2
$\begingroup$

Let $X$ be a compact Riemann surface with smooth boundary $\partial X$. Is it always possible to realize $X$ as a complex submanifold of $\mathbb CP^2$? In other words, is it true that there exists a Riemann surface $Y \subset \mathbb CP^2$ with boundary $\partial Y$ and a bijection $f \colon X \to Y$, $f(\partial X) = \partial Y$ such that $f$ and $f^{-1}$ are both analytic?

$\endgroup$
  • 1
    $\begingroup$ I'm not sure about the case with boundary, but without boundary the answer is no--you cannot even embed all elliptic curves. It turns out the answer is yes thought if you replace $2$ by $3$. I can't seem to find a reference, but I think it is due to Narashima (spelling?). $\endgroup$ – Alex Youcis Jul 30 '14 at 8:46
  • $\begingroup$ @AlexYoucis Then in general it isn't also true in the case with boundary. Is there a name for compact Riemann surfaces with boundary that can not be extended to a closed Riemann surface? Maybe something like "transcendent Riemann surface"? $\endgroup$ – Appliqué Jul 30 '14 at 9:12
  • $\begingroup$ Dear @Alex, I'm not sure what you mean, but every elliptic curve can be embedded into $\mathbb P^2$: any divisor of degree 3 on X is very ample and will do the job. $\endgroup$ – Georges Elencwajg Jul 30 '14 at 10:13
2
$\begingroup$

a) Every compact Riemann surface (without boundary) $X$ can be embedded into $\mathbb P^3$.
Indeed, if $X$ has genus $g$ and $x\in X$ is an arbitrary point, the divisor $D=(2g+1)\cdot x$ is very ample and embeds $X$ into $\mathbb P^{g+1}$.
If $g\geq 2$ it is easy to project the obtained embedded Riemann surface into a linear subspace $P^3\subset \mathbb P^{g+1}$ of dimension $3$ and thus obtain the composed embedding $X\hookrightarrow P^3\cong \mathbb P^3$.

b) However Riemann surface of genus $g\geq 2$ cannot in general be embedded in the plane $ \mathbb P^2$:
one obstruction is that the genus of a smooth plane curve of degree $d$ is $g=\frac {(d-1)(d-2)}{2}$ and , of course, most integers are not of this form!
For example Riemann surfaces of genus $2,4,5,7,8,9,...$ can never be embedded into $ \mathbb P^2$ .

$\endgroup$
  • $\begingroup$ Dear @Georges, the question wasn't about Riemann surfaces without boundary, see also my comment to Alex. Maybe it is always possible to embed in $\mathbb C P^2$ such compact Riemann surfaces with boundary that can not be extended to a closed Riemann surface? $\endgroup$ – Appliqué Jul 30 '14 at 10:21
  • $\begingroup$ Dear @Nimza: you are right. Let's say that the answer summarizes the case where the boundary is empty! But could you please explain exactly what the charts and change of coordinates for these Riemann surfaces with boundaries are? And what is a Riemann surface $Y\subset \mathbb P^2$ with boundary ? Is there a reference for those notions (I have never heard of them)? I suppose a Riemann surface with an open disk deleted is a Riemann surface with boundary ? $\endgroup$ – Georges Elencwajg Jul 30 '14 at 11:02
  • $\begingroup$ I use the following definitions: the "near-boundary charts" are $\{ z \in \mathbb C \colon |z| < 1, \; \Im z \leq 0 \}$ and transition functions are supposed to be extendable to biholomoprhic functions in some neighborhood of $\Im z = 0$. But I've seen the definitions where it is a $C^\infty$ $2$-manifold with boundary with a class of conformally equivalent metrics or with a smooth section $J$ of $\mathop{\mathrm{End}} TX$, satisfying $J^2 = -Id$. I don't know if first and two last definitions agree. $\endgroup$ – Appliqué Jul 30 '14 at 11:45
  • $\begingroup$ Thanks for the clarification, Nimza. I would guess (but do not guarantee) that both definitions are equivalent since almost complex structures are integrable in complex dimension one. More to the point: I know nothing about these structures in either definition! $\endgroup$ – Georges Elencwajg Jul 30 '14 at 11:45
  • $\begingroup$ (I think that the first definition may be more strict since extendibility may require the boundary to be more good than just $C^\infty$, e.g. real analytic, but I'm not sure) $\endgroup$ – Appliqué Jul 30 '14 at 11:54

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.