# RCHM × Hodge Conjecture — Research Round 3
## Relational Spectrum, Constraint Incidence Geometry, and Symmetry-Pinned Couplings

**Status:** Internal research note  
**Version:** Round 3 / v0.3  
**Date:** 2026-09-17  
**Methodological base:** Relational Constraint–Handoff Methodology (RCHM) v0.1  
**Previous rounds:** RCHM × Hodge Conjecture — Rounds 1–2

---

# 0. Round 3 Objective

Round 2 showed that constraint conjunction and constraint superposition are different operations.

For two obstruction channels

$$
\Phi_1,\Phi_2:V\to Q,
$$

the joint legal domain is

$$
\ker\Phi_1\cap\ker\Phi_2,
$$

whereas a superposed relation with coefficient $\lambda$ has legal domain

$$
\ker(\Phi_1+\lambda\Phi_2).
$$

Round 3 asks:

> Can the coupling coefficient itself be treated as a geometric variable, and can all legal relations be classified simultaneously?

The answer is yes.

The central construction of this round is the **constraint incidence variety**:

$$
\boxed{
\mathfrak I
=
\left\{
(v,[a]):
\sum_i a_i\Phi_i(v)=0
\right\}.
}
$$

This object simultaneously records:

1. which deformation states are legal for a fixed relation;
2. which relations legalize a fixed deformation state;
3. where relation-generated freedom jumps;
4. where exceptional coupling coefficients occur.

---

# 1. Coefficients Live in Projective Space

Let

$$
\gamma_1,\ldots,\gamma_m
$$

be classes or constraint channels.

A nonzero coefficient vector

$$
a=(a_1,\ldots,a_m)
$$

defines a combined object

$$
\gamma_a
=
\sum_{i=1}^m a_i\gamma_i.
$$

Scaling all coefficients by the same nonzero scalar does not change the condition that $\gamma_a$ be a Hodge class.

Therefore the natural coupling space is not affine space but

$$
\boxed{
\mathbb P^{m-1}.
}
$$

For $m=2$,

$$
[a:b]\in\mathbb P^1.
$$

On the affine chart $a\neq0$, write

$$
\lambda=\frac{b}{a}
$$

and therefore

$$
\gamma_\lambda
=
\gamma_1+\lambda\gamma_2.
$$

The points

$$
[1:0]
$$

and

$$
[0:1]
$$

represent the two pure, unmixed channels.

All other points are genuinely mixed relations.

---

# 2. Constraint Pencil

For two first-order obstruction maps

$$
\Phi_1,\Phi_2:V\to Q,
$$

define the constraint pencil

$$
\boxed{
\Psi(\lambda)
=
\Phi_1+\lambda\Phi_2.
}
$$

For fixed $\lambda$, the legal infinitesimal domain is

$$
L_\lambda
=
\ker\Psi(\lambda).
$$

The joint domain is

$$
L_{\mathrm{joint}}
=
\ker\Phi_1\cap\ker\Phi_2.
$$

Always,

$$
L_{\mathrm{joint}}
\subseteq
L_\lambda.
$$

Define the excess relation space

$$
\boxed{
E_\lambda
=
L_\lambda/L_{\mathrm{joint}}.
}
$$

Then:

$$
E_\lambda=0
$$

means the relation generates no additional legal direction, while

$$
E_\lambda\neq0
$$

means some legal directions exist only because of relational cancellation.

---

# 3. Relational Spectrum

Suppose

$$
\dim V<\infty,
\qquad
\dim Q<\infty.
$$

Choose bases so that

$$
\Psi(\lambda)
$$

is represented by a matrix pencil

$$
A+\lambda B.
$$

Let

$$
r_{\mathrm{gen}}
=
\max_{\lambda}
\operatorname{rank}(A+\lambda B).
$$

Define the first-order relational spectrum

$$
\boxed{
\Lambda_{\mathrm{tan}}
=
\left\{
[\lambda]\in\mathbb P^1:
\operatorname{rank}(A+\lambda B)
<
r_{\mathrm{gen}}
\right\}.
}
$$

Equivalently, $\Lambda_{\mathrm{tan}}$ is cut out by the simultaneous vanishing of the maximal minors associated with the generic rank.

Thus it is a determinantal discriminant.

For a generic one-parameter pencil, this set is finite unless the maximal minors vanish identically.

This gives a precise meaning to the phrase:

$$
\boxed{
\text{exceptional coupling coefficient}.
}
$$

It is a spectral point of the relation pencil.

---

# 4. Kloosterman 2025 as an Exact Calibration

Consider a smooth hypersurface

$$
X\subset\mathbb P^{2k+1}
$$

containing two $k$-planes

$$
\Pi_1,\Pi_2.
$$

Let

$$
c
=
k-\dim(\Pi_1\cap\Pi_2).
$$

Kloosterman studies

$$
\operatorname{NL}([\Pi_1]+\lambda[\Pi_2])
$$

versus

$$
\operatorname{NL}([\Pi_1],[\Pi_2]).
$$

For suitable constructed hypersurfaces and sufficiently large $k$, the generic mixed coefficient satisfies

$$
\boxed{
\operatorname{NL}([\Pi_1]+\lambda[\Pi_2])
=
\operatorname{NL}([\Pi_1],[\Pi_2])
}
$$

locally.

But the nonzero exceptional value can be identified exactly in two boundary regimes.

---

## 4.1 Cubic, codimension-three intersection

For

$$
d=3,
\qquad
c=3,
$$

Kloosterman's Theorem 1.2 gives

$$
S=\{0,1\}.
$$

Thus for the constructed examples, among nonzero mixed coefficients, the unique exceptional value is

$$
\boxed{
\lambda_*=1.
}
$$

At this value the special relation is

$$
[\Pi_1]+[\Pi_2].
$$

Section 6 constructs a new geometric carrier: a linear section of an iterated cone over the Plücker embedding

$$
G(2,5)\hookrightarrow\mathbb P^9.
$$

The limiting class satisfies

$$
[Y_0]
=
[\Pi_1]+[\Pi_2]+h^k,
$$

so the primitive class is the same as the primitive sum of the two planes.

Thus the exceptional spectral point carries a witness-transmutation branch.

---

## 4.2 Quartic, codimension-two intersection

For

$$
d=4,
\qquad
c=2,
$$

Kloosterman's Theorem 1.2 gives

$$
S=\{0,-1\}.
$$

Hence the unique nonzero mixed exceptional value is

$$
\boxed{
\lambda_*=-1.
}
$$

The special class is

$$
[\Pi_1]-[\Pi_2].
$$

A flat family of witnesses exists whose general member is a complete intersection of multidegree

$$
(2,2,1,\ldots,1),
$$

while

$$
[Y_0]_{\mathrm{prim}}
=
[\Pi_1]_{\mathrm{prim}}
-
[\Pi_2]_{\mathrm{prim}}.
$$

Again:

$$
\boxed{
\text{spectral exception}
\longleftrightarrow
\text{new witness carrier}.
}
$$

This correspondence is established in these examples, not asserted as a universal theorem.

---

# 5. Projective Completion of the Exceptional Set

The affine coordinate $\lambda$ misses the point

$$
\lambda=\infty,
$$

which corresponds to the pure class

$$
[\Pi_2].
$$

Thus in coefficient space

$$
\mathbb P^1,
$$

the trivial unmixed points are

$$
[1:0],
\qquad
[0:1].
$$

For the two calibrated cases the mixed special point is respectively

$$
[1:1]
$$

or

$$
[1:-1].
$$

So the projective coefficient line separates naturally into:

1. pure endpoints;
2. a generic mixed stratum;
3. a symmetry-selected exceptional mixed point.

---

# 6. Symmetry-Pinning Proposition

Suppose the base geometry admits an involution

$$
\sigma
$$

such that

$$
\sigma(\gamma_1)=\gamma_2,
\qquad
\sigma(\gamma_2)=\gamma_1.
$$

On coefficient space,

$$
\sigma:[a:b]\mapsto[b:a].
$$

In the affine coordinate

$$
\lambda=\frac{b}{a},
$$

this becomes

$$
\boxed{
\lambda\mapsto\lambda^{-1}.
}
$$

Assume:

1. the exceptional mixed set is invariant under $\sigma$;
2. there is exactly one exceptional mixed point.

Then this point must be fixed by inversion:

$$
\lambda_*=\lambda_*^{-1}.
$$

Hence

$$
\lambda_*^2=1,
$$

so

$$
\boxed{
\lambda_*\in\{+1,-1\}.
}
$$

This gives a simple RCHM explanation for why a unique symmetry-pinned mixed exceptional coefficient must occur on the symmetric or antisymmetric eigenline.

It does **not** determine which sign occurs.

The actual geometry selects the sign.

In the calibrated Hodge examples:

$$
(d,c)=(3,3)
\quad\Rightarrow\quad
\lambda_*=+1,
$$

whereas

$$
(d,c)=(4,2)
\quad\Rightarrow\quad
\lambda_*=-1.
$$

---

# 7. Symmetric and Antisymmetric Relation Modes

Under the swap involution, define

$$
\gamma_+
=
\gamma_1+\gamma_2,
$$

$$
\gamma_-
=
\gamma_1-\gamma_2.
$$

Then

$$
\sigma(\gamma_+)=+\gamma_+,
$$

$$
\sigma(\gamma_-)=-\gamma_-.
$$

Therefore the two fixed projective coupling points are precisely the two representation-theoretic eigenmodes of the relation.

This suggests the general heuristic:

$$
\boxed{
\text{symmetry can pin exceptional relational phases to irreducible symmetry modes}.
}
$$

For more than two constraints, the natural extension is to decompose the coefficient representation under the symmetry group acting on the constraint channels.

This is a research direction, not yet a theorem of the present framework.

---

# 8. Base-Point Dependence of the Spectrum

The exceptional set is not determined by coupling coefficients alone.

Let

$$
\mathcal M
$$

be the moduli/deformation space of geometric backgrounds $X$.

For each

$$
X\in\mathcal M,
$$

there is a constraint pencil

$$
\Psi_X(a).
$$

Define the total first-order relational discriminant

$$
\boxed{
\mathcal D_{\mathrm{tan}}
=
\left\{
(X,[a])\in
\mathcal M\times\mathbb P^{m-1}:
\operatorname{rank}\Psi_X(a)
<
r_{\mathrm{gen}}
\right\}.
}
$$

The fiber over $X$ is

$$
\Lambda_{\mathrm{tan}}(X)
=
\left\{
[a]:
(X,[a])\in\mathcal D_{\mathrm{tan}}
\right\}.
$$

Thus:

$$
\boxed{
\text{Relational spectrum is fibered over geometry.}
}
$$

A coefficient may be generic on one background and exceptional on another.

This explains why Fermat or split hypersurfaces can exhibit larger tangent spaces, reducibility, singularity or nonreduced behavior even when a very general point on the same joint Hodge locus behaves differently.

---

# 9. Evidence from Codimension-One Intersections

Kloosterman's 2025 codimension-one paper considers

$$
\dim(\Pi_1\cap\Pi_2)=k-1.
$$

For $k>1$ and $X$ very general on the joint Hodge locus,

$$
\operatorname{NL}([\Pi_1]+\lambda[\Pi_2])
=
\operatorname{NL}([\Pi_1],[\Pi_2])
$$

locally for

$$
\lambda\neq0,1.
$$

However, the paper also constructs a hypersurface where the joint locus is smooth while

$$
\operatorname{NL}([\Pi_1]+\lambda[\Pi_2])
$$

is singular for all

$$
\lambda\neq0,1,
$$

with an embedded component suggested as the explanation.

Therefore the same coefficient region can exhibit different scheme structure at different base points.

This requires at least two coordinates in the phase diagram:

$$
\boxed{
(X,[a]).
}
$$

---

# 10. Constraint Incidence Variety

Now move from coefficient-first analysis to a symmetric formulation.

Let

$$
V
$$

be the state/deformation space,

$$
Q
$$

the obstruction target,

and

$$
\Phi_i:V\to Q,
\qquad
i=1,\ldots,m.
$$

For each state $v\in V$, define the obstruction synthesis map

$$
\boxed{
\mathcal O_v:\mathbb K^m\to Q
}
$$

by

$$
\mathcal O_v(a_1,\ldots,a_m)
=
\sum_{i=1}^m a_i\Phi_i(v).
$$

Define the constraint incidence variety

$$
\boxed{
\mathfrak I
=
\left\{
(v,[a])
\in
V\times\mathbb P^{m-1}:
\mathcal O_v(a)=0
\right\}.
}
$$

This is the central construction of Round 3.

---

# 11. Two Projections, Two Interpretations

There are two natural projections.

## 11.1 Fix the relation

Project to coefficient space:

$$
\pi_A:
\mathfrak I
\to
\mathbb P^{m-1}.
$$

The fiber over $[a]$ is

$$
\pi_A^{-1}([a])
=
\left\{
v:
\sum_i a_i\Phi_i(v)=0
\right\}.
$$

Therefore

$$
\boxed{
\pi_A^{-1}([a])
=
\ker\Psi_a.
}
$$

This is the legal state domain for the chosen relation.

---

## 11.2 Fix the state

Project to state space:

$$
\pi_V:
\mathfrak I
\to
V.
$$

The fiber over $v$ is

$$
\pi_V^{-1}(v)
=
\mathbb P(\ker\mathcal O_v).
$$

Thus:

$$
\boxed{
\text{all relations that legalize }v
=
\mathbb P(\ker\mathcal O_v).
}
$$

This answers a fundamental RCHM question:

> Given a state that is individually illegal under several channels, do there exist coupling relations that make it legal?

The answer is determined exactly by the nullspace of $\mathcal O_v$.

---

# 12. Relational Nullspace Duality

The same incidence geometry can therefore be read in two directions:

$$
[a]
\longmapsto
\ker\Psi_a
$$

and

$$
v
\longmapsto
\mathbb P(\ker\mathcal O_v).
$$

This gives the Round 3 principle:

$$
\boxed{
\textbf{Relational Nullspace Duality}
}
$$

with interpretation:

$$
\boxed{
\text{a relation selects legal states,
while a state selects legal relations}.
}
$$

The coupling is not merely an external parameter.

It is part of the legality geometry.

---

# 13. Exact Two-Constraint Rank Classification

For

$$
m=2,
$$

write

$$
\mathcal O_v
=
\begin{bmatrix}
\Phi_1(v) & \Phi_2(v)
\end{bmatrix}.
$$

There are three qualitatively different cases.

---

## 13.1 Rank zero — universal relational legality

If

$$
\Phi_1(v)=0,
\qquad
\Phi_2(v)=0,
$$

then

$$
\operatorname{rank}\mathcal O_v=0.
$$

Hence

$$
\ker\mathcal O_v
=
\mathbb K^2
$$

and

$$
\pi_V^{-1}(v)
=
\mathbb P^1.
$$

Every coefficient relation is legal.

This is ordinary joint legality.

---

## 13.2 Rank one — relation-only legality

Suppose

$$
\operatorname{rank}\mathcal O_v=1.
$$

Then

$$
\dim\ker\mathcal O_v=1
$$

and therefore

$$
\pi_V^{-1}(v)
=
\mathbb P^0.
$$

There is exactly one projective coupling relation that legalizes $v$.

It is possible simultaneously that

$$
\Phi_1(v)\neq0,
\qquad
\Phi_2(v)\neq0.
$$

Then neither constraint is satisfied separately, but one unique relation cancels the two obstructions.

This is the purest form of **relationally generated legality**.

If

$$
\Phi_2(v)=\rho(v)\Phi_1(v)
$$

with both nonzero, then the legal coefficient satisfies

$$
a_1+\rho(v)a_2=0.
$$

On the chart $a_1=1$,

$$
\boxed{
\lambda(v)
=
-\frac{1}{\rho(v)}.
}
$$

This defines a local **compensation map** from cancellable states to coupling space.

---

## 13.3 Rank two — no legal relation

If

$$
\operatorname{rank}\mathcal O_v=2,
$$

then

$$
\ker\mathcal O_v=0.
$$

Thus

$$
\pi_V^{-1}(v)=\varnothing.
$$

No nonzero linear combination of the two constraints can legalize $v$.

This is a stronger form of incompatibility than merely failing each constraint separately.

---

# 14. Cancellable-State Locus

For two vector-valued obstruction channels, cancellation is possible only when

$$
\Phi_1(v)
$$

and

$$
\Phi_2(v)
$$

are linearly dependent in $Q$.

Define

$$
\boxed{
\mathcal K
=
\left\{
v\in V:
\operatorname{rank}
\begin{bmatrix}
\Phi_1(v)&\Phi_2(v)
\end{bmatrix}
\le1
\right\}.
}
$$

Then:

- rank $0$ points are jointly legal;
- rank $1$ points are relation-only legal;
- states outside $\mathcal K$ cannot be legalized by any nonzero two-channel superposition.

Thus the decomposition is

$$
\boxed{
V
=
V_{\mathrm{joint}}
\sqcup
V_{\mathrm{cancel}}
\sqcup
V_{\mathrm{incompatible}},
}
$$

where

$$
V_{\mathrm{joint}}
=
\left\{
v:
\operatorname{rank}\mathcal O_v=0
\right\},
$$

$$
V_{\mathrm{cancel}}
=
\left\{
v:
\operatorname{rank}\mathcal O_v=1
\right\},
$$

and

$$
V_{\mathrm{incompatible}}
=
\left\{
v:
\operatorname{rank}\mathcal O_v=2
\right\}.
$$

This provides a rigorous three-state replacement for the vague binary phrase “intersects / does not intersect”.

---

# 15. General $m$-Constraint Law

For

$$
m
$$

constraint channels, let

$$
r(v)
=
\operatorname{rank}\mathcal O_v.
$$

Then

$$
\dim\ker\mathcal O_v
=
m-r(v).
$$

Therefore the projective space of legal coupling relations has dimension

$$
\boxed{
\ell(v)
=
m-r(v)-1
}
$$

whenever the kernel is nonzero.

Interpretation:

$$
\ell(v)
$$

measures the number of independent relational freedoms capable of legalizing the state.

Thus:

$$
r(v)=m
$$

means no relation works;

$$
r(v)=m-1
$$

means a unique projective relation works;

smaller rank means a family of relations works.

This is a general relational-legality counting law.

---

# 16. Fiber-Dimension Jumps as Relational Phase Transitions

Consider again

$$
\pi_A:
\mathfrak I\to\mathbb P^{m-1}.
$$

Generically the fiber has some dimension

$$
d_{\mathrm{gen}}.
$$

Define the jump locus

$$
\boxed{
\mathcal J
=
\left\{
[a]:
\dim\pi_A^{-1}([a])
>
d_{\mathrm{gen}}
\right\}.
}
$$

This is the coefficient-space locus where a coupling suddenly legalizes extra state directions.

At first order, this is the relational spectrum.

Therefore:

$$
\boxed{
\text{exceptional coupling}
=
\text{fiber-dimension jump}.
}
$$

This gives a geometric interpretation of rank drop without depending on coordinates.

---

# 17. Three Different Spectra Must Be Separated

Hodge-locus examples show that one spectrum is not enough.

Define three diagnostic sets.

## 17.1 Tangent spectrum

$$
\Lambda_{\mathrm{tan}}
$$

contains coefficients where tangent dimension jumps.

---

## 17.2 Reduced-carrier spectrum

$$
\Lambda_{\mathrm{red}}
$$

contains coefficients where the reduced Hodge locus acquires additional geometric branches or carriers.

---

## 17.3 Scheme spectrum

$$
\Lambda_{\mathrm{sch}}
$$

contains coefficients where scheme structure changes, for example through:

- nonreduced structure;
- embedded components;
- multiplicity changes;
- singularity without reduced enlargement.

These sets need not coincide.

A tangent enlargement may arise from:

- a genuinely larger reduced component;
- intersection of several components;
- nonreduced thickening;
- embedded structure.

Therefore RCHM must record separately:

$$
\boxed{
\text{tangent phase},
\quad
\text{carrier phase},
\quad
\text{scheme phase}.
}
$$

---

# 18. Split Hypersurfaces as a Warning

In split hypersurfaces studied by Kloosterman, for the regimes

$$
(c,d)\in
\{(2,3),(3,3),(2,4)\},
$$

one has locally

$$
\operatorname{NL}([\Pi_1],[\Pi_2])_{\mathrm{red}}
\subsetneq
\operatorname{NL}([\Pi_1]+\lambda[\Pi_2])_{\mathrm{red}}
$$

for every nonzero

$$
\lambda.
$$

For sufficiently large $k$ in the cubic and quartic cases, the combined Hodge locus is reducible.

Thus an exceptional **background geometry** can make an entire coefficient region behave non-generically.

Hence the most faithful phase space is not merely

$$
\mathbb P^1,
$$

but

$$
\boxed{
\mathcal M\times\mathbb P^1.
}
$$

---

# 19. Revised Meaning of “Two Infinities Intersect”

The original motivating question can now be sharpened.

Suppose two large or unbounded domains generate obstruction channels

$$
\Phi_1,\Phi_2.
$$

Instead of asking only whether

$$
A\cap B
$$

is nonempty, ask:

1. what is the joint legal kernel?
2. what is the rank of the obstruction pair at each state?
3. does a null relation exist?
4. how many coupling relations legalize the state?
5. which coefficient values enlarge the legal state fiber?
6. does the enlarged fiber correspond to a new carrier, a singular intersection, or a nonreduced thickening?

The relevant object is therefore not merely an intersection but the incidence geometry

$$
\boxed{
\mathfrak I
=
\left\{
(v,[a]):
\mathcal O_v(a)=0
\right\}.
}
$$

This is the main conceptual closure of Round 3.

---

# 20. RCHM Upgrade Proposed by Round 3

The minimal RCHM working object should be enlarged from

$$
\mathfrak J
=
(A,B,C,f,g,R,\Gamma,\Sigma,\mathcal E,W)
$$

to include a coefficient/modulation space

$$
\mathcal P
$$

and an obstruction synthesis operator

$$
\mathcal O.
$$

Thus the extended object is

$$
\boxed{
\mathfrak J^+
=
(A_i,C,R,\Gamma,\Sigma,\mathcal E,W,\mathcal P,\mathcal O).
}
$$

Here

$$
\mathcal P
=
\mathbb P^{m-1}
$$

in a linear superposition problem.

The key derived objects are:

$$
\mathfrak I,
$$

$$
\mathcal D_{\mathrm{tan}},
$$

$$
\Lambda_{\mathrm{tan}}(X),
$$

$$
\mathbb P(\ker\mathcal O_v),
$$

and the carrier/scheme spectra.

---

# 21. New Principles

## Principle 1 — Projective Coupling Principle

If only the relative coefficients of a relation matter, the natural coupling domain is projective:

$$
\boxed{
a\sim ca,
\qquad
c\neq0.
}
$$

---

## Principle 2 — Relational Nullspace Principle

For a fixed state $v$, legal relations are exactly

$$
\boxed{
\mathbb P(\ker\mathcal O_v).
}
$$

---

## Principle 3 — Compensation Principle

A state can violate every component constraint individually and still satisfy one unique relation if its obstruction vectors are linearly dependent.

For two channels:

$$
\boxed{
\operatorname{rank}\mathcal O_v=1
}
$$

is the canonical relation-only legality state.

---

## Principle 4 — Spectral Jump Principle

Exceptional couplings are coefficient values where the legal-state fiber jumps:

$$
\boxed{
\dim\ker\Psi_a
>
\dim\ker\Psi_{\mathrm{generic}}.
}
$$

---

## Principle 5 — Geometry-Fibered Spectrum Principle

The relational spectrum depends on the background:

$$
\boxed{
\Lambda=\Lambda(X).
}
$$

Thus the global object is a discriminant in

$$
\mathcal M\times\mathcal P.
$$

---

## Principle 6 — Symmetry Pinning

If a swap symmetry exists and there is a unique mixed exceptional coupling, it must lie on a symmetry eigenline:

$$
\boxed{
\lambda=\pm1.
}
$$

---

# 22. Immediate Research Directions

Round 3 opens several concrete directions.

### Direction A — Compute the full discriminant

For explicit hypersurfaces, compute maximal minors of

$$
\Psi_X(\lambda)
$$

and obtain

$$
\Lambda_{\mathrm{tan}}(X)
$$

directly.

### Direction B — Compare spectra

Determine when

$$
\Lambda_{\mathrm{tan}},
\quad
\Lambda_{\mathrm{red}},
\quad
\Lambda_{\mathrm{sch}}
$$

agree or differ.

### Direction C — Multi-class extension

For

$$
m\ge3,
$$

study the determinantal locus

$$
\mathcal D
\subset
\mathbb P^{m-1}.
$$

Unlike the two-class case, it may have positive dimension.

### Direction D — Symmetry decomposition

If a group $G$ permutes the witness classes, decompose coefficient space into irreducible $G$-representations and test whether exceptional loci concentrate on symmetry modes.

### Direction E — Carrier map

Construct a partial map

$$
\operatorname{Carrier}:
\mathcal D_{\mathrm{red}}
\dashrightarrow
\{\text{geometric witness types}\}.
$$

This would convert a spectral point into a candidate witness geometry.

---

# 23. Round 3 Conclusion

The main result of Round 3 is that relational legality can be represented as an incidence geometry rather than a binary intersection.

The universal object is

$$
\boxed{
\mathfrak I
=
\left\{
(v,[a]):
\sum_i a_i\Phi_i(v)=0
\right\}.
}
$$

Its two fibers answer dual questions:

$$
\boxed{
[a]
\mapsto
\text{states legalized by the relation},
}
$$

and

$$
\boxed{
v
\mapsto
\text{relations capable of legalizing the state}.
}
$$

For two constraints the rank trichotomy is exact:

$$
\boxed{
\begin{array}{ccl}
\operatorname{rank}\mathcal O_v=0
&\Rightarrow&
\text{all relations legal},\\[1mm]
\operatorname{rank}\mathcal O_v=1
&\Rightarrow&
\text{exactly one projective relation legal},\\[1mm]
\operatorname{rank}\mathcal O_v=2
&\Rightarrow&
\text{no relation legal}.
\end{array}
}
$$

This provides a mathematically precise answer to the motivating question:

> If two domains are each “infinite” or individually incompatible, what can a legal relation between them generate?

It can generate a new legal state precisely when their obstruction vectors acquire a nontrivial null relation.

The coupling coefficients themselves form a geometric domain, and exceptional relations occur on its determinantal spectrum.

In Hodge-locus examples, the nonzero exceptional spectral points

$$
\lambda=+1
$$

and

$$
\lambda=-1
$$

are accompanied by concrete changes of witness carrier.

Thus the next-stage research object is no longer merely

$$
A\cap B,
$$

nor even

$$
\ker(\Phi_1+\lambda\Phi_2),
$$

but the full relation-state incidence geometry

$$
\boxed{
\mathfrak I
\subset
V\times\mathbb P^{m-1}.
}
$$

---

# References

1. Remke Kloosterman, **On a conjecture on Hodge loci of linear combinations of linear subvarieties**, Rendiconti del Circolo Matematico di Palermo Series 2, 74, Article 187 (2025).  
   DOI: https://doi.org/10.1007/s12215-025-01307-4

2. Remke Kloosterman, **Hodge loci associated with linear subspaces intersecting in codimension one**, Mathematische Nachrichten 298 (2025), 1220–1229.  
   DOI: https://doi.org/10.1002/mana.202400066  
   arXiv: https://arxiv.org/abs/2401.10775

3. Jorge Duque Franco and Roberto Villaflor Loyola, **On fake linear cycles inside Fermat varieties**, Algebra & Number Theory 17 (2023), 1847–1865.  
   DOI: https://doi.org/10.2140/ant.2023.17.1847  
   arXiv: https://arxiv.org/abs/2112.14818
