# Why is the central charge $c=9$ supersymmetry in the internal manifold?

In  (abstract [here]) (https://inspirehep.net/record/245643?ln=en) they say that, when compactifying any superstring theory, the six dimensional internal manifold must have $$N=2$$ supersymmetry with central charge $$c=9$$.
I have seen this stated before (e.g. in (abstract here)), but did not find any references justifying it.

 Gepner, Doron. "Space-time supersymmetry in compactified string theory and superconformal models." Current Physics–Sources and Comments. Vol. 4. Elsevier, 1989. 381-402. (Page 771)  Kazama, Yoichi, and Hisao Suzuki. "New N= 2 superconformal field theories and superstring compactification." Nuclear Physics B 321.1 (1989): 232-268. (Page 233)

## 1 Answer

The critical dimension of the superstring is $$d=10$$, fixed by the conformal anomaly. The total central charge for the world sheet conformal theory is therefore $$\begin{equation} c=d\left(1_\text{scalar}+\frac{1}{2}_\text{fermion}\right)=10 \times\frac{3}{2}=15. \end{equation}$$ From the spacetime point of view we want to compactify on a manifold $$\begin{equation} \mathcal{M}_{10}=\mathbb{R}^{1,3} \times M_6 \end{equation}$$ which amounts to studying a tensor product of conformal $$N=2$$ theories, in which central charges add. So the $$4$$ scalar fields and their superpartners in $$\mathbb{R}^{1,3}$$ contribute $$\begin{equation} c_{\mathbb{R}^{1,3}}=4\times 3/2=6 \end{equation}$$ and the remaining central charge of $$15-6=9$$ has to come from degrees of freedom in the internal manifold. That is, the theory of the internal manifold has to be described by a $$N=2$$ conformal theory of central charge 9.

Comment on $$N=2$$: If you start out writing down a generic worldsheet sigma-model for an internal space it is in general neither supersymmetric nor conformal. It can be shown that $$N=2$$ (or $$N=(2,2)$$ depending on notation) symmetry of the sigma model, corresponds to the internal space being a complex Kähler manifold. On top of that, conformal invariance requires it to be Ricci flat. These two conditions describe a Calabi-Yau manifold.

In other words, if you want to preserve SUSY by compactifying on a Calabi-Yau manifold, from the spacetime point of view, it corresponds to a superconformal $$N=2$$ theory on the worldsheet.

For technical details I would refer to Brian Greene's lectures (https://arxiv.org/pdf/hep-th/9702155.pdf) page 49, and references therein.

• Thanks, I get why $c=9$ now, but why $N=2$ SUSY on the worksheet? – Soap Sep 22 '19 at 16:33
• I added a comment to my answer. – Sparticle Sep 22 '19 at 20:08
• Two last questions: 1. From your comment, $N=2$ SUSY on the worldsheet implies that the internal manifold is Calabi-Yau. Do you know if the converse statement is also true? (Calabi-Yau implies $N=2$?) 2. Accepting that we need wordsheet $N=2$ SUSY, how does this mean that compactification "amounts to studying a tensor product of conformal $N=2$ theories"? – Soap Sep 23 '19 at 10:17
• Since the theories are $N=2$ superconformal, should,t you get $c_{\mathbb{R}^{1,3}} = 4\times (1+1/2+1/2)$ (because you have two fermions for each boson)? – Soap Oct 2 '19 at 12:57