Algebraic Actions of the Discrete Heisenberg Group

Transcription

1 Algebraic Actions of the Discrete Heisenberg Group Based on joint work with Doug Lind and Evgeny Verbitskiy Klaus Schmidt Vienna DAGT, Udine, July 2018 Klaus Schmidt Algebraic Actions 1 / 21

2 Toral automorphisms Let α be the automorphism of T n = R n /Z n given by a matrix A GL(n, Z). Then α is ergodic (with respect to Lebesgue measure) if and only if no eigenvalue of A is a root of unity, and expansive (or hyperbolic) if and only if no eigenvalue of A has absolute value 1. A point x T n is homoclinic under α if lim n α n x = 0. Hyperbolic toral automorphisms have nonzero homoclinic points, Markov partitions (Sinai, 68; Bowen, 70), hence both periodic and homoclinic specification, and the probability measures defined by partial or periodic orbits are dense in the set of all invariant probability measures. Nonexpansive ergodic toral automorphisms (assumed to be irreducible for simplicity) have no nonzero homoclinic points and no Markov partitions. However, they satisfy weak specification (Lind, 79), and the periodic orbit measures are again dense in the set of all invariant probability measures (Marcus, 80). Klaus Schmidt Algebraic Actions 2 / 21

3 Irreducible toral automorphisms Every irreducible matrix A GL(n, Z) is conjugate over Q to a companion matrix of the form A f = f 0 f 1 f 2 f n 2 f n 1, where f = t n f n 1 t n 1 f 0 is the (irreducible) characteristic polynomial of A. If (σx) k = x k+1 is the shift on T Z, then X f = ker f (σ) = {(x k ) T Z : x k+n f n 1 x k+n 1 f 0 x k = 0 for all k} is a closed, shift-invariant subgroup of T Z, and the restriction α f of σ to X f is conjugate to the toral automorphism defined by A f. In other words, every irreducible toral automorphism is finitely equivalent to a shift α f on X f T Z of the form ( ), where f is an irreducible monic polynomial with integer coefficients and f 0 = 1. ( ) Klaus Schmidt Algebraic Actions 3 / 21

4 Irreducible automorphisms of compact abelian groups Equation ( ) also makes sense if f is not monic: if f = f n t n f 0 is an integer polynomial with f 0 f n 0, then X f = ker f (σ) = {(x k ) T Z : f n x k+n f n 1 x k+n 1 f 0 x k = 0 for all k} is again a closed, shift-invariant subgroup of T Z. If f is irreducible, the restriction α f of σ to X f is irreducible in the sense that every closed, invariant subgroup Y X f is finite. Up to finite equivalence, every irreducible automorphism of a compact abelian group can be represented in this way. Basic dynamical properties of α f, like expansiveness or entropy, are again determined by the roots of f. In particular, α f is expansive and has nonzero homoclinic points if and only if f has no roots of absolute value 1. The entropy of α f is given by h(α f ) = log f n + {λ:f (λ)=0 and λ >0} log λ = where S = {s C : s = 1}. S log f (s) ds, Klaus Schmidt Algebraic Actions 4 / 21

5 Two elementary examples (1) Let f = 2. Then X f = {(x k ) T Z : 2x k = 0 for all k}. A homoclinic point v X f is given by { 1 if n = 0, v n = 0 otherwise. (2) Let f = 2t 3. Then X f = {(x k ) T Z : 2x k+1 = 3x k for all k}, α f is multiplication by 3/2 on T, and h(α f ) = log 3. The point v X f with { 2 n /3 n if n < 0, v n = 0 otherwise, is homoclinic. Klaus Schmidt Algebraic Actions 5 / 21

6 Two nonexpansive examples (1) Let A = ( ) GL(4, Z). The toral automorphism α defined by A is conjugate to α f with f = t 4 t 3 t 2 t + 1. Since f is irreducible, noncyclotomic, and has two roots of absolute value 1, α f is ergodic and nonexpansive, and has no nonzero homoclinic points. (2) Let f = 5t 2 6t + 5. Then X f = {(x k ) T Z : 5x k+2 6x k+1 + 5x k = 0 for all k}. The roots of f has absolute value 1, α f is nonexpansive, h(α f ) = log 5, and there are no nonzero homoclinic points. Klaus Schmidt Algebraic Actions 6 / 21

7 What are homoclinic points good for? If α f is expansive, all homoclinic points of α f decay exponentially. For every homoclinic point x X f we can thus define a shift-equivariant map ξ x : l (Z, Z) X f by setting ξ x (w) = n Z w nα n f x, w = (w n ) l (Z, Z). The map ξ x is weak -continuous on closed, bounded subsets of l (Z, Z). For sufficiently large N 1, ξ x ({0,..., N} Z ) = X f, so that α f is a continuous factor of a full shift. Some homoclinic points are particularly useful: a homoclinic point v X f is fundamental if every homoclinic point of α f lies in the subgroup of X f generated by the orbit {α n f v : n Z} of v under α f. If the homoclinic point v X f is fundamental, one can choose a closed, bounded, shift-invariant subset W l (Z, Z) which is a sofic shift, such that ξ v (W ) = X f and ξ v W : W X f is almost one-to-one. In other words, fundamental homoclinic points yield sofic partitions of (X f, α f ) (Vershik 92, Einsiedler-Schmidt 97, Kenyon-Vershik 98,... ). Klaus Schmidt Algebraic Actions 7 / 21

8 A formula for the fundamental homoclinic point of α f We write Ω f for the set of roots of f and consider the partial fraction decomposition 1 f (t) = 1 b ω f m ω Ω f t ω of 1/f with b ω C for every ω Ω f. Define w l (Z, R) by 1 wn f = m ω Ω b ω ω n 1 if n 1, f 1 b ω ω n 1 if n 0, f m ω Ω + f where Ω + f and Ω f denote the set of large, resp. small, roots of f. Then x = w (mod 1) lies in X f and is the fundamental homoclinic point of α f. For example, if f = t 2 t 1, then x X f is given by xn 1 = 5 ( 1 ) 5 n 2 (mod 1) if n 1, 1 5 ( 1+ ) 5 n (mod 1) if n 0. 2 Klaus Schmidt Algebraic Actions 8 / 21

9 Periodic points of α f For every k 1, the set of points Fix k (α f ) = {x T n : α k x = x} with period k under α f satisfies that Fix k (α f ) = f (ω). {ω C:ω k =1} In particular, if f is ergodic, Fix k (α f ) is finite for every k 1. Furthermore, 1 lim k k log Fix k(α f ) = lim = k S 1 log f (ω) k {ω C:ω k =1} log f (s) ds = h(α f ). If α f is expansive, this is obvious. If α f is nonexpansive, this requires a diophantine result due to Gelfond 32. Klaus Schmidt Algebraic Actions 9 / 21

10 What if α f is nonexpansive? If f has roots of absolute value 1, there are no nonzero homoclinic points, so that the approach just described is unavailable. However, by using either a geometric approach (Lind, Marcus, around 80) or an algebraic approach (Lindenstrauss-S 05, S 06 and 16) one can, to a limited extent, rescue some of the expansive results for nonexpansive automorphisms. What is interesting is that the number of unitary roots of f (i.e., the size of the unitary variety U(f ) of f ) seems to play no role in these partial results. For the corresponding Z d -actions by automorphisms of compact abelian groups, which are again determined by an integer polynomial f in d 2 variables, things are quite different. Here the size of the unitary variety U(f ) = {z = (z 1,..., z d ) C d : z i = 1 for i = 1,..., d and f (z) = 0} of f will play an important role. Klaus Schmidt Algebraic Actions 10 / 21

11 Algebraic Z d -actions Let d 1, and let α: n α n be a Z d -action by automorphisms of a compact abelian group X. If h(α) > 0, then X contains a closed, α-invariant subgroup Y such that (Y, α) is algebraically conjugate to the shift-action α f of Z d on a closed, shift-invariant subgroup X f T Zd of the form X f = ker f (σ) = { (x n ) T Zd : m Z d f mx m+n = 0 for every n Z d}, where f = m Z d f mu m is a polynomial in d variables u 1,..., u d with integer coefficients (here u m = u m 1 1 u m d d ). We write R d for the ring of all (Laurent) polynomials in d variables with integer coefficients and consider R d as a subring of the convolution algebra l 1 (Z d, R). Klaus Schmidt Algebraic Actions 11 / 21

12 Properties of (X f, α f ) Let d > 1 and 0 f R d. (X f, α f ) is ergodic. (X f, α f ) is expansive if and only if U(f ) = (S 90). (X f, α f ) is expansive if and only if the polynomial f is invertible in l 1 (Z d, R) (Wiener 32, Deninger-S 07). In this case 1/f (mod 1) lies in X f and is a fundamental homoclinic point of α f (here f = m Z d f mu m ). Example: A (Laurent) polynomial f = m Z d f mu m is lopsided if there exists an m Z d such that f m > n Z d {m} f n. If f is lopsided, then α f is expansive. In fact, α f is expansive if and only if the principal ideal f = R d f R d contains a lopsided polynomial (observation by Hanfeng Li). Klaus Schmidt Algebraic Actions 12 / 21

13 Entropy of (X f, α f ) Suppose that d > 1 and f R d. If f is nonzero, then (Lind-S-Ward 90). If f = 0, then h(α f ) =. h(α f ) = log f (s) ds S d If f 0, α f has positive entropy if and only if f is not a product of generalized cyclotomic polynomial (Lind-S-Ward 90). If f 0, h(α f ) coincides with the upper logarithmic growth rate of the number of connected components of points with finite orbits: h(α f ) = lim sup 1 Z d / log Fix (α f )/Fix (α f ), ( ) where Fix (α f ) is the connected component of the identity in the group Fix (α f ) of -periodic points in X f, and where the lim sup is taken over all finite-index subgroups Z d as = min { n : n {0}} (S 95). Klaus Schmidt Algebraic Actions 13 / 21

14 Homoclinic points of α f Suppose that d > 1 and f R d is nonzero and irreducible. By definition, every homoclinic point x = (x n ) X f T Zd satisfies that lim x n = 0. n We write αf (X f ) for the group of homoclinic points in X f. A homoclinic point x αf (X f ) is summable if n Z d x n <. The group of summable homoclinic points of α f is denoted by 1 α f (X f ). (a) If α f is expansive, then 1 α f (X f ) is countable and dense in X f, and αf (X f ) = 1 α f (X f ) (Lind-S 99). (b) If α f is nonexpansive, 1 α f (X f ) is either trivial or countable and dense in X f (Lind-S 99). (c) If U(f ) is not contained in a finite union of (d-1)-dimensional subgroups of S d, then αf (X f ) is uncountable (Linnell-Puls 01). (d) If 1 α f (X f ) {0} then 0 < h(α f ) <. (e) If 1 α f (X f ) {0}, then h(α f ) = lim 1 Z d / log Fix (α f )/Fix (α f ). ( ) Klaus Schmidt Algebraic Actions 14 / 21

15 Existence of summable homoclinic points Theorem (Lind-S-Verbitskiy 13) Suppose that d > 1 and f R d is nonzero and irreducible. Then 1 α f (X f ) {0} if and only if dim U(f ) d 2. Examples If f is asymmetric (i.e., if gcd(f, f ) = 1), then 1 α f (X f ) {0} (Lind-S-Verbitskiy 13). The harmonic system: Let f = 4 u v u 1 v 1 R 2. Then U(f ) = {(1, 1)}, and 0 = dim U(f ) = d 2. Hence 1 α f (X f ) {0} (S-Verbitskiy 09). Let f = 1 + u + v + w R 3. Then 1 = dim U(f ) = 3 2, so 1 α f (X f ) {0}. Let f = 3 u v u 1 v 1 R 2. then 1 = dim U(f ) > 2 2, so 1 α f (X f ) = {0}, but αf (X f ) is uncountable. For this example, the specification properties of α f are not understood. Klaus Schmidt Algebraic Actions 15 / 21

16 Principal Algebraic Actions of Discrete Groups Let Γ be a countable discrete group. We consider the shift actions λ and ρ of Γ on T Γ, where for every γ Γ and x = (x γ ) T Γ. (λ γ x) γ = x γ 1 γ, (ργ x) γ = x γ γ To get so-called principal actions, consider the integer group ring ZΓ of Γ. Write f ZΓ as a formal sum f = γ Γ f γ γ and set λ f = γ Γ f γλ γ, ρ f = γ Γ f γρ γ. Then X f := ker ρ f = { x T Γ : } f γρ γ x = 0 γ Γ is a closed, λ-invariant subgroup of T Γ (it is the kernel of right convolution by f = γ Γ f γ γ 1 ). We denote by α f the restriction of the left shift-action λ to X f and call (X f, α f ) the principal Γ-action defined by f. Klaus Schmidt Algebraic Actions 16 / 21

17 Principal Algebraic Actions: Expansiveness and Entropy There are very few general results about principal algebraic actions of discrete groups. Expansiveness: Let Γ be a discrete group and f ZΓ. Then α f is expansive if and only if f is invertible in l 1 (Γ) (Deninger-S, 07). Entropy: if Γ is sofic, f ZΓ, and ρ f is injective on l 2 (Γ), then h(α f ) = log det (f ), the Fuglede-Kadison determinant of f, acting by convolution on l 2 (Γ) (Sinai 59, Lind-S-Ward 90; Deninger 06, Deninger-S 07, Li 12, Li-Thom 14; Bowen, Hayes, Kerr, Li,... ). If ρ f in noninjective on l 2 (Γ), h(α f ) = (Chung-Li 15). If α f is expansive or, more generally, if α f has a nontrivial summable homoclinic point, then h(α f ) > 0. Klaus Schmidt Algebraic Actions 17 / 21

18 Principal Algebraic Actions of the Heisenberg Group Let H = generators x = {( 1 a b 0 1 c ( ) } : a, b, c Z be the discrete Heisenberg group, with ), y = ( ), and z = ( ), where yx = xyz. In this notation, every f ZH can be written as an integer polynomial f = k,l,m f k,l,mx k y l z m in the noncommuting variables x, y, z. Theorem (Einsiedler-Rindler, 01). Fix f ZH. The action α f is nonexpansive if and only if there exist an irreducible unitary representation π of H on a Hilbert space H and a unit vector v H such that π(f )v = 0. Problem: H is not a Type I group. So the set of irreducible unitary representations of H is complicated. Klaus Schmidt Algebraic Actions 18 / 21

19 An application of Allan s local principle Theorem (Göll-S-Verbitskiy, 16). Let f ZH. Then α f is nonexpansive if and only if π(f ) is invertible for every irreducible representation π of H which is induced from a one-dimensional representation of a subgroup of H (such representations are called monomial). The following examples are taken from Einsiedler-Rindler 01 and Göll-S-Verbitskiy, 14. f = 3 + x + y + z is invertible in l 1 (H), but not in l 1 (Z 3 ). f = 3 + x + y z is noninvertible in l 1 (H) as well as in l 1 (Z 3 ). f = 3 ± 2x ± y + z is invertible in l 1 (H), but not in l 1 (Z 3 ). Problem: Find an effective criterion for expansiveness of principal actions of H. Klaus Schmidt Algebraic Actions 19 / 21

20 Homoclinic and periodic points of nonexpansive actions For f Z[Z d ] we saw earlier that α f may be nonexpansive, but 1 α f (X f ) may nevertheless be nontrivial. The same can happen for f ZH. Examples (Göll-S-Verbitskiy, 16) f = 2 x y. f = 4 x x 1 y y 1. Currently it is not known which nonexpansive principal actions of H have summable homoclinic points. If α f is expansive, or if 1 α f (X f ) {0}, then the entropy of α f coincides with the logarithmic growth rate of the number of its periodic points: h(α f ) = lim 1 H/ log Fix (α f )/Fix (α f ). If α f is nonexpansive, then it is easy to see that h(α f ) lim sup 1 H/ log Fix (α f )/Fix (α f ). When does one have equality? Klaus Schmidt Algebraic Actions 20 / 21

21 Zero entropy? Which principal actions of H have zero entropy? Here is an example from Lind-S 15, based on Deninger 11: h(α y x+1/x ) = h(α x 2 yx 1) = h(α y 2 xy 1) = h(α y 2 yxz 1 1) = 0. Klaus Schmidt Algebraic Actions 21 / 21