Complex manifold with subvarieties but no submanifolds

I previously asked this question on MSE and offered a bounty but received no responses.

There are examples of compact complex manifolds with no positive-dimensional compact complex submanifolds. For example, generic tori of dimension greater than one have no compact complex submanifolds. The proof of this fact, see this answer for example, shows that these tori also have no positive-dimensional analytic subvarieties either (because analytic subvarieties also have a fundamental class).

My question is whether the non-existence of compact submanifolds always implies the non-existence of subvarieties.

Does there exist a compact complex manifold which has positive-dimensional analytic subvarieties, but no positive-dimensional compact complex submanifolds?

Note, any such example is necessarily non-projective.

• The following example does not quite work, but perhaps it can be made to work. Let $X$ be a generic Kaehler K3 surface. Let $n\geq 3$ be an integer. Consider the Douady space $Y=\text{Hilb}^n_{X/\mathbb{C}}$ parameterizing closed analytic subspaces of $X$ that are zero-dimensional of length $n$. There is a singular closed analytic hypersurface $E$ in $Y$, namely the locus parameterizing closed subspaces that have at least one point that is nonreduced. Sadly, the deepest stratum of the singular locus of $E$ is a manifold! What if we deform $(Y,E)$ with one multiple point? – Jason Starr May 2 '16 at 16:01
• My suggestion will not work. Ljudmila Kamenova reminded me that for a generic deformation of $Y$, the subvariety $E$ does not deform. – Jason Starr May 2 '16 at 16:19

There are surfaces of type $VII_0$ on which the only subvariety is a nodal rational curve (I. Nakamura, Invent. math. 78, 393-443 (1984), Theorem 1.7, with $n=0$).

• How does that answer the question? Is a complex submanifold necessarily a subvariety? – Igor Rivin May 3 '16 at 7:39
• @IgorRivin I'm surely embarrassing myself, but wouldn't any compact complex submanifold be of dimension $1$, and therefore a Riemann surface and algebraic? – Callus May 4 '16 at 13:19
• An analytic subvariety (as the OP presumably intended), also sometimes called a closed analytic subset, is a closed subset which is locally equal to the common zero locus of finitely many holomorphic functions. There is nothing algebraic about it. In particular, compact complex submanifolds are automatically analytic subvarieties. – YangMills May 4 '16 at 16:37
• Now I see the point of IgorRivin's question, so let me expand the answer I gave. On a compact complex surface, the only non-trivial closed analytic subsets are curves. On the surfaces considered by Nakamura, the nodal rational curve that he mentions is the unique curve. – inkspot May 5 '16 at 14:28
• Can you find an explicit example which is Kähler? – YangMills May 14 '16 at 1:48

The theorem that inkspot refers to in his answer is originally from Inoue's paper New Surfaces with No Meromorphic Functions, II which seems like a more complete reference for this question. In particular, Inoue gives an explicit example of a compact complex surface which has an analytic subvariety but no compact complex submanifolds.

If $$x$$ is a real quadratic irrationality (i.e. a real irrational solution of a real quadratic equation), denote it's conjugate by $$x'$$. Let $$M(x)$$ be the free $$\mathbb{Z}$$-module generated by $$1$$ and $$x$$, then set $$U(x) = \{\alpha \in \mathbb{Q}(x) \mid \alpha > 0, \alpha\cdot M(x) = M(x)\}$$ and $$U^+(x) = \{\alpha \in U(x) \mid \alpha\cdot\alpha' > 0\}$$. Both $$U(x)$$ and $$U^+(x)$$ are infinite cyclic groups and $$[U(x) : U^+(x)] = 1$$ or $$2$$.

If $$\omega$$ is a real quadratic irrationality such that $$\omega > 1 > \omega' > 0$$, then $$\omega$$ is a purely periodic modified continued fraction; that is, $$\omega = [[\overline{n_0, n_1, \dots, n_{r-1}}]]$$ where $$n_i \geq 2$$ for all $$i$$, $$n_j \geq 3$$ for at least one $$j$$, and $$r$$ is the smallest period. For every such $$\omega$$, Inoue constructs a compact complex surface $$S_{\omega}$$ which is now known as an Inoue-Hirzebruch surface.

There are compact subvarieties $$C$$ and $$D$$ of $$S_{\omega}$$ with irreducible components $$C_0, \dots, C_{r-1}$$ and $$D_0, \dots, D_{s-1}$$ respectively; here $$s$$ is the smallest period of the modified continued fraction expansion of another element $$\omega^*$$ related to $$\omega$$ (alternatively, $$s$$ can be determined from the modified continued fraction expansion of $$\frac{1}{\omega}$$). When $$r \geq 2$$, $$C$$ is a cycle of non-singular rational curves, and when $$r = 1$$, $$C$$ is a rational curve with one ordinary double point. Proposition $$5.4$$ shows that $$C_0, \dots, C_{r-1}, D_0, \dots, D_{s-1}$$ are the only irreducible curves in $$S_{\omega}$$.

In the case where $$[U(\omega) : U^+(\omega)] = 2$$, we have $$r = s$$. Furthermore, there is an involution $$\iota$$ such that $$\iota(C_i) = D_i$$ for $$i = 0, \dots, r - 1$$. The quotient of $$S_{\omega}$$ by $$\iota$$ is denoted $$\hat{S}_{\omega}$$ and is now known as a half Inoue surface. Note that the images of $$C_0, \dots, C_{r-1}$$ are the only irreducible curves in $$\hat{S}_{\omega}$$.

If we can find a real quadratic irrationality $$\omega$$ such that $$\omega > 1 > \omega' > 0$$, $$r = 1$$, and $$[U(\omega) : U^+(\omega)] = 2$$, then $$\hat{S}_{\omega}$$ is a compact complex surface containing a unique curve, namely a rational curve with one ordinary double point. In particular, it provides an example of a compact complex manifold with a subvariety but no compact complex submanifolds. One such $$\omega$$ was given in the paper (end of section 6).

Example. Take $$\omega = (3 + \sqrt{5})/2$$. Then $$[U(\omega) : U^+(\omega)] = 2$$ and $$\alpha_0 =\ \text{a generator of}\ U(\omega) = (1 + \sqrt{5})/2$$, $$\alpha = \alpha_0^2 = (3 + \sqrt{5})/2$$, $$\omega = [[\overline{3}]]$$, $$r = 1$$.

In this case, $$b_2(\hat{S}_{\omega}) = 1$$ and $$\hat{S}_{\omega}$$ contains exactly one curve $$\hat{C}$$. Moreover, $$\hat{C}$$ is a rational curve with one ordinary double point and $$(\hat{C})^2 = -1$$.

For those interested in the details, in addition to Inoue's paper, it may also be worth reading the earlier paper Hilbert modular surfaces by Hirzebruch. As mentioned in his paper, Inoue used some methods from Hirzebruch's paper (which gives some indication of why the resulting surfaces are jointly named).

• The surfaces $\hat{S}_\omega$ are actually called "half Inoue surfaces" – YangMills May 14 '16 at 1:47
• @YangMills: Fixed. – Michael Albanese May 17 '16 at 17:35