A clock with registers

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

This lesson builds the map of the counterexample and proves that it is continuously differentiable. The map has three kinds of coordinates. A circle coordinate $t$ is a clock: it advances by at most one per step, and it can slow down only near one reading. A sphere coordinate $z$, the main coordinate, usually squares: $z\mapsto z^2$. Further sphere coordinates $v_1,\ldots,v_q$, the registers, store small complex signals. Early in each clock period the map writes into the registers predictions of the phase that $z$ will have at a later readout; registers multiply their contents by a large gain $L$ at every step; at the readout the map adds the amplified signals to $z^2$. The signals are aligned with $z^2$, so they push $|z|$ up instead of disturbing it. The difficulty is regularity: a prediction $m$ steps ahead involves the phase $(z/|z|)^{2^{m+1}}$, whose derivative is of order $2^m$, and $m$ is unbounded. Weighting the $m$-th prediction by $L^{1-m}$ makes the series continuously differentiable; the gains restore its size before the readout. The next lesson analyses the orbits.

We use from the core course Real Analysis II (Lebl, Basic Analysis): the extreme value theorem (Theorem 7.5.6), differentiation of uniformly convergent sequences (Theorem 6.2.10), the chain rule (Theorem 8.3.7) and the criterion that a map is continuously differentiable when its partial derivatives exist and are continuous (Proposition 8.4.6). From Local tools for bundles and transport we use Theorem 3.1: for a closed set $K$ in a smooth manifold and an open neighbourhood $U$ of $K$ there is a smooth function with values in $[0,1]$, equal to one near $K$ and with support in $U$. We call such a function a cutoff for $(K,U)$.

1. The Riemann sphere and the manifold

The Riemann sphere $\Sigma=\mathbb C\cup\{\infty\}$ has two charts: the coordinate $z$ on $\mathbb C$, and the coordinate $w=1/z$ on $\Sigma\setminus\{0\}$, with $w(\infty)=0$. On their overlap $\mathbb C\setminus\{0\}$ the transition $w=1/z$ is smooth with smooth inverse, so $\Sigma$ is a compact smooth surface. A map of $\Sigma$ is smooth at $\infty$ when its expression in $w$ is smooth at $w=0$; for example $z\mapsto z^2$, with $\infty\mapsto\infty$, reads $w\mapsto w^2$ there.

We measure tangent vectors of $\Sigma$ by the round metric: a tangent vector with coordinate velocity $\dot z$ at a finite point $z$ has length $2|\dot z|/(1+|z|^2)$. In the chart $w$ the same vector has velocity $\dot w=-\dot z/z^2$ and $\frac{2|\dot w|}{1+|w|^2}=\frac{2|\dot z|/|z|^2}{(|z|^2+1)/|z|^2}=\frac{2|\dot z|}{1+|z|^2},$ so the formula is the same in both charts and defines a continuous norm on every tangent space, including at $\infty$. On $\mathbb R\times\Sigma$ we use the product norm $\|(\dot t,\xi)\|=(\dot t^2+\|\xi\|^2)^{1/2}$. For a $C^1$ map $\Psi$ between such spaces, $\|D\Psi(x)\|$ denotes the operator norm of its derivative at $x$ with respect to these norms. By the chain rule (Lebl, Theorem 8.3.7, applied in charts) $D(\Psi_2\circ\Psi_1)(x)=D\Psi_2(\Psi_1x)\,D\Psi_1(x)$, hence the inequality (1.1) holds: $\begin{aligned} &\|D(\Psi_2\circ\Psi_1)(x)\|\\ &\quad\leq\|D\Psi_2(\Psi_1x)\|\,\|D\Psi_1(x)\|. \end{aligned}$ On a compact subset of a chart domain these norms are comparable with the Euclidean norms of coordinate velocities: the factor $2/(1+|z|^2)$ is bounded above and below by positive constants there.

Fix an integer $q\geq2$ (the construction below works with $q=2$). The manifold of the counterexample is $\begin{gathered} M=(\mathbb R/100\mathbb Z)\times\Sigma\times\Sigma^q,\\ \text{coordinates }(t,z,v_1,\ldots,v_q), \end{gathered}$ a compact smooth manifold of dimension $2q+3$. We use $t$ also for a real lift of the clock.

2. The clock

Lemma 2.1 (the clock barrier). There is a smooth $100$-periodic function $b:\mathbb R\to[0,1]$ such that

(i) $b(t)=0$ exactly when $t\equiv50\pmod{100}$;

(ii) $b(t)=1$ when $|t-50|\geq2$, for $t\in[0,100]$;

(iii) $t+b(t)\leq50$ for $0\leq t\leq50$.

Proof. Let $\eta$ be a cutoff for $([-1,1],(-2,2))$ on $\mathbb R$, and on $[0,100]$ put $\begin{aligned} b(t)&=\eta(t-50)\,\frac{(t-50)^2}4\\ &\quad +1-\eta(t-50). \end{aligned}$ Near $t=0$ and $t=100$ we have $\eta(t-50)=0$, so $b=1$ there and the $100$-periodic extension is smooth. Write $x=t-50$. If $|x|\geq2$, then $\eta(x)=0$ and $b=1$; this is (ii). If $|x|\leq2$, then $b$ is a convex combination of $x^2/4\in[0,1]$ and $1$, so $b\in[0,1]$. If $\eta(x)<1$ then $b\geq1-\eta(x)>0$; if $\eta(x)=1$ then $b=x^2/4$, which vanishes only at $x=0$; this gives (i). For (iii) put $y=50-t\in[0,50]$; we need $b\leq y$. If $y\leq1$, then $\eta=1$ and $b=y^2/4\leq y$. If $1\leq y\leq2$, then $b\leq\max\{y^2/4,1\}\leq y$. If $y\geq2$, then $b=1\leq y$. $\square$

Let $e:\Sigma\to[0,1]$ be a cutoff for $(\{|z|\geq\tfrac34\}\cup\{\infty\},\ \{|z|>\tfrac12\}\cup\{\infty\})$; thus $e=0$ on $\{|z|\leq\tfrac12\}$ and $e=1$ on $\{|z|\geq\tfrac34\}\cup\{\infty\}$. The clock increment is $\begin{aligned} &h(t,z)\\ &\quad =b(t)+(1-b(t))\,e(z). \end{aligned}\tag{2.1}$ It is smooth and $100$-periodic in $t$, and $0\leq h\leq1$, $h\geq b$, with $h=1$ whenever $b(t)=1$ or $e(z)=1$. The clock can therefore slow down only near $t\equiv50$, and only while $|z|<\tfrac34$.

3. Cutoffs and register saturation

Fix the following smooth functions.

$\alpha:\mathbb R\to[0,1]$, a cutoff for $([9,11],(8,12))$: the preparation window.
$\beta_1,\ldots,\beta_q:\mathbb R\to[0,1]$, the readout windows, with supports in open intervals $J_\ell\subset(74,80)$ of length less than one, such that condition (3.1) holds: for every $t\in[76,77]$ some $\ell$ has $\beta_\ell(t)=1$. For $q=2$ take $J_1=(75.9,76.6)$, $J_2=(76.4,77.1)$ and let $\beta_\ell$ be a cutoff for $([76,76.5],J_1)$, respectively $([76.5,77],J_2)$. For larger $q$ put $\beta_\ell=0$ for $\ell\geq3$ (or use a finer cover). Write $\bar\beta_\ell(t)=\sum_{k\in\mathbb Z}\beta_\ell(t-100k)$ for the $100$-periodization; its translates have disjoint supports, so $0\leq\bar\beta_\ell\leq1$.
$\chi:\Sigma\to[0,1]$, a cutoff for $(\{\tfrac12\leq|z|\leq1\},\{\tfrac14<|z|<2\})$; it vanishes near $0$ and near $\infty$.

In the formulas below the functions $\beta_\ell$ on the real line, and not their periodizations, are used for predictions: only predicted readings in $J_\ell\subset(74,80)$ count.

The constant $L$ is chosen in Section 4, depending only on $h$. Once it is fixed, choose:

A smooth $p:[0,\infty)\to[0,3]$ with $p(r)=r$ for $0\leq r\leq1$, $p$ nondecreasing on $[0,3L]$, and $p(r)=0$ for all large $r$. To construct it, let $\psi$ be a cutoff on $\mathbb R$ for $((-\infty,1],(-\infty,2))$, put $p_0(r)=\int_0^r\psi$, which equals $r$ on $[0,1]$, is nondecreasing and is at most $2$; let $\phi$ be a cutoff for $([0,3L],(-1,3L+1))$ and put $p=\phi\,p_0$.
The saturation $S:\Sigma\to\mathbb C$ is given by (3.2): $\begin{gathered} S(v)=p(|v|)\frac v{|v|}\quad(0<|v|<\infty),\\ S(0)=S(\infty)=0. \end{gathered}$ It equals $v$ on the closed unit disc (so it is smooth near $0$), it is smooth where $0<|v|<\infty$, and it vanishes for $|v|\geq3L+1$ (so it is smooth near $\infty$). Moreover $|S|\leq3$.
A smooth $100$-periodic gain $a:\mathbb R\to[0,L]$ with $a=0$ on $[2,5]$ and $a=L$ on $[7,90]$: $L$ times the periodization of a cutoff for $([7,90],(6,96))$.

4. The prediction map and the gain

To predict later clock readings, we drop the readout additions and use the smooth map $\begin{gathered} G:\mathbb R\times\Sigma\to\mathbb R\times\Sigma,\\ G(t,z)=(t+h(t,z),\,z^2),\\ T_m=\mathrm{pr}_1\circ G^m\quad(m\geq1). \end{gathered}\tag{4.1}$ Here the clock is a real number, not a reading modulo $100$. Since $h$ is $100$-periodic, $G$ commutes with $(t,z)\mapsto(t+100,z)$, and so does the continuous function $x\mapsto\|DG(x)\|$. On the compact set $[0,100]\times\Sigma$ it is bounded (Lebl, Theorem 7.5.6); hence there is $B\geq1$ with $\|DG\|\leq B$ everywhere. By (1.1), and because the projection $\mathrm{pr}_1$ has derivative of norm at most one, $\begin{gathered} \|D(G^m)\|\leq B^m,\\ \|DT_m\|\leq B^m\\ (m\geq1). \end{gathered}\tag{4.2}$ Fix the gain $L>\max\{B,2\}.\tag{4.3}$ The bound $L>B$ will control the derivatives of the predicted clock readings and $L>2$ those of the predicted phases.

5. The inputs and the map

For $k\geq1$ let $\Phi_k(z)=\chi(z)\,(z/|z|)^k$ for $z\in\mathbb C\setminus\{0\}$, and $\Phi_k(0)=\Phi_k(\infty)=0$; these are smooth complex functions on $\Sigma$, vanishing near both poles. For $1\leq\ell\leq q$, $t\in\mathbb R$ and $z\in\Sigma$ define the input by (5.1): $\begin{aligned} U^\circ_\ell(t,z)&=\frac{\alpha(t)}{10}\sum_{m=1}^{\infty}L^{1-m}\,\beta_\ell\bigl(T_m(t,z)\bigr)\\ &\qquad\cdot\Phi_{2^{m+1}}(z), \end{aligned}$ and let $U_\ell(t,z)=\sum_{k\in\mathbb Z}U^\circ_\ell(t-100k,z)$ be its $100$-periodization. Since $\alpha$ is supported in $(8,12)$, at most one translate is nonzero at any $t$, and $U_\ell=U^\circ_\ell$ for $0\leq t<100$. The index $m$ counts iterates of $G$, not clock units.

Definition 5.1 (the map). Define $f:M\to M$, $f(t,z,v_1,\ldots,v_q)=(t',z',v_1',\ldots,v_q')$, by (5.2): $\begin{aligned} t'&=t+h(t,z)\pmod{100},\\ z'&=z^2+20\sum_{\ell=1}^q\bar\beta_\ell(t)\,S(v_\ell)\quad(z\neq\infty),\\ z'&=\infty\quad(z=\infty),\\ v_\ell'&=a(t)\,S(v_\ell)+U_\ell(t,z)\quad(1\leq\ell\leq q). \end{aligned}$ All right-hand sides use the old state. Every register output is a complex number, in the affine chart. When the clock reading lies in $[2,5]$ modulo $100$, both $a(t)$ and $U_\ell(t,z)$ vanish, so every register is reset to zero, whatever its value was.

How the timing is meant to work. Write $z_i$ for the main coordinate at step $i$. An input computed at step $i$ enters its register at step $i+1$. If it is read out at step $i+m$, it has received $m-1$ gains in between. If the gain is $L$ throughout and the register stays in the unit disc, where $S$ is the identity, the factor $L^{1-m}$ is exactly cancelled. If moreover $z$ squares exactly in between, its phase at step $i+m$ is $(z_i/|z_i|)^{2^m}$, so the phase of $z_{i+m}^2$ is $(z_i/|z_i|)^{2^{m+1}}$: the phase written into the input. Finally the clock moves by exactly one on $[70,82]$, which contains every readout support, whatever $z$ does. These are explanations, not proofs: the next lesson proves that the predicted and actual clocks agree, that each register stays in the unit disc until it is read, and only then uses the phase.

6. Continuous differentiability

Proposition 6.1. The inputs $U_\ell:(\mathbb R/100\mathbb Z)\times\Sigma\to\mathbb C$ are continuously differentiable, $|U_\ell|\leq L/(10(L-1))<\tfrac15$, and $f$ is a $C^1$ self-map of $M$.

Proof. Angular factors. The support of $\chi$ lies in the compact annulus $A=\{\tfrac14\leq|z|\leq2\}$, on which $u(z)=z/|z|$ is smooth; let $c_u$ and $c_\chi$ be the maxima of $\|Du\|$ and $\|D\chi\|$ on $A$ (norms from the round metric on the source and the modulus on $\mathbb C$). Since $D(u^k)=k\,u^{k-1}Du$ and $|u|=1$, $\begin{aligned} &|\Phi_k|\leq1,\\ &\|D\Phi_k\|\leq c_\chi\\ &\quad +k\,c_u\leq C_\chi(1+k) \end{aligned}\tag{6.1}$ with $C_\chi=\max\{c_\chi,c_u\}$, independent of $k$.

Summands. The $m$-th summand of (5.1), $\begin{aligned} &F_{\ell,m}(t,z)\\ &\quad =\frac{L^{1-m}}{10}\,\alpha(t)\,\beta_\ell\bigl(T_m(t,z)\bigr)\,\Phi_{2^{m+1}}(z), \end{aligned}$ is smooth on $\mathbb R\times\Sigma$. By the product rule its derivative is the sum of three terms, in which respectively $\alpha$, $\beta_\ell\circ T_m$ and $\Phi_{2^{m+1}}$ are differentiated. With $c_\alpha=\max|\alpha'|$, $c_\beta=\max_\ell\max|\beta_\ell'|$, (4.2) and (6.1), the bound (6.2) holds: $\begin{aligned} &|F_{\ell,m}|+\|DF_{\ell,m}\|\\ &\quad\leq C\,L^{1-m}\bigl(1+B^m+2^{m+1}\bigr) \end{aligned}$ for a constant $C$ independent of $\ell$ and $m$. By (4.3) the three series $\sum L^{1-m}$, $\sum L(B/L)^m$ and $\sum 2L(2/L)^m$ converge. Hence the series $\sum_mF_{\ell,m}$ and $\sum_mDF_{\ell,m}$ converge uniformly on $\mathbb R\times\Sigma$: their tails are bounded by tails of a convergent numerical series.

Differentiability of the sum. Let $K$ be a compact coordinate box inside a chart $(t,x_1,x_2)$ of $\mathbb R\times\Sigma$ (with $x_1+ix_2=z$ or $=w$). On $K$ the coordinate partial derivatives of a function are bounded by a constant $c_K$ times the norm of its derivative, so the series of partial derivatives $\sum_m\partial_jF_{\ell,m}$ converge uniformly on $K$. On each coordinate segment in $K$, Lebl's Theorem 6.2.10 shows that the partial derivative $\partial_jU^\circ_\ell$ exists and equals $\sum_m\partial_jF_{\ell,m}$. This limit is continuous: if $g_N\to g$ uniformly with $g_N$ continuous, then $\begin{aligned} &|g(x)-g(y)|\leq|g(x)-g_N(x)|\\ &\quad+|g_N(x)-g_N(y)|+|g_N(y)-g(y)|, \end{aligned}$ and the outer terms are uniformly small. By Lebl's Proposition 8.4.6, $U^\circ_\ell$ is continuously differentiable on the interior of $K$. Such boxes cover $\mathbb R\times\Sigma$, so $U^\circ_\ell$ is $C^1$.

Nothing in this argument depends on a selected prediction time: at a point where $\beta_\ell(T_m)=0$ for all $m$, every summand and its derivative vanish, and uniform convergence of the tails gives continuity even if nearby points have nonzero summands with indices tending to infinity.

Periodization and bound. $U^\circ_\ell$ vanishes, with its derivative, unless $t\in(8,12)$. The translates $U^\circ_\ell(\cdot-100k,\cdot)$ have disjoint supports and only one of them is nonzero near any point, so $U_\ell$ is $C^1$ and descends to $(\mathbb R/100\mathbb Z)\times\Sigma$. Since $\alpha,\beta_\ell,|\Phi_k|\leq1$, $\begin{aligned} |U_\ell|&\leq\frac1{10}\sum_{m\geq1}L^{1-m}\\ &=\frac{L}{10(L-1)}<\frac15, \end{aligned}\tag{6.3}$ using $L>2$.

The map. The clock component of (5.2) is smooth and well defined modulo $100$. Each register component is $C^1$, with values in the disc of radius $3L+\tfrac15$ of the affine chart; this includes source values $v_\ell=\infty$, because $S$ is smooth and vanishes near $\infty$. The main component is smooth at every finite $z$. Near $z=\infty$ put $C_0(t,\mathbf v)=20\sum_\ell\bar\beta_\ell(t)S(v_\ell)$, a smooth function with $|C_0|\leq60q$. In the coordinate $w=1/z$ at the source and $w'=1/z'$ at the target, $\begin{aligned} w'&=\frac1{w^{-2}+C_0}\\ &=\frac{w^2}{1+C_0\,w^2}. \end{aligned}\tag{6.4}$ For $|w|<(120q)^{-1/2}$ the denominator has modulus at least $1-60q|w|^2>\tfrac12$, uniformly in $t$ and the registers, so (6.4) extends smoothly to $w=0$ with value $0$, which is $z'=\infty$. Hence $f$ is $C^1$. $\square$

Remark 6.2 (why the series is infinite). The number of steps between preparation (clock near $10$) and a readout (clock near $76$) is not bounded: the clock passes $50$, where its increment $h$ can be as small as $b(t)$, when $|z|<\tfrac34$. An input built from finitely many predictions $m\leq m_0$ would miss every readout more than $m_0$ steps after preparation, and the main coordinate would then square without help. The weights $L^{1-m}$ let the series run over all delays while keeping its first derivatives summable. Nothing better than $C^1$ is claimed: the second derivatives of the summands grow like $L^{-m}(B^{2m}+4^m)$, and $L$ was only chosen larger than $B$ and $2$.

7. Exercises

7.1. Verify (6.4), and show that the denominator is at least $\tfrac12$ in modulus for $|w|<(120q)^{-1/2}$.

7.2. Why is the saturation $S$ needed in (5.2)? What would go wrong with the formulas $v_\ell'=a(t)v_\ell+U_\ell$ and $z'=z^2+20\sum_\ell\bar\beta_\ell(t)v_\ell$ on the sphere?

7.3. Show that $f$ is not injective. (Use that $S$ vanishes near $\infty$.)

7.4. Show that a register is reset to zero at the first step at which the clock reading lies in $[2,5]$ modulo $100$, and that it then stays zero until the clock reaches $8$.

8. Solutions

7.1. $z'=z^2+C_0$ with $z=1/w$ gives $1/w'=1/w^2+C_0=(1+C_0w^2)/w^2$, hence (6.4). For $|w|^2<1/(120q)$, $|C_0w^2|\leq60q|w|^2<\tfrac12$.

7.2. On the sphere, $a(t)v_\ell$ is undefined at $v_\ell=\infty$ when $a(t)=0$ (it would be $0\cdot\infty$), and $z^2+20\bar\beta_\ell(t)v_\ell$ is undefined when $z$ is finite and $v_\ell=\infty$, or makes $z'$ depend discontinuously on $v_\ell$. The saturation makes every term bounded, equal to the identity on the unit disc (where the registers will be shown to live until they are read), and zero near $\infty$.

7.3. Fix all coordinates except $v_1$, and let $v_1$ vary in $\{|v_1|>3L+1\}\cup\{\infty\}$. There $S(v_1)=0$, and $v_1$ enters (5.2) only through $S(v_1)$; so all these points have the same image.

7.4. At such a step $a(t)=0$ and $\alpha(t)=0$, so $U_\ell=0$ and $v_\ell'=0$. While the clock stays below $8$ (modulo $100$) the input remains zero, and $S(0)=0$ keeps the register at zero, whatever the gain.

7.5. For $g(z)=z^2$ at finite $z$, a tangent vector of round length $2|\dot z|/(1+|z|^2)$ is sent to $2z\dot z$, of round length $2|2z\dot z|/(1+|z|^4)$. The ratio is $2|z|(1+|z|^2)/(1+|z|^4)$, which is at most $2$ (it equals $2$ at $|z|=1$), since $2|z|(1+|z|^2)\leq2(1+|z|^4)$ is equivalent to $|z|+|z|^3\leq1+|z|^4$, that is $(1-|z|)(1-|z|^3)\geq0$. By symmetry in the chart $w$ the same bound holds near $\infty$. When $b(t)=1$ the clock component is $t+1$, whose derivative has norm one; so at such points $\|DG\|\leq2$, and larger values of $\|DG\|$ come from the variation of $b$ and $e$ in the clock increment.

References

[OpenAI-C1] OpenAI, A $C^1$ counterexample to the entropy conjecture, OpenAI Math Release preprint, 25 September 2026, Section 2. https://github.com/openai/math/tree/main/preprints/A-C1-Counterexample-to-the-Entropy-Conjecture-September-25-2026
[Lebl] J. Lebl, Basic Analysis I and II, version 6.3; the texts of the core courses Real Analysis I and II. https://www.jirka.org/ra/