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A mixing transformation for which Pinsker's conjecture fails
ADVANCES
IN
MATHEMATICS
A Mixing
10,
103-123
(1973)
Transformation Conjecture DONALD
Department
of Mathematics,
Stanford
for which Fails*>+
Pinsker’s
ORNSTEIN
S.
Urzivrrsity,
Stanford,
Califwnia
94305
The purpose of this paper is to construct a mixing transformation of nonzero entropy that is not the direct product of a transformation of O-entropy and a transformation of completely positive entropy (a K-automorphism). Pinsker [6] conjectured that every ergodic transformation of nonzero entropy was of the above form. This would have implied that the study of ergodic transformations could be reduced to the separate study of K-automorphisms and transformations of Oentropy. In [1] we exhibited a counterexample to Pinsker’s conjecture. The counterexample, however, was pathological in that its square was not ergodic. This might suggest that Pinsker’s conjecture could be saved if we assume some regularity such as continuous spectrum. W’e show here that Pinsker’s conjecture cannot be saved even if we assumemixing. We say that T is a K-automorphism if there is a finite partition P such that the TiP generate and nzzl VT=,T”P is trivial (that is, Vy==,T’P is the class of measurable sets generated by the T’P, n < i < CO, the only sets which are contained in the above classesfor all n have either measure 0 or 1). A special case is when the TiP generate and are independent. There is a beautiful theorem due to Roklin and Sinai [7] which says that T is a K-automorphism if and only if for every finite partition Q,, Q is not contained in V7LTi@ (or, equivalently, E(Q, T) + 0). This also implies that if T is a K-automorphism, any finite partition satisfies the condition of the previous paragraph. In constructing our example we take the example described in [3] of a K-automorphism that is not a Bernoulli shift. (We describe it again for the sake of completeness). We then take an example constructed in [4], of a mixing transformation that commutes only with its powers. We then take a skew product of these two transformations. * This research + Received July
was supported 1972.
in part
by NSF 103
Copyright All rights
q’ 1973 by Academic Press, Inc. of reproduction in any form reserved.
grant
GP 68064.
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1.1. T will be defined in stages, and at each stage we will extend the definition of T to a larger part of the measure space. P will have four sets: PO , P, , Pr , P, .
Dejkitions At stage n we will have the following situation: We will have a set F, , Ti, 0 < i < h(n) - 1 will be defined on F, , and the TiF, will all be disjoint. P will be defined on ~~~‘-‘T~F, . We thus have a e will call it the “gadget at stage n,” “gadget” in the terminology of [l]. W h(n)-lFn its top. (T is not defined on the top). we will call F, its base and T If the gadget G = uiziTiB, then we will call Y the height of G and denote it by h(G). Thus h(G,) = h(n). We will call 3 its base. We will define the name of a point x to be a sequence {nJ where ai is 0, e, f, or s according to whether Tix is in P,, , P, , Pr , or P, . If x is in a gadget G, th en only some of the 01~will be defined and we will call these the name of x in G. The k-name of x will denote the first K-term in the name of X. If G is a gadget, we will define a slice as follows: Partition the base B according to the name (in G) of the points in B. If J is an atom in this partition, we will call $$‘TiJ partitioned by P a slice. By the distribution of the R-names of the points in E’ we will mean the measure on sequences of 0, e, f, or s of length K that we get by normalizing E to have measure 1 and identifying each point in E with its k-name.
Construction
of T
(A) follows: equal uj”l”d”” tioned For
If the gadget at stage 2n is defined, we will get stage 2n Divide F, into f(2n) - 1 (f (2n) will equal 2n) disjoint measure F21L,i , 1 < i = F2n.i 7
+ 1 as sets of gadget partinot in
T(Th’2n)-1Fz,,i) = Fs,,,i,i+l
and T(Fzn,i,d = Fw.a Extend the definition j > 1 are in P, .
if
j # i, j #f(n).
of P so that Fzngisj , j < i are in Pf and Fzn,i,i ,
MIXING
This
105
TRANSFORMATION
defines the gadget at stage 2n + 1. F2,L+1 z u{(_21n)F21~,i,l. Also,
h(2n + I) = h(2n) +f(2n). We will begin our construction with G, of height h(2). (h(2) will be determined later). G, C P, (in fact, G, will turn out to equal I’,,). (B) Before going f r o m stage 212 + 1 to 2n + 2 we will describe an intermediate construction. Suppose we have two gadgets G, and G, where Gi , i = 1,2, is the union of TjJi , 0 < j < yi , partitioned by P. We also assume that the measure of Jl is the same as the measure of Jz and G, and G, are disjoint. We will now define T from TQ J1 onto Jz . This will give us a new gadget G, h G, , consisting of lJA1+rzTjJl , partitioned by P. T on TrlJl will be defined as follows: Partition J1 according to the v,-name of its points. Let I, be an atom in this partition and let I,’ = TQI, ((J&Tjl, is a slice). Partition J3 into sets E, in such a way that for each E, the gadget @,TjE, is isomorphic to G, and m(E,) = m(1,‘). I,’ will map onto E, . We will now use the above construction to define the gadget at stage 2n + 2, given the gadget at stage 2n + 1. Partition F,,t+, into 22n+1 disjoint sets of equal measure such that each is the base of a gadget isomorphic to the gadget at stage 272 + 1. Call these gadgets Gl,..., G(sZRfl’. Let s(2n + 1) b e a number to be determined later. For each Gi pick a collection of i . s(2n + 1) disjoint sets (not in the gadget at stage 2n + 1) all having the same measure as the base of Gi. These sets will be in P, . Define T so that it maps the first of these onto the second, the second onto the third, etc., and the last onto the base of Gi. Call the resulting gadget GL. Form (... (((Gl y G2) * G3) * G$) ... G@‘+‘)). Now take 22n+2s(2n + 1) intervals and map the top of G(2”“f1) onto the first of these, the first onto the second, etc. The resulting gadget will be G 20.t2 .
Choice of,f(n) and s(n) Let f(n) = n and let s(n) = lOOn ( we will and s(n) for n odd). We will have
(a) s(n) > 100 f f(i); i=l
(b)
S(YZ)< ($)10%(n).
only usef(n)
for n even
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(a) is obvious. We can choose h(2) so that (b) holds for all n. (First note that for all even n we can get (b’) s(n) < (*)10”+2h(n). To go from n to n + 2 we multiply the left side by something close to 1 and the right ’ 2o . 2”. Hence proper choice of h(2) gives us (b’) side by more than (2) for all even n. Thus (b) holds for all n). It is easily seen from (a) and (b) that the sup of the measure of the G, is finite and T is then defined on a measure space of finite measure whose measure could be taken to be 1. (At stage n, n odd, the measure of the P, that we add is less than 22n+2s(n)(h(n))-i u(GJ < (3)sn u(G,). Since f(a) < s(n) the measure of the Pr u P, added at stage n, n even, is less than s(n)(h(n))-i u(G,) < (&ion u(G,)). LEMMA 1. If we are given the name of x we can, for eachF, , determine which Tix are in F,, .
Proof. We will show that if we are given >h(n) consecutive terms in the name of x, we can tell which of the Tix’s are in F, . Obvious for n = 2. Simply take the first term in each group of h(2) consecutive 0, 1’s. If the lemma holds for n even, it ovbiously holds for n + Il. If it holds for n odd, we get it for n + 1 as follows: Find the term in F, with exactly s(n) s-terms in front of it. The first of these s’s will be in Fn+l . Remark. We could make our construction on (0, 1) and with a little care we could have P together with the F, generate all measurable sets. Then because of our lemma TiP would generate. This, however will not be necessary. Let G be a gadget with base J. We define a “rectangle” R in G as follows: Let E be a set in some Ti J, and let k > i be an integer smaller than the height of G. Then U~~~Ti+“E will be called a rectangle, and E its base. We will say that R is “pure below” if each of the sets T2E, -i < I< 0 is contained in some atom of P (which atom depends on 1 and E). DEFINITIONS.
LEMMA 2. If n is
MIXING
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TRANSFORMATION
G,, is equal to G, n G,,, = G,, and the union of bases of rectangles G,,, is equal to F, n G,,, = F, .
in
Proof. We will fix n and induct on m 3 n. Note that the lemma is obvious for n = m. We distinguish two cases: m even and m odd. If m is even, than it is obvious that the lemma holds for m + 1. To handle the case where m is odd it is enough to show that if the lemma holds for gadgets Gi and G2, then it holds for G’ * Gz. (G1 will be (-v* (G1 ‘f: G2) -I;-... J;)G’ and Ga will be G”+i). This is also obvious from the construction. LEMMA
3.
T is ergodic.
Proof. We will assume there is a set E, p(E) T(E) = E. We will now derive a contradiction of E). We will first introduce some more terminology. slice, C, in G,, we will call TiB (i < height of G,)
= ac, 0 < a: < 1, and (p(E) is the measure If B is the base of a a level in C.
(1) It is easy to see (by Lemma I) the following. Given E there is an n such that E intersects each level in a slice in G, (except for a collection the measure of whose union is less than E) in more than 1 - E of its measure or less than E of its measure. (2) The measure of the intersection of E with any two levels in the same slice is the same. (3) Let L, and L, be levels of different
slices in G,, . Then
We prove the above as follows: Given E we can find a K and E’ such that E’ C VK,TiP and j E - E’ 1 < E. For each m Lemma 2 implies that G,,, is the union of rectangles R, (in G,,,), each of which is isomorphic to G, and pure below. Therefore, if I > h(n) + K, TIE’ intersects each slice of R, in the same proportion (except for those Ri such that T-lR, is not defined in G,,,). If m is large enough, we can assume the measure of the union of the exceptional Ri is arbitrarily small and that TIE’ intersects each slice in G, in almost the same proportion. Since E’ could be taken as close to E as we want, T”E’ is arbitrarily close to E. This and (2) gives (3).
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LEMMA 4. Given an integer 1 and E > 0, there exists N such that all m > N (N even) have the following property: Let { Ji) be the collection of setsof the form TjF,,, and 0 < j < h(m). Except for a collection %?of Ji the measure of whose union is lessthan E we have that the distribution of l-names of points in uIr;‘Tj Ji is within E of the distribution of l-names of ponts in X.
Proof. Because T is ergodic we have the following: Given there is an M such that if f(m) > M, then all x, except for x of measure lessthan f, have the property that the distribution of in Ucm)Ti(x) is within 5 of the distribution of l-names in X. can be taken <(~/100)~, we get Lemma 4.
5 > 0, in a set l-names Since f
5. Given an integer 1 and E > 0, there exists N such that if n > N and n is odd we have the following: Let {Ji} be the collection Ti%, . Except for a collection of Ji the measure of whose union is lessthan E, the distribution of the l-names of points in Ji is within Eof the distribution of l-names of points in X. LEMMA
Proof. Pick m, even m > N, as in Lemma 4. n will be m + 1. It is obvious from Part A of the construction of T that F,,,, will be the union of disjoint sets Ki ,..., Kf(,,) such that Tf(m)-iKi = iK, is the base of a gadget, isomorphic to G, . If f(m) + 1 < Y < h(m + 1) f(m) - 1, then TTFm,, =
f(m) u
Trf&
=
i=l
f(m) u
Tr-fW+iiK,
i=l
We then have that the distribution of l-names of points in TTF,R,+,is the same as the distribution of l-names of points in Ufil=“4’Ti(TT-f(m))F,,,) and hence is within E of the distribution of l-names in X unless TT-fcnL)F, is in the exceptional set %?for Lemma 4. Since 1 is fixed and since (f (m)/f (m + 1)) + 0, we get Lemma 5. THEOREM.
T is a K-automorphism.
Proof. We must show the following: Given 1, there is an N such that ViTiP is E-independent of V,N+‘TiP for all r > 0. Lemma 5 says that ViTiP is E-independent of Q when Q is the partition of G, into levels. Take N = h(n). For each m > n define Qm as follows: Lemma 1 gives us a collection of disjoint rectangles
MIXING
TRANSFORMATION
109
in G,,, isomorphic to G,, . If Bi is the base of Ri, call TjB, , O
4
1.2, We will define an n-block in the P-name of x to be asequence of h(n) consecutive terms, the first of which is in F,, . Note that, because of Lemma 1, the n-blocks are uniquely determined by the P-name of x. The index of ai will be i. If ~1 > n, we will define the m-order of an n-block as follows: Each n-block a is contained in a unique m-block b. The m-order of a will be i if a is the i-th n-block contained in 6. (Note that if ?zis odd, then the n + l-order of an n-block is determined by the number of s’s in front of it). The next definitions will concern two points x and y and their Pnames (CQ>and {pi}. L et a be an n-block in {CQ)and b an n-block in {pi). Let 1a - b 1 denote the absolute value of the difference of the indices of the first terms of a and b. We will say that a and b are close if 1a - b i < Zr,.,,,f(k). (Note that if a is also close to an n-block b’, then b’ = b. This follows from (b) and (c) in “choice off(n) and s(n)” which imply Zk,,,f(k) < g-h(n).) Let a be an n-block in {ai). We will call CX~in a bad if olj = 0 and pj f 0. LEMMA 1. Let (OIJ and (pi} be the P-name of x and y. Let a and b be n-blocks in (CQ> and {&}, n odd. Assume a and b are close and have the same n + 1-oredr. Let a’ and b’ be the n + 1-blocks containing a and b. Then a’ is close to b’, and every n-block in a’ is close to an n-block in b’ of the same n + l-order.
Proof. The n-blocks in a (and b) are separated by a number of s-terms depending only on the n + l-order of the n blocks. LEMMA 2. Let {q} and (pi} be the P-names of x and y. Let a be an n-block in (OIJ, n even. Let a’ be an n - I -block in a, and assume that a’ is close to an n - l-block b’ whose n-order is dijerent from the n-order of a’.
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Then there is at most one other n to an n - l-block in {pi}.
l-block
a” in a such that a” is close
Proof. (1) It will be enough to show that if b is the n-block containing b’, then there is no other n - l-block in a that is close to an n - l-block in b. To see (l), note there is at most one n-block b different from b which contains an (n - I)-block 6’ close to an n - l-block in a, a’. Furthermore, a’ and 6’ must have different n-orders by Lemma 1. (2) Let ai’ and bi’ be the n - l-blocks in a and b whose n-orders differ from n-orders of a’ and b’ by i. Then ( ai’ - bi’ 1 3 i * s(n - 1) -Zk,,-, f (k). Thus ai’ and bi’ are not close by (a) in “Choice of f(n) and s(n).” (3)
1Ui’ - bi’ 1 < 2” * 2”s(n - 1) +
1
f(k)
< @(n -
1).
k
We get the first inequality as follows: When we change i by 1, the left side increases by at most 2?(n - l), the maximum length of a string of s. Also, i < 2”. We get the second inequality from (a) and (b) in “Choice off(n) and s(n).” (4) Because close to b,‘.
of (3), if ai ’ is close to any n -
l-block,
it must
be
LEMMA 3. There exists a sequence E, > c > 0 such that if (oli} and {pi} are the P-names of x and y, and if a is an n-block in {ai}, then either:
(1) There is an n-block in {/Ii} close to a, OY (2) There are more than Enh’(n), 01~in a that are very bad. (h’(n) is the number of O’s in an n-block). Proof. Lemma 3 is obviously true for n = 2 (Q, = (h(2))-l). We will induct on n. Assume that Lemma 3 is true for n - 1, n odd. Then by the induction hypothesis the n - l-block in a, a’ must be close to some (n - I)-block b’. Hence a must be close to the n-block containing b’. If n is even, we can assume that a contains some n - l-block, a”, that is close to an n - l-block b”. If a” and b” have the same n-order, then Lemma 1 implies that a is close to the n-block containing b”. If a” and b” have different orders, Lemma 2 and the induction hypothesis imply that the second alternative of Lemma 3 holds, with En
>1
c&l
-
2(&y-l).
MIXING
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We will call K consecutive terms in a sequence {ai} DEFINITIONS. a segment of length K. We will say that two segments of length K look alike if the i-th term of the first segment is the same as the i-th term of the second segment 1 < i < K (i.e., they are the same except for a shift of indices). We will say that two segments of length K dzjj$er by E if for EK of the i, the i-th terms in the two segments disagree. If c and d are segments, we will let (c - d) denote the difference between the indices of the first terms in c and d. LEMMA 4. Let {“J and {&) be n + l-block (n odd) and let c and Assume K > 3h(n). Let c and d be c and c dzjj‘er by <(l/IO) i and (F - 2) < $ h(n + 1). Then 1 (c -
the P-names of x and 2. Let a be an d be two segments of length K in a. segments of length K in {$> such that d and C? dz#er by <( 1/I O)E, and E) - (d - a)1 < 2 Zignf (i).
Proof. c and d must each contain at least two (n)-blocks, c1 , c2 and dI , d, . Since c and t differ by <(l/10)& c must contain two (n)-blocks i;l and cz such that (1)
I(c -
cJ
-
(C -
Cl)1 <
1
and
f(i)
I(c - c2) - (C - &)I < c f(i)
i
i< n
and [(cl - cg) - (c, - &)( < 2 x4G.n f (i) by Lemma 3. c1 - c2 = or - F, because any two possible spacings between consecutive (n)-blocks differ by more than two xiGrL,f(i). Thus (2) cr and c1 must have the same n + l-order. (I(c - cl) - (t - c,)I < xdGrL f (i)). Similarly d contains 2, and ;I, and (3) dI and d1 have the same n + l-order, and (4) I@
-
4)
-
(a
-
&)I
<
1 i
f(i).
Because of (2), (3), and the assumption that 1~ - a 1 < $ li(n + I), we have (5) c1 - dI = cI - d1 . Lemma 4 follows from (l), (4), and (5). 2, In this section we will construct a mixing transformation. (This is the same transformation as the one in [4]. However, it is described in a simpler way). It is convenient (but not necessary) first to prove the following theorem. THEOREM
formation 607/10/l-8
1. Let T be a l-l, invertible, measure-preserving qf (0, 1) onto itself, and let m be Lebesgue measure. If
trans(a)
every
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power of T is ergodic, and (b) there is a K such that iZi,,,m( T”A n B)
1. If T satisjies the hypothesis of Theorem 1, then T is weakly
mixing. Proof. Let T be the unitary operator on L, given by Tf (x) = f (T(x)). If T were not weakly mixing, then (by von Neumann’s theorem, see [8]) there would be a complex-valued function g and a complex number E such that Tg = olg. Since T is unitary, / 01/ = 1. Since T is ergodic, j g 1 is constant a.e. Since every power of T is ergodic, we have (1) Any set on which g is constant has measure 0. Let F be the set of x such that 0, > arg (g(x)) > 0, . Then, given c > 0, we can find arbitrarily large n such that TnF is the set of x, 0, + nol > arg (g(x)) > 0, + na, and j nol 1 < E. This and (1) imply that given E’ > 0, we can find arbitrarily large n such that 1 TnF - F 1 < E’. Since m(F) can be chosen as small as we want (by properly choosing 0, and @a), we have contradicted (b). Proof of Theorem 1. (1) We can pick a sequence of integers ni such that if C and D are intervals with rational end points, then
exists. (1) follows from a standard diagonal procedure since there are only a countable number of such C, D. (2) If T is not mixing, then ni can be chosen so that in addition to satisfying (1) there is one pair of intervals with rational end points C, , D, and lim,+mm[( TnC,) n DJ # m(C,) * m(Dl). (3) There is measure u on (0, 1) x (0, I) such that u is absolutely continuous with respect to Lebesgue measure on (0, 1) x (0, l), and if C and D are intervals with rational end points, then u(C x D) = lim,+mm[( TW’) n D]. (3) requires some proof. Order the pairs C, D and let F, be the field ofsetsin(O,l) X (0,l)g enerated by the first n, C x D. Define f, to be the function on (0, 1) x (0, 1) that is constant on each atom of F, and such that if C x D is in F, , then JjcxD f,dmdm = lim+,m[( T”C) n D]. 0 < fn < K by (b). The f, form a Martingale, and hence fn -+ f a.e. f will be the derivitive of u. (It is not really necessary to use the Martingale
MIXING
113
TRANSFORMATION
convergence theorem here. We could define u on the algebra generated by the C x D with rational end points, use (b) to show that u is countably additive when restricted to this algebra, and hence by a theorem in [5] can be extended to a measure on (0, 1) x (0, I)). (4) If A and B are any measurable sets in (0, I), then u(A >: B) = lim,,,m[( PiA) f7 B]. To see (4): (4) holds if A and B are each the union of a finite number of intervals with rational end points. Let A, and B,, be a sequence of such sets approaching A and B, respectively. u(A x B) = ;y* u(A. x B,), u(A4n X B,) = fim m[(PA,J However, ll+l
(b) implies
m[(PAn)
n
B,]
f7 B,].
that -
m[(P;i)
A
B]
< K(m j A, - A / . m(B) + m 1B, - B / m(A) + m j B, - B 1m 1A, - iz I), where
j A, -
A 1 denotes
the symmetric
difference
of A,t and A.
(5) T induces a transformation T on (0, 1) x (0, 1) as follows: T(x, y) = TX, Ty . u is invariant under T. To see (5), we need only check it on sets of the form A x B. u(A x B) = ;iz m[( PA) u(TA
x
TB) =
!im
1-1,
m[(PTL4)
n B], CI
TB] = l& m[T([PA]
n B)].
(b) Since T is weakly mixing, T is ergodic (with respect to Lebesgue measure on (0, 1) >( (0, 1)). S-mce ZLis invariant under T and since u is absolutely continuous with respect to Lebesgue measure, u must be a multiple of Lebesgue measure. This multiple is 1 since u[(O, I) x (0, l)] = 1. This gives We will defined in to a larger
a contradiction since (by (2)) u(C, x DJ # m(C,) * m(Di). now construct a certain mixing transformation T. T will be stages, and at each stage we will extend the definition of T part of the measure space.
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At stage n we will have a set F, , Ti, 0 < i < h(n) - 1 will be defined and the TiF, will be disjoint. Q will be defined on Utz)-‘TiF, . OnF,, We will call this gadget at stage n and denote it, as before, by G, . Q will have the property that each TiFn , 0 < i = h(n) - 1 will lie entirely in one atom of Q. We will define G n+l given G, as follows: We will be given positive integers p(n) and Z(n) and a sequence u~,~, 1 < i
2.
We can choose h(n), p(n), Z(n), and ai so that T is mixing.
The rest of this section will be devoted to a proof of this theorem. Lemma 1 will be the crucial lemma. Here we construct a sequence with certain properties. It is hard to do this explicitly but we will show by a probability argument that most sequences have this property. We will start with Lemma 0 which is a form of the law of large numbers. It is well known (and easy), and we include it only for the sake of completeness. LEMMA 0. Fix 01, 0 < OL< I, and give each sequence of zeros and ones of length n the measure a”( 1 - CX)~-~ where m is the number of zeros in the sequence. This puts a probability measure on the space of such sequences. Let P,(C) be the measure of the set of sequences of length n having more than (a + c)n zeros. Then for fixed E > 0,
+a (P,(E))l’” < 1.
MIXING
Proof.
Let T,,& be the probability y,ll
(1)
n(n -
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that there
are exactly
m zeros.
1) ... (w - m + 1) , N”‘(l _ (y)nH?l. m!
(2) If m > roln, Y > 1, then Y,~,+~/Y,,,< (1 - TCX)/~~. CX/( 1 - CY)< (l/y). This follows from (1). (3) If /3i > /3a > YCX,then
This follows from (2). (4)
This follows from (2). Lemma 0 follows from (3) and (4). LEMMA 1. Given E > 0 and an even positive integer K, then there is an L = L(K, E), and for all m large enough there is a sequence{a,}, i = l,..., m, of integers such that
(a) j Cjj+ka, / < K for all 1 < j < j + k < m. 1
(b) Let H(Z, k) be the number of j such that $hai = 1 where
(c) Except for a collection of
- Q(W) = I) < & .
(3) Now consider a fixed k < (1 - E)m. We can divide the pairs of integers (i, i + k) into two disjoint sets E(k) and El(k) in such a way that no integer occurs in more than one pair in E(k), and the same is
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true of El(k). Furthermore, / E(k)1 denotes the number and we omit the details).
1E(K)1 > 2 Em and 1 El(k)1 > a Em where of elements in E(K). (This is easy to see,
(4) There exists a y, 0 < y < 1, such that for all m large enough, all I and all k < (1 - E) m, we have Prob((number of i such that (i, i + k) E E(k) and s,~+~(w) - si(w) = I) > cl(K)-l 1 E(k)l} < ynl. This follows because of our choice of E(k) which makes all the s~+~(w) - s{(w), (i, i + k) E E(k) m . d ependent and makes 1 E(k)/ > Em, allowing us to apply (2) and Lemma 0. (5) Let ai = si(w) - si-i(w) (al(w) = si(w)). Then si+Jw) Si(W) = p+ki+raj(w). If m is large enough, then (4) tells us that the probability that {ui(w)} d oes not satisfy (b) is t2m * KY”. (To see (5), we note that (b) is automatically true for 1I / > K. Therefore, if (b) fails, (4) fails for one of the (1 - E)mK pairs k, I and E(K) or El(k).) (b) follows immediately from (5) since lim,+m2mKym = 0 implies that (b) holds on a set of measure close to 1. (6) Choose L so large that the measure of sequences of length L such that s<(w) - si+i(w), 1 < i 1 - * E. If m is large enough, the measure of the sequences si(w) having more than Em terms, such that the next L-consecutive terms fail to satisfy the above, is arbitrarily small. This and the last line of (5) prove Lemma 1. Remark. For each fixed k, the H(Z, k) give us a probability measure on the Z of absolute value 10n. (3) There is an odd integer n’ such that (3,a) Z(n) = 2[(n defined). (0)
%(n’)
< 4”
1)‘13 < 2-(1~4)nZz(n)
(assume
(n -
+ 1) < 4kdn’) . (3, 4 Mn’ + 1) > lO%L(K, E) * k(n + 1).
1)’
to
be
MIXING
117
TRANSFORMATION
( a,(,),,
will be an integer <4. We will leave its 13, p. 261. The above choice can be made as follows: We satisfy (3) for all previously chosen n. For all p(n) choose the ai,,& to satisfy (1). Choose I(n) = 2[(n so that (3 + +) h,(d)
definition
until later
assume p(n) and Z(n) large enough we can - I)‘]“. Choose p(n)
+ l(n)) < (4 -
9 h(n’).
(3,b) will follow because 1a,,n 1 < K = h(” - 1) < ,yz(n). If (3,b) holds and n’ is large enough, then (3,~) holds because h(n’ + l)lh(n’) ---f a (for odd n’). We also have that for all sufficiently large (n - l)‘, 2(n) = 2[(n -
1)‘]3 < $2-%&z - I)‘) < 223~lz(n).
The first few stages of the construction can be fairly arbitrary since we only need (l)-(3) to hold for all sufficiently large n. To start we can choose any K, Z(1) and h(1) > 101(l), h(1) > IOK. Now choose p(l) and ai,i to satisfy (l)-(3) (except for (3,a)). (1)
7 lim m[(PA)
M* m
n B] < 4m(A)
. m(B).
If we show that every power of T is ergodic, we can then apply Theorem 1 to show that T is mixing. (2) T is ergodic. If this were not true, there would be an invariant set A, m(A) < 8, and an n and i such that m( Tip,)
This implies that m( Ti(F,) contradicting m(A) < &.
n ,4) > ,a0 m( Tip,))
n A) > (9/10)m( Tj(F,))
for 1 < j < h(n),
(3) Every power of T is ergodic. If this were false, we could use (1) to show that there is a minimal set, A, invariant under T1. We would therefore get an (Y.such that the TjA are disjoint for 0
118
ORNSTEIN
S/l0 of both Ti(F,) and Tk(F,) is in th e same TjA. Since this is true for n and n + 1, we get that more than half of the h(n) + a,,n (1 < i < p(n)) must be congruent to 0 mod 01. Furthermore, the above statement must hold for all n large enough. By changing apcn),% , we change h(n + 1) and we can thus insure that for each integer 01< 4 there are infinitely many n such that more than half of the h(n) + ai,% are not congruent to 0 mod 01. We will now prove that Em m[(TMA) n B] < 4m(A) m(B). Fix M. Pick n such that h(n) < M < h(n + 1). Fix K, 0 < R < h(n) - 1. (A) Let $? be the collection of TiF,+,(O
MIXING
119
TRANSFORMATION
union of TkF, . Similarly B can be approximated very well by the union of T”F,,-, . Because of (A) and (B) we can break up most of the T’CF, into translates of F,,, or F,,, and TM applied to each of these translates intersects each T1F7,_1 in no more than four times what it ought to. 3.
Construction
of T,
We will now call the transformation constructed in Section 1, T, , and the one constructed in Section 2, T, . We will construct a transformation T, which will be a skew-product of Tl and T, . T, will be a mixing transformation that fails to satisfy Pinsker’s conjecture. T, will act on X, = X, x Xi as follows: Let (x, y) be a point in If x E 0, , x3 = x2 x x,. If XEQO, then T,(x, y) = ( Tzx, T,y). then T,(x, y) = (T,x, y). 1. The u-algebra of O-entropy E x X, , where E is any measurable
LEMMA
form
for T, consists of all sets of the set in X, .
First note that any partition consisting of sets of the form X1 have O-entropy relative to T, . To prove the lemma we will take any other partition, take the span with some partition whose sets have the form E x X, and show that this span has non-O-entropy relative to T, , thus getting a contradiction. Suppose our O-entropy partition had a set F, C X, not of the form E x X, . Let F, C Xi be the set of y such that (x, y) EF3 . We could find a set E, C X, and a set H, in X, such that x E E,, then I F, - H, j < E(1F,, - H, 1 denotes the measure of the symmetric difference of F, and Hi) and E has the property that if R is a partition consisting of three atoms, two of which have measure 0, thus getting a contradiction and proving the lemma. Pick K so that m( T,“(E,) n E2) > i (m(E,))2 for all 72> K, and if y(j) is any function satisfying j y(j) - y(lz)I > 4 K for J’ + K, then Proof.
E
x
1 lim - . E [t ‘l’;‘yq] n-,x n j=l
> &?qH,).
(This can be done because Tl is a K-automorphism).
120
ORNSTEIN
Define g(x, Z) to be the number of 0 < i < KZ such that Tzix E PO . Because of our construction (1) g(x, I) > QKZ. Note that Tf(x, y) = (TkK(x), Tpsz) (y)). For each z in Ez define Zi(z) to be the i-th z E T$‘E, . Define xi in E2 as zi = T-z@)K(z).
integer j, such that
(2) (We get (2) as follows: g(W,
3) = g(-&),
4,
I &w&>,
4 - g(-&),4I
> w
by (1). This and our choice of K gives (2).)
Because of our choice of K we have that for all M large enough (4) there is a set of z in E, of measure >&m(E,))2 such that the number of i, for which Zi(z) < M, is >*(m(E,))2M. (3) and (4) imply that E (it LEMMA
2.
T,"(R))
> &,(m(E,))4 * E(H,) * M.
T, is mixing.
Proof. Define ,$x, a) to be the number of 0 < i < n such that Tix E Q,, . Let E x F and E x F be rectangles in X2 x X, . T,“(E x F) consists of the points ( Tzn(x), T!‘“,“‘(y)) where x E E and y EF. Because of our choice of Q. , g(x, n) > B n. For each fixed TZnx, the set of possible 7 such that ( T2n~, p) E T3n(E x F) is Tf(“*n)F. Lemma 2 follows immediately from the fact that Tl and T2 are mixing and from Fubini’s theorem. 4. T3 doesnot satisfy Pinsker’s conjecture Throughout this section we will assume that T3 satisfies Pinsker’s conjecture, and will arrive at the end with a contradiction.
MIXING
TRANSFORMATION
121
If Pinsker’s conjecture held, there would be an u-algebra 0! invariant under T, which together with the u-algebra of O-entropy generates. Let P be a countable generator of finite entropy for the above o-algebra. Every point (x, y) in X, = X, x Xi has a Q, P-name under T, , icyi), (ri} and a Q, P-name under T, , {tiJ, [/3J. consisting By the O-part of {olJ, $,) we will mean the subsequence of those pi such that the 0~;of the same index is equal to 0. (The O-part is defined only up to some translate of the indices. If we talk about the index of a term in the O-part we will mean its old index). The O-part of {ui), @I,} is the P- name of y (or some Tliy) under Tl . We have a new measure-preserving automorphism of the o-algebra of X, induced by the identity on the a-algebra of O-entropy and TzL on 0’. This will be induced by a point transformation sending (x, y) to is the Q, P-name of x, y, then the Q, P-name of X, 7 (.?c,Y). If (q>, -(Yi> will be {ai>, {‘yi) where ri = yi+, Let g, be the mapping which takes the Q, P-name of (x, y), {a,$, {vi>, onto the 0, P-name (4, (Pi). (~((4, (rJ> = (&, (Pi:.) Because Q V p generates (under TJ we can find, for each E, a K and a partition P’ such that the symmetric difference between P and P’ has measure
being the Q, p-name of (x, y)). qK has the property that flj’ is determined by {cx~}~$+: and {~~}f$+g. It follows from the ergodic theorem that for a.e. (x, y) if we let (cYJ, (ri} be its Q, p-name and let F((cx(}, (rJ) = .(cx,~},{&f and ~,,((a~), of i such that pi f pi isI {Pi’)7 then the frequency We will now fix E = 10-5~ (g is defined in Section I .2, Lemma 4). This will give us a K (as above) (not to be confused with the K of Section 2; we will now call that K K,). Let the h(n) of Section 2 be h,(n). For each n there is an n’ defined in Section 2. Pick n so that K/h(n) < (l/10)6 . n and n’ (and E and K) will now be fixed for the rest of this paper. DEFINITIONS. Let {ai} be the Q-name of x under an n-block to be h,(n) consecutive terms starting (Note that any two n-blocks look alike). We will al t a3, a3 , a4 a 4-g+oup in (cY~>,if a, - a2 = a3 -
T, . We will define with an n (or Q,,). call four n-blocks a4 ; a, comes first,
122
ORNSTEIN
a2 second, as third, and a4 last. a2 is the first n-block following a, and a4 is the first n-block following a3 , and a3 , a4 are the first pair of n-blocks to the right of a2 satisfying all of the above. We also assume that a, , a2 , a3 , a4 all belong to the same n + l-block. We will also assume that [ a3 -
(L, is defined
a, j < L,(h,(n
in Section
-
l),
10-n-3) h(n) < lo-Vz,(n’
2, “Construction
of T,”
+ 1)
and satisfies
(3,~)).
LEMMA 1. Let (ai), (yi> be the (Q, Q-name of (x, y). Let (al , a,, a3 , a4) be a 4-group in (cQ}. Let k = a, - a2 = a3 - a4 . (1) Assume that the O-part of ~({a~}, {yi}) d oes not have an n’ + l-block whose jivst term has index dajjekring from the index of the jrst term of a, by <(l/lOOO)h,(n’ + 1). Let yi = yipk. Then v and y,, do not (l/lOO)Z agree on (al , a2 , a3 , 4 for {ai>, {YJ and {ai>, 173.
Proof. (1) The segment of {ai}, (y,} corresponding to a, looks like the segment of (01~1,{ri} corresponding to a2 . The above statement also holds if we replace a, by a3 and a2 by a4 . (This is obvious from the definition of 4-groups and the definition of {pi}). C2) WC{%>, {Yil> on the segments a, , a3 differs by
(3) If the conclusion of the lemma were false, then the segment corresponding to a, and the segment of ~({a~}, {yi}) Of df.49 {Yi>) corresponding to a2 differs by <(l/100)<. (3) also holds if we replace a, by a2 and a3 by a4 . (3) follows from (2). (3) implies that if we let c and d be the O-parts of ~({a~}, (70) corresponding to a, and a3 and if we let F and d be the O-parts of ~{(a~}, (?J) corresponding to a2 and a4, then c, C, d, d satisfy the hypothesis of Lemma 4 because of the definitions of 4-groups and conditions (a) and (6) in “The Construction of T2” in Section 2. (Assumption (1) implies that c and d are in the same n’ + l-blocks). However, the conclusion of Lemma 4, Section 1.2 is not satisfied because of the definition of Q2 which implies that the difference between the number of O’s between the first terms of a, and a2 and the number between the first terms of a3 and a4 is >2s(n’). This gives a contradiction. We will now use Lemma 1 to get a contradiction by showing that there must be some {ai}, {yi} satisfying the hypothesis of Lemma 1 but not its conclusion. We do this as follows:
MIXING
TRANSFORMATION
123
(1) Call (x, y) b a d i f w h en we take its Q, P-name {ai}, {ri} and if we let (ai}, {Pi} = p)(q), (ri} and {& {Pi’} = e((4, {A}), then either (i) /I,, # &’ or (ii) the first term of an n’ + l-block in the O-part of (ai}, {&} has index <(l/1000) h,(n’ + 1). (2) Take an (n + 1)-block, a, in the Q-name {ai} of x (all such (a -+ 1)-blocks look alike). Divide it into 4-groups of n-blocks. S/l0 of the n-blocks are in at least 1, and all are in at most four such 4-groups. (This holds because of the definition of 4-groups, condition (3,~) and (c) of Lemma 1). To each 4-group define a corresponding 4-set to be those (x, y) such that if we take the Q-name {ai} of X, then c+, lies in an 11+ l-block and lies in the above 4-groups in that (n + I)-block. (3) There must be at least one 4-group (ur , ua, aa , u4) whose corresponding 4-set, A, contains <(5/100@(A) bad points. (Since F and P)~ c-agree, and E < (l/1000), the measure of the points that are bad because of(i) is <(l/100). Th e measure of the bad points satisfying ecause for each X, y the frequency of T$(x, y) satisfying (ii) is <(2/1000) b (ii) is <(2/1000)). (4) Because of (3) and the ergodic theorem we have that for a.e. x, y we have th e f o 11owing: Let {ai}, {ri} be the Q, p-names of x, y. Let jji = yi&a, - a2 = A). There is one 4-group, a,, u2, u3, u4, looking like A, and y and P)~ applied to {a,}, {ri} and (ai}, {ri} 3E-agree on this 4-group. Furthermore, the O-part of ~({a~}, {ri}) does not have an m-block whose first term has index <(1/2000)h(n’ + 1). This contradicts Lemma 1 and proves our theorem.
REFERENCES I. D. S. ORNSTEIN, A K-automorphism with no square root and Pinsker’s conjecture, to appear. 2. D. S. ORNSTEIN, A K-automorphism that is not a Bernoulli shift, to appear. 3. D. S. ORNSTEIN, A mixing transformation that commutes only with its powers, Proc. Sixth Rerkeley Symp. Math. Stat. Prob., Vol. II, part 2, pp. 335-360, University of California Press, 1967. 4. M. S. PINSKER, Dynamical systems with completely positive and zero entropies, Dokl. Akad. Nauk SSSR 133 (I 960), 1025-1026. 5. V. A. ROKLIN AND J. G. SINAI, Construction and properties of invariant measurable partitions, Dokl. Akad. Nauk SSSR 141 (1961), 1038-1041. 6. P. R. HALMOS, “Lectures on Ergodic Theory,” Math. Sot. Japan, Tokyo (1956), 54-55.
IN
MATHEMATICS
A Mixing
10,
103-123
(1973)
Transformation Conjecture DONALD
Department
of Mathematics,
Stanford
for which Fails*>+
Pinsker’s
ORNSTEIN
S.
Urzivrrsity,
Stanford,
Califwnia
94305
The purpose of this paper is to construct a mixing transformation of nonzero entropy that is not the direct product of a transformation of O-entropy and a transformation of completely positive entropy (a K-automorphism). Pinsker [6] conjectured that every ergodic transformation of nonzero entropy was of the above form. This would have implied that the study of ergodic transformations could be reduced to the separate study of K-automorphisms and transformations of Oentropy. In [1] we exhibited a counterexample to Pinsker’s conjecture. The counterexample, however, was pathological in that its square was not ergodic. This might suggest that Pinsker’s conjecture could be saved if we assume some regularity such as continuous spectrum. W’e show here that Pinsker’s conjecture cannot be saved even if we assumemixing. We say that T is a K-automorphism if there is a finite partition P such that the TiP generate and nzzl VT=,T”P is trivial (that is, Vy==,T’P is the class of measurable sets generated by the T’P, n < i < CO, the only sets which are contained in the above classesfor all n have either measure 0 or 1). A special case is when the TiP generate and are independent. There is a beautiful theorem due to Roklin and Sinai [7] which says that T is a K-automorphism if and only if for every finite partition Q,, Q is not contained in V7LTi@ (or, equivalently, E(Q, T) + 0). This also implies that if T is a K-automorphism, any finite partition satisfies the condition of the previous paragraph. In constructing our example we take the example described in [3] of a K-automorphism that is not a Bernoulli shift. (We describe it again for the sake of completeness). We then take an example constructed in [4], of a mixing transformation that commutes only with its powers. We then take a skew product of these two transformations. * This research + Received July
was supported 1972.
in part
by NSF 103
Copyright All rights
q’ 1973 by Academic Press, Inc. of reproduction in any form reserved.
grant
GP 68064.
104
ORNSTEIN
1.1. T will be defined in stages, and at each stage we will extend the definition of T to a larger part of the measure space. P will have four sets: PO , P, , Pr , P, .
Dejkitions At stage n we will have the following situation: We will have a set F, , Ti, 0 < i < h(n) - 1 will be defined on F, , and the TiF, will all be disjoint. P will be defined on ~~~‘-‘T~F, . We thus have a e will call it the “gadget at stage n,” “gadget” in the terminology of [l]. W h(n)-lFn its top. (T is not defined on the top). we will call F, its base and T If the gadget G = uiziTiB, then we will call Y the height of G and denote it by h(G). Thus h(G,) = h(n). We will call 3 its base. We will define the name of a point x to be a sequence {nJ where ai is 0, e, f, or s according to whether Tix is in P,, , P, , Pr , or P, . If x is in a gadget G, th en only some of the 01~will be defined and we will call these the name of x in G. The k-name of x will denote the first K-term in the name of X. If G is a gadget, we will define a slice as follows: Partition the base B according to the name (in G) of the points in B. If J is an atom in this partition, we will call $$‘TiJ partitioned by P a slice. By the distribution of the R-names of the points in E’ we will mean the measure on sequences of 0, e, f, or s of length K that we get by normalizing E to have measure 1 and identifying each point in E with its k-name.
Construction
of T
(A) follows: equal uj”l”d”” tioned For
If the gadget at stage 2n is defined, we will get stage 2n Divide F, into f(2n) - 1 (f (2n) will equal 2n) disjoint measure F21L,i , 1 < i
+ 1 as sets of gadget partinot in
T(Th’2n)-1Fz,,i) = Fs,,,i,i+l
and T(Fzn,i,d = Fw.a Extend the definition j > 1 are in P, .
if
j # i, j #f(n).
of P so that Fzngisj , j < i are in Pf and Fzn,i,i ,
MIXING
This
105
TRANSFORMATION
defines the gadget at stage 2n + 1. F2,L+1 z u{(_21n)F21~,i,l. Also,
h(2n + I) = h(2n) +f(2n). We will begin our construction with G, of height h(2). (h(2) will be determined later). G, C P, (in fact, G, will turn out to equal I’,,). (B) Before going f r o m stage 212 + 1 to 2n + 2 we will describe an intermediate construction. Suppose we have two gadgets G, and G, where Gi , i = 1,2, is the union of TjJi , 0 < j < yi , partitioned by P. We also assume that the measure of Jl is the same as the measure of Jz and G, and G, are disjoint. We will now define T from TQ J1 onto Jz . This will give us a new gadget G, h G, , consisting of lJA1+rzTjJl , partitioned by P. T on TrlJl will be defined as follows: Partition J1 according to the v,-name of its points. Let I, be an atom in this partition and let I,’ = TQI, ((J&Tjl, is a slice). Partition J3 into sets E, in such a way that for each E, the gadget @,TjE, is isomorphic to G, and m(E,) = m(1,‘). I,’ will map onto E, . We will now use the above construction to define the gadget at stage 2n + 2, given the gadget at stage 2n + 1. Partition F,,t+, into 22n+1 disjoint sets of equal measure such that each is the base of a gadget isomorphic to the gadget at stage 272 + 1. Call these gadgets Gl,..., G(sZRfl’. Let s(2n + 1) b e a number to be determined later. For each Gi pick a collection of i . s(2n + 1) disjoint sets (not in the gadget at stage 2n + 1) all having the same measure as the base of Gi. These sets will be in P, . Define T so that it maps the first of these onto the second, the second onto the third, etc., and the last onto the base of Gi. Call the resulting gadget GL. Form (... (((Gl y G2) * G3) * G$) ... G@‘+‘)). Now take 22n+2s(2n + 1) intervals and map the top of G(2”“f1) onto the first of these, the first onto the second, etc. The resulting gadget will be G 20.t2 .
Choice of,f(n) and s(n) Let f(n) = n and let s(n) = lOOn ( we will and s(n) for n odd). We will have
(a) s(n) > 100 f f(i); i=l
(b)
S(YZ)< ($)10%(n).
only usef(n)
for n even
106
ORNSTEIN
(a) is obvious. We can choose h(2) so that (b) holds for all n. (First note that for all even n we can get (b’) s(n) < (*)10”+2h(n). To go from n to n + 2 we multiply the left side by something close to 1 and the right ’ 2o . 2”. Hence proper choice of h(2) gives us (b’) side by more than (2) for all even n. Thus (b) holds for all n). It is easily seen from (a) and (b) that the sup of the measure of the G, is finite and T is then defined on a measure space of finite measure whose measure could be taken to be 1. (At stage n, n odd, the measure of the P, that we add is less than 22n+2s(n)(h(n))-i u(GJ < (3)sn u(G,). Since f(a) < s(n) the measure of the Pr u P, added at stage n, n even, is less than s(n)(h(n))-i u(G,) < (&ion u(G,)). LEMMA 1. If we are given the name of x we can, for eachF, , determine which Tix are in F,, .
Proof. We will show that if we are given >h(n) consecutive terms in the name of x, we can tell which of the Tix’s are in F, . Obvious for n = 2. Simply take the first term in each group of h(2) consecutive 0, 1’s. If the lemma holds for n even, it ovbiously holds for n + Il. If it holds for n odd, we get it for n + 1 as follows: Find the term in F, with exactly s(n) s-terms in front of it. The first of these s’s will be in Fn+l . Remark. We could make our construction on (0, 1) and with a little care we could have P together with the F, generate all measurable sets. Then because of our lemma TiP would generate. This, however will not be necessary. Let G be a gadget with base J. We define a “rectangle” R in G as follows: Let E be a set in some Ti J, and let k > i be an integer smaller than the height of G. Then U~~~Ti+“E will be called a rectangle, and E its base. We will say that R is “pure below” if each of the sets T2E, -i < I< 0 is contained in some atom of P (which atom depends on 1 and E). DEFINITIONS.
LEMMA 2. If n is
MIXING
107
TRANSFORMATION
G,, is equal to G, n G,,, = G,, and the union of bases of rectangles G,,, is equal to F, n G,,, = F, .
in
Proof. We will fix n and induct on m 3 n. Note that the lemma is obvious for n = m. We distinguish two cases: m even and m odd. If m is even, than it is obvious that the lemma holds for m + 1. To handle the case where m is odd it is enough to show that if the lemma holds for gadgets Gi and G2, then it holds for G’ * Gz. (G1 will be (-v* (G1 ‘f: G2) -I;-... J;)G’ and Ga will be G”+i). This is also obvious from the construction. LEMMA
3.
T is ergodic.
Proof. We will assume there is a set E, p(E) T(E) = E. We will now derive a contradiction of E). We will first introduce some more terminology. slice, C, in G,, we will call TiB (i < height of G,)
= ac, 0 < a: < 1, and (p(E) is the measure If B is the base of a a level in C.
(1) It is easy to see (by Lemma I) the following. Given E there is an n such that E intersects each level in a slice in G, (except for a collection the measure of whose union is less than E) in more than 1 - E of its measure or less than E of its measure. (2) The measure of the intersection of E with any two levels in the same slice is the same. (3) Let L, and L, be levels of different
slices in G,, . Then
We prove the above as follows: Given E we can find a K and E’ such that E’ C VK,TiP and j E - E’ 1 < E. For each m Lemma 2 implies that G,,, is the union of rectangles R, (in G,,,), each of which is isomorphic to G, and pure below. Therefore, if I > h(n) + K, TIE’ intersects each slice of R, in the same proportion (except for those Ri such that T-lR, is not defined in G,,,). If m is large enough, we can assume the measure of the union of the exceptional Ri is arbitrarily small and that TIE’ intersects each slice in G, in almost the same proportion. Since E’ could be taken as close to E as we want, T”E’ is arbitrarily close to E. This and (2) gives (3).
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LEMMA 4. Given an integer 1 and E > 0, there exists N such that all m > N (N even) have the following property: Let { Ji) be the collection of setsof the form TjF,,, and 0 < j < h(m). Except for a collection %?of Ji the measure of whose union is lessthan E we have that the distribution of l-names of points in uIr;‘Tj Ji is within E of the distribution of l-names of ponts in X.
Proof. Because T is ergodic we have the following: Given there is an M such that if f(m) > M, then all x, except for x of measure lessthan f, have the property that the distribution of in Ucm)Ti(x) is within 5 of the distribution of l-names in X. can be taken <(~/100)~, we get Lemma 4.
5 > 0, in a set l-names Since f
5. Given an integer 1 and E > 0, there exists N such that if n > N and n is odd we have the following: Let {Ji} be the collection Ti%, . Except for a collection of Ji the measure of whose union is lessthan E, the distribution of the l-names of points in Ji is within Eof the distribution of l-names of points in X. LEMMA
Proof. Pick m, even m > N, as in Lemma 4. n will be m + 1. It is obvious from Part A of the construction of T that F,,,, will be the union of disjoint sets Ki ,..., Kf(,,) such that Tf(m)-iKi = iK, is the base of a gadget, isomorphic to G, . If f(m) + 1 < Y < h(m + 1) f(m) - 1, then TTFm,, =
f(m) u
Trf&
=
i=l
f(m) u
Tr-fW+iiK,
i=l
We then have that the distribution of l-names of points in TTF,R,+,is the same as the distribution of l-names of points in Ufil=“4’Ti(TT-f(m))F,,,) and hence is within E of the distribution of l-names in X unless TT-fcnL)F, is in the exceptional set %?for Lemma 4. Since 1 is fixed and since (f (m)/f (m + 1)) + 0, we get Lemma 5. THEOREM.
T is a K-automorphism.
Proof. We must show the following: Given 1, there is an N such that ViTiP is E-independent of V,N+‘TiP for all r > 0. Lemma 5 says that ViTiP is E-independent of Q when Q is the partition of G, into levels. Take N = h(n). For each m > n define Qm as follows: Lemma 1 gives us a collection of disjoint rectangles
MIXING
TRANSFORMATION
109
in G,,, isomorphic to G,, . If Bi is the base of Ri, call TjB, , O
4
1.2, We will define an n-block in the P-name of x to be asequence of h(n) consecutive terms, the first of which is in F,, . Note that, because of Lemma 1, the n-blocks are uniquely determined by the P-name of x. The index of ai will be i. If ~1 > n, we will define the m-order of an n-block as follows: Each n-block a is contained in a unique m-block b. The m-order of a will be i if a is the i-th n-block contained in 6. (Note that if ?zis odd, then the n + l-order of an n-block is determined by the number of s’s in front of it). The next definitions will concern two points x and y and their Pnames (CQ>and {pi}. L et a be an n-block in {CQ)and b an n-block in {pi). Let 1a - b 1 denote the absolute value of the difference of the indices of the first terms of a and b. We will say that a and b are close if 1a - b i < Zr,.,,,f(k). (Note that if a is also close to an n-block b’, then b’ = b. This follows from (b) and (c) in “choice off(n) and s(n)” which imply Zk,,,f(k) < g-h(n).) Let a be an n-block in {ai). We will call CX~in a bad if olj = 0 and pj f 0. LEMMA 1. Let (OIJ and (pi} be the P-name of x and y. Let a and b be n-blocks in (CQ> and {&}, n odd. Assume a and b are close and have the same n + 1-oredr. Let a’ and b’ be the n + 1-blocks containing a and b. Then a’ is close to b’, and every n-block in a’ is close to an n-block in b’ of the same n + l-order.
Proof. The n-blocks in a (and b) are separated by a number of s-terms depending only on the n + l-order of the n blocks. LEMMA 2. Let {q} and (pi} be the P-names of x and y. Let a be an n-block in (OIJ, n even. Let a’ be an n - I -block in a, and assume that a’ is close to an n - l-block b’ whose n-order is dijerent from the n-order of a’.
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Then there is at most one other n to an n - l-block in {pi}.
l-block
a” in a such that a” is close
Proof. (1) It will be enough to show that if b is the n-block containing b’, then there is no other n - l-block in a that is close to an n - l-block in b. To see (l), note there is at most one n-block b different from b which contains an (n - I)-block 6’ close to an n - l-block in a, a’. Furthermore, a’ and 6’ must have different n-orders by Lemma 1. (2) Let ai’ and bi’ be the n - l-blocks in a and b whose n-orders differ from n-orders of a’ and b’ by i. Then ( ai’ - bi’ 1 3 i * s(n - 1) -Zk,,-, f (k). Thus ai’ and bi’ are not close by (a) in “Choice of f(n) and s(n).” (3)
1Ui’ - bi’ 1 < 2” * 2”s(n - 1) +
1
f(k)
< @(n -
1).
k
We get the first inequality as follows: When we change i by 1, the left side increases by at most 2?(n - l), the maximum length of a string of s. Also, i < 2”. We get the second inequality from (a) and (b) in “Choice off(n) and s(n).” (4) Because close to b,‘.
of (3), if ai ’ is close to any n -
l-block,
it must
be
LEMMA 3. There exists a sequence E, > c > 0 such that if (oli} and {pi} are the P-names of x and y, and if a is an n-block in {ai}, then either:
(1) There is an n-block in {/Ii} close to a, OY (2) There are more than Enh’(n), 01~in a that are very bad. (h’(n) is the number of O’s in an n-block). Proof. Lemma 3 is obviously true for n = 2 (Q, = (h(2))-l). We will induct on n. Assume that Lemma 3 is true for n - 1, n odd. Then by the induction hypothesis the n - l-block in a, a’ must be close to some (n - I)-block b’. Hence a must be close to the n-block containing b’. If n is even, we can assume that a contains some n - l-block, a”, that is close to an n - l-block b”. If a” and b” have the same n-order, then Lemma 1 implies that a is close to the n-block containing b”. If a” and b” have different orders, Lemma 2 and the induction hypothesis imply that the second alternative of Lemma 3 holds, with En
>1
c&l
-
2(&y-l).
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We will call K consecutive terms in a sequence {ai} DEFINITIONS. a segment of length K. We will say that two segments of length K look alike if the i-th term of the first segment is the same as the i-th term of the second segment 1 < i < K (i.e., they are the same except for a shift of indices). We will say that two segments of length K dzjj$er by E if for EK of the i, the i-th terms in the two segments disagree. If c and d are segments, we will let (c - d) denote the difference between the indices of the first terms in c and d. LEMMA 4. Let {“J and {&) be n + l-block (n odd) and let c and Assume K > 3h(n). Let c and d be c and c dzjj‘er by <(l/IO) i and (F - 2) < $ h(n + 1). Then 1 (c -
the P-names of x and 2. Let a be an d be two segments of length K in a. segments of length K in {$> such that d and C? dz#er by <( 1/I O)E, and E) - (d - a)1 < 2 Zignf (i).
Proof. c and d must each contain at least two (n)-blocks, c1 , c2 and dI , d, . Since c and t differ by <(l/10)& c must contain two (n)-blocks i;l and cz such that (1)
I(c -
cJ
-
(C -
Cl)1 <
1
and
f(i)
I(c - c2) - (C - &)I < c f(i)
i
i< n
and [(cl - cg) - (c, - &)( < 2 x4G.n f (i) by Lemma 3. c1 - c2 = or - F, because any two possible spacings between consecutive (n)-blocks differ by more than two xiGrL,f(i). Thus (2) cr and c1 must have the same n + l-order. (I(c - cl) - (t - c,)I < xdGrL f (i)). Similarly d contains 2, and ;I, and (3) dI and d1 have the same n + l-order, and (4) I@
-
4)
-
(a
-
&)I
<
1 i
f(i).
Because of (2), (3), and the assumption that 1~ - a 1 < $ li(n + I), we have (5) c1 - dI = cI - d1 . Lemma 4 follows from (l), (4), and (5). 2, In this section we will construct a mixing transformation. (This is the same transformation as the one in [4]. However, it is described in a simpler way). It is convenient (but not necessary) first to prove the following theorem. THEOREM
formation 607/10/l-8
1. Let T be a l-l, invertible, measure-preserving qf (0, 1) onto itself, and let m be Lebesgue measure. If
trans(a)
every
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power of T is ergodic, and (b) there is a K such that iZi,,,m( T”A n B)
1. If T satisjies the hypothesis of Theorem 1, then T is weakly
mixing. Proof. Let T be the unitary operator on L, given by Tf (x) = f (T(x)). If T were not weakly mixing, then (by von Neumann’s theorem, see [8]) there would be a complex-valued function g and a complex number E such that Tg = olg. Since T is unitary, / 01/ = 1. Since T is ergodic, j g 1 is constant a.e. Since every power of T is ergodic, we have (1) Any set on which g is constant has measure 0. Let F be the set of x such that 0, > arg (g(x)) > 0, . Then, given c > 0, we can find arbitrarily large n such that TnF is the set of x, 0, + nol > arg (g(x)) > 0, + na, and j nol 1 < E. This and (1) imply that given E’ > 0, we can find arbitrarily large n such that 1 TnF - F 1 < E’. Since m(F) can be chosen as small as we want (by properly choosing 0, and @a), we have contradicted (b). Proof of Theorem 1. (1) We can pick a sequence of integers ni such that if C and D are intervals with rational end points, then
exists. (1) follows from a standard diagonal procedure since there are only a countable number of such C, D. (2) If T is not mixing, then ni can be chosen so that in addition to satisfying (1) there is one pair of intervals with rational end points C, , D, and lim,+mm[( TnC,) n DJ # m(C,) * m(Dl). (3) There is measure u on (0, 1) x (0, I) such that u is absolutely continuous with respect to Lebesgue measure on (0, 1) x (0, l), and if C and D are intervals with rational end points, then u(C x D) = lim,+mm[( TW’) n D]. (3) requires some proof. Order the pairs C, D and let F, be the field ofsetsin(O,l) X (0,l)g enerated by the first n, C x D. Define f, to be the function on (0, 1) x (0, 1) that is constant on each atom of F, and such that if C x D is in F, , then JjcxD f,dmdm = lim+,m[( T”C) n D]. 0 < fn < K by (b). The f, form a Martingale, and hence fn -+ f a.e. f will be the derivitive of u. (It is not really necessary to use the Martingale
MIXING
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TRANSFORMATION
convergence theorem here. We could define u on the algebra generated by the C x D with rational end points, use (b) to show that u is countably additive when restricted to this algebra, and hence by a theorem in [5] can be extended to a measure on (0, 1) x (0, I)). (4) If A and B are any measurable sets in (0, I), then u(A >: B) = lim,,,m[( PiA) f7 B]. To see (4): (4) holds if A and B are each the union of a finite number of intervals with rational end points. Let A, and B,, be a sequence of such sets approaching A and B, respectively. u(A x B) = ;y* u(A. x B,), u(A4n X B,) = fim m[(PA,J However, ll+l
(b) implies
m[(PAn)
n
B,]
f7 B,].
that -
m[(P;i)
A
B]
< K(m j A, - A / . m(B) + m 1B, - B / m(A) + m j B, - B 1m 1A, - iz I), where
j A, -
A 1 denotes
the symmetric
difference
of A,t and A.
(5) T induces a transformation T on (0, 1) x (0, 1) as follows: T(x, y) = TX, Ty . u is invariant under T. To see (5), we need only check it on sets of the form A x B. u(A x B) = ;iz m[( PA) u(TA
x
TB) =
!im
1-1,
m[(PTL4)
n B], CI
TB] = l& m[T([PA]
n B)].
(b) Since T is weakly mixing, T is ergodic (with respect to Lebesgue measure on (0, 1) >( (0, 1)). S-mce ZLis invariant under T and since u is absolutely continuous with respect to Lebesgue measure, u must be a multiple of Lebesgue measure. This multiple is 1 since u[(O, I) x (0, l)] = 1. This gives We will defined in to a larger
a contradiction since (by (2)) u(C, x DJ # m(C,) * m(Di). now construct a certain mixing transformation T. T will be stages, and at each stage we will extend the definition of T part of the measure space.
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At stage n we will have a set F, , Ti, 0 < i < h(n) - 1 will be defined and the TiF, will be disjoint. Q will be defined on Utz)-‘TiF, . OnF,, We will call this gadget at stage n and denote it, as before, by G, . Q will have the property that each TiFn , 0 < i = h(n) - 1 will lie entirely in one atom of Q. We will define G n+l given G, as follows: We will be given positive integers p(n) and Z(n) and a sequence u~,~, 1 < i
2.
We can choose h(n), p(n), Z(n), and ai so that T is mixing.
The rest of this section will be devoted to a proof of this theorem. Lemma 1 will be the crucial lemma. Here we construct a sequence with certain properties. It is hard to do this explicitly but we will show by a probability argument that most sequences have this property. We will start with Lemma 0 which is a form of the law of large numbers. It is well known (and easy), and we include it only for the sake of completeness. LEMMA 0. Fix 01, 0 < OL< I, and give each sequence of zeros and ones of length n the measure a”( 1 - CX)~-~ where m is the number of zeros in the sequence. This puts a probability measure on the space of such sequences. Let P,(C) be the measure of the set of sequences of length n having more than (a + c)n zeros. Then for fixed E > 0,
+a (P,(E))l’” < 1.
MIXING
Proof.
Let T,,& be the probability y,ll
(1)
n(n -
115
TRANSFORMATION
that there
are exactly
m zeros.
1) ... (w - m + 1) , N”‘(l _ (y)nH?l. m!
(2) If m > roln, Y > 1, then Y,~,+~/Y,,,< (1 - TCX)/~~. CX/( 1 - CY)< (l/y). This follows from (1). (3) If /3i > /3a > YCX,then
This follows from (2). (4)
This follows from (2). Lemma 0 follows from (3) and (4). LEMMA 1. Given E > 0 and an even positive integer K, then there is an L = L(K, E), and for all m large enough there is a sequence{a,}, i = l,..., m, of integers such that
(a) j Cjj+ka, / < K for all 1 < j < j + k < m. 1
(b) Let H(Z, k) be the number of j such that $hai = 1 where
(c) Except for a collection of
- Q(W) = I) < & .
(3) Now consider a fixed k < (1 - E)m. We can divide the pairs of integers (i, i + k) into two disjoint sets E(k) and El(k) in such a way that no integer occurs in more than one pair in E(k), and the same is
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true of El(k). Furthermore, / E(k)1 denotes the number and we omit the details).
1E(K)1 > 2 Em and 1 El(k)1 > a Em where of elements in E(K). (This is easy to see,
(4) There exists a y, 0 < y < 1, such that for all m large enough, all I and all k < (1 - E) m, we have Prob((number of i such that (i, i + k) E E(k) and s,~+~(w) - si(w) = I) > cl(K)-l 1 E(k)l} < ynl. This follows because of our choice of E(k) which makes all the s~+~(w) - s{(w), (i, i + k) E E(k) m . d ependent and makes 1 E(k)/ > Em, allowing us to apply (2) and Lemma 0. (5) Let ai = si(w) - si-i(w) (al(w) = si(w)). Then si+Jw) Si(W) = p+ki+raj(w). If m is large enough, then (4) tells us that the probability that {ui(w)} d oes not satisfy (b) is t2m * KY”. (To see (5), we note that (b) is automatically true for 1I / > K. Therefore, if (b) fails, (4) fails for one of the (1 - E)mK pairs k, I and E(K) or El(k).) (b) follows immediately from (5) since lim,+m2mKym = 0 implies that (b) holds on a set of measure close to 1. (6) Choose L so large that the measure of sequences of length L such that s<(w) - si+i(w), 1 < i
%(n’)
< 4”
1)‘13 < 2-(1~4)nZz(n)
(assume
(n -
+ 1) < 4kdn’) . (3, 4 Mn’ + 1) > lO%L(K, E) * k(n + 1).
1)’
to
be
MIXING
117
TRANSFORMATION
( a,(,),,
will be an integer <4. We will leave its 13, p. 261. The above choice can be made as follows: We satisfy (3) for all previously chosen n. For all p(n) choose the ai,,& to satisfy (1). Choose I(n) = 2[(n so that (3 + +) h,(d)
definition
until later
assume p(n) and Z(n) large enough we can - I)‘]“. Choose p(n)
+ l(n)) < (4 -
9 h(n’).
(3,b) will follow because 1a,,n 1 < K = h(” - 1) < ,yz(n). If (3,b) holds and n’ is large enough, then (3,~) holds because h(n’ + l)lh(n’) ---f a (for odd n’). We also have that for all sufficiently large (n - l)‘, 2(n) = 2[(n -
1)‘]3 < $2-%&z - I)‘) < 223~lz(n).
The first few stages of the construction can be fairly arbitrary since we only need (l)-(3) to hold for all sufficiently large n. To start we can choose any K, Z(1) and h(1) > 101(l), h(1) > IOK. Now choose p(l) and ai,i to satisfy (l)-(3) (except for (3,a)). (1)
7 lim m[(PA)
M* m
n B] < 4m(A)
. m(B).
If we show that every power of T is ergodic, we can then apply Theorem 1 to show that T is mixing. (2) T is ergodic. If this were not true, there would be an invariant set A, m(A) < 8, and an n and i such that m( Tip,)
This implies that m( Ti(F,) contradicting m(A) < &.
n ,4) > ,a0 m( Tip,))
n A) > (9/10)m( Tj(F,))
for 1 < j < h(n),
(3) Every power of T is ergodic. If this were false, we could use (1) to show that there is a minimal set, A, invariant under T1. We would therefore get an (Y.such that the TjA are disjoint for 0
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S/l0 of both Ti(F,) and Tk(F,) is in th e same TjA. Since this is true for n and n + 1, we get that more than half of the h(n) + a,,n (1 < i < p(n)) must be congruent to 0 mod 01. Furthermore, the above statement must hold for all n large enough. By changing apcn),% , we change h(n + 1) and we can thus insure that for each integer 01< 4 there are infinitely many n such that more than half of the h(n) + ai,% are not congruent to 0 mod 01. We will now prove that Em m[(TMA) n B] < 4m(A) m(B). Fix M. Pick n such that h(n) < M < h(n + 1). Fix K, 0 < R < h(n) - 1. (A) Let $? be the collection of TiF,+,(O
MIXING
119
TRANSFORMATION
union of TkF, . Similarly B can be approximated very well by the union of T”F,,-, . Because of (A) and (B) we can break up most of the T’CF, into translates of F,,, or F,,, and TM applied to each of these translates intersects each T1F7,_1 in no more than four times what it ought to. 3.
Construction
of T,
We will now call the transformation constructed in Section 1, T, , and the one constructed in Section 2, T, . We will construct a transformation T, which will be a skew-product of Tl and T, . T, will be a mixing transformation that fails to satisfy Pinsker’s conjecture. T, will act on X, = X, x Xi as follows: Let (x, y) be a point in If x E 0, , x3 = x2 x x,. If XEQO, then T,(x, y) = ( Tzx, T,y). then T,(x, y) = (T,x, y). 1. The u-algebra of O-entropy E x X, , where E is any measurable
LEMMA
form
for T, consists of all sets of the set in X, .
First note that any partition consisting of sets of the form X1 have O-entropy relative to T, . To prove the lemma we will take any other partition, take the span with some partition whose sets have the form E x X, and show that this span has non-O-entropy relative to T, , thus getting a contradiction. Suppose our O-entropy partition had a set F, C X, not of the form E x X, . Let F, C Xi be the set of y such that (x, y) EF3 . We could find a set E, C X, and a set H, in X, such that x E E,, then I F, - H, j < E(1F,, - H, 1 denotes the measure of the symmetric difference of F, and Hi) and E has the property that if R is a partition consisting of three atoms, two of which have measure
E
x
1 lim - . E [t ‘l’;‘yq] n-,x n j=l
> &?qH,).
(This can be done because Tl is a K-automorphism).
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Define g(x, Z) to be the number of 0 < i < KZ such that Tzix E PO . Because of our construction (1) g(x, I) > QKZ. Note that Tf(x, y) = (TkK(x), Tpsz) (y)). For each z in Ez define Zi(z) to be the i-th z E T$‘E, . Define xi in E2 as zi = T-z@)K(z).
integer j, such that
(2) (We get (2) as follows: g(W,
3) = g(-&),
4,
I &w&>,
4 - g(-&),4I
> w
by (1). This and our choice of K gives (2).)
Because of our choice of K we have that for all M large enough (4) there is a set of z in E, of measure >&m(E,))2 such that the number of i, for which Zi(z) < M, is >*(m(E,))2M. (3) and (4) imply that E (it LEMMA
2.
T,"(R))
> &,(m(E,))4 * E(H,) * M.
T, is mixing.
Proof. Define ,$x, a) to be the number of 0 < i < n such that Tix E Q,, . Let E x F and E x F be rectangles in X2 x X, . T,“(E x F) consists of the points ( Tzn(x), T!‘“,“‘(y)) where x E E and y EF. Because of our choice of Q. , g(x, n) > B n. For each fixed TZnx, the set of possible 7 such that ( T2n~, p) E T3n(E x F) is Tf(“*n)F. Lemma 2 follows immediately from the fact that Tl and T2 are mixing and from Fubini’s theorem. 4. T3 doesnot satisfy Pinsker’s conjecture Throughout this section we will assume that T3 satisfies Pinsker’s conjecture, and will arrive at the end with a contradiction.
MIXING
TRANSFORMATION
121
If Pinsker’s conjecture held, there would be an u-algebra 0! invariant under T, which together with the u-algebra of O-entropy generates. Let P be a countable generator of finite entropy for the above o-algebra. Every point (x, y) in X, = X, x Xi has a Q, P-name under T, , icyi), (ri} and a Q, P-name under T, , {tiJ, [/3J. consisting By the O-part of {olJ, $,) we will mean the subsequence of those pi such that the 0~;of the same index is equal to 0. (The O-part is defined only up to some translate of the indices. If we talk about the index of a term in the O-part we will mean its old index). The O-part of {ui), @I,} is the P- name of y (or some Tliy) under Tl . We have a new measure-preserving automorphism of the o-algebra of X, induced by the identity on the a-algebra of O-entropy and TzL on 0’. This will be induced by a point transformation sending (x, y) to is the Q, P-name of x, y, then the Q, P-name of X, 7 (.?c,Y). If (q>, -(Yi> will be {ai>, {‘yi) where ri = yi+, Let g, be the mapping which takes the Q, P-name of (x, y), {a,$, {vi>, onto the 0, P-name (4, (Pi). (~((4, (rJ> = (&, (Pi:.) Because Q V p generates (under TJ we can find, for each E, a K and a partition P’ such that the symmetric difference between P and P’ has measure
being the Q, p-name of (x, y)). qK has the property that flj’ is determined by {cx~}~$+: and {~~}f$+g. It follows from the ergodic theorem that for a.e. (x, y) if we let (cYJ, (ri} be its Q, p-name and let F((cx(}, (rJ) = .(cx,~},{&f and ~,,((a~), of i such that pi f pi is
T, . We will define with an n (or Q,,). call four n-blocks a4 ; a, comes first,
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a2 second, as third, and a4 last. a2 is the first n-block following a, and a4 is the first n-block following a3 , and a3 , a4 are the first pair of n-blocks to the right of a2 satisfying all of the above. We also assume that a, , a2 , a3 , a4 all belong to the same n + l-block. We will also assume that [ a3 -
(L, is defined
a, j < L,(h,(n
in Section
-
l),
10-n-3) h(n) < lo-Vz,(n’
2, “Construction
of T,”
+ 1)
and satisfies
(3,~)).
LEMMA 1. Let (ai), (yi> be the (Q, Q-name of (x, y). Let (al , a,, a3 , a4) be a 4-group in (cQ}. Let k = a, - a2 = a3 - a4 . (1) Assume that the O-part of ~({a~}, {yi}) d oes not have an n’ + l-block whose jivst term has index dajjekring from the index of the jrst term of a, by <(l/lOOO)h,(n’ + 1). Let yi = yipk. Then v and y,, do not (l/lOO)Z agree on (al , a2 , a3 , 4 for {ai>, {YJ and {ai>, 173.
Proof. (1) The segment of {ai}, (y,} corresponding to a, looks like the segment of (01~1,{ri} corresponding to a2 . The above statement also holds if we replace a, by a3 and a2 by a4 . (This is obvious from the definition of 4-groups and the definition of {pi}). C2) WC{%>, {Yil> on the segments a, , a3 differs by
(3) If the conclusion of the lemma were false, then the segment corresponding to a, and the segment of ~({a~}, {yi}) Of df.49 {Yi>) corresponding to a2 differs by <(l/100)<. (3) also holds if we replace a, by a2 and a3 by a4 . (3) follows from (2). (3) implies that if we let c and d be the O-parts of ~({a~}, (70) corresponding to a, and a3 and if we let F and d be the O-parts of ~{(a~}, (?J) corresponding to a2 and a4, then c, C, d, d satisfy the hypothesis of Lemma 4 because of the definitions of 4-groups and conditions (a) and (6) in “The Construction of T2” in Section 2. (Assumption (1) implies that c and d are in the same n’ + l-blocks). However, the conclusion of Lemma 4, Section 1.2 is not satisfied because of the definition of Q2 which implies that the difference between the number of O’s between the first terms of a, and a2 and the number between the first terms of a3 and a4 is >2s(n’). This gives a contradiction. We will now use Lemma 1 to get a contradiction by showing that there must be some {ai}, {yi} satisfying the hypothesis of Lemma 1 but not its conclusion. We do this as follows:
MIXING
TRANSFORMATION
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(1) Call (x, y) b a d i f w h en we take its Q, P-name {ai}, {ri} and if we let (ai}, {Pi} = p)(q), (ri} and {& {Pi’} = e((4, {A}), then either (i) /I,, # &’ or (ii) the first term of an n’ + l-block in the O-part of (ai}, {&} has index <(l/1000) h,(n’ + 1). (2) Take an (n + 1)-block, a, in the Q-name {ai} of x (all such (a -+ 1)-blocks look alike). Divide it into 4-groups of n-blocks. S/l0 of the n-blocks are in at least 1, and all are in at most four such 4-groups. (This holds because of the definition of 4-groups, condition (3,~) and (c) of Lemma 1). To each 4-group define a corresponding 4-set to be those (x, y) such that if we take the Q-name {ai} of X, then c+, lies in an 11+ l-block and lies in the above 4-groups in that (n + I)-block. (3) There must be at least one 4-group (ur , ua, aa , u4) whose corresponding 4-set, A, contains <(5/100@(A) bad points. (Since F and P)~ c-agree, and E < (l/1000), the measure of the points that are bad because of(i) is <(l/100). Th e measure of the bad points satisfying ecause for each X, y the frequency of T$(x, y) satisfying (ii) is <(2/1000) b (ii) is <(2/1000)). (4) Because of (3) and the ergodic theorem we have that for a.e. x, y we have th e f o 11owing: Let {ai}, {ri} be the Q, p-names of x, y. Let jji = yi&a, - a2 = A). There is one 4-group, a,, u2, u3, u4, looking like A, and y and P)~ applied to {a,}, {ri} and (ai}, {ri} 3E-agree on this 4-group. Furthermore, the O-part of ~({a~}, {ri}) does not have an m-block whose first term has index <(1/2000)h(n’ + 1). This contradicts Lemma 1 and proves our theorem.
REFERENCES I. D. S. ORNSTEIN, A K-automorphism with no square root and Pinsker’s conjecture, to appear. 2. D. S. ORNSTEIN, A K-automorphism that is not a Bernoulli shift, to appear. 3. D. S. ORNSTEIN, A mixing transformation that commutes only with its powers, Proc. Sixth Rerkeley Symp. Math. Stat. Prob., Vol. II, part 2, pp. 335-360, University of California Press, 1967. 4. M. S. PINSKER, Dynamical systems with completely positive and zero entropies, Dokl. Akad. Nauk SSSR 133 (I 960), 1025-1026. 5. V. A. ROKLIN AND J. G. SINAI, Construction and properties of invariant measurable partitions, Dokl. Akad. Nauk SSSR 141 (1961), 1038-1041. 6. P. R. HALMOS, “Lectures on Ergodic Theory,” Math. Sot. Japan, Tokyo (1956), 54-55.