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Let $\lambda_1, \lambda_2 \in \Rbb_{\neq 0}$, and
$\alpha_1 = e^{\lambda_1}$, $\alpha_2 =
e^{\lambda_2} \in \ol{\Qbb}$. Let $\beta =
\frac{\lambda_2}{\lambda_1} \in \ol{\Qbb}
\setminus \Qbb$. So we assumed that
Gelfond-Schneider is false. We aim for a
contradiction.

Let $T_0, T_1, S \in \Zbb_{> 0}$ with
\[
  L \defeq (T_0 + 1)(2T_1 + 1) = (2S + 1)^2
.\]
Consider the ``monomials''
\[
  X^\tau \exp(t\lambda_1 X)
\]
for $\tau = 0, \ldots, T_0$, $t = -T_1, \ldots,
T_1$ and the points $s_1 + \beta s_2$ for $s_1,
s_2 = -s, \ldots, s$.

\begin{notation*}
  $[-]_{\substack{\tau, t \\ s_1, s_2}}$ means a
  matrix with rows indexed by $\tau, t$ and
  columns indexed by $s_1, s_2$.
\end{notation*}

Let
\begin{align*}
  \Delta
  &= \det [(s_1 + \beta s_2)^\tau \cdot
  \exp(t \lambda_1(s_1 + \beta
  s_2))]_{\substack{\tau, t \\ s_1, s_2}} \\
  &= \det [(s_1 + \beta s_2)^\tau \alpha_1^{ts_1}
  \alpha_2^{ts_2}]_{\substack{\tau, t \\ s_1, s_2}}
\end{align*}
Steps:
\begin{enumerate}[(1)]
  \item Give an analytic upper bound on $\Delta$
  \item Give an arithmetic lower bound on $\Delta$
  \item ``zero estimate'' $\implies \Delta \neq
    0$.
\end{enumerate}
Steps (1) and (2) will be done in such a way that
together they will give $\Delta = 0$. Then this
will contradict (3).

We will alternate between viewing $(s_1 + \beta
s_2)^\tau \cdot \exp(t \lambda_1(s_1 + \beta
s_2))$ as a function of a single variable
(function of $s_1 + \beta s_2$) and thinking of it
as a function of two variables (function of $s_1$
and $s_2$).

\subsubsection*{Upper bound}

\begin{proposition*}
  For $n \in \Zbb_{> 0}$, there exists $c = c(n) >
  0$ such that the following holds:

  Let $L \in \Zbb_{> 0}$, $E \in \Rbb_{> 1}$. Let
  $f_1, \ldots, f_L : \Cbb^n \to \Cbb$ be
  \emph{analytic} functions (here, analytic means
  convergent power series on $\Cbb^n$). Let
  $\xi_1, \ldots, \xi_L \in \Cbb^n$. Let $r =
  \max_{\substack{s = 1, \ldots, L \\ j = 1,
  \ldots, n}}
  |\xi_{s, j}|$. Then
  \[
    \det[f_t(\xi_s)]_{\substack{t = 1, \ldots, L
    \\ s = 1, \ldots, L}}
    \le E^{-cL^{1 + \frac{1}{n}}} \cdot L! \cdot
    \prod_{t = 1}^{L} |f|_{Er}
  .\]
\end{proposition*}

\begin{notation*}
  $|f|_R = \max_{|x_1|, \ldots, |x_n| \le R} |f(x_1,
  \ldots, x_n)|$.
\end{notation*}

\begin{corollary*}
  With $\Delta, T_0, T_1, S, L$ as above, there
  exists $c, C > 0$ depending only on $\beta, z$,
  such that for all $E \in \Rbb_{\ge e}$:
  \[
    |\Delta| \le \exp(-cL^2 \log E + CL \cdot T_0
    \log (ES) + CLT_1ES)
  .\]
\end{corollary*}

\begin{proof}
  We take $n = 1$ and some $E \ge e$. We have
  $|s_1 + \beta s_2| < C_0 \cdot S$ with $C_0 =
  C_0(\beta)$.
  \[
    |z^\tau \exp(t\lambda_1 z)|
    < \exp(C_1 T_0 \cdot \log ES + C_1 T_1 ES)
  \]
  for $|z| < E \cdot C_0 \cdot S$, with $C_1 =
  C_1(\beta, \lambda_1)$.
\end{proof}

One possible choice of the parameters: $E = e$. $S
\sim L^{\half}$, $T_0 \sim L^{1 - \eps}$, $T_1
\sim L^\eps$. In this case:
\[
  |\Delta| = \exp(-cL^2)
.\]
(for large $L$).

\begin{lemma*}[Schwart's Lemma]
  \label{schwart}
  Let $f$ be a holomorphic function on $D_R$ the
  disc of radius $R$ with a zero of order $k$ at
  $0$. Then: for all $z \in D_R$:
  \[
    |f(z)| \le \frac{|z|^K \cdot |f|_R}{R^K}
  .\]
\end{lemma*}

\begin{proof}
  The maximum modulus principle for
  $\frac{f(z)}{z^K}$.
\end{proof}

\begin{proof}[[Proof of Proposition]]
  We apply \nameref{schwart} for
  \[
    f(z) = \det [f_t(z \cdot \xi_s)]_{\substack{t
    \\ s}}
  \]
  and $R = E$.
  Note: $|F|_E \le L! \cdot \prod_{t = 1}^{T}
  |f_t|_{Er}$.

  So the proposition follows if we show that $F$
  vanishes to order $cL^{1 + \frac{1}{n}}$ at $0$.
  We prove this. Enough to do it when each $f_t$
  is of the form $z_1^{a_1} \cdots z_n^{a_n}$ for
  some $a_1, \ldots, a_n \in \Zbb$ depending on
  $t$.

  This is because all $f_t$s are infinite linear
  combinations of such $f_t$s, and hence the
  determinant can be written as an infinite
  combination of special determinants. Furthermore
  we may assume that the $(a_1, \ldots, a_n)$ are
  distinct for different $t$s.

  Observe: $\det[f_t(z \cdot \xi_s)]_{\substack{t
  \\ s}} = z^{\sum \deg f_t} \cdot \det
  [f_t(\xi_s)]_{\substack{t \\ s}}$ if each $f_t$
  is of the special form.

  The number of monomials with degree $\le d$ is
  at most $d^n$. We take $d = \left\lfloor \left(
  \frac{L}{2} \right)^{\frac{1}{n}}
  \right\rfloor$. Then at least half of the $f_t$s
  have degree $\ge d$. So $\sum \deg f_t \ge
  \left( \frac{L}{2} \right) \cdot d \ge c \cdot
  L^{1 + \frac{1}{n}}$.
\end{proof}