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Which symplectic submanifolds can be realized in $\mathbb{R}^4$ with standard ($\text{d}\,\boldsymbol{p} \wedge \text{d}\,\boldsymbol{q}$) symplectic structure? It's easy to show that such submanifolds aren't compact. So, they are spheres with some handles and holes. Which relations between the numbers of holes and handles do exist?

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    $\begingroup$ Every algebraic curve in the complex projective plane, minus the points that lie on a complex projective line. So that gives any compact Riemann surface of genus $g$ minus a set of $d$ points so that $g=(d-1)(d-2)/2$. $\endgroup$
    – Ben McKay
    Apr 3, 2014 at 6:21
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    $\begingroup$ Moreover you may allow your curve to be singular at infinity, and/or the infinity line to have multiple contact with the curve. This should give any $g$ and $d$ provided $g$ is not too large w.r.t. $d$. $\endgroup$
    – abx
    Apr 3, 2014 at 7:13
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    $\begingroup$ If you keep track of the complex structure on an algebraic curve, not every set of $d$ points can be removed from a curve of genus $g$ by slicing with a projective line in an embedding in $\mathbb{CP}^2$. But symplectically, by Moser's lemma, you can move any discrete set of distinct points around anywhere, as you like. $\endgroup$
    – Ben McKay
    Apr 3, 2014 at 7:45
  • $\begingroup$ In fact, you can get any genus minus a single point: Just take a hyperelliptic curve $z^2 = P(w)$ where $P$ is a polynomial of degree $2g+1$ with no repeated roots. $\endgroup$ Apr 3, 2014 at 8:44
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    $\begingroup$ Just add a dimension: If you take the complex structure for which $z = p_1 + i q_1$ and $w = p_2 + i q_2$ are holomorphic coordinates, then $F = z= p_1 + i q_1$ is holomorphic, but $\{p_1,q_1\}\not=0$, when $\omega = dp_1\wedge dq_1 +dp_2\wedge dq_2 $. $\endgroup$ Apr 3, 2014 at 12:13

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