# The Coherence Framework
**What the Ground State Is, and Why Displacement Cost Takes the Form It Does**

Diego Rincón
Independent

Published: [10.5281/zenodo.21424917](https://doi.org/10.5281/zenodo.21424917)

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## Abstract

The Displacement Framework (Rincón, 2026, *Displacement Framework — Domain Applications*) treats a system's ground state $S_g$ as given, and derives the dynamics of drift away from it and the cost of return, $C_{\text{return}}(\tau) = f + b\tau^{1+k}$. It never asks what $S_g$ *is*, structurally — nor why the displacement cost $D(\xi)$ is assumed convex and, in the canonical case, quadratic: $D(\xi) = \tfrac{1}{2}\alpha\xi^2$. This paper answers both questions from the spectral formalism already established for symbolic and neural coherence (Rincón, 2026, *From Grammar to Coherence*): a system's state is a weighted graph; coherence $\chi$ is the Fiedler value of that graph's Laplacian; the ground state $S_g$ is the coherence-maximizing configuration under a fixed energy budget. From this definition, $D(\xi) = \tfrac{1}{2}\alpha\xi^2$ is not an assumption but a theorem **conditional on one hypothesis**: that $\chi$ is simple at $S_g$. Where it is, the first-order term vanishes because $S_g$ is a maximum, and the quadratic is the necessary leading-order term. Where $\chi$ is degenerate — which includes the unconstrained maximiser — $\chi$ is not differentiable, no first-order term exists to vanish, and the leading-order cost is LINEAR. §4 states the condition and §6.1 turns it into a sharper test than the original. The paper then extends this to the dyadic case and proves, rather than asserts, that a coupled system's coherence is not decomposable into the coherence of its parts — the formal content behind the claim that coherence cannot be achieved alone.

## 1. What the Displacement Framework leaves open

*Optimal Maintenance* (Rincón, 2026) sets displacement $\xi \geq 0$ as distance from $S_g$, lets it drift toward an attractor $S^*$ at rate $\delta$, and takes the cost of being displaced as

$$D(\xi) = \tfrac{1}{2}\alpha\xi^2, \quad \alpha > 0. \tag{1}$$

This is a reasonable assumption — convex costs are standard in control theory, and the quadratic case is the simplest convex form. But it is exactly that: an assumption, imported from convention rather than derived from what $S_g$ and $\xi$ actually are. The Displacement Framework's papers are, by design, about what happens once this structure is granted. None of them ask what makes a state a *ground* state in the first place, as opposed to just a labeled reference point.

## 2. Coherence as a graph property, not a metaphor

Represent a system's internal state as a weighted graph $G = (V, E, w)$: nodes are the system's components (neurons, sentences, oscillators, people — the specific instantiation depends on the domain, exactly as in the Displacement Framework's domain papers), and edge weights $w_{ij}$ represent coupling strength between components $i$ and $j$. This is not a new formalism invented for this paper — it is the same graph Laplacian construction already implemented for discourse coherence in Clarity (`lib/spectral-grammar.ts`) and proposed for inter-brain coupling in the Phase 2 hyperscanning protocol (`phase2-hyperscanning-protocol.md`).

The graph Laplacian is $L = D - A$, where $A$ is the weighted adjacency matrix and $D$ is the diagonal degree matrix. Define coherence as the Fiedler value:

$$\chi(G) = \lambda_2(L), \tag{2}$$

the **second-smallest** eigenvalue of $L$ — the Fiedler value, the standard measure of algebraic connectivity in spectral graph theory. (Corrected 2026-07-29: this read "the smallest non-zero eigenvalue", which contradicts §5. Two disconnected triangles have Laplacian spectrum $[0,0,3,3,3,3]$: the second-smallest is $0$, which is what §5 requires, while the smallest non-zero is $3$. The Fiedler value is the one that vanishes exactly when the graph disconnects, and that property is what the dyadic argument rests on.) A graph with $\chi = 0$ is disconnected; larger $\chi$ means the graph resists being split into weakly-connected pieces. This is the same $\chi$ already computed by Clarity for sentence graphs and proposed for neural graphs in Phase 2. Nothing here is domain-specific; the graph's nodes and edges change across applications, exactly as the Displacement Framework's $S_g$, $S^*$, and $\xi$ change across its 21 domain papers, but $\chi$ itself does not.

## 3. The ground state, defined rather than assumed

For a system with a fixed total edge-weight budget $W = \sum_{i<j} w_{ij}$ — a fixed amount of coupling capacity to allocate — define the ground state as the configuration that maximizes coherence subject to that budget:

$$S_g = \arg\max_{G : \sum w_{ij} = W} \chi(G). \tag{3}$$

This gives $S_g$ actual content: it is not a labeled target a system is presumed to want, it is the specific, computable graph configuration that makes the system maximally resistant to fragmentation for the coupling resources it has. A displaced state $S$ is, on this definition, simply a graph with the same node set and budget but a different, lower-$\chi$ configuration.

## 4. Recovering $D(\xi) = \tfrac{1}{2}\alpha\xi^2$ as a theorem

Let $\xi$ parameterize a one-parameter family of perturbations away from $S_g$ along some direction in the space of graphs with fixed budget $W$, and write $\chi(\xi) = \chi(G(\xi))$ with $G(0) = S_g$. Because $S_g$ is defined as a maximizer of $\chi$ (Eq. 3), the first-order condition holds:

$$\left.\frac{d\chi}{d\xi}\right|_{\xi=0} = 0. \tag{4}$$

Taylor-expanding coherence loss around the ground state:

$$\chi(0) - \chi(\xi) = -\tfrac{1}{2}\chi''(0)\,\xi^2 + O(\xi^3). \tag{5}$$

The linear term vanishes by Eq. 4 — this is not an assumption, it is a direct consequence of $S_g$ being a true maximum, not just a reference point. Since $S_g$ is a maximum (not a saddle or minimum), $\chi''(0) \leq 0$, so coherence loss is a non-negative quadratic form near $S_g$ to leading order. Identifying displacement cost with coherence loss, $D(\xi) \equiv \chi(0) - \chi(\xi)$, and setting $\alpha = -\chi''(0)$:

$$D(\xi) = \tfrac{1}{2}\alpha\xi^2 + O(\xi^3). \tag{6}$$

This is Eq. 1 from *Optimal Maintenance*, recovered rather than assumed — **provided $\chi$ is simple at $S_g$.**

### 4.1 The hypothesis the derivation needs, and where it fails

Eq. 4 requires $\chi$ to be differentiable at $S_g$, and Eq. 5 requires it twice differentiable. An eigenvalue of a symmetric matrix is differentiable at a point only where it is **simple**. Where $\lambda_2$ is degenerate, perturbation splits the eigenspace and $\lambda_2$ follows the lowest branch — a minimum of several smooth functions, which has a kink. There is then no first-order term to vanish, and the leading behaviour is linear.

This is not a remote edge case. It is the unconstrained maximiser. For the complete graph $K_n$ with equal weights — which maximises $\chi$ for a fixed total edge-weight budget, confirmed here against 4{,}000 random equal-budget configurations — the spectrum is $\{0, n^{(n-1)}\}$, so $\lambda_2$ has multiplicity $n-1$. Perturbing $K_5$ along budget-preserving directions with all weights non-negative:

| $\xi$ | $\chi(0)-\chi(\xi)$ | $d/\xi$ | $d/\xi^2$ |
|---|---|---|---|
| 0.2000 | 0.512545 | 2.5627 | 12.81 |
| 0.1000 | 0.256273 | 2.5627 | 25.63 |
| 0.0250 | 0.064068 | 2.5627 | 102.51 |
| 0.0125 | 0.032034 | 2.5627 | 205.02 |

$d/\xi$ is constant and $d/\xi^2$ doubles with each halving of $\xi$: the cost is **linear**, not quadratic. Six independent perturbation directions give the same result, as do $n = 4, 6, 8$.

Where $\lambda_2$ **is** simple the derivation goes through exactly as stated. Optimising edge weights on a path graph $P_5$ (whose $\lambda_2$ is simple) to its constrained maximiser and perturbing gives $d/\xi^2 = 0.361$ constant with $d/\xi \to 0$ — quadratic, as Eq. 6 claims.

So the correct statement is conditional: **the quadratic cost is the necessary leading-order behaviour at a coherence maximum where $\chi$ is simple.** Symmetry is what breaks it — the more symmetric the maximiser, the more degenerate $\lambda_2$, and the unconstrained case is maximally symmetric. Real systems whose topology is constrained by anatomy, task or history generally have simple $\lambda_2$, which is why the assumption has served the domain papers; but it is an assumption about the system, not a theorem about all of them.

> **Correction, 2026-07-29.** This section previously asserted the quadratic without qualification, calling it "the generic, unavoidable leading-order behavior of any system whose ground state is genuinely a coherence maximum". That is false at the framework's own canonical ground state, and the numbers above are the demonstration. The hypothesis was missing rather than the theorem being wrong.

## 5. The dyadic case: why coherence is not decomposable

Consider two systems, each with its own graph $G_A$ and $G_B$, now coupled by a set of cross-edges $E_{AB}$ with weights $w_{AB}$ — the joint system studied throughout this research program's coupling claims (*From Grammar to Coherence*'s dyadic-eigenspectrum consciousness argument; the Phase 2 hyperscanning protocol's joint-Laplacian metric; redtooth's live phase-coupling). Form the joint graph $G_{AB}$ on the combined node set, with the joint Laplacian

$$L_{AB} = \begin{pmatrix} L_A + D_{AB} & -W_{AB} \\ -W_{AB}^\top & L_B + D_{AB}^\top \end{pmatrix}, \tag{7}$$

where $W_{AB}$ is the cross-coupling weight matrix and $D_{AB}$ its induced degree correction. The joint coherence is $\chi(G_{AB}) = \lambda_1(L_{AB})$.

The claim to establish is that $\chi(G_{AB})$ is not, in general, any fixed function of $\chi(G_A)$ and $\chi(G_B)$ alone — that joint coherence carries information the two individual coherences cannot supply.

Take the test vector that is constant on each side and orthogonal to $\mathbf{1}$: $x_i = n_B$ for $i \in A$, $x_i = -n_A$ for $i \in B$. Only cross edges contribute to $x^\top L_{AB}\, x$, since $x$ is constant within each side. Evaluating the Rayleigh quotient gives

$$\chi(G_{AB}) \;\le\; \frac{(n_A + n_B)\, W_{\text{cross}}}{n_A\, n_B}, \qquad W_{\text{cross}} = \sum_{i \in A,\, j \in B} w_{ij}. \tag{8}$$

Joint coherence is bounded **above** by total cross-coupling, scaled by the partition sizes. When $W_{\text{cross}} = 0$ the bound is zero, so $\chi(G_{AB}) = 0$ regardless of how coherent $A$ and $B$ are individually: the joint graph is disconnected across the $A$–$B$ boundary and no amount of internal coherence on either side can compensate. Two maximally coherent, entirely uncoupled systems have zero joint coherence. That is the formal content behind "coherence cannot be achieved alone" (`about/page.tsx`) and the dyadic-eigenspectrum claim in *From Grammar to Coherence* — a theorem about graph Laplacians rather than a philosophical position.

Non-decomposability follows by counterexample, which is stronger than an inequality here because it exhibits the failure directly. Take $A$ and $B$ both triangles, so $\chi(G_A) = \chi(G_B) = 3$ throughout, and vary a single cross edge:

| $W_{\text{cross}}$ | $\chi(G_A)$ | $\chi(G_B)$ | $\chi(G_{AB})$ |
|---|---|---|---|
| 0.1 | 3.000 | 3.000 | 0.064 |
| 1.0 | 3.000 | 3.000 | 0.438 |
| 5.0 | 3.000 | 3.000 | 0.821 |

The individual coherences are identical in every row and the joint coherence is not. No function of $\chi(G_A)$ and $\chi(G_B)$ alone can reproduce the third column.

> **Correction, 2026-07-29.** Eq. 8 previously read $\lambda_1(L_{AB}) \ge \min(\chi(G_A), \chi(G_B))$, "with equality only when $W_{AB} = 0$." That is false, and false in the direction that matters: at $W_{AB} = 0$ the joint coherence is $0$ while $\min(\chi_A, \chi_B)$ is positive, so the boundary case the paragraph relies on is the inequality's maximal violation rather than its equality case. Checked numerically — two triangles, one cross edge — the old form fails at every coupling strength up to full connection. The prose was right and the formula was wrong; the bound above is the correct statement and carries the same conclusion. Verified on 400 random graph pairs with no violation. This also changes §6.2, which had inherited the wrong threshold.

## 6. Falsifiable predictions

Following the Displacement Framework's own stated practice (`bifurcation_analysis.pdf`: "propose falsifiable experiments"), this framework makes testable claims distinct from the displacement papers':

1. **A linear leading-order cost distinguishes two failures, and the multiplicity of $\lambda_2$ tells them apart.** Fitting the cost curve near a claimed $S_g$ and finding it linear rather than quadratic means either (a) the reference state is not a coherence maximum at all — it was mislabeled — or (b) it is a genuine maximum at which $\chi$ is degenerate, so §4.1 applies and linearity is *predicted*. These are distinguishable without any new data: compute the multiplicity of $\lambda_2$ at the claimed $S_g$. Simple and linear falsifies the labelling; degenerate and linear confirms §4.1; simple and quadratic confirms Eq. 6. Only *degenerate and quadratic* is unaccounted for, and observing it would falsify §4.1.

   This is a sharper test than the original, which read a linear cost as evidence of mislabelling alone and would therefore have mis-diagnosed the framework's own canonical ground state.
2. **Eq. 8 is directly testable on the ds007471 hyperscanning data already analyzed** (`ds007471-hyperscanning-note.md`), as a **comparison between conditions rather than against an absolute threshold**: real dyads should show higher joint-Laplacian coherence than the same people paired with strangers, because shuffling destroys $W_{\text{cross}}$ while leaving both individual coherences untouched. This is the test that was actually run, and it came back positive — $p = 0.0012$ paired, 25 of 29 real pairs showing a positive gap.

   The earlier wording asked whether joint coherence *exceeds the minimum of the two individual coherences*. That threshold does not follow from Eq. 8 and should not be used: inter-brain coupling is weak relative to within-brain coupling, so genuinely coupled dyads sit far below $\min(\chi_A, \chi_B)$ and would fail an absolute test while passing the comparative one. A framework should not be able to fail a test its own mathematics predicts it will fail.
3. **The Phase 2 hyperscanning protocol's primary metric (joint graph Laplacian eigenspectrum) is a direct empirical instance of Eq. 7–8** — this paper supplies its formal justification; that protocol supplies its test.

## 7. What this does and does not claim

This paper derives one specific thing: that a widely-used modeling assumption in the Displacement Framework (quadratic cost near the ground state) follows from a more primitive definition of coherence as spectral connectivity, and that the qualitative claim "coherence requires coupling" has a precise, checkable form. It does not claim that every system's ground state is in fact a coherence maximum in this technical sense — that remains an empirical question, domain by domain, exactly as the Displacement Framework's own papers treat their claims as falsifiable per-domain rather than assumed globally true.

## References

Rincón, D. (2026). *Displacement Framework — Domain Applications.* Zenodo. https://doi.org/10.5281/zenodo.20675507

Rincón, D. (2026). *From Grammar to Coherence: How Symbolic AI Produces Displacement States.* Zenodo. https://doi.org/10.5281/zenodo.21403447

Phronesis Research (2026). *Phase 2: Inter-Brain Coupling and the Dyadic-Coherence Claim.* `research/phase2-hyperscanning-protocol.md`.

Phronesis Research (2026). *Joint agency EEG hyperscanning analysis.* `research/ds007471-hyperscanning-note.md`.