Timeline for Decoupling of Holomorphic and Anti-holomorphic parts in 2D CFT
Current License: CC BY-SA 3.0
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| Jun 24, 2014 at 21:37 | comment | added | nervxxx | quantum hall droplet is a chiral CFT. From this example note that the decoupling of holomorphic and antiholomorphic part is only true for CFTs on the $\mathbb{C}$- plane. For a CFT on a plane with a boundary, or a defect, for examples, the Virasoro algebra does not decouple and you cannot solve for the holomorphic and antiholomoprhic parts separately. | |
| Jun 24, 2014 at 21:35 | comment | added | nervxxx | say, $l_n$, and then simply copy the result for the other set. But we must always remember that the full theory is comprised of the tensor products of both sets of generators, thus, we need to tensor the holomoprhic and antiholomoprhic parts together. The physical significance of these two parts has to do with time-reversal. In a time reversal symmetric system the left and right moving parts must pair up together to cancel out, to give no net chirality. But there are chiral (time reversal broken) CFTs which have only movers in one direction. for e.g., the theory on the boundary of a fractional | |
| Jun 24, 2014 at 21:31 | comment | added | nervxxx | Why are you talking about the Lorentz group only? Your question is about CFT, so, you want to look at the conformal group (which subsumes the Lorentz group). Now to study the conformal group we look at its Lie algebra. On the 2D plane this happens to be the Witt algebra (quantum version is the Virasoro algebra). In both cases the generators of the Lie algebra fall under two sets, $l_n$ and $\bar{l}_n$, and they obey the same commutation relations within each set, and they commute between sets: $[l_n, \bar{l}_m] = 0$. So, that means we only need to use representation theory on one set, | |
| Jun 24, 2014 at 21:14 | history | edited | user7757 | CC BY-SA 3.0 |
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| Jun 10, 2014 at 9:54 | history | answered | user7757 | CC BY-SA 3.0 |