# Factor of $1/2$ in the Sugawara construction

I'm trying to reproduce the Sugawara construction calculation using this reference (page 14).

The normal-ordering of two local operators is defined as

$$N(XY)(w)=\frac{1}{2\pi i} \oint_w \frac{dx}{x-w} X(z) Y(w). \tag{2.56}$$

Ok, that makes sense. The residue picks up the $$0$$-th term in the OPE.

Now we proceed to calculate $$X(z) N(YZ)(w)$$, which for some reason picks up two terms:

I don't understand how these two terms appear. The authors mention a "appropriate version of the Wick theorem" which to me looks like handwaving. WZW isn't a free theory and therefore the Wick theorem doesn't work.

On the other hand, the way I see it, we can use the $$XY$$ OPE to write the product $$XY$$ as a sum of local operators, and then we take the OPEs of all that operators with $$Z$$. Equivalently, we could have started with $$XZ$$ and obtained a different sum of local operators, but taking their OPEs with $$Y$$ is guaranteed to give the same result by associativity of OPE.

I've done the calculation on the whiteboard and obtained

$$J^a(z) N(J^b J^b)(w) = \left( k + h^{\vee} \right) \frac{J^a (w)}{(z-w)^2} + \dots,$$

which is the expected result expect for the factor of $$2$$.

The irony is in that that factor of $$2$$ is easily "explained" by the Wick theorem (there's 2 equivalent contractions), but I just can't see how that explanation can work in a nonlinear model like WZW, and also I can't see why my calculation is wrong.

Where am I wrong?

Update: example calculation for the Abelian case ($$f^{abc} = 0$$):

$$\oint_w \frac{dx}{2 \pi i} \frac{J(z) J(w) J(x)}{x-w} = \oint_w \frac{dx}{2 \pi i} \frac{1}{x-z} \left( \frac{k}{(z-w)^2} + \mathcal{O}(1) \right) J(x) = \frac{k J(w)}{(z-w)^2}$$

1. There is an implicit radial ordering $${\cal R}$$ assumed on the right-hand side of eq. (2.56) and on the left-hand side of eq. (2.58).
2. The contraction on the left-hand side of eq. (2.57) involves the whole normal-ordered product $$N(YZ)$$, not just $$Y$$.
4. The factor 2 in eq. (2.58) is correct. (To convince oneself of whether or not there is a factor 2, it might be helpful to first consider an abelian Kac-Moody current with $$f^{abc}=0$$.)
• I've done the calculation for $f^{abc}=0$ and I still don't see $2$ appearing. – Prof. Legolasov May 21 '19 at 8:08
• Also please explain 3. Which contraction is proportional to $1$ and why? Do you mean $\left< JJ \right>$? How can I show that it's equal to $1$? – Prof. Legolasov May 21 '19 at 8:29