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Let $X$ be a connected complex manifold (not necessarily compact). Let $C \subset X$ be a compact complex $k$-dimensional submanifold (for some $k>0$).

Is it true, in this generality, that the homology class $[C] \in H_{2k} (X,\mathbb{Z})$ is non trivial?

EDIT - some motivating observations: a first striking fact in the study of complex manifolds is that there is no analogue of Whitney embedding theorem for compact ones; indeed by the maximum modulus principle $\mathbb{C}^n$ has no compact complex submanifolds. I am not very familiar with complex manifolds of dimension $n>1$ (and of course in dimension $1$ this problem is not very interesting). The examples of compact complex submanifolds I have and can handle (as far as the above problem is concerned) are the following

  1. the first factor in the product $K \times X$ where $K$ is any compact complex manifold and $X$ any complex manifold
  2. the base of a vector bundle over a compact manifold $K$
  3. complex projective subspaces $\mathbb{CP}^k \subseteq \mathbb{CP}^n$

and in these cases it is easy to see that I get something which is non trivial in homology, by quite general facts not really related to complex geometry. Moreover I stumbled upon the fact that there exists many (non algebraic) 2-dimensional tori without compact complex (1-dimensional) submanifolds, as discussed for instance here. This has boosted my impression that if we manage to find a compact complex submanifold, then it must be very special indeed, in some sense. I would like to know if there is some counterexample to the sentence above, or if it can be proved by general methods in complex geometry. I am asking it in that generality also because I am not very familiar with Kähler or algebraic geometry, but of course I appreciate answers under the additional hypothesis that $X$ is compact/projective/Kähler/...

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    $\begingroup$ It's certainly true if you're Kahler, because then you can define the Kahler energy $\int_C \omega^k$, and you can check that this form is strictly positive on $C$ by the hypothesis that it's a complex submanifold. But if the homology class were trivial then the integral of any closed form over $C$ would be zero. I think it's still true in the level of generality you're looking for but don't have a proof off the top of my head. $\endgroup$ – user98602 Dec 2 '15 at 15:18
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As Mike Miller points out in the comments, if $X$ is a Kähler manifold (not necessarily compact), and $C$ is a $k$-dimensional compact complex submanifold, then $i_*[C] \in H_{2k}(X, \mathbb{Z})$ is non-trivial (here $i : C \to X$ is the inclusion map and $[C] \in H_{2k}(C, \mathbb{Z})$ is the fundamental class of $C$). To see this, let $\omega$ be the Kähler form, then $\int_C\omega^k = \operatorname{Vol}(C)$ by Wirtinger's Theorem (actually, Wirtinger's Theorem is much stronger than this). Now note that $\int_C\omega^k$ is actually a pairing of homology and cohomology classes, namely

$$\int_C\omega^k = \langle i_*[C], [\omega]^k\rangle.$$

Keep in mind, this is a pairing of real homology and cohomology classes, not integral ones. Although $i_*[C] \in H_{2k}(X, \mathbb{Z})$, we only have $[\omega] \in H^2(X, \mathbb{R})$ - provided $X$ is compact, finding a Kähler metric with $[\omega]$ integral is equivalent to $X$ being projective. We're identifying $i_*[C] \in H_{2k}(X, \mathbb{Z})$ with its image under the map $H_{2k}(X, \mathbb{Z}) \to H_{2k}(X, \mathbb{R})$ induced by the inclusion $\mathbb{Z} \to \mathbb{R}$.

If $i_*[C] \in H_{2k}(X, \mathbb{Z})$ were trivial, then its image in $H_{2k}(X, \mathbb{R})$ would also be trivial, in which case the pairing $\langle i_*[C], [\omega]^k\rangle$ would be zero. As $\operatorname{Vol}(C) > 0$, we therefore see that $i_*[C]$ is non-trivial.

A common misconception with this argument is that if a class in $H_{2k}(X, \mathbb{Z})$ is non-zero, then its image in $H_{2k}(X, \mathbb{R})$ will also be non-zero. At no point of the argument did I make such a claim, which is good because it is false: $H_{2k}(X, \mathbb{Z})$ may have torsion which will necessarily be mapped to zero in $H_{2k}(X, \mathbb{R})$.


As for the non-Kähler case, the result is no longer true. Let $X$ be the standard Hopf surface: $(\mathbb{C}^2\setminus\{(0,0)\})/\mathbb{Z}$ where the $\mathbb{Z}$-action is generated by the map $(z_1, z_2) \mapsto (2z_1, 2z_2)$. The image of $\mathbb{C}^*\times\{0\}$ under the natural projection $\pi : \mathbb{C}^2\setminus\{(0,0)\} \to X$ is

$$C := \{[(w, 0)] : w \in \mathbb{C}^*\} \cong \mathbb{C}^*/\mathbb{Z}$$

where the $\mathbb{Z}$-action is given by $w \mapsto 2w$. This is a one-dimensional compact complex submanifold of $X$, namely a torus. To see that the image of the fundamental class of $C$ is trivial in $H_2(X, \mathbb{Z})$, note that $X$ is diffeomorphic to $S^1\times S^3$, so by the Künneth Theorem, $H_2(X, \mathbb{Z}) = 0$.

Combining the considerations in the Kähler case, together with this example in the non-Kähler case, Donu Araparu gave a nice example of a non-compact complex surface which is not Kähler.

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  • $\begingroup$ If I understand correctly the wiki section about Hopf surfaces, your $X$ has the structure of a bundle over the projective line with elliptic curves as fibers; I find it a bit annoying that both the fiber and the zero section provide compact complex submanifolds which are trivial in homology! Anyway this is exactly what I was looking for, thank you. $\endgroup$ – Lor Dec 4 '15 at 9:32
  • $\begingroup$ Are there mild assumptions one can make that force this to be true? I don't really know what I'm looking for, but mostly something topological as opposed to a Kahler assumption. $\endgroup$ – user98602 Dec 4 '15 at 18:14
  • $\begingroup$ @MikeMiller: I'm not aware of any such assumption. The symplectic analogue of this is true right (compact symplectic submanifolds of a symplectic manifold are non-trivial in homology)? $\endgroup$ – Michael Albanese Dec 4 '15 at 19:32
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    $\begingroup$ @MichaelA: Yes, that's true. Maybe the correct story is that symplectic manifolds are the part of Kahler manifolds that "really" matter, at least for surfaces. $\endgroup$ – user98602 Dec 4 '15 at 19:40
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    $\begingroup$ @MikeMiller: I think that's the case. A complex submanifold of a Kähler manifold is Kähler, and hence a symplectic submanifold. $\endgroup$ – Michael Albanese Dec 4 '15 at 19:45

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