# Holomorphic/Symplectic embedding of Riemann surfaces

Let $$\Sigma_g$$ denote a Riemann surface and let $$X$$ denote the complex surface $$\Sigma_g \times \Sigma_g$$. Then can there exist holomorphic embeddings of $$\Sigma_l$$ into $$X$$ for $$l < g$$?

What about in the symplectic category i.e if $$\omega$$ denotes the area 1 form on $$\Sigma_g$$ and we equip $$X = \Sigma_g \times \Sigma_g$$ with the form $$\omega \oplus \omega$$. Then does there exist a symplectic embedding $$\Sigma_l$$ into $$X$$ for $$l < g$$?

• No for the holomorphic part. Composing with one of the two projections would give a holomorphic map $\Sigma _l\rightarrow \Sigma _g$; such a map must be trivial.
– abx
Aug 10, 2022 at 19:26
• Also there are no embeddings in the symplectic category. Consider the pullback map on singular cohomology. The pullback map on $H^1$ must have nonvanishing kernel, since the rank of the domain is larger than the rank of the target. But this nonzero element in $H^1$ has a nonzero cup-product against some element in $H^1$, since the cup-product pairing is nondegenerate. Since the pullback map on cohomology is a ring homomorphism, the pullback of the generator of $H^2$ is zero. Since this holds for both projections, the pullback of $\omega\oplus \omega$ gives the zero cohomology class. Aug 10, 2022 at 20:08

Lemma. For integers $$\ell < g$$, for every continuous map from $$\Sigma_\ell$$ to $$\Sigma_g$$, the pullback map on $$H^2$$ is the zero map.
Proof. The pullback map on $$H^1$$ has rank no greater than $$2\ell$$, since that is the rank of $$H^1(\Sigma_\ell)$$. Since $$H^1(\Sigma_g)$$ has rank $$2g>2\ell$$, there exists a nonzero element $$\alpha$$ in the kernel. Since the cup product pairing on $$H^1(\Sigma_g)$$ is nondegenerate, there is an element $$\beta$$ in $$H^1(\Sigma_g)$$ such that $$\alpha\cup \beta$$ is nonzero in $$H^2(\Sigma_g)$$. Since the pullback map on cohomology is a ring homomorphism, the pullback of $$\alpha\cup \beta$$ is zero. Since $$H^2(\Sigma_\ell)$$ is torsion-free, and since $$\alpha\cup \beta$$ is a nonzero multiple of the generator of $$H^2(\Sigma_g)$$, the pullback map on $$H^2$$ is zero. QED
Thus, for a generator $$\omega$$ of $$H^2(\Sigma_g,\mathbb{R})$$, for every continuous function from $$\Sigma_\ell$$ to $$\Sigma_g\times \Sigma_g$$, the pullback of $$\omega\oplus \omega$$ is zero in $$H^2(\Sigma_\ell,\mathbb{R})$$. Since a symplectic form on $$\Sigma_\ell$$ has nonzero cohomology class, there is no differentiable map from $$\Sigma_\ell$$ to $$\Sigma_g\times \Sigma_g$$ that pulls back $$\omega\oplus \omega$$ to a symplectic form on $$\Sigma_\ell$$.
As noted by @abx, there is no nonconstant holomorphic map from $$\Sigma_\ell$$ to $$\Sigma_g$$. The lemma shows that there is not even a map that pulls back $$\omega$$ to a differential form with nonzero cohomology class (every nonconstant holomorphic map pulls back $$\omega$$ to a differential form with nonzero cohomology class).