> Full-text translation from Chinese. The original remains authoritative for its research claims; this translation is not independent mathematical verification.

# ACTM Round 4
## Structural Classification of a Hypothetical MID=2 State: Projective Orbit Avoidance, Rank Splitting, and an Exact Three-Term Power Chain
### Structural Classification of a Hypothetical MID=2 State:
### Projective Orbit Avoidance, Rank Splitting, and an Exact Three-Term Power Chain

**Research program:** Algebraic Combinatorial Transcendence Methodology (ACTM)  
**Authors:** Neo.K; Aletheia (GPT-5.6 Sol)  
**Institution:** EveMissLab / 一言諾科技有限公司  
**Date:** 2026-09-30  
**Version:** Round 4 / v0.1  
**Document type:** Formal research notes / Four Exponentials hypothetical-state reduction  
**Status statement:** This round does not claim to prove the Four Exponentials Conjecture. The external core results used include Baker's theorem, the Five Exponentials Theorem, the Six Exponentials Theorem, and Roy–Waldschmidt's known result for the transcendence-degree-$1$ case. The principal work is to classify and normalize hypothetical counterexamples and extract new ACTM research interfaces.

---

## Abstract

ACTM Round 3 compressed the Four Exponentials Conjecture into a multiplier-intersection problem.

Fix:

$$
0\neq\lambda
\in
\mathcal L_{\mathrm{alg}},
$$

where:

$$
\mathcal L_{\mathrm{alg}}
=
\left\{
z\in\mathbb C:
e^z
\in
\overline{\mathbb Q}^{\times}
\right\},
$$

and define:

$$
\mathfrak M_\lambda
=
\left\{
c\in\mathbb C:
c\lambda
\in
\mathcal L_{\mathrm{alg}}
\right\}.
$$

For:

$$
b
\in
\mathfrak M_\lambda
\setminus
\mathbb Q,
$$

Define:

$$
W_\lambda(b)
=
\mathfrak M_\lambda
\cap
b^{-1}\mathfrak M_\lambda.
$$

The Six Exponentials Theorem gives:

$$
1
\le
\dim_{\mathbb Q}W_\lambda(b)
\le
2,
$$

whereas the Four Exponentials Conjecture is equivalent to:

$$
\dim_{\mathbb Q}W_\lambda(b)=1.
$$

Round 4 therefore assumes, for contradiction:

$$
\boxed{
\operatorname{MID}_\lambda(b)
=
\dim_{\mathbb Q}W_\lambda(b)
=
2.
}
$$

Take:

$$
a
\in
W_\lambda(b)
\setminus
\mathbb Q.
$$

then:

$$
\boxed{
\lambda,
\quad
a\lambda,
\quad
b\lambda,
\quad
ab\lambda
\in
\mathcal L_{\mathrm{alg}}.
}
$$

This round obtains the following necessary structure.

First, $a$ and $b$ are both transcendental, and the Five Exponentials Theorem further forces all rational Möbius transforms:

$$
g(a),
\qquad
g(b),
\qquad
g\in\operatorname{PGL}_2(\mathbb Q)
$$

to avoid $\mathcal L_{\mathrm{alg}}$ entirely:

$$
\boxed{
e^{g(a)},
\ e^{g(b)}
\text{ are both transcendental}.
}
$$

Second, the transcendence-degree-$1$ result of Roy–Waldschmidt forces:

$$
\boxed{
\operatorname{trdeg}_{\mathbb Q}
\mathbb Q(\lambda,a,b)
\ge2.
}
$$

Thus, any counterexample requires at least two genuinely independent transcendental parameters.

Third, define:

$$
\rho
=
\dim_{\mathbb Q}
\operatorname{span}_{\mathbb Q}
\{1,a,b,ab\}.
$$

then:

$$
\boxed{
\rho\in\{3,4\}.
}
$$

There is no $\rho\le2$.

If:

$$
\rho=3,
$$

there exists a rational bilinear relation unique up to scaling:

$$
r_0+r_1a+r_2b+r_3ab=0,
$$

Hence:

$$
b
=
g(a)
$$

for some:

$$
g\in\operatorname{PGL}_2(\mathbb Q).
$$

Through rational basis normalization, the entire counterexample can be put in the form:

$$
\boxed{
\lambda',
\quad
t\lambda',
\quad
t^2\lambda'
\in
\mathcal L_{\mathrm{alg}},
}
$$

where:

$$
t
$$

is transcendental.

Roy–Waldschmidt further forces:

$$
\boxed{
\operatorname{trdeg}_{\mathbb Q}
\mathbb Q(\lambda',t)=2,
}
$$

that is, $\lambda'$ and $t$ are algebraically independent.

And the Six Exponentials Theorem forces:

$$
\boxed{
t^3\lambda'
\notin
\mathcal L_{\mathrm{alg}}.
}
$$

Thus, a hypothetical counterexample of this type forms an exact truncated power chain:

$$
\boxed{
\lambda',
\ t\lambda',
\ t^2\lambda'
\in
\mathcal L_{\mathrm{alg}},
\qquad
t^3\lambda'
\notin
\mathcal L_{\mathrm{alg}}.
}
$$

If:

$$
\rho=4,
$$

Baker's theorem forces:

$$
1,a,b,ab
$$

to be linearly independent over $\overline{\mathbb Q}$ as well, so there is no nonzero algebraic bilinear relation:

$$
c_{00}
+
c_{10}a
+
c_{01}b
+
c_{11}ab
=
0.
$$

Round 4 therefore divides the hypothetical MID=$2$ state exactly into:

$$
\boxed{
\text{Type P: Projectively reducible power-chain state}
}
$$

and:

$$
\boxed{
\text{Type B: Genuinely bilinear state}.
}
$$

---

# 1. The Round 4 Assumptions

This round assumes that there exist:

$$
0\neq
\lambda
\in
\mathcal L_{\mathrm{alg}},
$$

and:

$$
a,b
\notin
\mathbb Q
$$

such that:

$$
\boxed{
\lambda,
\quad
a\lambda,
\quad
b\lambda,
\quad
ab\lambda
\in
\mathcal L_{\mathrm{alg}}.
}
$$

This is equivalent to a normalized Four Exponentials counterexample:

$$
Z
=
\lambda
\begin{pmatrix}
1&a\\
b&ab
\end{pmatrix},
$$

whose four entries are all algebraic logarithms.

At the same time:

$$
\det Z=0.
$$

---

# 2. The First Layer: $a$ and $b$ Must Be Transcendental

Round 2 proved:

$$
\mathfrak M_\lambda
\cap
\overline{\mathbb Q}
=
\mathbb Q.
$$

And:

$$
a,b
\in
\mathfrak M_\lambda.
$$

Also:

$$
a,b\notin\mathbb Q.
$$

Therefore:

$$
\boxed{
a,b
\notin
\overline{\mathbb Q}.
}
$$

That is:

$$
\boxed{
a,b
\text{ are both transcendental}.
}
$$

---

# 3. The First Classification of $ab$

Since:

$$
ab
\in
\mathfrak M_\lambda,
$$

If:

$$
ab
\in
\overline{\mathbb Q},
$$

Round 2 immediately gives:

$$
ab
\in
\mathbb Q.
$$

Thus:

$$
\boxed{
ab
\text{ is either rational or transcendental}.
}
$$

There is no:

$$
ab
\in
\overline{\mathbb Q}\setminus\mathbb Q.
$$

---

# 4. The First Additional Restriction from Five Exponentials

The Five Exponentials Theorem states:

If:

$$
x_1,x_2
$$

are linearly independent over $\mathbb Q$,

$$
y_1,y_2
$$

are also linearly independent over $\mathbb Q$,

then:

$$
e^{x_1y_1},
\quad
e^{x_1y_2},
\quad
e^{x_2y_1},
\quad
e^{x_2y_2},
\quad
e^{x_2/x_1}
$$

cannot all be algebraic.

For the current normalized state, take:

$$
x_1=1,
\qquad
x_2=b,
$$

and:

$$
y_1=\lambda,
\qquad
y_2=a\lambda.
$$

The first four exponentials are all algebraic.

The fifth must therefore be transcendental:

$$
\boxed{
e^b
\text{ is transcendental}.
}
$$

Swapping the row / column roles similarly gives:

$$
\boxed{
e^a
\text{ is transcendental}.
}
$$

Therefore:

$$
\boxed{
a,b
\notin
\mathcal L_{\mathrm{alg}}.
}
$$

This is stronger than '$a,b$ are transcendental'.

---

# 5. Rational basis invariance

Take:

$$
P
\in
\operatorname{GL}_2(\mathbb Q).
$$

Let:

$$
\begin{pmatrix}
x'_1\\
x'_2
\end{pmatrix}
=
P
\begin{pmatrix}
1\\
b
\end{pmatrix}.
$$

Since $P$ is invertible, $x'_1,x'_2$ remain linearly independent over $\mathbb Q$.

And:

$$
x'_iy_j
$$

are rational linear combinations of the original:

$$
x_ky_j
$$

products.

Since:

$$
\mathcal L_{\mathrm{alg}}
$$

is a $\mathbb Q$-vector space, all:

$$
x'_iy_j
$$

still belong to:

$$
\mathcal L_{\mathrm{alg}}.
$$

Thus, a rational row-basis change preserves the hypothetical-counterexample property.

Column-basis changes work in exactly the same way.

---

# 6. Projective Orbit Avoidance

Take any:

$$
g
\in
\operatorname{PGL}_2(\mathbb Q).
$$

One can choose:

$$
P
=
\begin{pmatrix}
p&q\\
r&s
\end{pmatrix}
\in
\operatorname{GL}_2(\mathbb Q)
$$

to represent:

$$
g(z)
=
\frac{r+sz}{p+qz}.
$$

Since $b$ is transcendental:

$$
p+qb\neq0.
$$

Thus:

$$
\frac{x'_2}{x'_1}
=
g(b).
$$

The four transformed matrix exponentials are still all algebraic.

The Five Exponentials Theorem therefore forces:

$$
\boxed{
e^{g(b)}
\text{ transcendental}.
}
$$

Likewise, for $a$:

$$
\boxed{
e^{g(a)}
\text{ transcendental}.
}
$$

We therefore obtain:

$$
\boxed{
\operatorname{PGL}_2(\mathbb Q)\cdot a
\quad\cap\quad
\mathcal L_{\mathrm{alg}}
=
\varnothing
}
$$

and:

$$
\boxed{
\operatorname{PGL}_2(\mathbb Q)\cdot b
\quad\cap\quad
\mathcal L_{\mathrm{alg}}
=
\varnothing.
}
$$

This round calls this property:

**Projective Logarithmic Avoidance, PLA.**

---

# 7. The Meaning of PLA

Ordinary Baker / Gelfond–Schneider-type information only tells us that particular specified numbers do not belong to the algebraic domain.

PLA is stronger:

> Not only can $a$ and $b$ themselves not be algebraic logarithms; their entire rational projective orbits must avoid the algebraic-logarithm locus.

That is:

$$
a
\mapsto
\frac{r+sa}{p+qa}
$$

are all excluded as rational Möbius transforms.

This is a necessary condition at the projective level.

---

# 8. Transcendence-degree floor

Roy–Waldschmidt's known result is:

If:

$$
x_1,x_2
$$

and:

$$
y_1,y_2
$$

are respectively $\mathbb Q$-linearly independent, and:

$$
\operatorname{trdeg}_{\mathbb Q}
\mathbb Q(x_1,x_2,y_1,y_2)=1,
$$

then the four:

$$
e^{x_i y_j}
$$

cannot all be algebraic.

In the normalized state:

$$
x_1=1,
\quad
x_2=b,
\quad
y_1=\lambda,
\quad
y_2=a\lambda.
$$

Therefore:

$$
\mathbb Q(x_1,x_2,y_1,y_2)
=
\mathbb Q(\lambda,a,b).
$$

If a hypothetical counterexample exists, it must satisfy:

$$
\boxed{
\operatorname{trdeg}_{\mathbb Q}
\mathbb Q(\lambda,a,b)
\ge2.
}
$$

---

# 9. ACTM Parameter Dimension Floor

Round 4 therefore defines:

$$
\operatorname{PDF}
(\lambda;a,b)
=
\operatorname{trdeg}_{\mathbb Q}
\mathbb Q(\lambda,a,b).
$$

For a hypothetical Four Exponentials counterexample:

$$
\boxed{
\operatorname{PDF}\ge2.
}
$$

That is:

> A one-parameter algebraic family cannot accommodate a counterexample.

Any counterexample requires at least two algebraically independent transcendental degrees of freedom.

---

# 10. Bilinear span rank

Define:

$$
\boxed{
\rho(a,b)
=
\dim_{\mathbb Q}
\operatorname{span}_{\mathbb Q}
\{1,a,b,ab\}.
}
$$

Since:

$$
1,a
$$

are linearly independent over $\mathbb Q$, so:

$$
\rho\ge2.
$$

Clearly:

$$
\rho\le4.
$$

The next step excludes:

$$
\rho=2.
$$

---

# 11. Proposition R4.1: $\rho$ Cannot Equal $2$

If:

$$
\rho=2,
$$

Since:

$$
1,a
$$

already form a basis, we can write:

$$
b=u+va,
$$

and:

$$
ab=s+ta,
$$

where:

$$
u,v,s,t
\in
\mathbb Q.
$$

If:

$$
v=0,
$$

then:

$$
b=u\in\mathbb Q,
$$

a contradiction.

Hence:

$$
v\neq0.
$$

Multiplying:

$$
b=u+va
$$

by $a$:

$$
ab=ua+va^2.
$$

and:

$$
ab=s+ta
$$

and comparing the two expressions:

$$
va^2+(u-t)a-s=0.
$$

This is a nonzero rational polynomial satisfied by $a$.

Therefore:

$$
a
\in
\overline{\mathbb Q},
$$

contradicting the transcendence of $a$.

Hence:

$$
\boxed{
\rho\neq2.
}
$$

Therefore:

$$
\boxed{
\rho\in\{3,4\}.
}
$$

---

# 12. The Action of Baker Rigidity on the Bilinear Span

Since:

$$
\lambda,
\quad
a\lambda,
\quad
b\lambda,
\quad
ab\lambda
\in
\mathcal L_{\mathrm{alg}},
$$

Baker's theorem tells us:

The $\mathbb Q$-linear rank of these four algebraic logarithms equals their $\overline{\mathbb Q}$-linear rank.

Since $\lambda\neq0$:

$$
\dim_{\mathbb Q}
\operatorname{span}
\{\lambda,a\lambda,b\lambda,ab\lambda\}
=
\rho(a,b).
$$

Therefore:

$$
\boxed{
\dim_{\overline{\mathbb Q}}
\operatorname{span}
\{1,a,b,ab\}
=
\rho(a,b).
}
$$

---

# 13. Rank split

Thus, a hypothetical counterexample must lie in one of two types:

## Type P

$$
\boxed{
\rho=3.
}
$$

There is a rational bilinear relation.

## Type B

$$
\boxed{
\rho=4.
}
$$

There is no algebraic bilinear relation.

This is the first main classification of Round 4.

---

# 14. Type P: A Rational Bilinear Relation

If:

$$
\rho=3,
$$

there exist, not all zero:

$$
r_0,r_1,r_2,r_3
\in
\mathbb Q
$$

such that:

$$
\boxed{
r_0
+
r_1a
+
r_2b
+
r_3ab
=
0.
}
$$

This can be rewritten:

$$
(r_2+r_3a)b
=
-(r_0+r_1a).
$$

Since $a$ is transcendental:

$$
r_2+r_3a\neq0.
$$

Hence:

$$
\boxed{
b
=
-\frac{r_0+r_1a}{r_2+r_3a}.
}
$$

---

# 15. Type P Is Equivalent to a Rational Möbius Relation

The preceding expression says:

$$
\boxed{
b=g(a)
}
$$

where:

$$
g
\in
\operatorname{PGL}_2(\mathbb Q).
$$

If the corresponding $2\times2$ rational matrix has zero determinant, $g$ degenerates to a rational constant, giving $b\in\mathbb Q$.

Its determinant must therefore be nonzero.

Thus, all bilinear dependence of Type P is actually:

$$
\boxed{
\text{rational projective equivalence between }a\text{ and }b.
}
$$

---

# 16. Projective Normalization of Type P

Since:

$$
b=g(a),
$$

Take:

$$
h=g^{-1}
\in
\operatorname{PGL}_2(\mathbb Q).
$$

Apply to the row vector:

$$
(1,b)
$$

the corresponding rational basis change so that the new row ratio becomes:

$$
h(b)=a.
$$

Let the first transformed row factor be:

$$
c\neq0.
$$

Then the new rank-one matrix can be written:

$$
Z'
=
\lambda'
\begin{pmatrix}
1&a\\
a&a^2
\end{pmatrix},
$$

where:

$$
\lambda'
=
c\lambda.
$$

Since:

$$
c
$$

is a rational linear combination of $1$ and $b$, we have:

$$
\lambda'
$$

as a rational linear combination of:

$$
\lambda
$$

and:

$$
b\lambda
$$

the second term above.

Therefore:

$$
\lambda'
\in
\mathcal L_{\mathrm{alg}}.
$$

---

# 17. The Three-Term Power Chain of Type P

Let:

$$
t=a.
$$

then:

$$
\boxed{
\lambda',
\quad
t\lambda',
\quad
t^2\lambda'
\in
\mathcal L_{\mathrm{alg}}.
}
$$

This is the canonical normal form of Type P.

Its exponential outputs:

$$
e^{\lambda'},
\quad
e^{t\lambda'},
\quad
e^{t^2\lambda'}
$$

are all algebraic.

---

# 18. In Type P, $t$ Must Be Transcendental

Since:

$$
t=a,
$$

Round 2 has already given:

$$
\boxed{
t
\text{ transcendental}.
}
$$

Five Exponentials further strengthens this to:

$$
\boxed{
e^t
\text{ transcendental}.
}
$$

More generally:

$$
\boxed{
e^{g(t)}
\text{ transcendental}
}
$$

for all:

$$
g
\in
\operatorname{PGL}_2(\mathbb Q)
$$

as stated above.

---

# 19. The Transcendence Degree of Type P

The transformed counterexample of Type P uses:

$$
x_1=1,
\qquad
x_2=t,
$$

and:

$$
y_1=\lambda',
\qquad
y_2=t\lambda'.
$$

Its four exponentials are all algebraic.

If:

$$
\operatorname{trdeg}_{\mathbb Q}
\mathbb Q(\lambda',t)=1,
$$

the known special case of Roy–Waldschmidt directly excludes such a counterexample.

On the other hand, this field is generated by two elements, so:

$$
\operatorname{trdeg}\le2.
$$

Hence:

$$
\boxed{
\operatorname{trdeg}_{\mathbb Q}
\mathbb Q(\lambda',t)=2.
}
$$

Therefore:

$$
\boxed{
\lambda'
\text{ and }
t
\text{ are algebraically independent}.
}
$$

---

# 20. Six Exponentials Closes off the Power Chain

Now take:

$$
x_1=1,
\qquad
x_2=t,
$$

and:

$$
y_1=\lambda',
\qquad
y_2=t\lambda',
\qquad
y_3=t^2\lambda'.
$$

Since $t$ is transcendental:

$$
1,t
$$

are linearly independent over $\mathbb Q$, and:

$$
1,t,t^2
$$

are also linearly independent over $\mathbb Q$.

Hence:

$$
y_1,y_2,y_3
$$

are linearly independent over $\mathbb Q$.

The six products are:

$$
\lambda',
\quad
t\lambda',
\quad
t^2\lambda',
$$

and:

$$
t\lambda',
\quad
t^2\lambda',
\quad
t^3\lambda'.
$$

The three distinct exponents involved in the first five products:

$$
\lambda',
\quad
t\lambda',
\quad
t^2\lambda'
$$

all belong to:

$$
\mathcal L_{\mathrm{alg}}.
$$

The Six Exponentials Theorem therefore forces:

$$
\boxed{
t^3\lambda'
\notin
\mathcal L_{\mathrm{alg}}.
}
$$

---

# 21. Exact Power-Chain Depth

For:

$$
0\neq\lambda
\in
\mathcal L_{\mathrm{alg}},
$$

and:

$$
t\in\mathbb C,
$$

Define:

$$
D_\lambda(t)
=
\sup
\left\{
N\in\mathbb N:
t^k\lambda
\in
\mathcal L_{\mathrm{alg}}
\text{ for all }
0\le k\le N
\right\}.
$$

A hypothetical Type P counterexample satisfies:

$$
\boxed{
D_{\lambda'}(t)=2.
}
$$

Since:

$$
k=0,1,2
$$

all lie in the locus, whereas:

$$
k=3
$$

must escape.

This is the exact chain boundary obtained in this round.

---

# 22. The Logarithmic Geometric Progression Form

Let:

$$
\mu=t\lambda',
$$

and:

$$
\nu=t^2\lambda'.
$$

then:

$$
\lambda',
\mu,\nu
\in
\mathcal L_{\mathrm{alg}},
$$

and:

$$
\boxed{
\mu^2
=
\lambda'\nu.
}
$$

Type P is therefore equivalent to:

> There exists a nontrivial three-term geometric progression in the algebraic logarithm locus.

Its common ratio:

$$
t
=
\frac{\mu}{\lambda'}
=
\frac{\nu}{\mu}
$$

is transcendental.

---

# 23. The Three-Term Log-GP Conjecture

ACTM thus extracts a subproblem narrower than the full Four Exponentials problem:

> If
>
> $$
> 0\neq
> \lambda,\mu,\nu
> \in
> \mathcal L_{\mathrm{alg}}
> $$
>
> and:
>
> $$
> \mu^2=\lambda\nu,
> $$
>
> must we have:
>
> $$
> \frac{\mu}{\lambda}
> \in
> \mathbb Q?
> $$

A positive answer would exclude all Type P counterexamples.

This round calls it:

$$
\boxed{
\text{Logarithmic Geometric Progression Conjecture}
}
$$

It is only a special subproblem of Four Exponentials; it is not claimed to be known or proved.

---

# 24. The Classical $2^\pi$ Example Falls Exactly into Type P

Take:

$$
\lambda=\log2,
\qquad
t=\pi.
$$

Suppose, for contradiction, that:

$$
2^\pi
$$

and:

$$
2^{\pi^2}
$$

are both algebraic; then:

$$
\log2,
\quad
\pi\log2,
\quad
\pi^2\log2
\in
\mathcal L_{\mathrm{alg}}.
$$

This is exactly a Type P chain:

$$
\lambda,
\quad
t\lambda,
\quad
t^2\lambda.
$$

The known Six Exponentials result can only additionally yield:

$$
\boxed{
2^{\pi^3}
\text{ must be transcendental}.
}
$$

The simultaneous algebraicity of the first two cannot currently be excluded.

This shows that Type P is not an artificially created side issue: it includes a classical open case of Four Exponentials.

---

# 25. Type B: $\rho=4$

Now consider:

$$
\boxed{
\rho(a,b)=4.
}
$$

then:

$$
1,a,b,ab
$$

are linearly independent over $\mathbb Q$.

Since multiplying these four by $\lambda$ gives algebraic logarithms, Baker rigidity immediately strengthens this to:

$$
\boxed{
1,a,b,ab
\text{ are also linearly independent over }
\overline{\mathbb Q}
\text{ as stated above}.
}
$$

---

# 26. Type B Has No Algebraic Bilinear Relation

Thus, there do not exist, not all zero:

$$
c_{00},
c_{10},
c_{01},
c_{11}
\in
\overline{\mathbb Q}
$$

such that:

$$
\boxed{
c_{00}
+
c_{10}a
+
c_{01}b
+
c_{11}ab
=
0.
}
$$

Thus:

$$
\boxed{
\text{Type B has no algebraic relation of bidegree }(1,1).
}
$$

If $a,b$ are nevertheless algebraically dependent, their minimal relation must occur at higher algebraic complexity.

---

# 27. The Essential Difference Between Type P and Type B

Type P:

$$
\boxed{
b=g(a),
\qquad
g\in\operatorname{PGL}_2(\mathbb Q).
}
$$

Thus, it can be compressed into a single projective ratio:

$$
t.
$$

Type B:

$$
\boxed{
1,a,b,ab
\text{ algebraically bilinearly independent}.
}
$$

There is no rational projective normalization that merges:

$$
a
$$

and:

$$
b
$$

into the same variable.

Type B is therefore genuinely two-directional.

---

# 28. The Matrix Normal Form of Type P

After normalization, Type P is:

$$
M_P
=
\lambda'
\begin{pmatrix}
1&t\\
t&t^2
\end{pmatrix}.
$$

This is a:

$$
\boxed{
\text{symmetric rank-one Hankel matrix}.
}
$$

Its determinant relation is:

$$
\boxed{
(\lambda')(t^2\lambda')
-
(t\lambda')^2
=
0.
}
$$

That is:

$$
\boxed{
\lambda'\nu-\mu^2=0.
}
$$

Type P therefore compresses the four-log problem of Four Exponentials into a three-log quadratic relation.

---

# 29. Rational Invisibility Remains

For:

$$
M_P
=
\lambda'
\begin{pmatrix}
1&t\\
t&t^2
\end{pmatrix},
$$

its rank-one factors can be written:

$$
M_P
=
\lambda'
\begin{pmatrix}
1\\
t
\end{pmatrix}
\begin{pmatrix}
1&t
\end{pmatrix}.
$$

Since:

$$
t\notin\mathbb Q,
$$

There is no:

$$
0\neq
v\in\mathbb Q^2
$$

such that:

$$
\begin{pmatrix}
1&t
\end{pmatrix}
v=0.
$$

Thus, there are no nonzero rational $w,v$ such that:

$$
w^TM_Pv=0.
$$

Type P is therefore still a state of the Matrix Coefficient Conjecture described as:

$$
\boxed{
\text{singular but rationally invisible}
}
$$

a state of this form.

---

# 30. The Position of the 2026 Matrix Coefficient Conjecture

The Matrix Coefficient Conjecture of Dasgupta–Kakde asserts:

If a square algebraic-logarithm matrix:

$$
M
$$

satisfies:

$$
\det M=0,
$$

then there exist:

$$
0\neq
w,v
\in
\mathbb Q^n
$$

such that:

$$
w^TMv=0.
$$

Its $n=2$ case is equivalent to the Four Exponentials Conjecture.

Thus, ACTM's Type P / Type B classification concerns the singular-rationally-invisible states this conjecture seeks to exclude.

The paper also proposes an auxiliary-polynomial / support-based proof strategy, but a key implication needed to complete the Matrix Coefficient Conjecture remains open.

---

# 31. The Additional Strength of a Rank-3 State

Type P requires more than:

$$
\lambda',
t\lambda',
t^2\lambda'
\in
\mathcal L_{\mathrm{alg}}.
$$

It simultaneously forces:

$$
\boxed{
\lambda',t
\text{ algebraically independent},
}
$$

$$
\boxed{
e^{g(t)}
\text{ transcendental for every }
g\in\operatorname{PGL}_2(\mathbb Q),
}
$$

and:

$$
\boxed{
t^3\lambda'
\notin
\mathcal L_{\mathrm{alg}}.
}
$$

Thus, a hypothetical Type P counterexample must be a very precise algebraic-logarithm chain 'of length $3$, but admitting no further extension'.

---

# 32. The Additional Strength of a Rank-4 State

Type B must simultaneously satisfy:

$$
\operatorname{trdeg}_{\mathbb Q}
\mathbb Q(\lambda,a,b)
\ge2,
$$

$$
1,a,b,ab
\text{ are }
\overline{\mathbb Q}\text{-linearly independent},
$$

and:

$$
\operatorname{PGL}_2(\mathbb Q)\cdot a
\cap
\mathcal L_{\mathrm{alg}}
=
\varnothing,
$$

$$
\operatorname{PGL}_2(\mathbb Q)\cdot b
\cap
\mathcal L_{\mathrm{alg}}
=
\varnothing.
$$

Yet it must still satisfy:

$$
\lambda,
a\lambda,
b\lambda,
ab\lambda
\in
\mathcal L_{\mathrm{alg}}.
$$

Thus, Type B is a:

$$
\boxed{
\text{highly rigid four-corner logarithmic rectangle}.
}
$$

---

# 33. Counterexample classification table

| Type | $\rho$ | Core relation | Normalizable form | Known additional restrictions |
|---|---:|---|---|---|
| Type P | $3$ | rational bilinear relation | $\lambda',t\lambda',t^2\lambda'$ | $\operatorname{trdeg}\mathbb Q(\lambda',t)=2$；$t^3\lambda'\notin\mathcal L_{\mathrm{alg}}$ |
| Type B | $4$ | No algebraic bilinear relation | Genuine $\lambda,a\lambda,b\lambda,ab\lambda$ rectangle | $\operatorname{trdeg}\mathbb Q(\lambda,a,b)\ge2$; PLA |

---

# 34. The Most Important Compression of This Round

Round 3:

$$
\boxed{
\operatorname{MID}=2
}
$$

still looks like an abstract intersection anomaly.

Round 4 splits it into:

$$
\boxed{
\operatorname{MID}=2
\Rightarrow
\text{Type P or Type B}.
}
$$

Type P is further compressed into:

$$
\boxed{
\lambda,
t\lambda,
t^2\lambda
\in
\mathcal L_{\mathrm{alg}},
\quad
t^3\lambda
\notin
\mathcal L_{\mathrm{alg}}.
}
$$

The first narrow frontier genuinely suitable for a concentrated attack has therefore formed.

---

# 35. Reconnecting ACTM to the Original Question

ACTM originally asked:

> Which algebraic structures have the combinatorial potential to generate transcendence under some generating mechanism?

Round 4 shows that the reverse direction also holds:

> If hypothetical 'excessive algebraization' actually occurs, it cannot occur arbitrarily: it is forced into very special generating chains and matrix shapes.

That is:

$$
\boxed{
\text{Transcendence Generation}
}
$$

and:

$$
\boxed{
\text{Impossible Algebraic Compression}
}
$$

are actually two sides of the same question.

---

# 36. Round 4 Verdict

**PASS — Hypothetical MID=2 State Classified into Two Structural Types**

This round completes:

1. Prove that $a,b$ must be transcendental.
2. Prove that $e^a,e^b$ must be transcendental.
3. Generalize to rational projective orbit avoidance:
   $$
   e^{g(a)},e^{g(b)}
   \text{ transcendental}.
   $$
4. Obtain the counterexample parameter floor:
   $$
   \operatorname{trdeg}\mathbb Q(\lambda,a,b)\ge2.
   $$
5. Define the bilinear span rank:
   $$
   \rho
   =
   \dim_{\mathbb Q}
   \operatorname{span}\{1,a,b,ab\}.
   $$
6. Exclude:
   $$
   \rho\le2.
   $$
7. Obtain:
   $$
   \rho\in\{3,4\}.
   $$
8. Type P：
   $$
   \rho=3
   $$
   is equivalent to a rational Möbius relation.
9. Normalize Type P to:
   $$
   \lambda,t\lambda,t^2\lambda
   \in\mathcal L_{\mathrm{alg}}.
   $$
10. Prove that, in Type P:
    $$
    \operatorname{trdeg}\mathbb Q(\lambda,t)=2.
    $$
11. Use Six Exponentials to obtain:
    $$
    t^3\lambda
    \notin
    \mathcal L_{\mathrm{alg}}.
    $$
12. Define Exact Power-Chain Depth:
    $$
    D_\lambda(t)=2.
    $$
13. Recast Type P as a three-log geometric-progression problem.
14. Type B：
    $$
    \rho=4
    $$
    gives linear independence of $1,a,b,ab$ over $\overline{\mathbb Q}$.
15. Obtain the complete Type P / Type B dichotomy.

---

# 37. The Formal Entry Point to Round 5

Round 5 proposes first tackling the narrower:

$$
\boxed{
\text{Type P}
}
$$

rather than directly addressing Type B.

The basic inputs become three algebraic logarithms:

$$
\lambda,
\quad
\mu,
\quad
\nu
\in
\mathcal L_{\mathrm{alg}}
$$

satisfying:

$$
\boxed{
\mu^2=\lambda\nu.
}
$$

and:

$$
\lambda,\mu,\nu
$$

linearly independent over $\mathbb Q$.

Define:

$$
t
=
\frac{\mu}{\lambda}
=
\frac{\nu}{\mu}.
$$

The known requirements for a hypothetical Type P state are:

$$
t
\text{ transcendental},
$$

$$
\operatorname{trdeg}_{\mathbb Q}
\mathbb Q(\lambda,t)=2,
$$

$$
e^{g(t)}
\text{ transcendental}
$$

for all:

$$
g
\in
\operatorname{PGL}_2(\mathbb Q),
$$

and:

$$
t^3\lambda
\notin
\mathcal L_{\mathrm{alg}}.
$$

The question for Round 5 becomes:

> Can the fact that 'three algebraic logarithms form an exact geometric progression' itself yield new rational visibility, height obstructions, period obstructions, or approximation obstructions?

That is, formally study:

$$
\boxed{
\lambda\nu-\mu^2=0
}
$$

this minimal quadratic relation.

---

# 38. The One-Sentence Conclusion

> **If the one-dimensional gap of Four Exponentials actually exists, it is not arbitrary: a hypothetical counterexample can either be compressed by rational projective normalization into an exact three-term algebraic-logarithm geometric chain, whose fourth term must immediately escape, or its four corner multipliers are already completely independent at the algebraic bilinear level.**

The narrowest first type is:

$$
\boxed{
\lambda,
\quad
t\lambda,
\quad
t^2\lambda
\in
\mathcal L_{\mathrm{alg}},
}
$$

However:

$$
\boxed{
t^3\lambda
\notin
\mathcal L_{\mathrm{alg}},
}
$$

At the same time:

$$
\boxed{
\operatorname{trdeg}_{\mathbb Q}
\mathbb Q(\lambda,t)=2.
}
$$

This is the formal frontier of ACTM Round 4.

---

# References

1. Damien Roy and Michel Waldschmidt, **Quadratic Relations between Logarithms of Algebraic Numbers**, *Proceedings of the Japan Academy, Series A*, 71 (1995), 151–153.  
   This proves the Four Exponentials Conjecture when the field generated by the four parameters has transcendence degree $1$.

2. Michel Waldschmidt, **The Four Exponentials Problem and Schanuel’s Conjecture**, in *Mathematics Going Forward*, Lecture Notes in Mathematics 2313, Springer, 2023, pp. 579–592.

3. Michel Waldschmidt, expository accounts of the **Five Exponentials Theorem** and **Six Exponentials Theorem**, including *Legacy of Srinivasa Ramanujan* lecture material.

4. Samit Dasgupta and Mahesh Kakde, **Ranks of Matrices of Logarithms of Algebraic Numbers II: The Matrix Coefficient Conjecture**, *Advances in Mathematics* 487 (2026), Article 110753.  
   DOI: 10.1016/j.aim.2025.110753

5. Samit Dasgupta, **Ranks of Matrices of Logarithms of Algebraic Numbers I: The theorems of Baker and Waldschmidt–Masser**, *Essential Number Theory* 2 (2023).

6. Alan Baker, classical theorem on linear independence of logarithms of algebraic numbers over algebraic coefficients.

---

**End of ACTM Round 4**