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

# ACTM Round 1
## Exact Freedom Conversion of Algebraic Seeds Under the Exponential Map
### Exact Freedom and Relation Transport of Algebraic Seeds Under Exponentiation

**Research programme:** 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 1 / v0.1  
**Document type:** Formal research note / calibration round  
**Novelty statement:** The main results of this round are structural consequences of the Lindemann–Weierstrass theorem and their ACTM reformulation; no new transcendental-number theorem is claimed. At present, the new aspects lie only in the organisation of the research, candidate quantities, and subsequent research questions.

---

## Abstract

The first round of ACTM does not directly tackle the general complex inputs of Schanuel's conjecture, but restricts attention to the cleanest calibration domain:

$$
A=(a_1,\ldots,a_n)\in\overline{\mathbb Q}^{\,n},
$$

and studies the coordinatewise exponential map

$$
\exp(A)
=
(e^{a_1},\ldots,e^{a_n}).
$$

This round obtains an exact benchmark of great importance to ACTM: if

$$
r(A)
=
\operatorname{ldim}_{\mathbb Q}(a_1,\ldots,a_n),
$$

then the Lindemann–Weierstrass theorem implies

$$
\boxed{
\operatorname{trdeg}_{\mathbb Q}
\mathbb Q(e^{a_1},\ldots,e^{a_n})
=
r(A).
}
$$

Thus, in the purely algebraic seed domain, the exponential map converts each $\mathbb Q$-linear degree of freedom into exactly one degree of algebraic independence.

Furthermore, define the integer relation lattice

$$
L(A)
=
\left\{
m\in\mathbb Z^n:
m\cdot A=0
\right\}.
$$

Then all algebraic relations of $\exp(A)$ in the Laurent polynomial ring are generated precisely by

$$
X^m-1,
\qquad
m\in L(A)
$$

In other words, on algebraic inputs, the exponential function produces no additional hidden algebraic relations; integer additive relations on the input side are converted exactly into multiplicative binomial relations on the output side.

This round therefore establishes ACTM's first calibration principle:

$$
\boxed{
\text{Additive Linear Freedom}
\longleftrightarrow
\text{Multiplicative Algebraic Freedom}
}
$$

and its first exact relation-transport law:

$$
\boxed{
\text{Input Relation Lattice}
\longleftrightarrow
\text{Output Lattice Ideal}.
}
$$

---

# 1. Research Questions for This Round

The original question of ACTM is:

> Which algebraic inputs have combinatorial potential such that a non-algebraic generating operator produces new transcendental degrees of freedom?

The first round selects the most classical operator:

$$
\mathcal O=\exp.
$$

and imposes the restriction:

$$
A\subseteq\overline{\mathbb Q}.
$$

The research objective is not to prove Lindemann–Weierstrass again, but to answer three ACTM questions:

1. Which kind of freedom in the algebraic input controls the transcendental freedom of the output?
2. How are relations in the input transformed after applying $\exp$?
3. Apart from relations directly induced by the input relations, does the output have any additional hidden algebraic relations?

---

# 2. Basic Setup

Let

$$
A=(a_1,\ldots,a_n)
\in
\overline{\mathbb Q}^{\,n}.
$$

Define its $\mathbb Q$-linear rank:

$$
r(A)
=
\operatorname{ldim}_{\mathbb Q}
\operatorname{span}_{\mathbb Q}
\{a_1,\ldots,a_n\}.
$$

For brevity, write below:

$$
r=r(A).
$$

Define the exponential image:

$$
E(A)
=
(e^{a_1},\ldots,e^{a_n})
\in
(\mathbb C^\times)^n.
$$

The first candidate measure of combinatorial potential in ACTM is defined as:

$$
\boxed{
\operatorname{CP}_{\exp}^{\mathrm{alg}}(A)
:=
r(A).
}
$$

For now, this definition applies only to purely algebraic seeds.

---

# 3. Calibration Theorem: Lindemann–Weierstrass

The classical theorem used in this round can be stated as follows:

> If
>
> $$
> \beta_1,\ldots,\beta_r
> $$
>
> are $\mathbb Q$-linearly independent algebraic numbers, then
>
> $$
> e^{\beta_1},\ldots,e^{\beta_r}
> $$
>
> are algebraically independent over $\mathbb Q$.

An equivalent formulation is:

If

$$
\lambda_1,\ldots,\lambda_s
$$

are distinct algebraic numbers, and

$$
c_1,\ldots,c_s
\in
\overline{\mathbb Q}
$$

are not all zero, then

$$
c_1e^{\lambda_1}
+\cdots+
c_se^{\lambda_s}
\neq0.
$$

The second formulation is particularly useful for the "complete relation transport" of this round.

---

# 4. Proposition R1.1: Exact Freedom Conversion

## Proposition

For any

$$
A=(a_1,\ldots,a_n)
\in
\overline{\mathbb Q}^{\,n},
$$

we have

$$
\boxed{
\operatorname{trdeg}_{\mathbb Q}
\mathbb Q
(e^{a_1},\ldots,e^{a_n})
=
r(A).
}
$$

Since all entries of $A$ itself are algebraic, this is also equivalent to:

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

---

## Proof

Take

$$
\beta_1,\ldots,\beta_r
$$

as a basis for

$$
\operatorname{span}_{\mathbb Q}
\{a_1,\ldots,a_n\}
$$

over $\mathbb Q$.

Since each $a_i$ is algebraic, each $\beta_j$ is also algebraic.

By Lindemann–Weierstrass:

$$
e^{\beta_1},\ldots,e^{\beta_r}
$$

are algebraically independent, so

$$
\operatorname{trdeg}_{\mathbb Q}
\mathbb Q
(e^{\beta_1},\ldots,e^{\beta_r})
=
r.
$$

On the other hand, each $a_i$ can be written as

$$
a_i
=
\sum_{j=1}^{r}
q_{ij}\beta_j,
\qquad
q_{ij}\in\mathbb Q.
$$

After clearing denominators, there exists a positive integer $N$ such that:

$$
N a_i
=
\sum_{j=1}^{r}
m_{ij}\beta_j,
\qquad
m_{ij}\in\mathbb Z.
$$

Therefore:

$$
(e^{a_i})^N
=
\prod_{j=1}^{r}
(e^{\beta_j})^{m_{ij}}.
$$

Thus each $e^{a_i}$, relative to

$$
\mathbb Q
(e^{\beta_1},\ldots,e^{\beta_r})
$$

is algebraic.

In the reverse direction, each $\beta_j$ can also be written as a rational linear combination of the $a_i$, so there likewise exists an integer $M$ such that:

$$
(e^{\beta_j})^M
$$

belongs to

$$
\mathbb Q
(e^{a_1},\ldots,e^{a_n}).
$$

Thus the two function fields differ only by algebraic extensions and have the same transcendence degree:

$$
\operatorname{trdeg}_{\mathbb Q}
\mathbb Q
(e^{a_1},\ldots,e^{a_n})
=
r.
$$

This completes the proof.

---

# 5. The First ACTM Conclusion: Combinatorial Potential Is Linear Rank, Not Numerical Size

For purely algebraic inputs, under the exponential operator:

$$
\boxed{
\operatorname{CP}_{\exp}^{\mathrm{alg}}(A)
=
\operatorname{ldim}_{\mathbb Q}(A)
}
$$

and:

$$
\boxed{
\Delta_{\exp}(A)
=
\operatorname{CP}_{\exp}^{\mathrm{alg}}(A).
}
$$

Thus the first round answers:

> Which algebraic numbers have "combinatorial potential"?

At least for $\exp$, what matters is not:

- whether $a_i$ is large;
- whether its decimal expansion is complicated;
- how high its algebraic degree is by itself;
- whether it looks irregular;

but:

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

---

# 6. The Integer Relation Lattice

Define:

$$
L(A)
=
\ker
\left(
\mathbb Z^n
\longrightarrow
\overline{\mathbb Q}
\right),
$$

where the map is:

$$
m=(m_1,\ldots,m_n)
\longmapsto
m_1a_1+\cdots+m_na_n.
$$

Thus:

$$
L(A)
=
\left\{
m\in\mathbb Z^n:
m\cdot A=0
\right\}.
$$

By the rank–nullity relation:

$$
\boxed{
\operatorname{rank}_{\mathbb Z}L(A)
=
n-r(A).
}
$$

This lattice completely describes the integer additive relations among the input coordinates.

---

# 7. Direct Relation Transport by the Exponential Map

If

$$
m\in L(A),
$$

then:

$$
m\cdot A=0.
$$

Therefore:

$$
e^{m\cdot A}=1.
$$

that is:

$$
\prod_{i=1}^{n}
(e^{a_i})^{m_i}
=
1.
$$

Using

$$
X_i=e^{a_i},
$$

as notation, we obtain:

$$
X^m=1.
$$

Thus the additive relation in the input:

$$
\sum_i m_i a_i=0
$$

is converted into the multiplicative relation in the output:

$$
\prod_i X_i^{m_i}=1.
$$

This is a direct manifestation in this problem of the group-homomorphism structure

$$
\boxed{
(\mathbb C,+)
\xrightarrow{\exp}
(\mathbb C^\times,\cdot)
}
$$

displayed above.

---

# 8. Proposition R1.2: Complete Relation Transport

Let the Laurent polynomial ring be:

$$
R
=
\overline{\mathbb Q}
[X_1^{\pm1},\ldots,X_n^{\pm1}].
$$

Define the evaluation map:

$$
\operatorname{ev}_A:
R
\longrightarrow
\mathbb C
$$

by:

$$
X_i
\longmapsto
e^{a_i}.
$$

Define the output relation ideal:

$$
J_{\exp}(A)
=
\ker(\operatorname{ev}_A).
$$

---

## Proposition

We have:

$$
\boxed{
J_{\exp}(A)
=
\left\langle
X^m-1:
m\in L(A)
\right\rangle.
}
$$

In other words:

> All Laurent algebraic relations of $\exp(A)$ are generated by the integer linear relations of the input $A$.

---

# 9. Proof of Proposition R1.2

Take any:

$$
P(X)
=
\sum_{u\in F}
c_uX^u
\in
R,
$$

where $F\subset\mathbb Z^n$ is finite, and:

$$
c_u\in\overline{\mathbb Q}.
$$

Suppose:

$$
P(e^{a_1},\ldots,e^{a_n})=0.
$$

Then:

$$
\sum_{u\in F}
c_u e^{u\cdot A}
=
0.
$$

This is because:

$$
u\cdot A
$$

is still an algebraic number.

Now define an equivalence relation on $F$:

$$
u\sim v
\iff
u\cdot A=v\cdot A.
$$

Equivalently:

$$
u-v\in L(A).
$$

Grouping all terms with the same exponent

$$
\lambda=u\cdot A
$$

gives:

$$
\sum_{\lambda}
C_\lambda e^\lambda
=
0,
$$

where:

$$
C_\lambda
=
\sum_{u:\,u\cdot A=\lambda}
c_u
\in
\overline{\mathbb Q}.
$$

Distinct $\lambda$ are distinct algebraic numbers.

By the linear-independence formulation of Lindemann–Weierstrass:

$$
C_\lambda=0
$$

holds for every $\lambda$.

Thus the sum of the coefficients in each equivalence class is zero.

Fix a representative $u_0$ of that class. Then:

$$
\sum_{u\sim u_0}
c_uX^u
=
\sum_{u\neq u_0}
c_u
\left(
X^u-X^{u_0}
\right).
$$

Moreover:

$$
X^u-X^{u_0}
=
X^{u_0}
\left(
X^{u-u_0}-1
\right),
$$

and:

$$
u-u_0\in L(A).
$$

Thus each group belongs to:

$$
\left\langle
X^m-1:
m\in L(A)
\right\rangle.
$$

Hence:

$$
P
\in
\left\langle
X^m-1:
m\in L(A)
\right\rangle.
$$

The reverse direction is immediate.

This completes the proof.

---

# 10. Geometric Version

Let:

$$
x_A
=
(e^{a_1},\ldots,e^{a_n})
\in
\mathbb G_m^n.
$$

Consider its closure over

$$
\overline{\mathbb Q}
$$

in the Zariski topology:

$$
T_A
=
\overline{\{x_A\}}^{\,\mathrm{Zar}}_{\overline{\mathbb Q}}.
$$

By Proposition R1.2, $T_A$ is defined by:

$$
X^m=1,
\qquad
m\in L(A)
$$

and is therefore an algebraic subtorus, with:

$$
\boxed{
\dim T_A
=
n-\operatorname{rank}_{\mathbb Z}L(A)
=
r(A).
}
$$

Thus:

$$
\boxed{
\dim
\overline{\{\exp(A)\}}^{\,\mathrm{Zar}}_{\overline{\mathbb Q}}
=
\operatorname{ldim}_{\mathbb Q}(A).
}
$$

---

# 11. The Exact Conservation Identities of Round 1

Input freedom:

$$
F_{\mathrm{in}}
=
r(A).
$$

Number of independent integer constraints on the input:

$$
C_{\mathrm{in}}
=
n-r(A).
$$

Output Zariski dimension:

$$
F_{\mathrm{out}}
=
\dim T_A
=
r(A).
$$

Output algebraic codimension:

$$
C_{\mathrm{out}}
=
n-\dim T_A
=
n-r(A).
$$

Thus:

$$
\boxed{
F_{\mathrm{in}}
=
F_{\mathrm{out}}
}
$$

and:

$$
\boxed{
C_{\mathrm{in}}
=
C_{\mathrm{out}}.
}
$$

This is ACTM's first exact conservation law for relation transport on purely algebraic exponential inputs.

---

# 12. Some Benchmark Cases

## 12.1 $A=(1)$

$$
r(A)=1,
\qquad
L(A)=\{0\}.
$$

Therefore:

$$
\operatorname{trdeg}_{\mathbb Q}\mathbb Q(e)=1.
$$

---

## 12.2 $A=(1,2)$

$$
r(A)=1,
$$

and:

$$
L(A)=\mathbb Z(2,-1).
$$

Thus the output:

$$
(e,e^2)
$$

satisfies:

$$
X_1^2-X_2=0,
$$

and has no additional independent algebraic relations.

---

## 12.3 $A=(1,\sqrt2)$

$$
r(A)=2,
\qquad
L(A)=\{0\}.
$$

Therefore:

$$
e,
\quad
e^{\sqrt2}
$$

are algebraically independent:

$$
\operatorname{trdeg}_{\mathbb Q}
\mathbb Q(e,e^{\sqrt2})
=
2.
$$

---

## 12.4 $A=(1,\sqrt2,2\sqrt2)$

$$
r(A)=2,
$$

and:

$$
(0,2,-1)\in L(A).
$$

Therefore:

$$
X_2^2-X_3=0
$$

is an output relation, while:

$$
e,
\quad
e^{\sqrt2}
$$

retain two algebraically independent directions.

---

## 12.5 $A=(0,1)$

$$
r(A)=1.
$$

From:

$$
(1,0)\in L(A)
$$

we obtain:

$$
X_1-1=0.
$$

A zero seed supplies no transcendental degree of freedom.

---

## 12.6 $A=(1,-1)$

$$
r(A)=1,
$$

and:

$$
(1,1)\in L(A).
$$

Therefore:

$$
X_1X_2-1=0.
$$

Additive cancellation on the input side is translated exactly into multiplicative cancellation on the output side.

---

# 13. The Value of Schanuel Predimension on Algebraic Seeds

Define:

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

By R1.1:

$$
\boxed{
\delta(A)=0
}
$$

For all:

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

this identity holds.

Thus:

> All purely algebraic seeds lie on the zero boundary of Schanuel predimension.

This supplies a fixed reference surface for subsequent research:

$$
\boxed{
\text{Algebraic Domain}
\Rightarrow
\delta=0.
}
$$

---

# 14. ACTM's First Structural Interpretation

This round shows that, on purely algebraic seeds, the exponential function is not:

$$
\text{algebraic}
\rightarrow
\text{mysteriously transcendental}.
$$

A more precise description is:

$$
\boxed{
\text{additive linear freedom}
\rightarrow
\text{algebraic transcendence freedom}.
}
$$

At the same time:

$$
\boxed{
\text{integer additive relations}
\rightarrow
\text{multiplicative lattice relations}.
}
$$

Thus the main phenomenon observed in this round is not "freedom appearing from nothing", but:

$$
\boxed{
\text{freedom-type conversion}.
}
$$

---

# 15. ACTM's First Candidate Scale

Define the calibration scale:

$$
\boxed{
\operatorname{ACP}_{\exp}(A)
=
\operatorname{ldim}_{\mathbb Q}(A),
\qquad
A\in\overline{\mathbb Q}^{\,n}.
}
$$

Here ACP stands for:

**Algebraic Combinatorial Potential**.

In the calibration domain of this round:

$$
\boxed{
\operatorname{ACP}_{\exp}(A)
=
\operatorname{trdeg}_{\mathbb Q}\mathbb Q(e^A)
=
\dim T_A.
}
$$

This is not a claim to a new mathematical invariant, but a boundary condition that all subsequent ACTM generalisations must satisfy.

---

# 16. The First Falsifiable Direction of Generalisation

Round 1 supports, but does not prove, the following more general principle:

> For transcendental operators with a group-homomorphism or uniformisation structure, the capacity to generate transcendence may be controlled by "the number of independent input directions not absorbed by the operator kernel and algebraic subgroups".

Abstractly:

$$
\boxed{
\text{Transcendence Potential}
\sim
\text{Free Directions Modulo Structural Kernel}.
}
$$

In the setting:

$$
\overline{\mathbb Q}^{\,n}
\xrightarrow{\exp}
\mathbb G_m^n
$$

it passes the first-round calibration.

However, it may fail for other operators and must therefore be tested in a counterexample-first manner.

---

# 17. The Boundaries of Round 1

## 17.1 The Number of Coordinates Is Not the Number of Degrees of Freedom

For example:

$$
A=(1,2,\ldots,N)
$$

has $N$ coordinates, but:

$$
r(A)=1.
$$

Thus:

$$
\operatorname{trdeg}_{\mathbb Q}
\mathbb Q(e,e^2,\ldots,e^N)
=
1.
$$

---

## 17.2 No Direct Extension to Non-Algebraic Inputs

If:

$$
A\not\subseteq\overline{\mathbb Q},
$$

then R1.1 is no longer guaranteed by Lindemann–Weierstrass.

Thus one cannot directly assert that:

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

holds for every complex tuple.

---

## 17.3 The General Complex Domain Reintroduces the Exponential Kernel

In general:

$$
e^z=1
$$

allows:

$$
z\in2\pi i\mathbb Z.
$$

Whereas in the purely algebraic seed domain:

$$
\overline{\mathbb Q}
\cap
2\pi i\mathbb Z
=
\{0\},
$$

so the kernel produces no nonzero algebraic direction of degeneration.

This is an important reason why Round 1 is particularly clean.

---

# 18. Connection to the Method of "Compressing $\pi$"

The earlier method pointed out:

$$
\text{Object}
\neq
\text{Representation}
\neq
\text{Expansion}.
$$

The corresponding version of Round 1 for the problem of transcendence generation is:

$$
\boxed{
\text{Output Number}
\neq
\text{Generating Seed}
\neq
\text{Relation Structure}.
}
$$

Thus ACTM does not study:

$$
e^{\sqrt2}
$$

as an isolated output, but places it back into:

$$
(1,\sqrt2)
\xrightarrow{\exp}
(e,e^{\sqrt2}),
$$

and studies how:

$$
\operatorname{ldim}_{\mathbb Q}(1,\sqrt2)=2
$$

is mapped exactly to:

$$
\operatorname{trdeg}_{\mathbb Q}
\mathbb Q(e,e^{\sqrt2})=2.
$$

---

# 19. Round 1 Assessment

**PASS — Calibration Successful**

This round completes:

1. Exact input freedom:
   $$
   r(A)=\operatorname{ldim}_{\mathbb Q}(A).
   $$

2. Exact output transcendental freedom:
   $$
   \operatorname{trdeg}_{\mathbb Q}\mathbb Q(e^A).
   $$

3. A proof that the two are equal.

4. Construction of the complete input relation lattice:
   $$
   L(A).
   $$

5. Construction of the complete output lattice ideal:
   $$
   J_{\exp}(A).
   $$

6. Exclusion of additional hidden algebraic relations.

7. Construction of the Zariski geometric version:
   $$
   \dim T_A=r(A).
   $$

8. Calibration of Schanuel predimension:
   $$
   \delta(A)=0.
   $$

---

# 20. The Formal Entry Point for Round 2

The next round does not jump directly to general Schanuel.

It is proposed to study a two-layer combination:

$$
\boxed{
(\alpha,\beta)
\mapsto
\alpha^\beta
=
e^{\beta\log\alpha}.
}
$$

Reasons:

1. It connects to Gelfond–Schneider;
2. It combines multiplication, logarithms, and exponentiation;
3. Branch structure will appear;
4. It can test whether "linear rank" still suffices as a measure of combinatorial potential;
5. It will very likely force ACTM to introduce a second freedom axis;
6. It is the smallest nontrivial step from a single operator to a composite generator.

---

# 21. One-Sentence Conclusion

> **For purely algebraic tuples, the exponential function translates the input's $\mathbb Q$-linear freedom exactly into algebraic-independence freedom in the output, and translates all integer additive relations exactly into multiplicative lattice relations.**

Core identity:

$$
\boxed{
\operatorname{ldim}_{\mathbb Q}(A)
=
\operatorname{trdeg}_{\mathbb Q}\mathbb Q(e^A)
=
\dim
\overline{\{e^A\}}^{\,\mathrm{Zar}}_{\overline{\mathbb Q}}.
}
$$

and:

$$
\boxed{
L(A)
\longleftrightarrow
J_{\exp}(A).
}
$$

This is ACTM's first formal benchmark.

---

# References

1. Lindemann theorem, *Encyclopedia of Mathematics*.  
   https://encyclopediaofmath.org/wiki/Lindemann_theorem

2. Yu. V. Nesterenko, “Lindemann–Weierstrass theorem,” *Moscow University Mathematics Bulletin* 76(6), 239–243, 2021.  
   DOI: 10.3103/S0027132221060073

3. Jean-Paul Bezivin and Philippe Robba, “A new p-adic method for proving irrationality and transcendence results,” *Annals of Mathematics* 129(1), 151–160, 1989.  
   DOI: 10.2307/1971488

4. Manuel Eberl, *Hermite–Lindemann–Weierstrass Transcendence Theorem*, Archive of Formal Proofs.  
   https://isa-afp.org/browser_info/current/AFP/Hermite_Lindemann/Hermite_Lindemann.html

5. Vahagn Aslanyan, “Adequate predimension inequalities in differential fields,” *Annals of Pure and Applied Logic* 173(1), 103030, 2022.  
   DOI: 10.1016/j.apal.2021.103030

6. Daniel Bertrand and Anand Pillay, “A Lindemann-Weierstrass theorem for semi-abelian varieties over function fields,” *Journal of the American Mathematical Society* 23(2), 491–533, 2010.  
   DOI: 10.1090/S0894-0347-09-00653-5

---

**End of ACTM Round 1**