# Decoupling of Holomorphic and Anti-holomorphic parts in 2D CFT

This maybe a very naive question.

I have just started studying CFT, and I am confused by why we have two separate parts of everything in CFT (operator algebras and hilbert space), the holomorphic and anti-holomorphic, which are decoupled from each other. We initially introduced $z$ and $\bar{z}$ as two independent variables instead of say $t$ and $x$ in two dimensions. But now, we have got two isomorphic parts, the holomorphic and antiholorphic (they may given by $z \to \bar{z}$ and $h \to \bar{h}$), then what extra info. does the anti-holomorphic parts provide? And how is all the information contained in just one independent coordinate $z$? Also a physical theory should be the tensor product of the verma modules? Why do we need both the parts, and what is the physical significance of each.

• As far as I understand, the decoupling of holomorphic and antiholomorphic parts may not hold always. However, many of the important examples of 2d conformal field theories have this property. For these theories, holomorphic and antiholomorphic parts can be studied independently and later combined to give the full theory. – user10001 Jun 7 '14 at 4:57
• To understand the physical meaning consider a gas of particle on a circle. Assume that the particles which are moving clockwise do not interact with the particle moving anticlockwise, so that these two subsystems are decoupled from each other. If we want to compute (say) the partition function of the whole system, we can calculate it independently for both right and left moving parts and then take their product. Thus decoupling simplifies things a lot. – user10001 Jun 7 '14 at 4:57

Let us for simplicity consider a bosonic string $X:\Sigma\to M$ in the matter sector only (as opposed to the ghost sector). Here $M$ is a target manifold. The worldsheet $\Sigma$ is a Lorentzian manifold of real dimension 2.

Locally in a neighborhood $U\subseteq \Sigma$ of the worldsheet, we may work in a so-called conformal gauge, which means to choose a worldsheet metric equal to the flat 2-dimensional Minkowski metric

$$\tag{1M} \mathrm{d}\sigma^1 ~\mathrm{d}\sigma^1-\mathrm{d}\sigma^0 ~\mathrm{d}\sigma^0~=~-\mathrm{d}\sigma^{+} ~\mathrm{d}\sigma^{-}.$$

Here

$$\tag{2M} \sigma^{\pm}~=~\sigma^0\pm\sigma^1$$

We will assume that it is possible to Wick rotate to an Euclidean signature. We identify the local Wick-rotated coordinates (of Euclidean signature) with a single independent complex coordinate $z=x+iy\in \mathbb{C}$. Normally, one assumes that the Wick-rotated worldsheet $\Sigma$ globally form a Riemann surface.

We stress that the complex conjugate variable $\bar{z}=x-iy$ is fundamentally not an independent variable, although one may in certain calculations get away with treating it as an independent variable. For a very similar discussion of dependence of a complex variable and its complex conjugate, see this Phys.SE post.

The Lagrangian density in such local coordinates becomes

$$\tag{3M} {\cal L} ~\sim~ \partial_{+} X^{\mu}~ \partial_{-}X_{\mu}.$$

This means that the classical equation of motion is just the wave equation

$$\tag{4M} \partial_{+}\partial_{-} X^{\mu}~\sim~ \Box X^{\mu}~=~0.$$

The full solution to the wave equation (4M) is left- and right-movers

$$\tag{5M} X^{\mu} = X_L^{\mu}(\sigma^{+}) + X_R^{\mu}(\sigma^{-}).$$

If we Wick-rotate the worldsheet metric to the Euclidean signature, then the left- and right-movers in eq. (5M) become holomorphic and antiholomorphic parts, respectively:

$$\tag{5E} X^{\mu} = f^{\mu}(z) + g^{\mu}(\bar{z}).$$

Note that in order to perform the Wick rotation in the worldsheet $\Sigma$, it in general becomes necessary to consider a complexification of the target space $M$. The Wick-rotated classical equation of motion (4M) is just Laplace's equation

$$\tag{4E} \partial\bar{\partial} X^{\mu}~\sim~ \Delta X^{\mu}~=~0,$$

with general complex solution (5E).

The splitting (5E) in holomorphic and antiholomorphic parts for the classical solutions carries manifestly over to the operator formalism (in contrast to the path integral formalism, where the splitting is not manifest). Both sectors in eq. (5E) are needed in a full description to meet pertinent physical requirements.

The splitting (5E) in holomorphic and antiholomorphic parts is also encoded in the CFT representation theory for the partition function $Z$ and the $n$-point correlation functions in terms of a tensor product of conformal blocks. An additional important requirement is modular invariance.

• Thanks for your answer. The point I am trying to take is the holomorphic and the anti-holomorphic sectors are identical in every aspect, so I am puzzled how working with just one variable gives you the whole answer. Just replace $z$ by $\bar{z}$ and $h$ by $\bar{h}$. I am trying to think of this as writing a coupled system in 2D into normal modes, but the two normal modes are physically distinguishable, but here there is no such thing, as the hilbert spaces seem to be isomorphic. – user7757 Jun 6 '14 at 16:31
• Also why are the two parts called the left and right chiral components? I don't see any association with helicity. – user7757 Jun 6 '14 at 17:20
• Could you have a look at my answer and check if it is right? – user7757 Jun 11 '14 at 10:44

Usually for $D \ne 2$, the vectors are irreducible under the lorentz group. But the lorentz group in the case $D=2$ is $SO(1,1)$ which is abelian. The irreducible representations of an abelian group at 1 D. (See this wiki article). Hence the vectors in 2D are irreducible and the $z$ and $\bar{z}$ components don't mix.

• 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, – nervxxx Jun 24 '14 at 21:31
• 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 – nervxxx Jun 24 '14 at 21:35
• 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. – nervxxx Jun 24 '14 at 21:37