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Let $X$ be a K3 surface and $\omega$ be a nowhere vanishing 2-form on $X$. Suppose $Y\subset X$ be a smooth real surface. How can one see that $\omega|_Y=0$ implies $Y$ is a complex submanifold (a complex curve)?

The other direction "If $Y$ is a complex a complex curve in $X$, then $\omega|_Y=0$." is relatively easy to see by taking a local coordinate.

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1 Answer 1

In local holomorphic coordinates $(z,w)=(x,y,u,v)$, write the 2-form as $\omega=f(z,w) dz \wedge dw$. (I assume you meant a holomorphic 2-form.) If we have a real 2-plane $P \subset T_x X$, and if $dx \wedge dy \ne 0$ on $P$, then on $P$, $dw = a \, dz + b \, d\bar{z}$, and $\omega = f b dz \wedge d\bar{z}$ vanishes if and only if $P$ is complex linear. Similarly if $du \wedge dv \ne 0$ on $P$. So we can assume that $dx\wedge dy=0$ and $du \wedge dv=0$ on $P$. Therefore there is a linear relation among $dx$ and $dy$ on $P$. Rotating the variables $z$ and $w$, by replacing $z$ by $Z=e^{i \theta_0} z$ for some constant $\theta_0$, and similarly for $w$, we can arrange that these linear relations have the form $dy=p \, dx$ and $dv=q \, du$ for some real numbers $p$ and $q$. These are two linear relations among the 4 variables, so these must be all of the relations that cut out $P$. So on $P$, $dx \wedge du \ne 0$. Therefore on $P$, $\omega=f \, (dx+ip \, dx) \wedge (du + i q \, du) = f(1+ip)(1+iq) dx \wedge du \ne 0$. Therefore all tangent planes to $Y$ are complex lines.

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