Is there a complex surface into which every Riemann surface embeds?

Every Riemann surface can be embedded in some complex projective space. In fact, every Riemann surface $\Sigma$ admits an embedding $\varphi : \Sigma \to \mathbb{CP}^3$. It follows from the degree-genus formula that the same is not true if we replace $\mathbb{CP}^3$ with $\mathbb{CP}^2$; for example, no Riemann surface of genus two can be embedded in $\mathbb{CP}^2$.

Said another way, there exists a compact complex manifold of dimension three into which every compact Riemann surface embeds (namely $\mathbb{CP}^3$), but the natural two-dimensional candidate (namely $\mathbb{CP}^2$) does not have this property. So my question is

Is there a compact complex surface into which every compact Riemann surface embeds?

The only other surface I have checked is $\mathbb{CP}^1\times\mathbb{CP}^1$. My hope was that every Riemann surface $\Sigma$ admitted an embedding $\varphi : \Sigma \to \mathbb{CP}^3$ with $\varphi(\Sigma)$ contained in the image of the Segre embedding $\mathbb{CP}^1\times\mathbb{CP}^1 \to \mathbb{CP}^3$. Many more Riemann surfaces embed in $\mathbb{CP}^1\times\mathbb{CP}^1$ than in $\mathbb{CP}^2$ - in particular, every genus can be realised. However, not all Riemann surfaces can be embedded in $\mathbb{CP}^1\times\mathbb{CP}^1$: any genus three Riemann surface which embeds in $\mathbb{CP}^1\times\mathbb{CP}^1$ must be hyperelliptic, but one can show using a dimension counting argument that there exist non-hyperelliptic genus three Riemann surfaces.

• I feel like one should be aiming to disprove this; it feels like it would be extraordinarily unlikely.
– user98602
Commented Oct 20, 2015 at 3:15
• My first guess would be to use a Chow ring argument to show that this fails for all minimal models, and then try too bootstrap off of this using the fact that we have control over the blow ups of the minimal surfaces' Chow rings. Interesting question though. I believe that I discussed this with a friend a while ago, and we concluded no, but I don't remember why. I'll have to ask him. EDIT: I guess this only works for the algebraic analogue of your question? Commented Oct 20, 2015 at 4:24
• Like Mike Miller, I would bet the farm that this isn't true. One idea is to take a general genus $g$ curve in a surface $X$ and show that its space of deformations inside $X$ has dimension less than $3g-3$ (for some value of $g$ say). The tangent space of the space of deformations is global sections of the normal bundle, so Riemann--Roch is available; I couldn't get the details to work out, but there may be a way to do it. Commented Oct 20, 2015 at 15:50
• Call me uninformed, but what's wrong with just taking the disjoint union of all Riemann surfaces? Isn't that a complex surface, or is there some implicit connectedness assumption you are using? (If there is, then I don't see how you can get any surface which properly embeds a sphere without being a sphere itself.) Commented Oct 20, 2015 at 23:17
• @MarioCarneiro: A compact manifold can only have finitely many connected components. I think you are confusing the notions Riemann surface and complex surface. A Riemann surface is a one-dimensional complex manifold, whereas a complex surface is a two-dimensional complex manifold. Commented Oct 20, 2015 at 23:46

This question was asked and answered on MathOverflow. The accepted answer by Olivier Benoist has been replicated below.

The answer is negative. Suppose for contradiction that $$S$$ is such a surface, and let me first assume that it is smooth and projective.

Fix $$g\geq 24$$. Then the coarse moduli space of genus $$g$$ curves $$M_g$$ is of general type (this is due to Harris, Mumford and Eisenbud, see for instance [The Kodaira dimension of the moduli space of curves of genus $$\geq 23$$]), hence a fortiori not uniruled. Let $$H$$ be the Hilbert scheme of smooth curves of genus $$g$$ in $$S$$ and let $$(H_i)_{i\in I}$$ be its irreducible components: the index set $$I$$ is countable. By hypothesis, the classifying morphism $$H\to M_g$$ is surjective at the level of $$\mathbb{C}$$-points. By a Baire category argument, there exists $$i\in I$$ such that $$H_i\to M_g$$ is dominant.

Suppose that the natural morphism $$H_i\to \operatorname{Pic}(S)$$ is constant. Then $$H_i$$ is an open subset of a linear system on $$S$$, hence is covered by (open subsets of) rational curves. By our choice of $$g$$, $$H_i\to M_g$$ cannot be dominant, which is a contradiction. Thus, $$H_i\to \operatorname{Pic}(S)$$ cannot be constant, which proves that $$\operatorname{Pic}^0(S)$$ cannot be trivial. Equivalently, the Albanese variety $$A$$ of $$S$$ is not trivial.

Consider the Albanese morphism $$a:S\to A$$. The curves embedded in $$S$$ are either contracted by $$a$$ or have a non-trivial morphism to $$A$$. Those that are contracted by $$a$$ form a bounded family, hence have bounded genus. Curves $$C$$ that have a non-trivial morphism to $$A$$ are such that there is a non-trivial morphism $$\operatorname{Jac}(C)\to A$$, but this is impossible if $$\operatorname{Jac}(C)$$ is simple of dimension $$>\dim(A)$$. Consequently, a smooth curve with simple jacobian that has high enough genus cannot be embedded in $$S$$. This concludes because a very general curve of genus $$g$$ has simple jacobian (there even exist hyperelliptic such curves by [Zarhin, Hyperelliptic jacobians without complex multiplication]).

As pointed out in the comments, the argument needs to be modified if $$S$$ is non-algebraic or singular. I explain now these additional arguments.

If $$S$$ is smooth but non-algebraic, we can use the following (probably overkill) variant. We can consider the open space $$H$$ of the Douady space of $$S$$ parametrizing smooth connected curves in $$S$$. It is a countable union of quasiprojective varieties by [Fujiki, Countability of the Douady space of a complex space] and [Fujiki, Projectivity of the space of divisors on a normal compact complex space]. It has an analytic morphism to the analytic space $$\operatorname{Pic}(S)$$ parametrizing line bundles on $$S$$ [Grothendieck, Techniques de construction en géométrie analytique IX §3]. The dimension of $$\operatorname{Pic}(S)$$ is finite equal to $$h^1(S,\mathcal{O}_S)$$. The dimension of the fibers of $$H\to\operatorname{Pic}(S)$$ is at most $$1$$. Indeed, otherwise, we would have a linear system of dimension $$>1$$ on $$S$$ consisting generically of smooth connected curves, hence a dominant rational map $$S\dashrightarrow\mathbb{P}^2$$, showing that $$S$$ is of algebraic dimension $$2$$, hence that $$S$$ is algebraic. It follows that every connected component of $$H$$ has dimension $$\leq h^1(S,\mathcal{O}_S)+1$$. Now choose $$g$$ such that $$\dim(M_g)>h^1(S,\mathcal{O}_S)+1$$, and let $$H_g$$ be the union of connected components of $$H$$ parametrizing genus $$g$$ curves. A Baire category argument applied to the image of $$H_g\to M_g$$ shows that there are genus $$g$$ curves that do not embed in $$S$$, as wanted.

Finally, if $$S$$ is singular, consider a desingularization $$\tilde{S}\to S$$. The hypothesis on S implies that every smooth projective curve may be embedded in $$\tilde{S}$$, with the exception of at most a finite number of isomorphism classes of curves (namely, the curves that are connected components of the locus over which $$\tilde{S}\to S$$ is not an isomorphism). The arguments above apply as well in this situation.