# What is the difference between $N=(2,2)$ with $N=(2,2)^*$?

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What is the difference between $N=(2,2)$ with $N=(2,2)^*$? In some sense, i heard, they are totally different theory. I heard from breaking of $N=(4,4)$ supersymmetry it comes $N=(2,2)^*$. What is the crucial difference between $N=(2,2)$ with $N=(2,2)^*$?

This post imported from StackExchange Physics at 2014-10-01 04:28 (UTC), posted by SE-user phy_math
Please consider adding relevant links and explanations to your question, it is currently quite unclear what the $N = (k,l)$ are that you are talking about.

This post imported from StackExchange Physics at 2014-10-01 04:28 (UTC), posted by SE-user ACuriousMind

Why is there a flag on ACuriousMind's comment? It seems like a legitimate concern.

I got rid of it by adding a +1. :-)

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I am assuming that you are talking about two-dimensional supersymmetry. While I haven't seen the usage $(2,2)^*$ -- my guess is that it refers to a theory with higher supersymmetry, say $(4,4)$ that is broken to $(2,2)$ by adding some terms that explicitly break it to $(4.4)$. In other words, the spectrum of fields is that of $(4,4)$ but the interactions are $(2,2)$. These are special cases of $(2,2)$ theories since not all $(2,2)$ theories arise in this way.

Remark:  Many $(2,2)$ (resp. $(4,4)$)) theories arise via dimensional reduction of four-dimensional $\mathcal{N}=1$ (resp. $\mathcal{N}=2$) supersymmetric theories.

answered Oct 1, 2014 by (1,545 points)
edited Oct 2, 2014 by suresh

Some explicit example of breaking of $(4,4)$ to $(2,2)$ which I think is called $(2,2)^*$. The $(4,4)$ vector multiplet is made of  one $(2,2)$ vector multiplet and one chiral multiplet. It is possible to break $(4,4)$ to $(2,2)$ by giving a mass to the chiral multiplet.

Hi please explain the notation used. What are the 2 numbers in the bracket

The numbers in the brackets denote the number of supersymmetries (left,right).

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