Shortcut-Collatz Incidence

For every nonzero integer n, a two-generator projective-incidence question built from two fixed Möbius maps has a solution exactly when n reaches 1 under the shortcut Collatz map.

The two maps are the inverse branches of the shortcut Collatz map; only the target row depends on n, and words that follow an illegal branch leave the integers forever. A second construction compiles every generic two-generator incidence instance into two rational 3 × 3 matrices of rank two that are mortal exactly when the instance has a solution. Together they place the reachability question for each given integer inside the rank-(2, 2) part of M₃(2).

Neither construction proves undecidability, and nothing here bears on the Collatz conjecture. Both constructions are machine-checked; their composition is not packaged as one formal reduction.

Known Stuff#

Shortcut-Collatz reachability#

The shortcut Collatz map is

T (x) = {\begin{cases} x / 2, & x even, \\ (3 x + 1) / 2, & x odd. \end{cases}

Write x ⇝ 1 when some iterate of x equals 1. This is a question about one given integer; the Collatz conjecture is the statement that every positive integer reaches 1.

Projective incidence#

A two-generator projective-incidence instance consists of invertible rational 2 × 2 matrices G, H, a nonzero row r^T, and a nonzero column c. With X_w the product along a word w over {G, H}, it asks whether

\exists w \in {G, H}^{*} : r^{T} X_{w} c = 0 .

The empty word is allowed. On the projective line the question is whether the point c can be moved into the point ker r^T by the Möbius maps of G and H. Multiplying a generator by a nonzero scalar changes no answer. Two scalars of an ordered instance matter below:

κ_{1} = r^{T} H^{- 1} c, κ_{2} = r^{T} H^{- 1} G H^{- 1} c .

The instance is generic when both are nonzero; this excludes at most two source points c. Rescaling G and H can then make both equal to one, and GPI₂ denotes the problem restricted to instances normalized in this way.

New Stuff#

Result#

A forward Collatz step chooses its branch by parity, while a matrix word chooses its letters freely. Running the map backward removes the mismatch. From 1, the two inverse branches z ↦ 2z and z ↦ (2z − 1)/3 generate every integer that reaches 1, and a word that uses the odd branch where it is not integral acquires a factor 3 in its denominator that never disappears. The two branches are Möbius maps, so incidence with the point n asks exactly whether n is a predecessor of 1.

Incidence tests one coefficient, and mortality needs a whole zero product. The compiler uses two rank-two cuts, each factoring through a plane, and chooses the four products between adjacent letters to be G, H, the identity, and a rank-one test. Two planes in ℚ³ share a line, and the scalar κ₂ makes the four maps agree on it, so three dimensions suffice.

Theorem (shortcut-Collatz incidence). For every nonzero integer n, put

A = [\begin{matrix} 2 & 0 \\ 0 & 1 \end{matrix}], B = [\begin{matrix} 2 & - 1 \\ 0 & 3 \end{matrix}], r_{n}^{T} = (1, - n), c = {(1, 1)}^{T} .

Some word w over {A, B}, possibly empty, has r_n^TX_wc = 0 exactly when n ⇝ 1. After the rescaling H_n = λ_nA, G_n = (λ_n²/μ_n)B of the normalization, the instance lies in GPI₂.

Theorem (rank-two incidence compiler). For every generic two-generator incidence instance there are rational 3 × 3 matrices P₀, P₁, each of rank exactly two, with

Mortal {P_{0}, P_{1}} \Leftrightarrow \exists w : r^{T} X_{w} c = 0 .

The empty word is necessary in the first theorem: n = 1 reaches 1 in no steps.

Inverse branches#

The projective actions of A and B are the inverse branches

a (z) = 2 z, b (z) = \frac{2 z - 1}{3},

and c represents the point 1. Reading a word from right to left applies the branches to 1 and produces a candidate predecessor x(w); matrix multiplication tracks it with a scale factor,

X_{w} c = s (w) {(x (w), 1)}^{T}, r_{n}^{T} X_{w} c = s (w) (x (w) - n),

where s(w) is 3 to the number of letters B in w. The coefficient vanishes exactly when x(w) = n.

Malformed words#

The branch b is a genuine inverse step at an integer z exactly when 3 divides 2z − 1; its value is then an odd integer mapped to z by T, and a(z) is always an even integer mapped to z. When 3 does not divide 2z − 1, the result has 3-adic valuation −1. Doubling preserves a negative 3-adic valuation, and b lowers it further, since the numerator 2z − 1 then has valuation 0. No later letter returns to an integer.

Every word therefore either spells a legal backward run from 1 or leaves the integers permanently, and only the first kind can produce the integer n. The words of the first kind produce exactly the integers that reach 1.

Normalization#

For the ordering G = B, H = A, the chart scalars are

κ_{1} = λ_{n} = \frac{1}{2} - n, κ_{2} = μ_{n} = - 3 n,

both nonzero for nonzero integers n. The rescaled generators H_n = λ_nA and G_n = (λ_n²/μ_n)B have the same Möbius maps and chart scalars κ₁ = κ₂ = 1, and every word keeps its zero or nonzero coefficient. The compiler below needs only that κ₁ and κ₂ are nonzero.

From incidence to two 3 × 3 matrices#

Let (G, H, r^T, c) be generic, and put

v_{1} = H^{- 1} c, v_{0} = H^{- 1} G H^{- 1} c, Θ = κ_{2}^{- 1} v_{1} r^{T} .

Then r^Tv₁ = κ₁ and r^Tv₀ = κ₂, so v₀, v₁ are nonzero, and the rank-one test Θ satisfies Θv₀ = v₁.

Two planes sharing a line#

In ℚ³ take the planes Π₀ = span{e₁, e₂} and Π₁ = span{e₁, e₃}, which share the line ℚe₁, with embeddings ι₀(x, y) = (x, y, 0) and ι₁(x, y) = (x, 0, y). Each generator will have the form P_t = ι_tO_t with O_t: ℚ³ → ℚ². In the language of interface compression, both generators are cuts, there are no transitions, and the bridge between adjacent letters P_tP_s is O_tι_s, a map from plane s to plane t. The design prescribes these four bridges:

Adjacent letters	Bridge	Role
`P`₀`P`₀	Θ	rank-one test
`P`₀`P`₁	`I`	change of plane
`P`₁`P`₀	`H`	generator
`P`₁`P`₁	`G`	generator

A single linear map O_t must produce both bridges out of plane t, from ι₀ and from ι₁, and the two embeddings agree on the shared line. So the two bridges into P_t must agree on a distinguished vector of each plane. The vectors v₀ for plane 0 and v₁ for plane 1 work:

Θ v_{0} = I v_{1}, H v_{0} = G v_{1} .

Change coordinates in plane s by the frame F_s = [v_s | Jv_s], where J is the quarter turn; det F_s = |v_s|² ≠ 0 over ℚ. The adapted bridges Ê_ts = F_t⁻¹E_tsF_s, where E_ts is the tabulated bridge, then agree on e₁: Ê_t0e₁ = Ê_t1e₁. Put

O_{t} = [{\hat{E}}_{t 0} e_{1} | {\hat{E}}_{t 0} e_{2} | {\hat{E}}_{t 1} e_{2}], P_{t} = ι_{t} O_{t} .

Then O_tι_s = Ê_ts for all t, s. Each P_t factors through ℚ², so its rank is at most two, and projecting Ê₀₁ = F₀⁻¹F₁ and Ê₁₀ = F₁⁻¹HF₀, both invertible, out of P₀ and P₁ shows that the rank is exactly two. Frames cancel along any path, so adapted and tabulated bridges have the same zero products.

For the forward direction, map each letter G of w to P₁ and each letter H to P₀P₁. The physical word P₀P₀P₁φ(w)P₀P₀ has bridge product ΘX_wHΘ, and

Θ X_{w} H Θ = κ_{2}^{- 1} (r^{T} X_{w} c) Θ,

which is zero when w solves the incidence problem.

Every zero product#

By interface compression, a zero product of P₀, P₁ gives a zero product of bridges along a path of planes. Cut that product at every test Θ:

B_{0} Θ B_{1} Θ \dots Θ B_{m} = (B_{0} v_{1}) [\prod_{1 \leq k < m} κ_{2}^{- 1} r^{T} B_{k} v_{1}] (κ_{2}^{- 1} r^{T} B_{m}),

where each block B_k is a product of the invertible bridges I, G, H. With no test the product is invertible, and otherwise the outer factors are nonzero. A test comes from the letters P₀P₀, so the bridge before it comes from P₀P₀ or P₁P₀: every inner block is empty or ends in H. An empty block contributes κ₁/κ₂ ≠ 0. A block ending in H is X_wH after the identity bridges are deleted, and contributes κ₂⁻¹r^TX_wc. A zero product therefore needs a word w with r^TX_wc = 0.

The hypothesis κ₁ ≠ 0 excludes a spurious zero: Θ² = (κ₁/κ₂)Θ. Multiplying P₀ and P₁ by nonzero integers clears denominators without changing ranks or zero products.

What follows#

For each nonzero integer n, composing the two theorems and clearing denominators gives two integer 3 × 3 matrices of rank two that are mortal exactly when n ⇝ 1. A decision procedure for M₃(2), or for GPI₂, would therefore decide, for a given integer, whether it reaches 1 under the shortcut Collatz map, a question for which no algorithm is known. That is the only direction established.

Neither theorem decides GPI₂ or M₃(2), proves either undecidable, or reduces them back to Collatz reachability. The valuation guard shows the same phenomenon for the other open rank profile: M₃(2) already contains orbit questions about rational numbers whose status is unknown.

Bookkeeping#

Verification and provenance#

Inverse language

reachesOne_iff_shortcutCollatz identifies the backward closure of 1 with shortcut-Collatz reachability.

Malformed words

predecessorState_reaches_or_negative proves, for every word, that it spells a legal backward run or reaches negative 3-adic valuation.

Normalized incidence

normalizedScalars and exists_normalizedIncidence_zero_iff prove the normalization and the final equivalence for every nonzero integer.

Rank-two compiler

isMortal_adaptedGenerator_iff and adaptedGenerator_rank, with rawEdge_agrees, adaptedEdge_compatible, adaptedEdge_pathProduct_eq_zero_iff, and the path grammar pathBlocks_closed.

Transitive axioms

The publication-facing declarations use only propext, Classical.choice, and Quot.sound.

Formal scope#

Lean checks the shortcut-step equivalence, the backward semantics of every word, the 3-adic trap, the raw incidence identity, the normalization, and the final equivalence for every nonzero integer. Lean separately checks the compiler for every generic instance, including both ranks and every zero product. Scaling generators by nonzero constants preserves mortality by isMortal_smulMatrix_iff.

The composition of the two theorems with denominator clearing is not a single Lean declaration, and the map from n to the matrices is not packaged as a primitive-recursive reduction; the formulas above make it explicitly computable.

References#

GPT-5.6 Sol, elicited by @eternalism_4eva, Interface Compression, 2026. Cuts, bridges, and paths.
GPT-5.6 Sol, elicited by @eternalism_4eva, A Valuation Guard in M₃(2), 2026. The rank profile (3, 2).