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Lets $E_{\tau}^{\rho}$ be the elliptic curve with complex structure given by $\tau$ in upper half plane and complexified Kahler form $\rho \frac{dz\wedge d\bar{z}}{2}$.( $\rho$ is in upper half plane too)

Then mirror symmetry says that mirror to $E_{i}^{\rho}$ in A-side is $E^{i}_{\rho}$ in B-side.(see the paper of Polishchuk and Zaslow)

then what is the mirror for general $E_{\tau}^{\rho}$ in A-side (i.e. when we change the complex structure on A-side from the one given by $i$ to something else)??

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up vote 3 down vote accepted

The mirror of $E^\rho_\tau$ is $E^\tau_\rho$, as you may have guessed. The reason this is not discussed in, say, Polishchuk/Zaslow is that the derived category does not depend on the symplectic structure, and the Fukaya category does not depend on the complex structure, so for their purposes the parameter $\tau$ is irrelevant.

Btw, another article on "Mirror symmetry and elliptic curves" that you might find interesting is due to Dijkgraaf's, with title as indicated; it is contained in a 1995 volume on "The moduli space of curves" (see e.g.

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At the first glance this seems to be the answer, but I think it is not. Here is the reason: Consider $E_{\tau}^{\rho}$ in the A-side while $\rho$ is fixed and $\tau$ is moving. As $\tau$ moves the special Lagrangians deform and the map $m_2$ of $A_{\infty}$ structure change (due to the change of area of holomorphic triangles) so the A-side deforms. But from your claim the B-model is $E_{\rho}^{\tau}$ and so the complex structure is fixed and nothing changes. – Mohammad F. Tehrani Nov 10 '10 at 2:08
Infact there is a lack of symmetry in the homological mirror symmetry which I can not digest: The A-side (Special Lagrangians) depends on both complex structure and symplectic structure but the B-side depends only on complex structure. I wish some body can explain it to me. – Mohammad F. Tehrani Nov 10 '10 at 2:10
@Mohammad: The special Lagrangians change as the complex structure changes, but the Lagrangians do not change. The Fukaya category has objects Lagrangians, not special Lagrangians. – Kevin H. Lin Nov 10 '10 at 5:22
Mohammad, Kevin is correct. A special Lagrangian (if one exists, which is rarely known) gives a preferred representative within a Hamiltonian isotopy classes of Lagrangians. Hamiltonian-isotopic Lagrangians are quasi-isomorphic objects of the Fukaya category. – Tim Perutz Nov 10 '10 at 16:27
Sorry Mohammed, you are right. What I meant is that the choice of complex structure and symplectic structure get interchanged. However, $\rho$ does not correspond to a choice of symplectic structure on the (fixed) differentiable manifold $E$, but a choice of symplectic structure on $\mathbb R^2$. I will correct my answer later. – Arend Bayer Nov 10 '10 at 18:24

I think the answer is: $E_{\tau}^{\rho/\tau_{im}} \leftrightarrow E_{\rho}^{\tau/ \rho_{im}}$ where $\tau_{im}$ is the imaginary part of $\tau$ (a positive number).

Infact the answer of AByer is correct if write everything in the coordinate $(x,y)$ for the point $x+i\tau=X+iY$ on $E_{\tau}$ and not the $(X,Y)$.

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