Here is a conjecture in transcendental number theory.
Conjecture 1 (Schanuel, 1966). Let $z_1, \dotsc, z_n \in \mathbb{C}$ be a linearly independent over $\mathbb{Q}$. Then $$ \operatorname{trdeg}_ \mathbb{Q} \mathbb{Q}(z_1, \dotsc, z_n, e^{z_1}, \dotsc, e^{z_n}) \ge n. $$
This implies that stuff like $e^{\sqrt{2}}$ or $\pi$ are transcendental. The case when $z_1, \dotsc, z_n \in \bar{\mathbb{Q}}$ is known, and is called the Lindemann–Weierstrass theorem.
Theorem 2 (Ax, 1971). Let $f_1, \dotsc, f_n \in \mathbb{C}[[t_1, \dotsc, t_m]]$ be elements that are $\mathbb{Q}$-linearly independent when we throw away the constant terms. Then we have $$ \operatorname{trdeg}_ \mathbb{C} \mathbb{C}(f_1, \dotsc, f_n, e^{f_1}, \dotsc, e^{f_n}) \ge n + \operatorname{rk} \Bigl(\frac{\partial f_i}{\partial t_j}\Bigr), $$ where we calculate the rank in the fraction field of $\mathbb{C}[[t_1, \dotsc, t_m]]$.
The geometric interpretation#
Assume that $f_1, \dotsc, f_n$ all have positive radii of convergence. Then these functions can be combined into holomorphic map $$ f \colon B(0,\epsilon)^m \to \mathbb{C}^n $$ for small enough $\epsilon$. Since complex analytic spaces have a well-behaved dimension theory, we can ask what its dimension is.
Theorem 3 (V3). Let $V \subseteq \mathbb{C}^n \times (\mathbb{C}^\times)^n$ be an irreducible subvariety, and let $U$ be an irreducible component of $V \cap D_n$. If $\pi_m(U)$ is not contained in a coset of a proper subtorus of $(\mathbb{C}^\times)^n$, then $$ \dim_\mathbb{C} V \ge \dim_\mathbb{C} U + n. $$
The proof#
We will use o-minimality of the structure $\mathbb{R}_ {\mathrm{an},\mathrm{exp}}$. We will also be using induction on the lexicographical ordering of $(n, \dim V - \dim U, n - \dim U)$.
Definition 4. For $X \subseteq \mathbb{C}^n \times (\mathbb{C}^n)^\times$ an irreducible analytic set, we denote by $\pi_a(X) \subseteq X^\mathrm{lin} \subseteq \mathbb{C}^n$ the smallest affine $\mathbb{C}$-linear subspace containing $\pi_a(X)$.
We also define, for $0 \le d \le n$ and $L \subseteq \mathbb{C}^n$ a $\mathbb{C}$-linear subspace, the open subset $$ G_d(X, L) \subseteq X $$ consisting of those points $x \in X$ with the property that
- $X$ is smooth at $x$ of dimension $d$,
- for the irreducible component $X_0 \subseteq X$ containing $x$, the affine linear subspace $X_0^\mathrm{lin}$ is a translate of $L$.
We now consider the subset $$ F = \lbrace (\underline{z}, \underline{w}) : 0 \le \operatorname{Im}(z_i) \le 2 \pi \rbrace \subseteq \mathbb{C}^n \times (\mathbb{C}^\times)^n, $$ so that $D_n \cap F$ is some sort of “$(\mathbb{C}^\times)^n$ with cuts.” We now consider the set $$ I = \lbrace \ell \in i\mathbb{R}^n : G_{\dim U}((V+\ell) \cap (D_n \cap F), U^\mathrm{lin}) \neq \emptyset \rbrace, $$ where the translate $V + \ell$ means adding $\ell$ on the additive part.
The point is that for $\ell \in 2\pi i \mathbb{Z}$, the condition is that $$ G_{\dim U}(V \cap D_n \cap (F-\ell), U^\mathrm{lin}) = G_{\dim U}(V \cap D_n, U^\mathrm{lin}) \cap (F-\ell) \neq \emptyset. $$ Because $U$ is irreducible, we see that $\ell \in 2 \pi i \mathbb{Z}$ is in $I$ if $U \cap (F - \ell) \neq \emptyset$.
Claim 5. This subset $I \subseteq i\mathbb{R}^n$ is definable.
To see this, we first note that $D_n \cap F$ is definable, even if $D_n$ isn’t. Now under the isomorphism $D_n \cap F = (i[0, 2\pi] + \mathbb{R})^n$ the equation cutting out $V$ pulls back to a nice definable holomorphic function, and detecting whether they cut out a $d$-dimensional submanifold with given linear span is can be encoded in first-order logic.
Claim 6. The set $I \cap 2\pi i \mathbb{Z}^n$ is infinite.
If $I \cap 2 \pi i \mathbb{Z}^n$ is finite, then the $\ell \in 2 \pi i \mathbb{Z}^n$ such that $U \cap (F-\ell) \neq \emptyset$ is also finite. This means that $U$ is definable in $\mathbb{C}^n \times (\mathbb{C}^\times)^n$. But $U$ is a closed analytic subset, and so if it is also definable then it is an positive-dimensional algebraic subvariety by definable Chow. Now we get a contradiction because there are no positive-dimensional varieties contained in $D_n$.
Claim 7. The size of $I \cap 2 \pi i \lbrace -N, -N+1, \dotsc, N \rbrace^n$ is at least $N + O(1)$.
Again, we use the fact that if $U \cap (F-\ell) \neq \emptyset$ then $\ell \in I$. We now consider the imaginary part of $\pi_a(U)$, as a subset of $\mathbb{R}^n$. Because $U$ is connected, we can now draw a lattice and see this.
Theorem 8 (Pila–Wilkie). If $I \subseteq i\mathbb{R}^n$ is a definable set with the property that $I \cap 2 \pi i \lbrace -N, \dotsc, N \rbrace^n$ has at least $N + O(1)$ elements, there exists a semialgebraic curve $C_\mathbb{R} \subseteq I$ containing a point $\ell \in C_\mathbb{R} \cap 2 \pi i \mathbb{Z}^n$ at which $C_\mathbb{R}$ is smooth.
Let $C$ be the complex algebraic curve corresponding to $C_\mathbb{R} \subseteq I$. By definition, for each point $c \in C_\mathbb{R}$ there is an irreducible component $$ W_c \subseteq (V+c) \cap (D_n \cap F) $$ that has dimension equal to $\dim U$ satisfying the property that $W_c^\mathrm{lin}$ is a translate of $U^\mathrm{lin}$.
- Case 1: there are infinitely many components $W_c$ as $c \in C_\mathbb{R}$ vary.
Note that $V+C$ (or rather its closure) is another algebraic subvariety over $\mathbb{C}^n \times (\mathbb{C}^\times)^n$. Because each $W_c$ is contained an irreducible component of $(V+C) \cap D_n$, there exists some irreducible component $W \subseteq (V+C) \cap D_n$ that contains infinitely many $W_c$. This implies that $$ \dim W \ge \dim U + 1. $$ On the other hand, $U^\mathrm{lin}$ is not contained in a $\mathbb{Q}$-coefficient proper $\mathbb{C}$-linear subspace of $\mathbb{C}^n$, and so neither is $W^\mathrm{lin}$. Applying the induction hypothesis to $V+C$ and $W$, we conclude that $$ \dim V + 1 \ge \dim V+C \ge \dim W + n \ge \dim U + n + 1. $$
- Case 2: there are only finitely many components $W_c$.
In this case, there is some component $W$ such that $W = W_c$ for infinitely many $c \in C_\mathbb{R}$. So by analyticity (assuming that $C$ is connected) $W$ is contained in $V+c$ for all $c \in C$.
If $V+c$ depends on $c$, we can apply the induction hypothesis to $W$ and $\bigcap_{c \in C} (V+c)$, because $W^\mathrm{lin} = U^\mathrm{lin}$, and obtain $$ \dim U = \dim W \ge \dim \bigcap_{c \in C} (V+c) + n \ge \dim V + n. $$ So we may as well assume that $V+c$ is independent of $c \in C$. In particular, $V$ is stable under the translation of the $\mathbb{C}$-vector space spanned by $C-C$.
Claim 9 (confusion). We may as well assume that the $\mathbb{C}$-vector space spanned by $C-C$ contains an element of $\mathbb{Q}^n$.
This allows us to make a change of coordinates so that $$ V = \mathbb{C} \times V_0, \quad V_0 \subseteq \mathbb{C}^{n-1} \times \mathbb{C}^\times \times (\mathbb{C}^\times)^{n-1}. $$ In this case, we can write $$ U = \bigcup_{z \in D_1} \lbrace z \rbrace \times U_z. $$