Invariant measures for bipermutative cellular automata

A `right-sided, nearest neighbour cellular automaton' (RNNCA) is a continuous transformation F:A^Z-->A^Z determined by a local rule f:A^{0,1}-->A so that, for any a in A^Z and any z in Z, F(a)_z = f(a_{z},a_{z+1}) . We say that F is `bipermutative' if, for any choice of a in A, the map g:A-->A defined by g(b) = f(a,b) is bijective, and also, for any choice of b in A, the map h:A-->A defined by h(a)=f(a,b) is bijective. We characterize the invariant measures of bipermutative RNNCA. First we introduce the equivalent notion of a `quasigroup CA', to expedite the construction of examples. Then we characterize F-invariant measures when A is a (nonabelian) group, and f(a,b) = a*b. Then we show that, if F is any bipermutative RNNCA, and mu is F-invariant, then F must be mu-almost everywhere K-to-1, for some constant K . We use this to characterize invariant measures when A^Z is a `group shift' and F is an `endomorphic CA'.


Introduction
If A is a (discretely topologized) finite set, then A Z is compact in the Tychonoff topology. Let σ : A Z −→A Z be the shift map: σ(a) = [b z | z∈Z ], where b z = a z−1 , ∀z ∈ Z. A cellular automaton (CA) is a continuous map Φ : A Z −→A Z which commutes with σ. Equivalently, Φ is defined by a local rule φ : A [−ℓ...r] −→A (for some ℓ, r ≥ 0) so that, for any a ∈ A Z and any z ∈ Z, Φ(a) z = φ(a z−ℓ , . . . , a z+r ). We say Φ is right-permutative if, for any fixed a ∈ A [−ℓ...r) , the map A ∋ b → φ(a, b) ∈ A is bijective. Likewise, Φ is 2 Quasigroup Cellular Automata A quasigroup [9] is a finite set A equipped with a binary operation ' * ' which has the leftand right-cancellation properties. In other words, for any a, b, c ∈ A, If we identify A with [1..N] in some arbitrary way, then the 'multiplication table' for * is the N × N matrix M * = [m i,j ] N i,j=1 where m i,j = i * j. We say M * is a Latin square [4] if every column and every row of M * contains each element of [1..N] exactly once. It follows: (A, * ) is a quasigroup ⇐⇒ M * is a Latin square .
Note that the operator ' * ' is not necessarily associative. Indeed, it is easy to show: ' * ' is associative ⇐⇒ (A, * ) is a group .
A quasigroup cellular automaton (QGCA) is a right-sided, nearest neighbour cellular automaton Φ : A Z −→A Z with local rule φ : A {0,1} −→A given: φ(a 0 , a 1 ) = a 0 * a 1 , where ' * ' is a quasigroup operation. For example, any Ledrappier automaton is a QGCA. It follows: The obvious generalization of Proposition 3 fails for arbitrary quasigroup CA. If B ⊂ A, then we call B a subquasigroup (and write 'B ≺ A') if B is closed under the ' * ' operation.
This suggests that the correct generalization of Proposition 3 is: If µ is a Φ-invariant and σ-ergodic measure, and h µ (Φ) > 0, then µ is the uniform measure on B Z , for some B ≺ A.
Conjecture 7 is false, as we will show with Example (12b) of §3.
Unilateral vs. Bilateral Cellular Automata: Any right-sided CA Φ : A Z −→A Z induces a unilateral CA Φ : A N −→A N with the same local rule. Any Φ-invariant measure on A Z projects to a Φ-invariant measure on A N ; conversely, any Φ-and σ-invariant measure on A N extends to a unique (Φ, σ)-invariant measure on A Z . In what follows, we will abuse notation and write Φ as Φ. Thus, Conjecture 7 is equivalent to: Dual Cellular Automata: There is a well-known conjugacy between any right-permutative unilateral CA and a full shift. Define Ξ : Let (A, * ) be a quasigroup. The dual quasigroup is the set A equipped with binary operator * defined: a * b = c, where c is the unique element in A such that a * c = b.
Lemma 9 Let (A, * ) be a quasigroup and let Φ be the corresponding QGCA. Then: (a) (A, * ) is a quasigroup, and Φ is a QGCA. The dual of * is * ; the dual of Φ is Φ.
(b) Ξ is a topological conjugacy from the dynamical system (A N , σ) to the system (A N , Φ), so that we have the following commuting cube:

PSfrag replacements
Let µ be a measure on A Z , and let µ = Ξ(µ). Then: Thus, Conjecture 7 is equivalent to: If µ is a Φ-ergodic and σ-invariant measure, and h µ (σ) > 0, then µ is the uniform measure on B N , for some B ≺ A.
It is Conjecture 7 which we'll refute in §3.

Multiplication CA on Nonabelian Groups
Let N = {0, 1, 2, 3, . . .}, and let µ be a σ-invariant measure on A N . Let N = {1, 2, 3, . . .} For any a ∈ A N , let µ a be the conditional measure induced by a on the zeroth coordinate. That is, for any b ∈ A, (where x ∈ A N is a µ-random sequence) Let µ be the projection of µ onto A N . Then we have the following disintegration [11]: Suppose A is a finite (possibly nonabelian) group, and let C ≺ A be a subgroup. We call µ a C-measure if, for ∀ µ a ∈ A N , supp (µ a ) is a right coset of C, and µ a is uniformly distributed on this coset. It follows: Lemma 10 (a) If µ is a C-measure, then h(µ, σ) = log 2 |C|. (c) µ is an A-measure ⇐⇒ µ is the uniform measure on A N . ✷ Let Φ : A N −→A N be the nearest neighbour multiplication CA, having local map φ(a 0 , a 1 ) = a 0 · a 1 . This type of CA was previously studied in [7,10]. Our goal is to prove: Theorem 11 If µ is σ-invariant and Φ-ergodic, then µ is a C-measure for some C ≺ A. ✷ Example 12: (a) Let C ≺ A be any subgroup, and let µ be the uniform measure on C N . Then µ is a C-measure (for any a ∈ A N , µ a is uniform on C), and µ is σ-invariant and Φ-ergodic.
(b) Let Q = {±1, ±i, ±j, ±k} be the Quaternion group [2, §1.5], and let Φ Q : Q N −→Q N be the nearest neighbour multiplication CA. It follows: Let µ Q be the probability measure on Q N assigning probability 1/3 to each of p, Φ Q (p) and Φ 2 Q (p). Then µ Q is σ-invariant and Φ Q -ergodic. Now, let C be any other group, and let A = C × Q. Identify C with C × {1} ≺ A; then C is a normal subgroup of A, and Q = A/C. The cosets of C all have the form C × {q} for some q ∈ Q. There is a natural identification A N ∼ = C N × Q N , given: Let µ C be the uniform Bernoulli measure on C N , and let µ = µ C ⊗ µ Q .
Proof: Theorem 11 says µ must be a C-measure for some subgroup C ≺ A. But if B is any proper subgroup, then h(µ, σ) > h max ≥ log 2 |B|, so Lemma 10(a) says C can't be B. Thus, C = A. Then Lemma 10(c) says that µ is the uniform measure.
Proof: Suppose not. Let .

Lemma 21
If ν is invariant under scalar multiplication by c, then, for ∀ µ a ∈ A N , µ a is invariant under left multiplication by c.
Claim 1: Proof: Generalize the reasoning behind equation (2) Proof: For any measurable subset U ⊂ A N , Claim 3: For any measurable subset W ⊂ A N , .

Proof of Theorem 11
Let C be the set of all c ∈ A so that there is some a ∈ A N and b ∈ A with both [b, a] and [(cb), a] being (Φ, µ)-generic.

Degree of QGCA relative to invariant measures
If µ is a Φ-invariant measure, then Φ is K-to-1 (µ-ae) if there is a subset U ⊂ A Z such that: 3. µ-almost every element u ∈ U has exactly K preimages in U -ie. U ∩ Φ −1 {u} = K.
We will generalize the methods of [1] to prove: Theorem 22 Let Φ : A Z −→A Z be a quasigroup CA, and let µ be a measure which is Φinvariant and σ-ergodic. Let |A| = N. Then there is some K ∈ [1.
.N] so that Example 23: Let λ be the uniform Bernoulli measure on A Z . Then λ is invariant for any QGCA, h λ (Φ) = log 2 (N), and Φ is N-to-1 (λ-ae). Indeed, λ is the only (Φ, σ)-invariant measure with entropy log 2 (N). Thus, Proposition 3 is proved in [1] by first proving a special case of Theorem 22 (when Φ is a Ledrappier CA) and then showing that K = N.
Let µ be a measure on A Z . If q is any partition of A Z , and S is any sigma-algebra, define Let p 0 be the partition of A Z generated by zero-coordinate cylinder sets, and let p [ℓ,n] = n m=ℓ σ −m (p 0 ). Thus,

Proof:
Let x ∈ A Z be an unknown sequence. Because Φ is bipermutative, complete information about (Φ t (x)) [−r,r] (for t ∈ [0..T )) is sufficient to reconstruct x [−T −r,T +r] , and vice versa. In other words, we have an equality of partitions: Letting T →∞, we get an equality of sigma-algebras: Applying Φ −1 to everything yields: Now, Φ is bipermutative, so if we have complete knowledge of Φ(x), then we can reconstruct x from knowledge only of x 0 . Thus, H µ p [−r,r] |B 1 = H µ (p 0 |B 1 ).
. Hence, the sets F (x) (for x ∈ A Z ) are the 'minimal elements' of the sigma algebra B 1 .
The conditional expectation operator E µ [• |B 1 ] defines 'fibre' measures µ x (for ∀ µ x ∈ A Z ) having three properties: (F2) For any fixed x ∈ A Z , µ x is a probability measure on A Z , and supp (µ x ) = F (x).
(F3) For any fixed measurable U ⊂ A Z , the function A Z ∋ x → µ x (U) ∈ R is B 1measurable. Hence, µ x = µ y for any y ∈ F (x).
Our goal is to show that there is some constant K and, for ∀ µ x ∈ A Z , there is a subset E ⊂ F (x) of cardinality K so that µ x is uniformly distributed on E.

Thus, we must show that
as desired. Here (E) is the defining property of conditional expectation, (I) is because µ is σ-invariant, and (S) is the substitution For any x ∈ A Z , let η(x) = µ x {x}. Thus, if y is an unknown, µ-random sequence, then η(x) represents the conditional probability that y = x, given that Φ(y) = Φ(x).

Here, (c2) follows from Claim 2 below, and (26b) is by Corollary 26(b). ✸ [Claim 1]
Proof: Φ is bipermutative, so if y ∈ F (x), then y is entirely determined by y 0 . Thus, Our goal is to show that H = 1 K for some K. Suppose |A| = N, and identify A with the group Z /N in an arbitrary way. Define τ : A Z −→A Z as follows. For any x ∈ A Z , τ (x) = y, where y is the unique element in F (x) such that y 0 = x 0 + 1 (mod N). Existence/uniqueness of y follows from bipermutativity.
Lemma 28 µ n is absolutely continuous relative to µ.
Proof: Lemma 28 means that a statement which is true for ∀ µ x is also true for ∀ µn x. Now apply Lemma 26(c). ✷
Corollary 31 For ∀ µ x ∈ A Z , µ x is equidistributed on E(x). If card [E(x)] = K, then µ x assigns mass 1 K to each element in E(x). In particular, η(x) = 1 K .

Proof of Theorem 22:
is a set of cardinality K, by definition of U. ✷

Endomorphic Cellular Automata
A group shift is a sequence space A Z equipped with a topological group structure such that σ is a group automorphism. Equivalently, the multiplication operation • on A Z is defined by some local multiplication map ψ : .r] −→A so that, if a, b ∈ A Z and c = a • b, then c 0 = ψ(a −ℓ , . . . , a r ; b −ℓ , . . . , b r ). The most obvious group shift is a product group, where A is a finite group and multiplication on A Z is defined componentwise. However, this is not the only group shift [5].
An endomorphic cellular automaton (ECA) is a cellular automaton Φ : A Z −→A Z which is also a group endomorphism of A Z . For example, it is easy to verify: Proposition 33 Let (A, +) be an additive abelian group. Let A Z be the product group. Let Φ : A Z −→A Z be a right-sided, nearest neighbour CA, with local map φ : A {0,1} −→A. Then: (a) Φ is an ECA iff φ(a 0 , a 1 ) = φ 0 (a 0 ) + φ 1 (a 1 ), where φ 0 , φ 1 are endomorphisms of A.
(b) Φ is bipermutative iff φ 0 and φ 1 are automorphisms of A. ✷ We will now apply the results of §4 to bipermutative ECA, to prove: Theorem 34 Let A Z be a group shift and let Φ : A Z −→A Z be a bipermutative ECA. Suppose ker(Φ) contains no nontrivial σ-invariant subgroups. If µ is Φ-invariant and totally σ-ergodic, and h µ (Φ) > 0, then µ = λ. ✷ Lemma 35 Let Φ : A Z −→A Z be a bipermutative ECA on a group shift. Let K = ker(Φ).
(a) For any x ∈ A Z , F (x) = x • K.
(b) Let e ∈ A Z be the identity element. Then e is a constant sequence -ie. there is some e ∈ A so that e = (...., e, e, e, ....).
(c) K is σ-invariant. Also, if k ∈ K, then k is entirely determined by k 0 . It follows that every element of K is P -periodic, for some P < |A|.
(f) Any σ-invariant subgroup J ≺ K is thus a disjoint union of periodic σ-orbits, which corresponds to a disjoint union of ρ-orbits in A.  (c) The map ρ : A−→A from Lemma 35(e) is a group automorphism. To be precise, suppose Φ has local map φ(a 0 , a 1 ) = φ 0 (a 0 ) + φ 1 (a 1 ), where φ 0 and φ 1 are automorphisms of A, as in Proposition 33(b).
(e) In particular, A has no nontrivial ρ-invariant subgroups ⇐⇒ K has no nontrivial σ-invariant subgroups .
Proof: We need only verify the claim in (b) that ζ is a group homomorphism. To see this, suppose k = ζ(a) and k ′ = ζ(a ′ ). Let j = k + k ′ and let i = ζ(a + a ′ ); we want to show j = i. From Lemma 35(c), it suffices to show that i 0 = j 0 . But the operation on K is componentwise addition. Thus, j 0 = k 0 + k ′ 0 = a + a ′ = i 0 . Hence, ζ is a homomorphism; being bijective, ζ is thus an isomorphism. All other claims follow. ✷ Let η be as in §4, and for any k ∈ K,
Proof: By Lemma 35(e), find P ∈ N so that σ P (j) = j. But then Lemma 37(b) says that σ P (E j ) = E j . But µ is σ P -ergodic, so this means that µ(E j ) = 1. ✸ [Claim 1] Proof: Thus, we want to show that, for ∀ µ x ∈ A Z , and all k ∈ K, x ∈ E k ⇐⇒ k ∈ J . Observe that µ   k∈K\J E k   = 0 (by definition of J ) and µ j∈J E j = 1 (by Claim 1).
Claim 3: If k ∈ K, then k ∈ J ⇐⇒ U • k ⊂ U, modulo a set of measure zero .

Proof of Theorem 34
If h µ (Φ) > 0, then Corollary 39 says |J | > 1, so J is a nontrivial σ-invariant subgroup of K. Thus, J = K, which means |J | = |K| = |A|, where the second equality is by Lemma 35(d). Thus, h µ (Φ) = log |A|. Thus, h µ (σ) = log |A|, which means µ must be the uniform measure. ✷ Lemmas 35(g) and 36(e) provide conditions under which K has no σ-invariant subgroups. For example, suppose p ∈ N is prime, and let A = (Z /p ) N for some N > 0. Then A is a vector space over the field Z /p , and ρ : A−→A is a group automorphism iff ρ is a Z /plinear automorphism. Thus, ρ can be described by an N × N matrix M of coefficients in Z /p . Furthermore, B ⊂ A is a (ρ-invariant) subgroup iff B is a (ρ-invariant) subspace. The structure of ρ-invariant subspaces in A is described by the rational canonical form of ρ; this is an N × N matrix M, similar to M, having the block-diagonal form Thus, ρ = −φ 0 is simple. Hence, if µ is Φ-invariant and totally σ-ergodic, and h µ (Φ) > 0, then µ is the uniform measure.

Conclusion
We have characterized the invariant measures for several natural families of bipermutative cellular automata. Many questions remain unanswered. For example, in §3 and §5, we exploited an algebraic structure on A Z to study the Φ-invariant measures. What other algebraic properties of the quasigroup structure of A can be exploited in this way? Also, if Φ is a QGCA on A Z , then the system (A Z , Φ, λ) is measurably isomorphic to the uniform Bernoulli shift (A N , σ, λ) [12]. Theorem 22 suggests that, if µ is any positiveentropy, Φ-invariant measure, then the system (A Z , Φ, µ) is isomorphic to (K Z , σ, κ), where K is an alphabet of K letters and κ is the uniform Bernoulli measure on K N . Is this true?
Finally, Example 12b refuted Conjecture 7, but did so by using a structural decomposition A = C×Q to get an invariant measure without full support. This leaves us with the following: Conjecture: Let (A, * ) be a quasigroup and let Φ : A Z −→A Z be the corresponding QGCA. Let µ be a Φ-invariant and σ-ergodic measure.