It is well known that $\mathbb{C}^n$ does not admit any compact complex submanifold, I was wondering if this can happen for compact manifolds, i.e., does there exist an example of compact complex manifold with no compact submanifold?
Complex manifolds without compact submanifolds
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0yes, I mean embedded complex submanifolds of positive dimension. โ 2011-10-04
2 Answers
Yes, there exists a compact holomorphic torus $X$ of dimension $2$ with no compact holomorphic subvariety of dimension 1 (actually there exist many such tori).
Of course the only meromorphic functions on $X$ are constants: $\mathcal M (X)=\mathbb C$ .
Else the zero set or the pole set of a non-constant meromorphic function $f\in \mathcal M (X)\setminus \mathbb C$ would furnish holomorphic subvarieties of dimension $1$.
Needless to say these examples are not algebraic, that is are not abelian varieties. Indeed an algebraic variety $V$ over a field $k$ has many rational functions, since $trdeg_k (Rat(V))=dim (V)$.
Bibliography There is a discussion of 2-dimensional tori in Shafarevich's Basic Algebraic Geometry, Vol. 2, Chap. VIII, ยง1.4. You will see calculations there which show that most 2-dimensional tori contain no holomorphic curve.
Edit (later) Here is an article relevant to Mariano's question in his comment below.
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0@Mariano: Oops, sorry Mariano, I hadn't seen your comment before now (I don't get notified because I left no email address).I have added an edit addressing your question. โ 2011-10-31
Indeed, any compact submanifold in a compact complex manifold is a finite set or the whole manifold.
This can be proved by the fact: every holomorphic function on a connected compact manifold is constant.
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0Huh. That's interesting. Beware of non-Stein manifolds bearing gifts. โ 2011-11-08