Operator weights in 2-D CFTs -- what is happening with the degrees of freedom?

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This may be overtly obvious or simple and I'm being very dense, but it's something that has been bothering me. I am confused about how correlation functions of generic spin operators work in 2-d CFTs. There is a formula that is quoted in many texts (e.g. diFrancesco's text, see section 5.1-5.2) for the four point function of operators with weights $(h_i, \bar{h}_i)$:

$\langle \phi_1(z_1,\bar{z}_1) \phi_2(z_2,\bar{z}_2) \phi_3(z_3,\bar{z}_3) \phi_4(z_4,\bar{z}_4)\rangle = f(\eta,\bar{\eta}) \prod z_{ij}^{h/3-h_i-h_j} \bar{z}_{ij}^{\bar{h}/3-\bar{h}_i - \bar{h}_j}$

where the product is supposed to be for $i<j$ up to 4 and $h = \sum_i h_i$. The function $f$ is allowed to depend arbitrarily on the conformally invariant cross ratio $\eta$ and its conjugate. Here lies the crux of my problem:

If we consider any tensor; for example, $D^{\mu \nu \sigma}$, which denotes the four point function of a spin-3 current and three scalars, then it is clear that under the change to the complex coordinates, we will have a tensor $D^{abc}$ where $a,b,c$ may denote either $z$ or $\bar{z}$. If the current is symmetric, then of course we may eliminate all but 4 degrees of freedom. The issue now is the following: how does the above four point function capture all of these independent components? Setting $(h_1,\bar{h}_1) = (h_1, h_1-3)$, and the rest scalars so that $\bar{h}=h$, we clearly only have an expression for one of the components. Which one is it? What happens to the other degrees of freedom? Must we assume tracelessness?

asked Jun 22, 2014
edited Jun 22, 2014

1 Answer

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The vector representation of $SO(2)$ is reducible as the OP implicitly points out. Thus, one first decomposes $D^{\mu\nu\sigma}$ into irreducible components and then computes the four-point function for each irreducible component. There are as many functions of the cross-ratio as there are irreducible components. Of course, each irreducible component will have a different set of weights that one can easily compute. Let $\Delta=(h+\bar{h})$ denote the scaling dimension of the field and $s=(h-\bar{h})$ the spin (basically the $SO(2)$ charge) of the irreducible component. For instance, $D^{zz\bar{z}}$ has spin $1$ and  thus will  have conformal weight $(\Delta+1)/2,(\Delta-1)/2)$ and so on. One needs information about the scaling dimension of $D^{\mu\nu\sigma}$ to determine the conformal weights.

A classic example is that of the energy momentum tensor: $T^{zz}$ has weight $(2,0)$ while $T^{\bar{z}z}$ has weight $(1,1)$ and $T^{\bar{z}\bar{z}}$ has weight $(0,2)$. (There is some looseness in my statement as it is not a true primary due to the conformal anomaly but nevertheless illustrates my point.)

answered Jun 23, 2014 by (1,545 points)
edited Jun 23, 2014 by suresh
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Thanks for the answer! Could you clarify why the different irreducible components would correspond to different weights? Is this due to the transformation property of the components? Are the weights of these components always determined by how many z vs. zbar indices there are?

I had some incorrect statements in my original answer and I have corrected them. Hope that helps.

@Galois If you consider left arnd right movers, then $(L_0-\bar{L}_0)$ is the generator of rotations while $(L_0+\bar{L}_0)$ generates dilatations.

Absolutely right about the global conformal group. Rotations are also conformal transformations and so are a  part of $SL(2,C)$. Recall that $SL(2,C)$ is generated by $(L_0,L_{\pm1},\bar{L}_0,\bar{L}_{\pm1})$ (see di Francesco et al or Ginsparg's Les Houches lectures posted not the arXiv for details, if necessary). Irreps are labelled by the eigenvalues of $(L_0,\bar{L_0})$ on the highest weight state.

Yes.

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I'd like to clarify another point: even if D itself involves a current that is spin 3 (has 3 indices), the irred. components of D themselves have different spin due to having different weights? Is that correct?

@anonymous Yes.

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