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

# ACTM Round 2
## Algebraic Scalars, Logarithmic Lifts, and the Exact Escape Boundary
### Algebraic Scalars, Logarithmic Lifts, and the Exact Escape Boundary

**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 2 / v0.1  
**Document type:** Formal research notes / structural calibration round  
**Novelty statement:** The mathematical core of this round rests on Gelfond–Schneider and Baker's theory of linear forms. The principal additions are ACTM's structural reformulation, scalar escape boundary, matrix-based language of relation transport, and subsequent falsifiable research quantities; no claim is made to reprove or replace established classical theorems.

---

## Abstract

ACTM Round 1 studied the transformation of degrees of freedom for purely algebraic seeds under the exponential map:

$$
A\in\overline{\mathbb Q}^{\,n}
\quad\longmapsto\quad
e^A,
$$

and obtained:

$$
\operatorname{ldim}_{\mathbb Q}(A)
=
\operatorname{trdeg}_{\mathbb Q}\mathbb Q(e^A).
$$

Round 2 lifts the question one level to study:

$$
(\alpha,\beta)
\longmapsto
\alpha^\beta
=
e^{\beta\lambda},
\qquad
e^\lambda=\alpha,
$$

where:

$$
\alpha\in\overline{\mathbb Q}^{\times},
\qquad
\beta\in\overline{\mathbb Q}.
$$

The core transformation is no longer 'algebraic inputs entering the exponential directly', but:

$$
\boxed{
\text{Algebraic Point}
\rightarrow
\text{Logarithmic Lift}
\rightarrow
\text{Algebraic Scalar Action}
\rightarrow
\text{Exponential Descent}
}
$$

This round obtains an exact scalar boundary. Let:

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

If:

$$
0\neq\lambda\in\mathcal L_{\mathrm{alg}},
\qquad
\beta\in\overline{\mathbb Q},
$$

then:

$$
\boxed{
\beta\lambda\in\mathcal L_{\mathrm{alg}}
\iff
\beta\in\mathbb Q.
}
$$

Equivalently:

$$
\boxed{
e^{\beta\lambda}\in\overline{\mathbb Q}
\iff
\beta\in\mathbb Q.
}
$$

The nontrivial direction is precisely the Gelfond–Schneider theorem.

Thus, although $\mathcal L_{\mathrm{alg}}$ is a $\mathbb Q$-vector space, its algebraic scalar stabilizer ring is exactly:

$$
\boxed{
\operatorname{Stab}_{\overline{\mathbb Q}}
(\mathcal L_{\mathrm{alg}})
=
\mathbb Q.
}
$$

Baker's theorem also allows this result to be expressed in matrix form: if:

$$
\lambda_1,\ldots,\lambda_r
\in
\mathcal L_{\mathrm{alg}}
$$

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

$$
B\in M_{s\times r}(\overline{\mathbb Q}),
$$

then:

$$
B\lambda\in\mathcal L_{\mathrm{alg}}^s
$$

if and only if:

$$
B\in M_{s\times r}(\mathbb Q).
$$

This establishes ACTM's second exact boundary:

$$
\boxed{
\text{Rational Coefficients}
=
\text{Algebraic Descent-Preserving Region}
}
$$

whereas:

$$
\boxed{
\text{Algebraic Irrational Coefficients}
=
\text{Forced Escape Region}.
}
$$

---

# 1. Reversing the Question in Round 2

The traditional form of Gelfond–Schneider is:

If:

$$
\alpha\in\overline{\mathbb Q},
\qquad
\alpha\neq0,1,
$$

and:

$$
\beta\in\overline{\mathbb Q}\setminus\mathbb Q,
$$

then any appropriately defined:

$$
\alpha^\beta
$$

is transcendental.

ACTM does not first interpret this as:

> Why is a particular number $\alpha^\beta$ transcendental?

Instead, it asks:

> After an algebraic input $\alpha$ undergoes a logarithmic lift, which algebraic scalar actions can still descend to algebraic points, and which must escape?

Thus, the core research object is not the individual value:

$$
\alpha^\beta
$$

but:

$$
\boxed{
\mathcal L_{\mathrm{alg}}
\subset\mathbb C.
}
$$

---

# 2. The Algebraic Logarithm Locus

Define:

$$
\mathcal L_{\mathrm{alg}}
=
\exp^{-1}
\left(
\overline{\mathbb Q}^{\times}
\right).
$$

That is:

$$
\lambda\in\mathcal L_{\mathrm{alg}}
\iff
e^\lambda
\in
\overline{\mathbb Q}^{\times}.
$$

The elements here are generally themselves transcendental.

For example:

$$
\log 2,
\qquad
i\pi,
\qquad
2\pi i
$$

all belong to:

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

---

# 3. Proposition R2.1: $\mathcal L_{\mathrm{alg}}$ Is a $\mathbb Q$-Vector Space

## Addition

If:

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

then:

$$
e^{\lambda+\mu}
=
e^\lambda e^\mu
\in
\overline{\mathbb Q}^{\times}.
$$

Therefore:

$$
\lambda+\mu
\in
\mathcal L_{\mathrm{alg}}.
$$

---

## Additive Inverses

$$
e^{-\lambda}
=
(e^\lambda)^{-1}
\in
\overline{\mathbb Q}^{\times}.
$$

Thus:

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

---

## Rational Scalars

Let:

$$
q=\frac{p}{n}\in\mathbb Q,
\qquad
n>0.
$$

then:

$$
\left(e^{q\lambda}\right)^n
=
e^{p\lambda}
=
(e^\lambda)^p
\in
\overline{\mathbb Q}.
$$

Therefore:

$$
e^{q\lambda}
$$

is algebraic; hence:

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

Thus:

$$
\boxed{
\mathcal L_{\mathrm{alg}}
\text{ is a }
\mathbb Q\text{-vector space}.
}
$$

---

# 4. The First Exact Escape Boundary

Fix:

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

Let:

$$
\alpha=e^\lambda
\in
\overline{\mathbb Q}^{\times}.
$$

Consider:

$$
\beta\lambda,
\qquad
\beta\in\overline{\mathbb Q}.
$$

---

# 5. Proposition R2.2: Scalar Descent Rigidity

## Proposition

For any:

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

and:

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

we have:

$$
\boxed{
\beta\lambda\in\mathcal L_{\mathrm{alg}}
\iff
\beta\in\mathbb Q.
}
$$

Equivalently:

$$
\boxed{
e^{\beta\lambda}
\in
\overline{\mathbb Q}^{\times}
\iff
\beta\in\mathbb Q.
}
$$

---

## Proof: The Rational Direction

If:

$$
\beta\in\mathbb Q,
$$

by the previous section:

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

---

## Proof: The Algebraic Irrational Direction

If:

$$
\beta
\in
\overline{\mathbb Q}\setminus\mathbb Q,
$$

and:

$$
\lambda
$$

is a nonzero algebraic logarithm, the Gelfond–Schneider theorem gives:

$$
e^{\beta\lambda}
$$

as a transcendental number.

Therefore:

$$
\beta\lambda
\notin
\mathcal L_{\mathrm{alg}}.
$$

This completes the proof.

---

# 6. ACTM's First 'If and Only If' Escape Law

Round 1 tells us that:

$$
\mathbb Q\text{-linear rank}
$$

controls the transcendental degrees of freedom of algebraic seeds after $\exp$.

Round 2 now obtains a sharper result:

$$
\boxed{
\beta\in\mathbb Q
\Rightarrow
\text{descent stays algebraic},
}
$$

whereas:

$$
\boxed{
\beta\in
\overline{\mathbb Q}\setminus\mathbb Q
\Rightarrow
\text{forced transcendental escape}.
}
$$

This is not a probabilistic tendency.

It is an exact boundary.

---

# 7. The Algebraic Scalar Stabilizer Ring

Define:

$$
\operatorname{Stab}_{\overline{\mathbb Q}}
(\mathcal L_{\mathrm{alg}})
=
\left\{
\beta\in\overline{\mathbb Q}:
\beta\mathcal L_{\mathrm{alg}}
\subseteq
\mathcal L_{\mathrm{alg}}
\right\}.
$$

By R2.2:

$$
\boxed{
\operatorname{Stab}_{\overline{\mathbb Q}}
(\mathcal L_{\mathrm{alg}})
=
\mathbb Q.
}
$$

Therefore:

> The algebraic logarithm locus is not an $\overline{\mathbb Q}$-vector space.

It naturally admits only:

$$
\mathbb Q
$$

as scalar actions.

This is the core structural fact of ACTM Round 2.

---

# 8. Baker Rigidity

Baker's theory gives a more general result.

If:

$$
\lambda_1,\ldots,\lambda_r
\in
\mathcal L_{\mathrm{alg}}
$$

are linearly independent over $\mathbb Q$, they remain linearly independent over:

$$
\overline{\mathbb Q}
$$

that field.

Therefore:

$$
\boxed{
\operatorname{ldim}_{\mathbb Q}
(\lambda_1,\ldots,\lambda_r)
=
\operatorname{ldim}_{\overline{\mathbb Q}}
(\lambda_1,\ldots,\lambda_r).
}
$$

for every finite algebraic-logarithm tuple.

---

# 9. Scalar-Extension Defect

ACTM defines the working quantity:

$$
\operatorname{SED}(Z)
=
\operatorname{ldim}_{\mathbb Q}(Z)
-
\operatorname{ldim}_{\overline{\mathbb Q}}(Z),
$$

called:

**Scalar-Extension Defect**.

In general:

$$
\operatorname{SED}(Z)\ge0.
$$

Baker rigidity tells us that:

If:

$$
Z\subset\mathcal L_{\mathrm{alg}},
$$

then:

$$
\boxed{
\operatorname{SED}(Z)=0.
}
$$

Therefore:

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

can be understood as a $\mathbb Q$-vector space with extremely strong rigidity under algebraic scalar extension.

---

# 10. Reformulating Gelfond–Schneider via SED

Let:

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

and:

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

Consider:

$$
Z=(\lambda,\beta\lambda).
$$

Since $\beta$ is irrational:

$$
\operatorname{ldim}_{\mathbb Q}
(\lambda,\beta\lambda)=2.
$$

However, since:

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

Thus:

$$
\operatorname{ldim}_{\overline{\mathbb Q}}
(\lambda,\beta\lambda)=1.
$$

Hence:

$$
\boxed{
\operatorname{SED}
(\lambda,\beta\lambda)
=
1.
}
$$

If:

$$
\beta\lambda
\in
\mathcal L_{\mathrm{alg}},
$$

then the entire tuple:

$$
(\lambda,\beta\lambda)
$$

lies in:

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

contradicting the Baker-rigidity requirement:

$$
\operatorname{SED}=0
$$

as stated above.

Thus:

$$
\boxed{
\beta\lambda
\notin
\mathcal L_{\mathrm{alg}}.
}
$$

That is:

$$
e^{\beta\lambda}
$$

is transcendental.

This is a structural reformulation of Gelfond–Schneider within Baker's theory.

---

# 11. ACTM's Interpretation: Where the Actual Conflict Lies

On the traditional surface:

$$
\alpha^\beta
$$

looks like 'raising an algebraic number to an algebraic irrational power'.

ACTM expands it as:

$$
\boxed{
\alpha
\rightarrow
\lambda
\rightarrow
\beta\lambda
\rightarrow
e^{\beta\lambda}.
}
$$

where:

$$
e^\lambda=\alpha.
$$

The actual conflict is not that 'powers are strange'.

Rather:

$$
\boxed{
\mathbb Q\text{-rank}
\neq
\overline{\mathbb Q}\text{-rank}
}
$$

is forcibly created on a special locus that does not permit this defect.

The transcendental output is therefore:

$$
\boxed{
\text{rank incompatibility}
\Rightarrow
\text{failure of algebraic descent}.
}
$$

---

# 12. The Matrix Form

Round 2 next generalizes the individual scalar:

$$
\beta
$$

to an algebraic matrix.

Let:

$$
\lambda
=
(\lambda_1,\ldots,\lambda_r)^T,
$$

where:

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

that are linearly independent over:

$$
\mathbb Q
$$

that field.

By Baker:

$$
\lambda_i
$$

are also linearly independent over:

$$
\overline{\mathbb Q}
$$

that field.

Take:

$$
B
=
(b_{ji})
\in
M_{s\times r}
(\overline{\mathbb Q}).
$$

Define:

$$
\mu
=
B\lambda.
$$

That is:

$$
\mu_j
=
\sum_{i=1}^{r}
b_{ji}\lambda_i.
$$

---

# 13. Proposition R2.3: Algebraic Matrix Descent Rigidity

## Proposition

Under the conditions above:

$$
\boxed{
B\lambda
\in
\mathcal L_{\mathrm{alg}}^s
\iff
B\in M_{s\times r}(\mathbb Q).
}
$$

---

# 14. The Rational Direction of R2.3

If:

$$
B\in M_{s\times r}(\mathbb Q),
$$

then each:

$$
\mu_j
=
\sum_i
b_{ji}\lambda_i
$$

is a rational linear combination in:

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

that space.

Since:

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

is a $\mathbb Q$-vector space:

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

Hence:

$$
B\lambda
\in
\mathcal L_{\mathrm{alg}}^s.
$$

---

# 15. The Converse Direction of R2.3

Suppose:

$$
B\lambda
\in
\mathcal L_{\mathrm{alg}}^s.
$$

Choose an arbitrary row $j$:

$$
b_j
=
(b_{j1},\ldots,b_{jr}).
$$

If:

$$
b_j
\notin
\mathbb Q^r,
$$

then:

$$
\mu_j
=
b_j\cdot\lambda
$$

does not lie in:

$$
\operatorname{span}_{\mathbb Q}
\{\lambda_1,\ldots,\lambda_r\}
$$

that space.

Therefore:

$$
\lambda_1,\ldots,\lambda_r,\mu_j
$$

are linearly independent over $\mathbb Q$.

However:

$$
\mu_j
=
\sum_i b_{ji}\lambda_i
$$

is an $\overline{\mathbb Q}$-linear combination, so:

$$
\lambda_1,\ldots,\lambda_r,\mu_j
$$

are linearly dependent over $\overline{\mathbb Q}$.

On the other hand, all the elements in the assumption belong to:

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

This violates Baker rigidity.

Every row must therefore satisfy:

$$
b_j\in\mathbb Q^r.
$$

Therefore:

$$
B\in M_{s\times r}(\mathbb Q).
$$

This completes the proof.

---

# 16. The Single-Row Version: Multi-Base Gelfond–Baker Escape

Take:

$$
b
=
(b_1,\ldots,b_r)
\in
\overline{\mathbb Q}^{\,r}.
$$

Let:

$$
\mu
=
\sum_{i=1}^{r}
b_i\lambda_i.
$$

If:

$$
b\notin\mathbb Q^r,
$$

R2.3 immediately gives:

$$
\boxed{
e^\mu
=
\exp
\left(
\sum_i b_i\lambda_i
\right)
\notin
\overline{\mathbb Q}.
}
$$

If:

$$
\alpha_i=e^{\lambda_i},
$$

we can write this formally as:

$$
e^\mu
=
\prod_{i=1}^{r}
\alpha_i^{b_i},
$$

where the branches are fixed by the selected $\lambda_i$.

Therefore:

$$
\boxed{
b\notin\mathbb Q^r
\Rightarrow
\prod_i
\alpha_i^{b_i}
\text{ transcendental}
}
$$

under the condition that the logarithmic lifts $\lambda_i$ are $\mathbb Q$-independent.

---

# 17. The Exact Coefficient Boundary

Thus, for a fixed independent logarithmic basis:

$$
\lambda_1,\ldots,\lambda_r,
$$

there is an exact boundary:

$$
\boxed{
b\in\mathbb Q^r
\iff
e^{b\cdot\lambda}
\in
\overline{\mathbb Q}.
}
$$

and:

$$
\boxed{
b
\in
\overline{\mathbb Q}^{\,r}
\setminus
\mathbb Q^r
\iff
e^{b\cdot\lambda}
\text{ transcendental}.
}
$$

This is the most important multidimensional version of Round 2.

---

# 18. Example 1: The Gelfond–Schneider Constant

Let:

$$
\alpha=2,
\qquad
\lambda=\log2,
\qquad
\beta=\sqrt2.
$$

Then:

$$
\sqrt2\notin\mathbb Q.
$$

Thus:

$$
\sqrt2\log2
\notin
\mathcal L_{\mathrm{alg}},
$$

Therefore:

$$
\boxed{
2^{\sqrt2}
}
$$

is transcendental.

---

# 19. Example 2: Rational Scalars Do Not Escape

Again, take:

$$
\lambda=\log2.
$$

If:

$$
\beta=\frac32,
$$

then:

$$
e^{\frac32\lambda}
=
2^{3/2}
=
2\sqrt2
\in
\overline{\mathbb Q}.
$$

Thus:

$$
\frac32\lambda
\in
\mathcal L_{\mathrm{alg}}.
$$

---

# 20. Example 3: Multi-Base Escape

Take:

$$
\lambda_1=\log2,
\qquad
\lambda_2=\log3.
$$

Since:

$$
2^m3^n=1,
\qquad
m,n\in\mathbb Z
$$

is possible only when:

$$
m=n=0,
$$

Hence:

$$
\lambda_1,\lambda_2
$$

are linearly independent over $\mathbb Q$.

Let:

$$
b=
\left(
\sqrt2,
\frac13
\right).
$$

Since:

$$
b\notin\mathbb Q^2,
$$

Thus:

$$
\boxed{
2^{\sqrt2}3^{1/3}
}
$$

is transcendental.

---

# 21. Example 4: Multi-Base Algebraic Descent

Again:

$$
\lambda_1=\log2,
\qquad
\lambda_2=\log3.
$$

Take:

$$
b=
\left(
\frac12,
\frac13
\right)
\in
\mathbb Q^2.
$$

then:

$$
e^{b\cdot\lambda}
=
2^{1/2}3^{1/3}
$$

is algebraic.

Thus, the Round 2 boundary is not:

$$
\text{whether the coefficients are integers},
$$

but:

$$
\boxed{
\text{whether the coefficients belong to }\mathbb Q.
}
$$

---

# 22. Branches Are No Longer Treated as Noise

For fixed:

$$
\alpha\in\overline{\mathbb Q}^{\times},
$$

select a logarithm:

$$
\lambda_0.
$$

All branches are:

$$
\lambda_k
=
\lambda_0+2\pi i k,
\qquad
k\in\mathbb Z.
$$

Define:

$$
\tau=2\pi i.
$$

Then:

$$
e^\tau=1,
$$

Thus:

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

Therefore:

$$
\boxed{
\lambda_k
=
\lambda_0+k\tau
}
$$

is merely an action by:

$$
\mathbb Z\subset\mathbb Q
$$

these coefficients.

In other words:

> Branch changes themselves belong to the algebraic descent-preserving region.

They are not the source of transcendental escape.

---

# 23. Separating Branches from Algebraic Irrational Scalars

Now let:

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

Every branch output:

$$
x_k
=
e^{\beta\lambda_k}
$$

is transcendental.

And:

$$
x_k
=
e^{\beta\lambda_0}
e^{2\pi i k\beta}.
$$

Let:

$$
q_\beta
=
e^{2\pi i\beta}.
$$

Then:

$$
\boxed{
x_k
=
x_0q_\beta^k.
}
$$

Applying R2.2 to:

$$
\tau=2\pi i,
$$

gives:

$$
q_\beta
$$

as another transcendental number.

---

# 24. The Relation Structure of a Branch Family

Although each:

$$
x_k
$$

is transcendental, the entire branch family is not 'random'.

Since:

$$
x_k=x_0q_\beta^k.
$$

it always satisfies:

$$
\boxed{
x_{k+1}x_{k-1}
=
x_k^2.
}
$$

More generally:

$$
x_{k+a}x_{k+b}
=
x_kx_{k+a+b}.
$$

Therefore:

$$
\boxed{
\text{Every value transcendental}
\not\Rightarrow
\text{values algebraically independent}.
}
$$

This is an important warning ACTM must retain.

---

# 25. Roots of Unity and Branch Rank

Consider:

$$
\lambda_0,
\qquad
\tau=2\pi i.
$$

If:

$$
\alpha=e^{\lambda_0}
$$

is a root of unity, there exists:

$$
m\neq0
$$

such that:

$$
\alpha^m=1.
$$

Therefore:

$$
m\lambda_0
\in
2\pi i\mathbb Z,
$$

Thus:

$$
\lambda_0
$$

and:

$$
\tau
$$

are linearly dependent over $\mathbb Q$.

If $\alpha$ is not a root of unity, they are linearly independent over $\mathbb Q$.

Thus, the branch lift itself carries a computable rank:

$$
\rho_{\mathrm{branch}}
=
\operatorname{ldim}_{\mathbb Q}
\{\lambda_0,2\pi i\}
\in
\{1,2\}.
$$

---

# 26. The New Axis of Degrees of Freedom in Round 2

The main axis of Round 1 is:

$$
\boxed{
\text{Direction Rank}
=
\operatorname{ldim}_{\mathbb Q}.
}
$$

Round 2 adds:

$$
\boxed{
\text{Scalar-Extension Defect}
=
\operatorname{SED}.
}
$$

ACTM's minimal two-axis model therefore becomes:

$$
\boxed{
(\text{Direction Rank},
\text{Scalar-Extension Defect})
}
$$

The first axis asks:

> How many independent directions are there?

The second axis asks:

> Do these directions undergo additional linear collapse when extending from $\mathbb Q$ to $\overline{\mathbb Q}$?

Within the algebraic logarithm locus:

$$
\operatorname{SED}=0.
$$

Thus, no construction with:

$$
\operatorname{SED}>0
$$

can remain entirely within that locus.

---

# 27. Coefficient Novelty

For a matrix:

$$
B\in M_{s\times r}(\overline{\mathbb Q}),
$$

we can define the working quantity:

$$
\kappa(B)
=
\dim_{\mathbb Q}
\operatorname{span}_{\mathbb Q}
\left(
\mathbb Q^r
\cup
\{b_1,\ldots,b_s\}
\right)
-r,
$$

where:

$$
b_j
$$

is the row of $B$ indexed by $j$.

It measures how many new $\mathbb Q$-directions the algebraic coefficient rows add relative to the rational coefficient space.

If:

$$
\kappa(B)=0,
$$

then:

$$
B
$$

is actually a rational matrix.

If:

$$
\kappa(B)>0,
$$

it is impossible for all:

$$
e^{(B\lambda)_j}
$$

to be algebraic.

---

# 28. Caution: $\kappa(B)$ Is Not Yet a Transcendence-Degree Lower Bound

ACTM must not directly claim:

$$
\operatorname{trdeg}
\mathbb Q
\left(
e^{B\lambda}
\right)
\ge
\kappa(B).
$$

Baker's theory suffices to rule out certain simultaneous algebraic descents, but does not automatically give this full algebraic-independence lower bound.

Therefore:

$$
\boxed{
\kappa(B)
}
$$

is currently only a candidate 'coefficient novelty' quantity.

Connecting it to:

$$
\operatorname{trdeg}
$$

is the genuinely difficult subsequent question.

---

# 29. The First Schanuel-Type Gap Appears Here

For a single output:

$$
e^{b\cdot\lambda},
$$

Round 2 can determine whether:

$$
b\in\mathbb Q^r
$$

or:

$$
b\notin\mathbb Q^r
$$

gives the exact algebraic / transcendental boundary.

However, when considering multiple:

$$
b_1,\ldots,b_s,
$$

we begin to ask:

> How many outputs are individually transcendental?

and:

> How many outputs are algebraically independent of one another?

These questions are not the same.

This naturally gives:

$$
\boxed{
\text{Individual Escape}
\quad\neq\quad
\text{Joint Algebraic Independence}.
}
$$

This is the point where ACTM gradually approaches Schanuel.

---

# 30. The Interface with Six Exponentials / Four Exponentials

The classical six exponentials theorem considers:

$$
x_1,x_2
$$

linearly independent over $\mathbb Q$, and:

$$
y_1,y_2,y_3
$$

linearly independent over $\mathbb Q$.

Then among the six:

$$
e^{x_i y_j}
$$

at least one is transcendental.

The adjacent four exponentials conjecture predicts an analogous result for:

$$
2\times2
$$

matrices, but remains an unresolved direction.

This shows that ACTM's matrix escape is not an artificial construction:

$$
\boxed{
\text{rank}
+
\text{matrix coupling}
+
\text{simultaneous descent}
}
$$

is already one of the central frontiers of transcendence theory.

---

# 31. The Core Transformation of ACTM Round 2

Round 1:

$$
\boxed{
\text{Linear Rank}
\rightarrow
\text{Transcendence Degree}.
}
$$

Round 2:

$$
\boxed{
\text{Algebraic Scalar Action}
\rightarrow
\text{Rank Defect}
\rightarrow
\text{Descent Obstruction}.
}
$$

Together, the two rounds:

$$
\boxed{
\text{Seed Directions}
+
\text{Coefficient Field}
+
\text{Operator Structure}
}
$$

begin to replace mere:

$$
\text{output classification}.
$$

---

# 32. Answering the Original Methodological Question

The original question is:

> Which ordinary algebraic objects have the combinatorial potential to become transcendental numbers in the traditional sense after being combined?

Round 2 gives an exact answer:

First, do not look only at algebraic values:

$$
\alpha_i.
$$

Lift them to logarithmic generators:

$$
\lambda_i
\in
\log\alpha_i.
$$

If these generators are independent over:

$$
\mathbb Q
$$

that field, then:

- Rational coefficient combinations can descend to algebraic outputs;
- Algebraic irrational coefficient combinations must escape algebraic outputs.

Thus, one form of 'combinatorial potential' is:

$$
\boxed{
\text{independent logarithmic directions}
+
\text{non-rational algebraic coefficients}.
}
$$

---

# 33. Round 2 Verdict

**PASS — Exact Scalar Escape Boundary Established**

This round completes:

1. Define the algebraic logarithm locus:
   $$
   \mathcal L_{\mathrm{alg}}.
   $$

2. Prove it is a $\mathbb Q$-vector space.

3. Use Gelfond–Schneider to obtain:
   $$
   \beta\lambda\in\mathcal L_{\mathrm{alg}}
   \iff
   \beta\in\mathbb Q.
   $$

4. Obtain:
   $$
   \operatorname{Stab}_{\overline{\mathbb Q}}
   (\mathcal L_{\mathrm{alg}})
   =
   \mathbb Q.
   $$

5. Introduce Scalar-Extension Defect:
   $$
   \operatorname{SED}.
   $$

6. Use Baker's theory to obtain:
   $$
   Z\subset\mathcal L_{\mathrm{alg}}
   \Rightarrow
   \operatorname{SED}(Z)=0.
   $$

7. Obtain matrix descent rigidity:
   $$
   B\lambda\in
   \mathcal L_{\mathrm{alg}}^s
   \iff
   B\in M_{s\times r}(\mathbb Q).
   $$

8. Absorb branch freedom into the:
   $$
   2\pi i
   $$
   generator, making branch shifts rational / integer transforms.

9. Separate:
   $$
   \text{branch freedom}
   $$
   from:
   $$
   \text{transcendence-generating scalar freedom}.
   $$

10. Find the next genuine frontier:
   $$
   \text{individual escape}
   \rightarrow
   \text{joint algebraic independence}.
   $$

---

# 34. The Formal Entry Point to Round 3

The most natural entry point for the next round is no longer a single:

$$
\alpha^\beta.
$$

but an exponential matrix:

$$
\boxed{
M_{ij}
=
e^{x_i y_j}.
}
$$

Study:

$$
\boxed{
\text{row rank}
+
\text{column rank}
+
\text{algebraic descent capacity}.
}
$$

Priorities:

### R3-A

Reconstruct the ACTM rank picture of the six exponentials theorem:

$$
2\times3.
$$

### R3-B

Analyze why:

$$
2\times2
$$

becomes the unresolved boundary of the four exponentials conjecture.

### R3-C

Establish a candidate quantity for the degrees of freedom of:

$$
\operatorname{MatrixEscape}(X,Y)
$$

without presupposing equality with transcendence degree.

### R3-D

Find which structural constraint is genuinely added between:

$$
2\times2
\rightarrow
2\times3
$$

rather than viewing it merely as 'two more numbers'.

This will be ACTM's first genuine entry into:

$$
\boxed{
\text{simultaneous transcendence}
}
$$

rather than single-point transcendence.

---

# 35. The One-Sentence Conclusion

> **Round 2 shows that the algebraic logarithm locus is a rigid space closed under $\mathbb Q$ but not under algebraic irrational scalars; applying coefficients in $\overline{\mathbb Q}\setminus\mathbb Q$ to independent logarithmic directions creates a rank incompatibility that prevents descent to algebraic exponential values.**

Core formulas:

$$
\boxed{
\mathcal L_{\mathrm{alg}}
=
\exp^{-1}
(\overline{\mathbb Q}^{\times})
}
$$

$$
\boxed{
\operatorname{Stab}_{\overline{\mathbb Q}}
(\mathcal L_{\mathrm{alg}})
=
\mathbb Q
}
$$

and:

$$
\boxed{
B\lambda\in
\mathcal L_{\mathrm{alg}}^s
\iff
B\in
M_{s\times r}(\mathbb Q)
}
$$

hold under the condition that $\lambda_1,\ldots,\lambda_r$ are $\mathbb Q$-independent algebraic logarithms.

This is ACTM's second formal benchmark.

---

# References

1. **Gel'fond–Schneider method**, Encyclopedia of Mathematics.  
   https://encyclopediaofmath.org/wiki/Gel%27fond%E2%80%93Schneider_method

2. **Schneider method**, Encyclopedia of Mathematics.  
   https://encyclopediaofmath.org/wiki/Schneider_method

3. **Transcendental number**, Encyclopedia of Mathematics.  
   https://encyclopediaofmath.org/wiki/Transcendental_number

4. **Gel'fond–Baker method**, Encyclopedia of Mathematics.  
   https://encyclopediaofmath.org/wiki/Gel%27fond-Baker_method

5. Alan Baker, *Transcendental Number Theory*, Cambridge University Press.

---

**End of ACTM Round 2**