# How can we see that a 4D N = 2 sigma model will yield a 3D N = 4 sigma model when compactified on a circle?

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I have a question about sigma models in 3D.

If we have $\mathcal{N}=2$ field theory on $\mathbb{R}^4$ and compactify it on $\mathbb{R}^3 \times S^1_R$ (in which $S^1_R$ is a circle of radius $R$) we get a 3D effective field theory whose Lagrangian is dependent on $R$. If we change variables of Lagrangian in suitable way and impose the preservation of SUSY (8 real supercharges), then we get $\mathcal{N}=4$ sigma model whose target space is a Hyperkähler manifold. My question is:

How we can prove this rigorously or using theorems of supersymmetry? Is there any reference other than Gauge Dynamics & Compactification To Three Dimensions that explains this more carefully?

This post imported from StackExchange Physics at 2015-05-22 20:56 (UTC), posted by SE-user QGravity

edited May 22, 2015

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Let me start by some general considerations. In a theory with massless scalars it is possible that these scalars acquire non-trivial expectation values. The space of the possible expectation values is the moduli space $M$ of vacua of the theory. The kinetic term of the scalars in the low-energy effective action around a given vacua, or equivalently the two points functions of the scalar fluctuations around the vacua, defines a natural metric on the moduli space $M$. If the massless scalars are the only massless degrees of freedom then the low energy description of the theory is the sigma model of target $M$. But if there exists other massless degrees of freedom the low energy description is in general more complicated.

For a $\mathcal{N}=2$ $4d$ gauge theory, the low energy description at a generic point of the moduli space of vacua  is an abelian gauge theory. In particular if the abelian gauge group is non trivial it is something more complicated than a sigma model with values in the moduli space of vacua.

After compactification on a circle we obtain a $\mathcal{N}=4$ $3d$ gauge theory. At low energy, at a generic point of the moduli space of vacua, we obtain again an abelian gauge theory. The key point is that in three dimensions an abelian gauge field is dual to a scalar field. Thus all the (bosonic) massless degrees of freedom can be seen as scalars and so the low energy effective description of the theory is the sigma model of target the moduli space $M$ of these scalars.  This $3d$ sigma model has $\mathcal{N}=4$ supersymmetries ($8$ real supercharges). This implies that $M$ is naturally Hyperkähler. The easiest way  to see that is maybe to reduce to two dimensions: it is classical that a $2d$ sigma model has $\mathcal{N}=(4,4)$ supersymmetries if and only if the target is Hyperkähler (for more informations and references see the answer to this question: http://physicsoverflow.org/23966/why-are-complex-structures-important-in-physics ). The idea is that the $\mathcal{N}=(4,4)$ $2d$ supersymmetric algebra has a $SO(4) \sim SU(2) \times SU(2)$ $R$-symmetry rotating the four supersymmetries which also rotates  three complex structures $I,J,K$  on $M$.

answered May 23, 2015 by (5,140 points)

@40227 Thank you!  So as I understand, $\mathcal{N}=(4,4)$ has $8$ real supercharges ($4$ chiral and $4$ anti-chiral) which sit in an irreducible rep of Lorentz algebra in $2D$ which is $\mathfrak{so} (1,1) \oplus \mathfrak{s}_o$ in which $\mathfrak{s}_o$ is odd part of the algebra. I have another questions. Does target space have the same structure when we compactify from 3D to 2D? I mean when we compactify from 4D to 3D, the moduli space is essentially a torus fibration over the Coulomb branch of $\mathcal{N}=2$ theory and we know that the Coulomb branch has a rigid special Kähler structure. So when we reduce to 2D, why does the same Hyperkähler target space (moduli space) of 3D theory arise?

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