Field notes

The one-shot Prisoner's Dilemma proves cooperation cannot survive: defecting pays more no matter what your partner does, so two rational players both defect. This lattice adds the thing the proof leaves out — structure. Each cell plays the dilemma with its eight neighbours, then copies whichever neighbour scored best (Nowak & May, 1992).

A lone cooperator on a defector frontier is exploited and lost. But inside a cluster, a cooperator's neighbours are mostly other cooperators, so the colony's interior out-earns the defectors around it and the cluster grows. Cooperation survives as geography, not goodwill — and the marked threshold is where a cooperative front stops gaining ground, not where cooperation dies. The payoff panel shows the whole threshold as the comparison it is: a front cell scores 6·M[C][C], the defector facing it scores 3·M[D][C] + 6·M[D][D], and the front advances only while the first is larger — which is exactly T < 2(R−P) = 1.80.

At or above 1.80 the front stops; it is not destroyed. A 12×12 cooperator seed at T = 1.80 sits at its own 144 cells forever. What happens to a ragged world past the threshold is not one story and not monotonic: the same 40-defector field ends 98% cooperative at T = 1.70, 3% at 1.80, permanently frozen at 56% in 66 colonies anywhere from 1.90 to 2.10, and 1% above 2.15. Coexistence and Barren load that identical field — only the dial differs.

Rounds per meeting is the shadow of the future. At one round tit-for-tat is just a cooperator: its row in the payoff table is C's row, digit for digit. Give it memory and it stops being suckered after the first betrayal, and its threshold moves out to 2K(R−P) — at K = 6 that is 10.80, far past the top of this dial, so nothing you can set here will stop it. Note too that once the defectors are gone C and tit-for-tat earn exactly the same against each other, so the border between them freezes wherever it happens to lie: that seam is a tie, not a result.

Past the notch at T = 2 the game is no longer a textbook Prisoner's Dilemma — 2R > T + S fails, so a cooperating pair no longer makes the most between them. The lattice still runs; the story it tells is about a different game.