added the copy of d at g 1 but d also occurs at some g 2 g 1. Then g 1 T g 2 T since d 1 does not occur in µ, so we have overlapping T blocks. Now if we have any α with d occurring at g 1 and g 2 with overlapping T blocks (i.e., g 1 T g 2 T ), then g 2 g 1 T T 1 g 1 T. If g 2 / g 1 T then we should have α(g 2 ) = from the occurrence of d at g 1, whereas α(g 2 ) = 1 from the occurrence at g 2 since e T. Hence g 2 g 1 T. From the properties of T we must have g 2 T g 1 T ; hence there is h g 2 T \g 1 T. But now h g 1 T 2 g 1 T so we should have α(h) = from the occurrence of d at g 1, whereas α(h) = 1 from the occurrence of d at g 2. (claim 2) 5

6 Let k from Lemma?? be such that the sets {s k n T : n Z} are pairwise disjoint, and let c i(), c i(1), c d(), c d(1), c e, c a, c a 1, c b, c b 1 be, respectively, s k,..., s 9k in G so that they yield pairwise disjoint T translates. Define patterns u i(), u i(1), u d(), u d(1), u e, u a, u a 1, u b, u b 1 to be equal to d on T 1 and on all the other listed translates of T T except that the corresponding c should have value 1 (e.g., u a (c a ) = 1). Hence, each such pattern encodes one of i(), i(1), d(), d(1), e, a, a 1, b, or b 1. None of these patterns occur in µ, and any two occurrences of any of these patterns do not overlap on the T block of all 1 s. Also, if we replace disjoint T 1 blocks in µ by copies of these patterns, the T block of 1 s in each can occur only at the T block in d. Now let T 1 = 9 hi k i= T be the union of all of the above T translates (the domain of all of the above patterns). We want to extend T 1 so that we can encode links to neighboring patterns. Let n be such that H T 1 F <n, and let T (n) 1 = T 1 F <n for n > n. For a pattern u with domain T 1 and p F <n \ T 1 we define the pattern u p on T (n+n ) 1 to be equal to u on T 1, 1 on p and elsewhere. Note that since T 1 F <n, if d µ u p then d u p. Let C(n) = {p F <n \ T 1 : d d p }. note that if h F 2n+2n then h T (n+n ) 1 T 1 T (n+n ) 1 =. We want to choose n large enough so that we can encode two elements of F 2n+2n in T (n+n ) 1 using elements of C(n). We claim that C(n) F <n \T 1. To see this, observe that if p odd k < n ( F k \ T 1 ) or p even k < n ( F k \ T 1 ) than d d p since F T, and at least one of these two sets has size at least 1 2 F <n \ T 1. Also, F 2n+2n 2 ( F 2n +2 F n 1 F n 1 ) 2 F 2n +2 2 F <n 4. Hence we just need n such that ( F <n T 1 ) F 2n +2 2 F <n F <n T 1 F 2n +2 2 F <n 4. This is be true for sufficiently large n since F <n and the exponential on the left hand side dominates any polynomial. 6

7 Choose n with this property and let ñ = 2n + 2n. For each pair of elements (h, h 1 ) in F 2n +2n we can choose a distinct p(h, h 1 ) C(n ). Let g (n) = sñ n, and for each of the 9 patterns u introduced above and n 1 let u n = u p(s 1 ñ(n 1) s ñn,s 1 ñn s ñ(n+1)) (and u = u p(e,sñ) ). These are patterns with domain T 2 = T 1 F <(n +n ), and {g (n) T 2 : n N} are pairwise disjoint. Let g 1 (n) = g (2n). Let µ = µ n N (g (2n) u 2n i() g (2n + 1) u 2n+1 i(m(n)) ), i.e., we encode the positive part of the minimal element m into odd g translates of T 1, clear even translates, and encode links to the previous and next translate. Note that the subflow generated by µ is still free (although it may not be minimal). Let {π j : 2 B j 2} enumerate all of the block codes, with B j G finite, and let ϕ j : 2 G 2 G be the induced mapping. We can assume that the B j s are increasing and that each B j is symmetric and contains e. For a finite T G we define the B-saturation of T, [T ] B = T B; note that when ϕ is induced by the block code π, then ϕ(α) T is completely determined by α [T ] B. Choose a sufficiently quickly increasing sequence {q n : n N} so that (with q 1 = ) H j n, m 2ñq n 1 For j N and i 2 define: and let ( T 1 B j B 1 j T 1 1 F m T 1 B j B 1 v j i = (µ u d(i) ) T 2B j p j i = µ v j i g 1(q 2j+i ) v j i j ) T1 1 F <2ñqn. P = {p j i : j N, i 2}. Note that the two copies of v j i are disjoint since g 1(q 2j+i ) F 2ñq 2j+i is disjoint from T 1 B j Bj 1 T1 1. which Claim 3. If ϕ j carries each p j i to a shift of some p j i, then ϕ j (p j i ) = a shift of p j i for i 2. 7

8 Proof: Suppose not, and let ϕ j (p j i ) = h p j i with (j, i ) (j, i ) (so 2j + i 2j + i ). If 2j + i < 2j + i, then there are j 1, i 1, j 2, i 2 with 2j 1 + i 1 < 2j 2 + i 2 and 2j + i 2j 2 + i 2 with ϕ j (p j 1 i1 ) = h p j 2 i2 ; otherwise j 1 = j, i 1 = i, j 2 = j, i 2 = i, h = h satisfy these conditions. Let n 1 = 2j 1 + i 1 < n 2 = 2j 2 + i 2. We have that h p j 2 i2 contains occurrences of u d(i 2 ) at h and at h g 1(q n2 ) and hence occurrences of the pattern d at h T 1 and h g 1 (q n2 ) T 1. The values here are determined by the values of p j 1 i1 at h T 1 B j and h g 1 (q n2 ) T 1 B j. Since the pattern d does not occur in µ, and µ is recurrent, we can not have these occurrences induced by a T 1 b j block of µ. Hence each of the two sets h T 1 B j and h g 1 (q n2 ) T 1 B j must intersect one of T 1 B j and g 1 (n 1 ) T 1 B j. We have four possibilities: 1. If h T 1 B j and h g 1 (q n2 ) T 1 B j both meet T 1 B j, then ht 1 b 1 = t 2 b 2 and hg 1 (q n2 )t 3 b 3 = t 4 b 4 for some t 1,... t 4 T 1 and b 1 1,..., b 4 B j, so g 1 (q n2 ) = h 1 t 4 b 4 b 1 = t 1 b 1 b 1 2 t 1 2 t 4b 4 b 1 g 1 (q n2 ) T 1 B j Bj 1 T 1 1 T 1 B j B 1 j T If h T 1 B j and h g 1 (q n2 ) T 1 B j both meet g 1 (q n1 ) T 1 B j, then ht 1 b 1 = g 1 (q n1 )t 2 b 2 and hg 1 (q n2 )t 3 b 3 = g 1 (q n1 )t 4 b 4 : g 1 (q n2 ) = h 1 g 1 (q n1 )t 4 b 4 b 1 = t 1 b 1 b 1 2 t 1 2 g 1(q n1 ) 1 g 1 (q n1 )t 4 b 4 b 1 = t 1 b 1 b 1 2 t 1 2 t 4b 4 b 1 g 1 (q n2 ) T 1 B j Bj 1 T1 1 T 1 B j B 1 T If h T 1 B j meets T 1 B j and h g 1 (q n2 ) T 1 B j meets g 1 (q n1 ) T 1 B j, then ht 1 b 1 = t 2 b 2 and hg 1 (q n2 )t 3 b 3 = g n 1 1 t 4b 4 : g 1 (q n2 ) = h 1 g 1 (Q n1 )t 4 b 4 b 1 = t 1 b 1 b 1 2 t 1 2 g 1(q n1 )t 4 b 4 b 1 g 1 (q n2 ) T 1 B j Bj 1 T1 1 F 2ñn 1 T 1 B j Bj 1 j T If h T 1 B j meets g 1 (q n1 ) T 1 B j and h g 1 (q n2 ) T 1 B j meets T 1 B j, 8

9 then ht 1 b 1 = g 1 (q n1 )t 2 b 2 and hg 1 (q n2 )t 3 b 3 = t 4 b 4 : g 1 (q n2 ) = h 1 t 4 b 4 b 1 = t 1 b 1 b 1 2 t 1 2 g 1(q n1 ) 1 t 4 b 4 b 1 g 1 (q n2 ) T 1 B j Bj 1 T1 1 F 2ñn 1 T 1 B j Bj 1 T1 1. All of these are ruled out by the conditions on q n2 since j n 2 and n 1 < n 2. (claim 3) Let F 2 = {w i : i N}. For a word w and j N we define ρ j (w) for w = q q 1 q n a reduced word in a, a 1, b, b 1 (and q = e for w = e) we define the pattern ρ j (w) by ρ j (w) = g 1 (j) u 2j i() n k= g 1(j+1+k) u 2(j+1+k) q j g 1 (j+1+n+1) u 2(j+1+n+1) i(). We next define a function k n k from N to N as follows. Partition N into disjoint infinite sets N j for j N, and let e(n) = (n) be the first coordinate of the pairing function n = (n), (n) 1, so that e(n) : n N enumerates each natural number infinitely often. Define µ k = µ ρ 1+k2 (e), and for each j and k let { i j k = least i such that ρ l (w i ) ϕ j (µ k ) for some l is no such i exists { i j least i such that {k N j : i j k = i} is infinite, if such exists = otherwise. Note that for B j s potentially inducing isomorphisms of the subflows we will use, there will always be a unique i satisfying the conditions for i j k, and there will always be only finitely many i j k s for a given j (since occurrences of some ρ(w) will have to be induced by the B j saturation of an occurrence of ρ(e)), so i j will be defined. Now let Ñ j = {k j l : l N} = {k N j : i j k = ij } be an increasing enumeration of Ñ j (and simply enumerate N j when i j = ). We can now define n k for k N as { e(l) if k = k j l for some j and l n k = k otherwise. 9

10 The point is that for appropriate j, the set {n k : k Ñj} will enumerate each element of N infinitely often. Finally, for x 2 F 2 we define the G-subflow X x = ϕ(x) as the orbit closure of the set A x given by A x = P {µ u d(w i x(w nk )) ρ1+k2 (w i ) : i, k N}. The map ϕ : x 2 F 2 X x can be implemented in a Borel way as in [?]. Claim 4. The subflow ϕ(x) = X x is free. Proof: Suppose we have α ϕ(x) with g α = α for some g e. Since O µ is free, we can assume if α / O µ. Hence there is some pattern u α with u µ. Let T = dom(u). We have α = lim α i where α i = h i β i with β i A x. We can assume that α i T = u for all i. Since u does not occur in µ, T must intersect some occurrence of one of the finitely many u n patterns in each α i. By passing to a subsequence, we may assume that it is the same pattern in each case, with the same intersection (since there are only finitely many possibilities). That is, one of the u n patterns (which we will call u ) occurs at the same position h, i.e., α i h T 1 = h u for all i. Multiplying each α i by h 1 (and replacing g with h 1 gh ) we may assume in fact that α i T 1 = u for all i. Since the d 1 pattern of T 1 s occurs in elements of A x only at elements g (k) S, we have that for each i there is a k i N with h i g (k i ) T = T, and since g T T for g e we must have h i = g (k i ) 1. Thus, α = lim g (k i ) 1 β i with each β i A x. The T 2 block in α i must then be of the form u k i, so if we consider the two F ñ links encoded there, they must be s 1 ñ(k i 1) s ñk i and s 1 sñ(ki ñk i +1)), and these must be the same for all i. Choosing i large enough so that the corresponding T 2 translates stabilize, the same holds for the links in the translates, so the limit exists in the definition of the following sequence in F Z : g (n) = lim i g (k i + n) (we may assume that the k i s are unbounded since otherwise α will have a unique occurrence of a u d(i) block, and hence can not be a fixed point). Passing to a subsequence we can assume that all of the k i s are even or all are odd; we assume they are all even (the following sequence may be shifted by 1 if they are odd). We then have that the g (2i + 1) T 1 blocks in α must be either u i() or u i(1), so we may define the sequence γ 2 Z by { if α g (2i + 1) T 1 = u γ(i) = i() 1 if α g (2i + 1) T 1 = u i(1). 1

11 Each finite subsequence occurring in γ is induced by a subsequence in some β i, and hence must be a subsequence of our minimal element m encoded at these positions. Thus, γ O m must be a minimal free element. We similarly see that since g α = α, by considering i so the that the limit stabilizes at g T, we must have that g carries the g (k i ) translate of T to some other g (k i ) translate, so g = g (k i )g (k i ) 1 for some k i N for each sufficiently large i, and we may assume k i > k i by using g 1 if necessary. But now the g (k i ) T 2 blocks of β i must be the same as the g (k i ) T 2 blocks, and so (for large enough i), we must have the same links encoded at each such translate. Hence g is actually periodic with period p = k i k i for some sufficiently large i. Squaring g, we can take p = 2q to be even, and we see γ must actually be periodic under the shift by q. This contradicts that γ is aperiodic, and hence we could not have had g α = α. (claim 4) We now check that ϕ witnesses that E(F 2, 2) B E. Suppose first that x, y 2 F 2 with x E(F 2, 2) y; we show that (X x, S) = (X y, S). It will be sufficient to show this when y = a x or y = b x, for a and b generating F 2. Consider the case y = a x (the case of y = b x is identical). We have: A x = P {µ u d(w i x(w nk )) ρ1+k2 (w i ) : i, k N} A y = P {µ u d(w i y(w nk )) ρ1+k2 (w i ) : i, k N} = P {µ u d(w i a x(w nk )) ρ1+k2 (w i ) : i, k N} = P {µ u d(w i x(w nk )) ρ1+k2 (w i a 1 ) : i, k N}. We can define a block code π : 2 B 2 such the induced map f π is an isomorphism of X x and X y as follows. The idea is to ensure that each element µ u d(w i x(w nk )) ρ1+k2 (w i ) is mapped to µ u d(w i x(w nk )) ρ1+k2 (w i a 1 ) (and likewise for their shifts) and all other sequences are left unchanged. This will provide an S-invariant homeomorphism of A x with A y and hence an isomorphism of X x with X y. It suffices to map the patterns of the forms shown in Figure?? (where u v w on the left abbreviates a pattern of the form a 1 2 a 1 1 u s p(a2, ) v p(a 1,b 1 ) t p(,b2) b 1 b 2 w with u, s, v, t, w patterns on T 1, and u v w on the right abbreviates a 1 2 a 1 1 u s p(a2, ) v p(a 1,b 1 ) t p(,b2) b 1 b 2 w ), and all other patterns should be left unchanged. None of the listed patterns overlaps a shift of another and there are only finitely many, so this can be ensured by choosing the block code π appropriately; taking B = (F 2ñ T 2 T 2 ) T 2 (since only the central T 2 block is changed in each of the patterns shown) is sufficient. 11

12 u i() u e u i() u i() u a 1 u i() u i() u a g 1 u i() u i() u e u i() u a u a u i() u a u i() u i() u b u a u i() u b u i() u i() u b 1 u a u i() u b 1 u i() u i() u a 1 u i() u i() u a 1 u a 1 u i() u b u i() u i() u b u a 1 u i() u b 1 u i() u i() u b 1 u a 1 u i() Figure 1: Patterns to be mapped by the block code Finally, suppose x, y A are such that X x = Xy via the function ϕ = ϕ j induced by the block code π j : 2 B j 2. We will show that x E(F 2, 2) y. Note that the points of A x (and shifts thereof) are the isolated points in X x, and so ϕ must map send each point in A x to a shift of a point in A y. We first check that ϕ must send each point in P to a shift of a point in P. Note that each p j i P differs from µ in only finitely many places, whereas the other points of A y contain infinitely many copies of the pattern ( d which does not occur ) in µ. Suppose we have ϕ(p j i ) = β = h µ u d(w i x(w nk )) ρ1+k2 (w i ) for some i and k. Then one of these u patterns must be induced by a T 1 B j block in µ. This block recurs uniformly in µ and hence in p j i, and hence the pattern d 1 must recur uniformly in β. Hence there would be a finite T such that g t T T T (β(gtt ) = d 1 (t )), i.e., d 1 β gt T for all g G. Since d 1 occurs in β only at hg (k) for k, we have that for all g G there is k such that hg (k) gt T. But we can find some k such that T T F <ñk. But now if we let g = hg ( k ) we have that for any k, g ( k ) 1 g (k) / F <ñ(k+k), so hg (k ) 1 g (k) / hg (k ) 1 F <ñk, i.e., for all k we have g (k) / hg k T T, a contradiction. Thus ϕ sends every point in P to the shift of another point in P. Since the same holds for ϕ 1, we must have that ϕ gives a bijection between the closure of P in X x and the closure of P in X y. Since the conditions of claim?? are satisfied, we must have that ϕ j (p j i ) = p j i for i {, 1}. This implies that points in A x containing the initial T 1 B j block of µ u d(i) must maps to (shifts of) points with the same initial T 1 block u d(i). Let h and 12

13 h 1 be the respective shiftings; note that both must be in T 1 B j Bj 1. Choose k such that g 1 (1 + k 2 ) T 1 T 1 B j Bj 1 = for all k k. Consider now the points α k = µ u d(x(w nk )) ρ1+k2 (e) in A x for k k. Each must be mapped to the shift by some h x(wnk ) of a point µ u d(w i(k) y(w nj(k) )) ρ1+j(k)2 (w i(k) ) in A y, with x(w nk ) = w i(k) y(w nj(k) ). We claim that j(k) = k for sufficiently large k. To see this, consider α k. Note that µ u d(i) = lim j p j i (since we took the B j s to be increasing), so ϕ(µ u d(i) ) = lim j ϕ(p j i ) = lim j h i v k(j) i(j) (with k(j) ) = h i (µ u d(i) ). Hence, ϕ(α k ) = h x(wnk ) (µ u d(x(w nk )) ) except on the block g 1(1 + k 2 ) T 2 B j. In particular, g 1 (1 + j(k) 2 ) T 2 g 1 (1 + k 2 ) T 2 B j. Since g 1 (j) T 2 will be disjoint from T 2 B j for all but finitely many j, we must have that j(k) 2 k 2 = (j(k) + k) j(k) k k j(k) k is bounded, and hence j(k) = k for all but finitely many k. Choose k 1 k so that j(k) = k for all k k 1. Note also that the above shows that there is a bound on the length of w i(k), so that there are only finitely many possible w ik s. We now have i(k) = i j k from the definition of i j k earlier. Since there are only finitely many i(k) s, we have that i j is defined. Hence for all but finitely many k Ñj we have i(k) = i j k = i j, and {n k : k Ñj } enumerates N infinitely often. Hence the set {n k : k N j i(k) = i j j(k) = k} includes all of N, so for all n N we have x(w n ) = w i j y(w n ). Hence x = w i j y, so xe(f, 2)y and we are done. 5 Questions Question 1. What is the complexity of isomorphism of free or arbitrary G-subflows for G locally finite? Question 2. What is the complexity of isomorphism of minimal G-subflows for G not locally finite? References [1] J.D. Clemens, Isomorphism of subshifts is a universal countable Borel equivalence relation, Israel Journal of Mathematics, vol. 17 (29), pp

14 [2] S. Gao, S. Jackson and B. Seward, A coloring property for countable groups, to appear in the Mathematical Proceedings of the Cambridge Philosophical Society. [3] S. Gao, S. Jackson and B. Seward, Group colorings and Bernoulli subflows, in preparation. [4] E. Glasner and V.V. Uspenskij, Effective minimal subflows of Bernoulli flows, Proceedings of the American Mathematical Society, vol. 137, no. 9 (September 29), pp [5] S. Jackson, A.S. Kechris, and A. Louveau, Countable Borel equivalence relations. J. Math. Logic, vol. 2, no. 1, (22), pp [6] A.S. Kechris, Classical Descriptive Set Theory, Graduate Texts in Mathematics 156, Springer-Verlag,