# SLAGs on elliptic curves are only lines?

Let $E=\mathbb{C}/(\mathbb{Z}+\tau\mathbb{Z})$ be an alliptic curve. Define a holomorphic 1-form $dz=dx+idy$ and a Kahler form $\omega=dx\wedge dy$. How can one prove that special Lagrangians on $E$ are all given by a line?

By special Lagrangian, I mean a submanifold $L$ of $E$ with $\omega|_L=0$ and $Im(dz)|_L=C \cdot\mathrm{vol}$ for some constant $C$. $\mathrm{vol}$ is the Riemannian volume form induced by the metric associated to $\omega$.

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Note that you have a mistake in your definition: The condition should be $\mathrm{Im}(dz)_{| L}=C\cdot \mathrm{vol}$ instead of $\mathrm{Im}(\omega)_{| L}=C\cdot \mathrm{vol}$. If you write out what the condition $\mathrm{Im}(dz)_{| L}=C\cdot \mathrm{vol}$ means, you'll see that it simply says that the slope of the curve in the $xy$-plane is constant. – Robert Bryant Apr 19 '13 at 14:46
@Robert Thank you for the correction. $Im(dz)=dy|_L$ and $C\mathrm{vol}=Cdy|_{L}$, but does this imply a constant slope? – James Apr 19 '13 at 15:39
@James: Your interpretation of $C\mathrm{vol}$ is not correct. For an oriented curve $L$ parametrized in the form $\bigl(x(t),y(t)\bigr)$, one has $\mathrm{vol} = \sqrt{x'(t)^2+y'(t)^2}\ dt$, so the equation you are looking at is $y'(t)\ dt = C \sqrt{x'(t)^2+y'(t)^2}\ dt$, and this does indeed imply that the ratio $\bigl[x'(t):y'(t)\bigr]$ is constant. – Robert Bryant Apr 19 '13 at 16:45