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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?

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Do you mean proper, non-empty submanifold? –  Rasmus Oct 4 '11 at 19:44
@Lucas: I suppose you are excluding the case of a point (or a set of points)? Such a thing is trivially a compact manifold... –  Zhen Lin Oct 4 '11 at 21:11
yes, I mean embedded complex submanifolds of positive dimension. –  Lucas Kaufmann Oct 4 '11 at 21:51

2 Answers 2

up vote 5 down vote accepted

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, §1.4. 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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I should probably go read in the book, I know... but do all non-algebraic $2$-tori have this property of not containing compact holomorphic curves? –  Mariano Suárez-Alvarez Oct 4 '11 at 22:57
@Mariano: No, there are non-algebraic tori with non-zero Picard number which admit closed complex subvarieties. IIRC this is shown in Birkenhake & Lange's book (either the actual one, or "Complex tori" which is milling about on the internet somewhere (edit: section 1.7 of "Complex tori" seems to be where the money is)). –  Gunnar Magnusson Oct 31 '11 at 7:24
@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. –  Georges Elencwajg Oct 31 '11 at 8:47

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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Any compact submanifold in a non-compact complex manifold is a finite set of points. There are many compact complex manifold with lots of subvarieties, like projective manifolds. For an explicit example, $\mathbb P^2$ contains $\mathbb P^1$ as a submanifold. –  Gunnar Magnusson Oct 31 '11 at 7:27
Thank you for your correction ! I know where I've made a mistake in my brain. The non-compactness is essential. –  user18537 Oct 31 '11 at 10:17
@Gunnar Magnusson: Unfortunately, non-compactness is not essential. Indeed, you can change your counterexample a little to make a non-compact complex manifold having proper compact submanifolds which is not finite. The real essential factor is the dimension. We have in dimension 1, any connected complex manifold possesses only finite complex proper compact submanifold. In dimension greater than 1, this is not correct. –  user18537 Nov 7 '11 at 11:19
The maximum principle says that any compact subvariety of a variety that can be embedded in $\mathbb C^N$ is a finite set of points. Now, perhaps one can find a non-compact manifold with a proper compact subvariety, but one would have to start by giving an example of a non-Stein manifold which does not lie in $\mathbb C^N$. In any case this would require more than a slight modification of my example, but I'd be very interested in seeing it if it exists. –  Gunnar Magnusson Nov 7 '11 at 12:24
@Gunnar Magnusson: Let $M=\mathbb{C}\mathbb{P}^2-\{[0,0,1]\}$, $M$ is a complex manifold. $\{[z_1,z_2,0]: z_i\in\mathbb{C},|z_1|^2+|z_2|^2\neq0\}$ is a compact complex submanifold in $M$, which is equivalent to $\mathbb{C}\mathbb{P}^1$. But $M$ is not compact. –  user18537 Nov 8 '11 at 6:28

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