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In my understanding, mirror symmetry in physics originates from representation of the $N=2$ superconformal algebra. Why do we need precisely 2 supersymmetries (why not 1 or 4)?

Moreover, a chiral (anti-chiral) field is defined as a state that is annihilated by $G^+_{-1/2}$ ($G^-_{-1/2}$), where $G^+_{-1/2}$ and $G^-_{-1/2}$ are coefficients of the Fourier mode expansion of some anti-commuting current $G^+(z)$ and $G^-(z)$ of conformal weight $3/2$. How should I understand this chiral (anti-chiral) field?

In $N=(2,2)$ superconformal algebra, there are four rings: $(c,c),(a,a),(a,c),(c,a)$. It is known that the first two are charge conjugate, but what does theism mean? Right and left-moving, and chiral and anti-chiral rings... these all confuse me.

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  1. It was shown by Zumino (Supersymmetry and Kahler Manifolds Phys.Lett. B87 (1979) 203 ) that the supersymmetric non-linear sigma model in four-dimensions (with target $M$) necessarily requires the manifold, $M$, to be Kahler. A dimensional reduction of this model leads to a two-dimensional nonlinear sigma model with $(2,2)$ supersymmetry. (See also: B. Zumino, “Supersymmetric sigma-models in two-dimensions,”)

  2. Consistency of string propagation on $M$ requires it to be Ricci-flat (this is a result due to Friedan). A six-dimensional compact manifold that is Kahler and Ricci-flat is a Calabi-Yau threefold.

These are the circle of ideas that eventually lead to mirror symmetry. Brian Greene's TASI lectures as well as Nick Warner's ICTP lectures on "N=2 Supersymmetric Integrable Models and Topological Field Theories"are two other references that might be of interest to you.

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    $\begingroup$ Note one can define the A model even if the manifold is just Kahler. $\endgroup$ – Ryan Thorngren Mar 1 '14 at 23:59
  • $\begingroup$ @RyanThorngren I was discussing the untwisted theories. You are right that the topological A-model doesn't need point 2 mentioned above. There are many bells and whistles that can be added to the above story and I wanted to make my answer as simple as possible. $\endgroup$ – suresh Mar 2 '14 at 0:06
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    $\begingroup$ Is $N=2$ SCFT the same as $N=(2,2)$ field theory? I think the former give the latter. $\endgroup$ – Mathematician Mar 11 '14 at 23:25
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    $\begingroup$ $N=2$ SCFT usually refers to a chiral half. $(2,2)$ says that there are two left-moving and two right-moving supersymmetries. Thus $(2,2)$ has two copies of the $N=2$ SCFT. $\endgroup$ – suresh Mar 12 '14 at 3:17

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