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Motivating example : Using complex analysis, one can show that for some lattice $\Omega$ in $\mathbb{C}$, the field of meromorphic functions on the torus $\mathbb{C}/\Omega$ is generated by the associated Weierstrass elliptic function and its derivative, $\wp$ and $\wp'$. Moreover these functions satisfy the differential equation $$ \wp'^2=4\wp^3-g_2\wp-g_3 $$ where $g_2,g_3$ are the associated modular invariants : this relation actually gives an isomorphism between the torus $\mathbb{C}/\Omega$ and a smooth cubic in $\mathbb{P}^2(\mathbb{C})$. Thus any complex 1-torus can be embedded into the complex projective plane.

Question : Does the study of the field of meromorphic functions on some arbitrary (compact) complex manifold give us any information as to whether or not the given manifold can be embedded into some projective space ? For starters a generalization to $n$-dimensional tori seems reasonable (edit : apparently it isn't !).

More generally, having "many" such functions could be a necessary condition. In our case we have the happy accident of the Weirestrass function but I suspect the general case to be much messier. Any enlightenment would be appreciated.

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    $\begingroup$ Looking first at other compact Riemann surfaces is easier : $m$ generators of the function field makes it an algebraic variety and gives an embedding in $\mathbb{P}^m$. Complex tori of dimension $\ge 2$ aren't always algebraic varieties, if it is not then you can't embed it biholomorphically into $\mathbb{P}^n$ $\endgroup$ – reuns Nov 1 '17 at 14:43
  • $\begingroup$ @reuns Thanks ! That's a surprise. Would you care to explain why these tori fail to satisfy Chow's theorem (aren't algebraic)? $\endgroup$ – Bass Nov 1 '17 at 14:50
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    $\begingroup$ Here it gives an argument in term of homology classes of $\mathbb{C}^2/\Lambda$ $\endgroup$ – reuns Nov 1 '17 at 15:03
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Your guess is absolutely correct.

For an arbitrary compact complex manifold $X$ let $\mathcal{M}(X)$ be the field of meromorphic functions on $X$. Consider its transcendence degree $a(X):=\mathrm{tr}\deg_{\mathbb{C}} \mathcal{M}(X)$. This quantity is called the algebraic dimension of $X$.

It is a standard fact from algebraic geometry that if $X$ is algebraic and normal, then $a(X) = \dim X$. This fails for arbitrary complex manifolds and the degree of failure somehow measures non-algebraicity of the manifold.

Let me prove the following simple lemma:

Lemma: For any compact $X$ one has $0 \le a(X) \le \dim X$. Particarly, $a(X)$ is always finite.

Proof: Assume that $X$ is $n$-dimensional. Let $f_1, \ldots, f_{n+1}$ be a collection of different non-constant meromorphic functions on $X$. We want to find an algebraic relations between them. The proof goes in two steps.

Step 1. For the dimension reasons one can (algebraically) perturb all the functions until they will be in a general position.

More precisely, I claim that one can find such constants $\lambda_1, \ldots, \lambda_{n+1} \in \mathbb{C}$ that all the functions $\tilde{f}_j := f_j -\lambda_j$ do not vanish simultaneously.The argument is the following:

For each $f_j$ put $V_j(\lambda):=f_j^{-1}(\lambda)$. Each $V_j$ is an analytical subvariety of codimension $1$ in $X$. Find the smallest $k$, such that $\mathrm{dim} \bigcap_{j=1}^k V_j(0)> n-k$. Then $V':=\bigcap_{j=1}^{k-1}V_j(0)$ is a subvariety in $X$ of dimension $n-k+1$. Consider the set $f_k(V') \subset \mathbb{CP}^1$. Since $f_k$ is analytical and $X$ is compact, it is either a discrete union of points or the whole $\mathbb{CP}^1$.

In the first case one can always find a $\lambda \in \mathbb{C} \subset \mathbb{CP^1}$, such that $V' \cap V_k(\lambda) = \varnothing$. Then replacing $f_k$ with $\tilde{f}_k =f_k - \lambda$ we get sure that $\dim \bigcap_{j=1}^{k-1}V_{j}(0) \cap \tilde{f_k}^{-1}(0) = -1 < n-k$.

In the second case for a general $t \in \mathbb{CP}^1$ one has $\dim V' = 1 + \dim V' \cap f_k^{-1}(t)$, hence for a certain $\lambda \in \mathbb{C}$ $$\dim (V' \cap V_k(\lambda))= n-k,$$ so we again replace $f_k$ with $\tilde{f}_k = f_k - \lambda$.

Iterating the process one gets sure that $\dim \bigcap_{j=1}^{n+1}V_j(0)< n-(n+1)$, hence $\bigcap V_j(0)$ is empty.

Step 2. Starting from now we might assume that all the $f_j$ do not vanish simultaneously. Indeed, any algebraic relation between $\tilde{f}_j$ will give us an algebraic relation between $f_j$.

Consider the map $F \colon X \to \mathbb{CP}^{n+1}$ given by $z \mapsto [1: f_1(z): \ldots :f_{n+1}(z)]$. By the step 2 it is a well-defined holomorphic map. Since $X$ is compact, $Y := F(X)$ is a closed subvariety in $\mathbb{CP}^{n+1}$. Due to Chow's lemma any such subvariety is algebraic, i.e. is defined by a system of homogenous polynomials. By the construction, $f_1, \ldots, f_{n+1}$ are the roots of any of this polynomials. This finishes the proof.


Let me also make a couple of additional remarks.

First if all, it is easy to see from the proof of the lemma, that if $a(X) = n$, then $\dim Y = \dim X$ and after passing to a finite covering of $Y$ we get that $X$ is bimeromorphic to a certain projective variety. Such manifolds are called Moishezon. There is a deep theorem (due to Moishezon, as well) that a complex manifold is projective iff it is both Moishezon and Kähler.

Generalizing this, notice that if $\pi \colon B \to Y$ is the normalization, one gets a meromorphic map $\phi \colon X \dashrightarrow B$ from $X$ to a normal projective variety of dimension $\dim B = a(X)$. Such a pair $(B, \phi)$ is called the algebraic reduction of $X$. It is characterized by the following property: $\phi^* \colon \mathcal{M}(B) \to \mathcal{M}(X)$ is an isomorphism. It is not unique, however for any two algebraic reductions $(B, \phi)$ and $(B', \phi')$ the varieties $B$ and $B'$ are normal projective and has isomorphic function fields. It is a general result that such varieties are always bimeromorphic.

Finally, it is necessary to mention that the estimate from the lemma cannot be improved. It is a nice exercise to explicitly construct, for a given $k<n$, an $n$-dimensional complex torus whith algebraic dimension being equal to $k$. Some of the examples are discussed in "Basic algebraic geometry:II" by Igor Shafarevich. You can also find there another (less geometrical and more analytical) proof of the Lemma above.

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  • $\begingroup$ Thank you so much for this. $\endgroup$ – Bass Nov 7 '17 at 23:57

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