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Mirror Symmetry of Elliptic Curve

6

1

$begingroup$

I'm a little bit unsure about the mirror symmetry statement for elliptic curves; specifically, how the flipping of the Kähler and complex moduli works. Perhaps I should say at the outset, the reason I've been thinking about this, is that I'm doing a computation involving a torus with parameter $tau in mathbb{H}$, and my answer in invariant under $tau to tau+1$, but not $tau to -1/tau$. So I'm thinking that maybe I'm only using the Kähler structure, not the complex structure.

Of course, the complex structure is given by $tau in mathbb{H}/rm{PSL}(2, mathbb{Z})$. I think for the Kähler structure, we choose a class $[omega] in H^{2}(X,mathbb{C})$, which we can parameterize by Kähler parameter

$$t=t_{1} + i , t_{2}, t=frac{1}{2pi i } int_{X} [omega]$$

We identify $t_{2}>0$ with the area of the torus. So unlike the complex structure, Kähler structures related by $rm{PSL}(2,mathbb{Z})$, aren't necessarily identical, correct? After all, one will have small area, the other large.

So I'm confused about how the mirror symmetry acts. If mirror symmetry is some mysterious equivalence of the torus under interchange of the complex and Kähler moduli, doesn't that imply that the space of equivalent Kähler structures is also $mathbb{H}/rm{PSL}(2, mathbb{Z})$. Is this correct?

Thank you.

algebraic-geometry complex-geometry symplectic-geometry mirror-symmetry

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edited Mar 24 at 11:31

Andrews

1,3012423

asked Apr 20 '16 at 7:01

BenightedBenighted

1,054510

$endgroup$

add a comment | 

6

1

$begingroup$

I'm a little bit unsure about the mirror symmetry statement for elliptic curves; specifically, how the flipping of the Kähler and complex moduli works. Perhaps I should say at the outset, the reason I've been thinking about this, is that I'm doing a computation involving a torus with parameter $tau in mathbb{H}$, and my answer in invariant under $tau to tau+1$, but not $tau to -1/tau$. So I'm thinking that maybe I'm only using the Kähler structure, not the complex structure.

Of course, the complex structure is given by $tau in mathbb{H}/rm{PSL}(2, mathbb{Z})$. I think for the Kähler structure, we choose a class $[omega] in H^{2}(X,mathbb{C})$, which we can parameterize by Kähler parameter

$$t=t_{1} + i , t_{2}, t=frac{1}{2pi i } int_{X} [omega]$$

We identify $t_{2}>0$ with the area of the torus. So unlike the complex structure, Kähler structures related by $rm{PSL}(2,mathbb{Z})$, aren't necessarily identical, correct? After all, one will have small area, the other large.

So I'm confused about how the mirror symmetry acts. If mirror symmetry is some mysterious equivalence of the torus under interchange of the complex and Kähler moduli, doesn't that imply that the space of equivalent Kähler structures is also $mathbb{H}/rm{PSL}(2, mathbb{Z})$. Is this correct?

Thank you.

algebraic-geometry complex-geometry symplectic-geometry mirror-symmetry

share|cite|improve this question

edited Mar 24 at 11:31

Andrews

1,3012423

asked Apr 20 '16 at 7:01

BenightedBenighted

1,054510

$endgroup$

add a comment | 

$begingroup$

I'm a little bit unsure about the mirror symmetry statement for elliptic curves; specifically, how the flipping of the Kähler and complex moduli works. Perhaps I should say at the outset, the reason I've been thinking about this, is that I'm doing a computation involving a torus with parameter $tau in mathbb{H}$, and my answer in invariant under $tau to tau+1$, but not $tau to -1/tau$. So I'm thinking that maybe I'm only using the Kähler structure, not the complex structure.

Of course, the complex structure is given by $tau in mathbb{H}/rm{PSL}(2, mathbb{Z})$. I think for the Kähler structure, we choose a class $[omega] in H^{2}(X,mathbb{C})$, which we can parameterize by Kähler parameter

$$t=t_{1} + i , t_{2}, t=frac{1}{2pi i } int_{X} [omega]$$

We identify $t_{2}>0$ with the area of the torus. So unlike the complex structure, Kähler structures related by $rm{PSL}(2,mathbb{Z})$, aren't necessarily identical, correct? After all, one will have small area, the other large.

So I'm confused about how the mirror symmetry acts. If mirror symmetry is some mysterious equivalence of the torus under interchange of the complex and Kähler moduli, doesn't that imply that the space of equivalent Kähler structures is also $mathbb{H}/rm{PSL}(2, mathbb{Z})$. Is this correct?

Thank you.

algebraic-geometry complex-geometry symplectic-geometry mirror-symmetry

share|cite|improve this question

edited Mar 24 at 11:31

Andrews

1,3012423

asked Apr 20 '16 at 7:01

BenightedBenighted

1,054510

$endgroup$

share|cite|improve this question

edited Mar 24 at 11:31

Andrews

1,3012423

asked Apr 20 '16 at 7:01

BenightedBenighted

1,054510

share|cite|improve this question

edited Mar 24 at 11:31

Andrews

1,3012423

asked Apr 20 '16 at 7:01

BenightedBenighted

1,054510

edited Mar 24 at 11:31

Andrews

1,3012423

edited Mar 24 at 11:31

Andrews

1,3012423

asked Apr 20 '16 at 7:01

BenightedBenighted

1,054510

asked Apr 20 '16 at 7:01

BenightedBenighted

1,054510

1 Answer
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$begingroup$

I know this question is now rather old, but I think it's worth putting up the reference. The place you'll want to look at is Section 2 of Dijkgraaf's Mirror Symmetry and Elliptic Curves. In summary, the extra symmetries you see on the complex moduli suggest an unexpected symmetry in the A-model structures of the mirror Elliptic curve. This is related to the prediction that generating series of Gromov-Witten invariants are something like modular forms. As you point out, that is not a symmetry you see on the classical level by comparing the areas of the Elliptic curves.

Of course, if you are using something which is not invariant under the modular group, then you have something which does not only depend on the complex structure of your Elliptic curve, but something more.

share|cite|improve this answer

edited Sep 27 '18 at 15:11

answered Sep 27 '18 at 15:03

tprincetprince

963

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add a comment | 

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2

$begingroup$

I know this question is now rather old, but I think it's worth putting up the reference. The place you'll want to look at is Section 2 of Dijkgraaf's Mirror Symmetry and Elliptic Curves. In summary, the extra symmetries you see on the complex moduli suggest an unexpected symmetry in the A-model structures of the mirror Elliptic curve. This is related to the prediction that generating series of Gromov-Witten invariants are something like modular forms. As you point out, that is not a symmetry you see on the classical level by comparing the areas of the Elliptic curves.

Of course, if you are using something which is not invariant under the modular group, then you have something which does not only depend on the complex structure of your Elliptic curve, but something more.

share|cite|improve this answer

edited Sep 27 '18 at 15:11

answered Sep 27 '18 at 15:03

tprincetprince

963

$endgroup$

add a comment | 

2

$begingroup$

I know this question is now rather old, but I think it's worth putting up the reference. The place you'll want to look at is Section 2 of Dijkgraaf's Mirror Symmetry and Elliptic Curves. In summary, the extra symmetries you see on the complex moduli suggest an unexpected symmetry in the A-model structures of the mirror Elliptic curve. This is related to the prediction that generating series of Gromov-Witten invariants are something like modular forms. As you point out, that is not a symmetry you see on the classical level by comparing the areas of the Elliptic curves.

Of course, if you are using something which is not invariant under the modular group, then you have something which does not only depend on the complex structure of your Elliptic curve, but something more.

share|cite|improve this answer

edited Sep 27 '18 at 15:11

answered Sep 27 '18 at 15:03

tprincetprince

963

$endgroup$

add a comment | 

$begingroup$

I know this question is now rather old, but I think it's worth putting up the reference. The place you'll want to look at is Section 2 of Dijkgraaf's Mirror Symmetry and Elliptic Curves. In summary, the extra symmetries you see on the complex moduli suggest an unexpected symmetry in the A-model structures of the mirror Elliptic curve. This is related to the prediction that generating series of Gromov-Witten invariants are something like modular forms. As you point out, that is not a symmetry you see on the classical level by comparing the areas of the Elliptic curves.

Of course, if you are using something which is not invariant under the modular group, then you have something which does not only depend on the complex structure of your Elliptic curve, but something more.

share|cite|improve this answer

edited Sep 27 '18 at 15:11

answered Sep 27 '18 at 15:03

tprincetprince

963

$endgroup$

share|cite|improve this answer

edited Sep 27 '18 at 15:11

answered Sep 27 '18 at 15:03

tprincetprince

963

answered Sep 27 '18 at 15:03

tprincetprince

963

answered Sep 27 '18 at 15:03

tprincetprince

963