The Eisenstein Lattice

The Collatz map speaks two languages: halving ($\div 2$) and tripling ($\times 3 + 1$). Both primes have a clean description in the Eisenstein integers $\mathbb{Z}[\omega]$, where $\omega = e^{2\pi i/3}$ is a primitive cube root of unity: 3 is the one prime that ramifies there, and 2 is the smallest prime that stays prime.

The Eisenstein integers form a triangular lattice in the complex plane. Every Collatz orbit can be drawn as a walk on this lattice, which turns questions about convergence into geometric questions about where the walk goes. This page is an exploration: a way of looking, not a set of results about every orbit.

The Two Primes

The Eisenstein integers are complex numbers $a + b\omega$ where $a, b \in \mathbb{Z}$ and $\omega^2 + \omega + 1 = 0$. Their norm is:

$$N(a + b\omega) = a^2 - ab + b^2$$

The ring has 6 units: $\{\pm 1, \pm\omega, \pm\omega^2\}$.

The two Collatz primes have distinct algebraic characters in $\mathbb{Z}[\omega]$:

Prime	Behavior	Factorization	Norm
2	Inert (stays prime)	$(2)$ remains prime	$N(2) = 4$
3	Ramifies (squares)	$3 = -(1+2\omega)^2$	$N(1+2\omega) = 3$

Halving is division by the inert prime 2. Tripling is multiplication by $3 = -(1+2\omega)^2$, up to sign the square of the ramified prime. The norms $N(2) = 4$ and $N(1+2\omega) = 3$ are in the ratio $4/3$, the same number as in the average-drift heuristic (two halvings per tripling). I use that as a way of writing the heuristic, not as an explanation of it.

Orbits as Lattice Walks

Each Syracuse step (triple once, then halve $\alpha$ times) is recorded as a displacement vector on the Eisenstein lattice:

$$\vec{v}_i = \alpha_i + \omega$$

where $\alpha_i = v_2(3x_i + 1)$ is the number of halvings after the $i$-th tripling. In Eisenstein coordinates, this is the lattice point $(\alpha_i, 1)$ with norm:

$$N(\alpha_i, 1) = \alpha_i^2 - \alpha_i + 1$$

$\alpha$	Vector $\alpha + \omega$	Angle	Norm	Meaning
1	$1 + \omega$	60$^\circ$	1 (unit)	Growth: $x$ becomes about $\tfrac32 x$
2	$2 + \omega$	30$^\circ$	3	Shrinks: about $\tfrac34 x$
3	$3 + \omega$	19$^\circ$	7	Shrinks: about $\tfrac38 x$
4	$4 + \omega$	14$^\circ$	13	Shrinks: about $\tfrac{3}{16} x$

As $\alpha$ grows, the displacement vector rotates toward the real axis — more horizontal, more contractive. The full orbit is the sum of all displacement vectors, landing at the Eisenstein lattice point $(h, s)$ where $h$ = total halvings and $s$ = total triplings.

n (odd, up to 99,999) = Grid Vectors

h = 70s = 41h/s = 1.7073N(h,s) = 3711above geodesic by 5.017

Each Syracuse step adds a vector α_i + ω to the lattice walk. Green segments end above the geodesic (the value is back at or below its start, give or take the +1 corrections); red segments end below it (the value is above its start). Dot size reflects α (halvings per tripling).

The Geodesic

The geodesic is the line $h = s \cdot \log_2 3$ on the lattice. Which side of it the walk is on records where the orbit stands relative to its starting value:

• Above the geodesic ($h/s > \log_2 3$): more halvings so far than the triplings need — the value is back at or below its start, give or take the $+1$ corrections
• Below the geodesic ($h/s < \log_2 3$): not enough halvings yet to compensate — the value is above its start

Every orbit that reaches 1 ends above the geodesic. This is an identity, not a finding:

$$h - s \cdot \log_2 3 = \log_2(x_0) + \lvert\varepsilon\rvert > 0$$

The excess above the geodesic equals the initial bit-length $\log_2(x_0)$ plus the size of the cumulative $+1$ correction $\varepsilon$ (the same $\varepsilon$ as in the conservation law).

For the orbit of 27: 38 of the 41 points of the walk (93%) lie below the geodesic, where the value is above 27. The walk first gets above the line at step 37, and it ends above it with the steps $53 \to 5 \to 1$, which have $\alpha = 5$ and $\alpha = 4$.

The Endgame

In a sample of orbits, large $\alpha$ values sit late. For the odd starts from 3 to 5999 (orbits with at least five odd steps), the mean relative position of each kind of step is:

$\alpha$	Mean position in orbit	Interpretation
1	0.46	Slightly early — these are the growth steps
2	0.45	Slightly early
3	0.55	Slightly late
$\geq 4$	0.70	Late

Where each α value appears in orbits (odd starts 3–5999). α ≥ 4 steps sit late on average (mean position about 0.7 in this sample). About a quarter of them are the final step into 1, which always has α ≥ 4; without those the mean is 0.60. Triangles mark the mean position of each group.

Much of this is built in rather than discovered. The last step of every orbit that reaches 1 comes from 5, 21, 85, … and so always has $\alpha \geq 4$; with that final step left out, the mean for $\alpha \geq 4$ falls to 0.60, and over the odd starts below $10^5$ it is 0.65 (0.57 without the final step). It is not that small numbers are more divisible: the share of odd $x$ with $v_2(3x+1) \geq 4$ is $1/8$ at every size. Likewise every orbit that reaches 1 ends above the geodesic, by the identity above. Whether any walk stays below the geodesic forever is open: such a walk would belong to a number that never falls below its start, a counterexample to the conjecture.

The Eigenvalue Connection

The matrix on the transfer operator page has three eigenvalues satisfying:

$$\lambda^3 = \frac{4}{3} = \frac{N(2)}{N(1+2\omega)}$$

The equation comes from the trivial cycle $1 \to 4 \to 2 \to 1$: two halvings of weight 2 and one tripling of weight $1/3$ (see that page). The numbers 4 and 3 are also the norms of the Eisenstein primes 2 and $1+2\omega$, so it can be written as a ratio of Eisenstein norms; nothing about $\mathbb{Z}[\omega]$ was used to get it. The three eigenvalues are:

$$\lambda_k = \left(\frac{4}{3}\right)^{1/3} \cdot \omega^k, \quad k = 0, 1, 2$$

They lie on a circle of radius $(4/3)^{1/3} \approx 1.1006$, equally spaced at $0°$, $120°$, $240°$, as the three cube roots of any positive number are. The three directions are the Eisenstein units $1, \omega, \omega^2$.

The radius exceeds 1 exactly because $4 > 3$:

$$(4/3)^{1/3} > 1 \iff N(2) > N(1+2\omega) \iff 4 > 3$$

It is the same inequality, $2^2 > 3$, that appears in the average-drift heuristic, because the trivial cycle has the average number of halvings, two. It is a statement about the trivial cycle, not about every orbit, and the same matrix for $5x+1$ also has radius above 1.

The Three Layers

Three ways of looking, side by side:

Layer	Question	Tool	What it shows
Thermodynamics	Does energy dissipate on average?	Criticality $\mu = 3/4$	Yes — $E[\alpha] = 2 > \log_2 3$
Spectral	What does the cut-off matrix show?	Transfer matrix $\lambda^3 = 4/3$	The trivial cycle: two halvings, one tripling
Geometric	Where do the large steps sit?	Eisenstein lattice walk	Late in the sample — $\alpha \geq 4$ at mean position 0.70 (0.60 without the forced last step)

The first two come back to one arithmetic fact, $2^2 > 3$: an odd step is followed by two halvings on average, and by exactly two around the trivial cycle. In $\mathbb{Z}[\omega]$ the same inequality reads $N(2) > N(1+2\omega)$, which is a way of writing it, not a reason for it. 4 is greater than 3 is the heuristic for why orbits shrink on average. It does not show that every orbit does.

One lap of the trivial cycle $\{1, 2, 4\}$ is the step $2 + \omega$, the lattice point $(2, 1)$, whose norm is 3: it is the ramified prime $1+2\omega$ times the unit $-\omega$. That is a pleasant coincidence of the encoding, not a structural fact: the norm of $\alpha + \omega$ is $\alpha^2 - \alpha + 1$, which equals 3 at $\alpha = 2$.

Related

• The Transfer Operator — the matrix whose three eigenvalues are the cube roots of $4/3$
• The Collatz Zoo — the $nx+c$ family and the thermodynamic analogy
• The Hidden Rotation — the base-6 circle: an irrational rotation plus a small perturbation
• Finite Fuel — a heuristic picture, not a proof, of why positive integers might not keep climbing
• abc Conjecture — how close powers of 2 and 3 can be