The record is everything that has happened, written down event by event. Each entry is a posting, and a posting is never erased. To follow one pair, picture four on/off switches, two for each station; one setting of all four is a configuration, and there are 2 × 2 × 2 × 2 = 16 of them, written 00|00 to 11|11. The paper calls such a set of switches a cube (three switches give the eight corners of an ordinary cube), and each station’s two switches a block. In the record’s account the two stations are the two blocks of one cube; for two separate laboratories that is composition, which is owed (section 06).
An act is what a station does to get a +1 or a −1 (in the textbook, a measurement with two outcomes), and each outcome has a weight, the textbook’s probability. Some acts only read: their result was already sitting in the configuration. Others write: for at least some configurations, their result is not there until the act is done. As a drawing of that real difference: a reading act looks at a box already ticked on the card; a writing act asks a question whose answer is not on every card.
If every act only reads, the configuration itself is the card of section 02, and the sum cannot pass 2, whatever the state. That is Bell’s inequality; the paper behind this page, a preprint posted on 14 November 2026 and not yet peer reviewed (section 09), proves it again in the record’s terms (Theorem 27a). The loophole-free tests pass 2, with each setting drawn at random and the stations too far apart for either choice to reach the other in time. So not every result is read off an entry already there. The record’s account of what is left: some act writes its result.
Acts that write can reach 2√2 exactly, on one cube: three of the four correlations come out exactly 1/√2 ≈ 0.707 and the fourth, the one subtracted, exactly −1/√2, so S = 2√2 = 2.828427… (Theorem 28). Figure 1 runs these acts; the construction follows for readers who want it.
The construction, step by step
The bound of 2 on values written in advance holds for any two stations that meet the conditions of section 02; what follows about acts that write is proved on one cube, or on one state, with the textbook’s squared weights (section 06), and not for two separate laboratories.
A state gives each of the 16 configurations a number. Taken together, the 16 numbers form an arrow in sixteen dimensions, of length one. The weight of an outcome is the squared length of the part of that arrow the outcome picks out: the textbook’s probability (section 06 says what the square rests on).
An act’s two results pick out two parts of the state, and an act read twice gives the same result both times. An act that only reads splits the state along whole configurations, so its result was already sitting in the configuration. An act that writes splits it, at least in part, into parts spread over several configurations, for example the sum and the difference of the two settings of one switch, so some configurations do not hold its result. Written as a 16 × 16 table, a reading act multiplies each configuration by +1 or −1, and a writing act moves configurations, for example by flipping a switch: that is how the figures draw them.
Let the state weigh equally the four configurations in which the two stations’ switches match: 00|00, 01|01, 10|10 and 11|11 (in the textbook’s words, a maximally entangled state of the two blocks). Give the first station two acts: flip its first switch; or flip its second switch while carrying the first switch’s sign, a factor −1 whenever the first switch is on. Give the second station the same two acts on its own switches, and let it mix them in equal measure: one setting is their sum divided by √2, the other their difference divided by √2. Three of the four correlations come out exactly 1/√2 ≈ 0.707 and the fourth, E(1,1), exactly −1/√2, so
S = 3/√2 + 1/√2 = 4/√2 = 2√2 = 2.828427…exactly, on this cube of 2 + 2 switches; the paper proves it for every cube of n + n switches with n ≥ 2 (Theorem 28). Figure 1’s “Open the matrices” shows the four acts as 16 × 16 tables.