A C^1 Counterexample to the Entropy Conjecture

Proof. We first estimate the input series in the fixed product Riemannian metric, then pass to source coordinates. The chain rule and \|DG\|\le B give

The second bound uses the real projection \operatorname{pr}_1, whose derivative has norm one everywhere. In particular, it requires no choice of output chart for G^m.

For an integer k\ge1, let

These are smooth complex-valued functions. The support of \chi lies in a fixed compact annulus, on which the angular map z\mapsto z/|z| and its first derivative are bounded. The product and chain rules therefore give a constant C_\chi, independent of k, such that

The extensions vanish on fixed neighborhoods of both poles.

Write the mth summand, including its prefactor, as

Its differential is

Since there are finitely many \beta_\ell, their first derivatives have a common bound. Equations (11) and (12) yield a constant C, independent of \ell and m, with

The prediction and angular derivatives occur in separate product-rule terms; their bounds are added. Each of the three resulting geometric series converges by (6).

For completeness, cover [0,100]\times\Sigma by finitely many relatively compact source coordinate neighborhoods V_j whose closures lie in larger coordinate charts W_j\subset\mathbb R\times\Sigma. If \kappa_j is the chart on W_j, the norm of D\kappa_j^{-1} on \kappa_j(\overline{V_j}) is bounded by a constant C_j independent of m. Thus the coordinate derivatives of F_{\ell,m}\circ\kappa_j^{-1} satisfy (13) with the additional fixed factor C_j. The series of functions and of all their first coordinate derivatives consequently converge uniformly on every such neighborhood. The usual differentiation theorem for series now applies: on a smaller coordinate ball, integrate the first derivatives of the partial sums along line segments and pass to the uniform limit. This identifies the derivative of the limit with the continuous limit of the derivatives. Hence each U_\ell is C^1 on the fundamental strip.

This proof also covers data with arbitrarily late predicted visits. At a datum for which \beta_\ell(T_m)=0 for every m, each summand and its first derivative vanish: a differentiable nonnegative function has derivative zero at any zero. Uniform convergence of the derivative tails ensures continuity even when nearby nonzero summands have indices tending to infinity. No regularity of a selected hitting time is needed.

Because \alpha vanishes near 0 and 100, the input and all its first derivatives vanish on neighborhoods of the periodic seams. The periodic extension is therefore C^1. Because \chi vanishes near zero and infinity, the same holds in fixed neighborhoods of the two poles, so the angular expression introduces no coordinate singularity. We also have the uniform bound

The clock component of (10) is smooth and well-defined modulo 100. Each register component is C^1 and takes its values in the fixed disk of radius 3L+L/(10(L-1)) in the affine chart. This includes all source values v_\ell=\infty, since S is smooth and zero near infinity.

It remains to check the main component at z=\infty. Set

This is a smooth function of all its variables. In the source inverse coordinate w=1/z and the corresponding inverse coordinate for the output, the main component is

For |w|<[2(60q+1)]^{-1/2} the denominator has modulus greater than 1/2, uniformly in t and the registers. Formula (15) therefore extends smoothly to w=0, with value zero. At all finite z, including z=0, the main component is already smooth in the affine chart. These checks establish the asserted global C^1 regularity of f.□

Proof. Monotonicity gives a finite limit t_*, and hence h(t_n,z_n)=t_{n+1}-t_n\to0. Since 0\le b(t_n)\le h(t_n,z_n), continuity and the zero set of b imply t_*=50+100k. For all sufficiently large n the clock therefore avoids every readout support, so z_{n+1}=z_n^2. Moreover, |z_n|\ge3/4, including z_n=\infty, would give e(z_n)=1 and h(t_n,z_n)=1. Thus eventually z_n is finite with |z_n|<3/4. Starting at such a time N after all readouts have ceased,

The registers may have arbitrary behavior; all their values are included in the definition of Z_0.□

Proof. Let r be the first index in the passage with s_r\ge2. The preceding reading is below 2, so 2\le s_r<3. At time r one has a(s_r)=0 and \alpha(s_r)=0, whence U_\ell(s_r,z_r)=0 and

This conclusion holds even if any earlier register was at infinity. Until preparation starts, the inputs remain zero, and S(0)=0 keeps all registers zero.

On (8,12) the clock speed is exactly one. Its readings in that interval therefore belong to a single translate of the integers, which gives |I|\le5. The first reading at least 9 belongs to [9,10) by (17); there \alpha=1. Similarly, the first reading at least 76 belongs to [76,77), where the plateau-covering property of the \beta_\ell supplies a coefficient equal to one. All these crossings occur in the passage because s_n eventually reaches 100.

Suppose now that i\in I and |z_i|\le1/2. As long as s_n\in[8,50] and |z_n|\le1/2, there is no readout, e(z_n)=0, and

The clock cannot decrease, and |z_{n+1}|\le1/4\le1/2. These two conditions thus persist by induction from time i, keeping s_n\le50 forever. This contradicts the assumed complete passage. This also excludes the endpoint |z_i|=1/2 and proves (18).□

Proof. Fix i\in I, and let r_{70} and r_{82} be the first indices after i with s_{r_{70}}\ge70 and s_{r_{82}}\ge82, respectively. Before r_{70} all readout coefficients vanish. The actual pair (s_n,z_n) therefore follows G exactly from time i through time r_{70}. At the latter time s_{r_{70}}\in[70,71). Throughout [70,82] one has b=1, so both the actual and predicted clock increments are one, independently of their respective main coordinates. Starting with the equal clocks at time r_{70}, induction gives

including after any main-coordinate additions. The predicted clock is nondecreasing at all later times. Once it reaches 82, it cannot return to any support of the nonperiodic \beta_\ell, all of which lie in (74,80).

The support of a fixed \beta_\ell lies in an interval of length less than one. If \beta_\ell(s_j)>0, the next clock reading is s_j+1, strictly beyond that support interval. Monotonicity then rules out any later positive value of this \beta_\ell; the same argument rules out two positive values in either order. Equation (20) identifies every possible predicted hit with an actual hit in this passage, and later predicted hits are impossible. This proves (19).□

Proof. Put i_0=\min I and, for each i\in I, define

The phases are defined by (18). Lemma 3.4 and (9) give the exact input

The input at time i first enters v_{i+1,\ell}; it receives j-i-1 subsequent factors of L before v_{j,\ell} is read. Thus its coefficient would be L^{1+i-j}L^{j-i-1}=1 if the register remained in the region where S is the identity. We verify that region condition directly.

For every integer u with i_0\le u\le j, define the candidate value

The triangle inequality and |I|\le5 give

By Lemma 3.3, v_{i_0,\ell}=0=w_{i_0}. For i_0\le u<j the clock lies between the preparation and readout intervals, so a(s_u)=L. The input is zero unless u\in I, and otherwise it is L^{1+u-j}c_u by (22). Consequently

If v_{u,\ell}=w_u, the bound (23) implies S(v_{u,\ell})=w_u, so the actual register update gives v_{u+1,\ell}=w_{u+1}. Induction proves v_{u,\ell}=w_u all the way through u=j. Every preparation index precedes j, hence w_j=\sum_{i\in I}c_i, which proves (21) and the stated bounds.

This induction concerns only the particular register whose readout is at j. Another register can already have been read and can subsequently leave the unit disk. By Lemma 3.4, its readout coefficient is zero for the rest of this passage. Registers do not enter one another’s updates, so such later nonlinear behavior has no effect on the induction just proved.□

Proof. Set \theta=z_{i_0}/|z_{i_0}|. Before the first readout, the main coordinate squares exactly, so it stays nonzero and its phase at time n is \theta^{2^{n-i_0}}. In particular this holds at every preparation index i\in I, since all preparation precedes all readouts.

Proceed chronologically through the readout indices. Suppose the phase at time j is \theta^{2^{j-i_0}} and z_j\ne0. For every i\in I the phase appearing in (21) is

which is the phase of z_j^2. Lemma 3.5 applies separately to each \ell with \beta_\ell(s_j)>0. The other channels have zero coefficient in the main update. Thus that update is exactly

One factor of \beta_\ell(s_j) comes from the input and the other from the readout in (10); both are essential to this coefficient. Every added coefficient is nonnegative, and the baseline |z_j|^2 is strictly positive. This proves nonvanishing, phase doubling, and the square lower bound at this readout, even if all its inputs vanish. Between readouts there is exact squaring. Induction therefore proves (24) through the end of the passage, including simultaneous readouts of several channels and successive readouts of different channels.

If some preparation value already has |z_i|>1, take N=i. Otherwise 1/2<|z_i|\le1 for every i\in I. Choose i_* with \alpha(s_{i_*})=1, supplied by Lemma 3.3. The prescribed cutoff satisfies \chi(z_{i_*})=1, including when |z_{i_*}|=1. The same lemma supplies a readout j_* and a channel \ell_* with s_{j_*}\in[76,77] and \beta_{\ell_*}(s_{j_*})=1. In (25), the single pair (\ell_*,i_*) therefore adds 2 to the radius, and all other terms have the same phase. Hence

Taking N=j_*+1 proves the claim; this index is still in the passage because s_{j_*+1}\le78<100.□

Proof of Proposition 3.1. If the lift is bounded, Lemma 3.2 proves the asserted coordinate limits. In a product metric the distance to Z_0 consequently tends to zero, irrespective of the registers.

If z_0=\infty, it remains at infinity, all inputs vanish, and the clock advances by one at every step. A translate of the integer lattice meets the reset interval [2,5] modulo 100. At such a step a=0, so every register becomes zero, regardless of its initial value, and stays zero. The orbit then lies in Z_\infty.

It remains to consider an unbounded lift with finite z_0. Choose any complete passage and an index N with |z_N|>1 given by Lemma 3.6. Its square lower bound continues through the rest of that passage. We spell out why it also holds at every later time. After the current readout region (74,80), all readout multipliers vanish until the next period’s region (174,180) in the same translated coordinate. In particular, the main variable squares exactly across the period boundary and throughout the next reset and preparation intervals. An old register, however large, cannot alter z there. It is erased by the next reset before any new preparation input is entered.

Every later period is a complete passage by the unbounded-lift assumption. Lemma 3.6 therefore applies afresh from its first preparation index through its end. The intervening indices, before that first preparation, have no readout and give exact squaring. These intervals cover every time after N, and hence

This argument does not require \chi to vanish immediately after the radius exceeds one. New inputs in later periods still have nonnegative weights \alpha\chi, satisfy the register bound, and give the aligned additions in (25). Thus they preserve (26) even while \chi remains nonzero.

It follows that z_n\to\infty in \Sigma. Since \chi vanishes in a neighborhood of infinity, there is a time after which \chi(z_n)=0 at every step. All inputs U_\ell(t_n,z_n) are then zero. A later reset, which exists because the clock lift is unbounded, makes every register zero; the recurrence and S(0)=0 keep them zero forever. The distance to Z_\infty therefore tends to zero.

We have proved the assertion in a product metric. All compatible metrics on the compact space M are uniformly equivalent, so the same convergence in distance holds for any such metric. The conclusion is pointwise in the initial state; no uniform entry time is asserted.□

Proof. Write s_X(n,\varepsilon) and s_Z(n,\varepsilon) for the separated-set cardinalities of F and F|_Z, respectively. Fix \varepsilon>0 and an integer m\geq1, and define

This metric induces the original topology of X. Choose a largest (m,\varepsilon/4)-separated subset of Z, of cardinality P=P(m)=s_Z(m,\varepsilon/4). Its open d_m-balls of radius \varepsilon/2, taken in X, have union V containing Z: otherwise a point of Z at d_m-distance greater than \varepsilon/4 from every center could be added to the separated set. Also choose Q=Q(m) open d_m-balls of radius \varepsilon/2 covering X.

We claim that for every 0<\tau<1 there is an integer J\geq1 such that

Indeed, since Z is compact and V is an open neighborhood of Z, the attraction hypothesis implies that F^{mi}x\in V for all sufficiently large i, with the threshold allowed to depend on x. Consequently, for each x there is an integer k_x\geq1 for which

Let I_x=\{0\leq i<k_x:F^{mi}x\in V\}. The open neighborhood

of x preserves all these visits: every y\in O_x has fewer than \tau k_x indices outside V among its first k_x samples. Choose a finite cover O_{x_1},\ldots,O_{x_a} of X, and put J=\max_{1\leq b\leq a}k_{x_b}.

Starting at x, choose a member of this cover containing the current point and follow the corresponding number k_{x_b} of iterates of F^m. Restart the same procedure at the next point, concatenating these segments for as long as the selected segment fits within the first r samples. The final remainder has length \ell<J. The completed segments contribute at most \tau(r-\ell) samples outside V, and the remainder contributes at most \ell. This proves (27), including when no segment fits. In particular, this compactness argument bounds a proportion of exceptional samples without bounding their last occurrence.

Now encode the first r blocks of length m of an orbit. Record the subset of indices i for which F^{mi}x\notin V. At every other index choose one of the P balls covering V that contains F^{mi}x; at an exceptional index choose one of the Q balls covering X. Choose the first available ball in a fixed ordering of each cover. By (27), the number of possible descriptions is at most

Here 2^r bounds the number of subsets, and P,Q\geq1 allow the displayed overcount even if \tau r+J\geq r. Two points with the same description lie in the same radius-\varepsilon/2 ball at each block, so their distance is less than \varepsilon at every time 0,\ldots,mr-1. Thus an (mr,\varepsilon)-separated set has at most the cardinality in (28).

For arbitrary n, use r=\lceil n/m\rceil and s_X(n,\varepsilon)\leq s_X(mr,\varepsilon). Taking the orbit-length limit with m and \tau fixed gives

First let \tau\downarrow0 with m fixed. Next let m\to\infty. The hypothesis h_{\mathrm{top}}(F|_Z)=0 implies

because every fixed-scale entropy rate is nonnegative and bounded above by the topological entropy. The right-hand side therefore tends to zero. This holds for every \varepsilon>0, proving the lemma.□

Proof. Fix a compatible metric d_\Sigma on \Sigma and use its maximum product metric on \Sigma^q. Fix \varepsilon>0. Uniform continuity of the polar map

allows a partition of [0,A] into H sufficiently short ordered intervals and a partition of \mathbb S^1 into K sufficiently short arcs, such that two points have d_\Sigma-distance less than \varepsilon whenever their radii lie in the same interval and their angular coordinates lie in the same arc. Use half-open intervals and arcs, with endpoints assigned once, so that the labels are single-valued. Let b:[0,A]\to\{1,\ldots,H\} be the resulting nondecreasing radius label.

For a radius r\in[0,A] and an integer N\geq1, consider its label vector

Every coordinate of this vector is nondecreasing as a function of the initial radius r. Its distinct values therefore form a chain in the coordinatewise order on \{1,\ldots,H\}^N. Along such a chain the sum of the coordinates strictly increases whenever the vector changes, and this sum lies between N and NH. Hence there are at most

radius itineraries. This orders itineraries by their initial radii; no monotonicity in time is being asserted.

Each point v\in D_A can be written v=ru with u\in\mathbb S^1; choose u=1 when v=0. Its iterates have the representations T^jv=R^j(r)u with that same u, including after a radius becomes zero. Thus one angular label, together with the radius itinerary, gives at most K[1+N(H-1)] descriptions of length-N orbits in D_A. Points with the same description stay within \varepsilon throughout those N iterates.

Finally cover the whole compact space \Sigma^q by C sets of diameter less than \varepsilon, and assign each initial point one cover label. For n\geq2, encode time zero by this label and encode each coordinate of its first image in D_A^q for the remaining N=n-1 times by the preceding construction. Points sharing a complete description stay within \varepsilon in the maximum product metric for all n times. Consequently

The finite initial cover accounts for all preimages of the disk, including \infty. The bound grows polynomially in n, so its exponential rate is zero at every \varepsilon>0.□

Proof. Recall the compact forward-invariant sets of Proposition 3.1:

On Z_0, the clock and main coordinate are fixed and the registers evolve independently by T(v)=LS(v). By the choice of S in Section 2, T(\Sigma)\subseteq D_{3L}, and on that disk its radius map is

This map is continuous, nondecreasing, fixes zero, and takes values in [0,3L]. Lemma 4.2 therefore gives h_{\mathrm{top}}(f|_{Z_0})=0.

On Z_\infty the only evolving coordinate is the clock, which advances by 1 modulo 100. Thus (f|_{Z_\infty})^{100} is the identity. For n\geq100 the orbit metric d_n on this set equals d_{100}, so its separated-set cardinalities are bounded independently of n at every fixed scale. In particular, h_{\mathrm{top}}(f|_{Z_\infty})=0.

For Z=Z_0\cup Z_\infty, splitting a separated set between these two invariant subsets gives

Both terms have zero exponential rate, hence h_{\mathrm{top}}(f|_Z)=0. Proposition 3.1 states that every orbit of f approaches this exact set Z. Lemma 4.1 now proves h_{\mathrm{top}}(f)=0.□

Proof. Let \pi:M\to\Sigma be projection to z, and put g(z)=z^2, with g(\infty)=\infty. Define

Since |S|\le3, we have |C|\le60q. For 0\le s\le1, set

when z is finite, and H_s(t,\infty,v)=\infty. This is a continuous homotopy from g\pi to \pi f. Indeed, in the inverse coordinate w=1/z at infinity, the output has inverse coordinate

For all s,t,v, the denominator is nonzero on one fixed neighborhood of w=0, since |sC|\le60q. This verifies continuity jointly in every variable, including s.

By homotopy invariance and functoriality of singular homology [2, Theorem 2.10],

The oriented sphere has H_2(\Sigma;\R)\cong\R, and the action of g on this group is multiplication by its degree. The regular value 1 has the two preimages 1 and -1. At either preimage the derivative is complex multiplication by 2z, with positive real determinant |2z|^2=4. Both local degrees are +1, so \deg g=2 [2, Proposition 2.30 and Lemma 2.49].

The projection \pi has a section obtained by fixing the clock and all register coordinates. Therefore \pi_*:H_2(M;\R)\to H_2(\Sigma;\R) is surjective. Write A=f_*|_{H_2(M;\R)} and P=\pi_*. Equation (31) gives P(A-2I)=0, with P\ne0. If A-2I were invertible, this equality would imply P=0. Thus 2 is an eigenvalue of A.□

Proof of Theorem 1.1. Proposition 2.1 constructs a C^1 self-map of the stated compact smooth manifold. Proposition 4.3 gives \htop(f)=0, and Proposition 5.1 supplies the eigenvalue 2 on H_2(M;\R). This group is an invariant summand of H_*(M;\R), so \rho(f_*)\ge2. Finally, S vanishes on a neighborhood of infinity. Holding all other coordinates fixed and varying one register within that neighborhood leaves every output in (10) unchanged. Hence f is noninvertible.□