3
$\begingroup$

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 $l < g$?

$\endgroup$
2
  • 4
    $\begingroup$ 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. $\endgroup$
    – abx
    Aug 10, 2022 at 19:26
  • $\begingroup$ 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. $\endgroup$ Aug 10, 2022 at 20:08

1 Answer 1

3
$\begingroup$

I am just posting my comment as one answer.

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.