Three-Point Functions for MN/SN Orbifolds¶with?= 4 Supersymmetry

By O. Lunin, Samir D. Mathur. 2001

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Abstract: The D1–D5 system is believed to have an “orbifold point” in its moduli space where its low energy theory is a ?=4 supersymmetric sigma model with target space MN/SN, where M is T4 or K3. We study correlation functions of chiral operators in CFTs arising from such a theory. We construct a basic class of chiral operators from twist fields of the symmetric group and the generators of the superconformal algebra. We find explicitly the 3-point functions for these chiral fields at large N; these expressions are “universal” in that they are independent of the choice of M. We observe that the result is a significantly simpler expression than the corresponding expression for the bosonic theory based on the same orbifold target space.

Subfields: Nuclear and High Energy Physics

View Paper: https://www.semanticscholar.org/paper/4e7176ac20943b6ccb3779c6aa7f8fac88a41fd3

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In Nuclear and High Energy Physics

Citations in the corpus

Emergent spacetime and holographic CFTs (2011, background): “e statements about the spectrum have to be understood in the NS sector which is dual to AdS3 in global coordinates. 13 The correlators also enjoy a large Nexpansion implying factorization at N→∞[36], [37], [38] and consequently we expect a simple structure for the multi-particle spectrum. These low-”
Microscopic formulation of black holes in string theory (2002, methodology): “cross-section calculation. This factor ﬁnally cancels oﬀ in the calculation as we will see in Section 8.6. Higher point correlations functions in the orbifold conformal ﬁeld theory were determined in [151, 152] using general methods of computing correlations functions of twist ﬁelds on symmetric pro”
Tensionless string spectra on AdS3 (2018, background): “ycle twisted sector is given by h 0 = (2B+F) L2 − 1 48L Lodd . (A.5) For example, for the case of T4 for which B= F= 4, this leads to h 0 = 1 4L (L2 −1), in agreement, for example, with eq. (2.11) in [28]. Incidentally, we also ﬁnd the same result for F= 8 and B= 2, h 0 = L2 −1 4L (Lodd) . (A.6) For”