# Does differentiation and (conformal) normal ordering commute?

In Polchinski's String theory Volume I, chapter 2, he writes down the following OPE, $$X^\mu (z_1,\bar{z}_1)X^\nu(z_2,\bar{z}_2) = -\frac{\alpha'}{2}\eta^{\mu\nu}\ln|z_{12}|^2 + :X^\nu X^\mu(z_2,\bar{z}_2): + \sum_{k=1}^{\infty}\frac{1}{k!} \bigg[ (z_{12})^k :X^\nu\partial^kX^\mu(z_2,\bar{z}_2): + \ (\bar{z}_{12})^k :X^\nu\bar{\partial}^kX^\mu(z_2,\bar{z}_2): \bigg]$$

which he gets by taylor expanding $:X^\mu(z_1,\bar{z}_1)X^\nu(z_2,\bar{z}_2):$ since it can be written as a sum of holomorphic and antiholomorphic functions (since it is harmonic in $(z_1,\bar{z}_1)$).

But if we call the harmonic function $\ f(z_1,\bar{z}_1;z_2,\bar{z}_2) = :X^\mu(z_1,\bar{z}_1)X^\nu(z_2,\bar{z}_2):$, and taylor expand $\ f$ in $(z_1,\bar{z}_1)$ around $(z_1,\bar{z}_1) = (z_2,\bar{z}_2)$, then we get terms that go like ~ $\partial^k f$ ~ $\partial^k :X^\mu(z_1,\bar{z}_1)X^\nu(z_2,\bar{z}_2):$ evaluated at $(z_2,\bar{z}_2)$. Here the derivatives are outside the normal ordered expression while in Polchinski's expression, they are inside the normal ordered expression. Why are they equal? What am I misunderstanding?

Reference for the expression : Eq. 2.2.4 Pg 38, Polchinski Chapter 2 (hardcover) (2004 reprint I think).

Edit 1: $z_{12} = z_1 - z_2$ Edit 2: All partial derivatives are with respect to $z_1,\bar{z}_1$.

## 1 Answer

Yes, differentiation and normal ordering commute. You can prove this using the definition (2.2.7) of the normal ordering: $$:\mathcal{F}: = \exp \left( \frac{\alpha '}{4} \int d^2z_1 d^2z_2 \ln |z_{12}|^2 \frac{\delta}{\delta X^{\mu}(z_1 , \bar{z}_1)} \frac{\delta}{\delta X^{\nu}(z_2 , \bar{z}_2)} \right) \mathcal{F}$$

Using this expression, the commutativity of the normal order and differentiation boils down to commutativity of functional derivative and differentiation. Let us consider any functional $F$ of $X^{\mu}(z )$. Then $$\frac{\delta F[X^{\mu}(z )]}{\delta X^{\mu}(z_1)} = \lim\limits_{\epsilon \rightarrow 0} \frac{F[X^{\mu}(z) + \epsilon \delta(z-z_1) ]-F[X^{\mu}(z)]}{\epsilon}$$ (see for instance eq A.28 here). Then apply this for $F[X^{\mu}(z , \bar{z})] = \partial X^{\mu}(z)$: $$\frac{\delta \partial X^{\mu}(z)}{\delta X^{\mu}(z_1)} = \delta ' (z-z_1)$$ On the other hand, $$\partial \frac{\delta X^{\mu}(z)}{\delta X^{\mu}(z_1)} = \delta ' (z-z_1)$$ since $\frac{\delta X^{\mu}(z)}{\delta X^{\mu}(z_1)} = \delta (z-z_1)$. The two expressions are equal, which proves the claim.

• If so, you can consider upvoting and/or accepting the answer. (I just mention this because I see you are a newcomer on this site). – Antoine Mar 15 '17 at 15:24
• Yea, I did upvote it. But apparently it doesn't become visible until I have 15 reputation points or something(although it gets recorded). So let's hope for both our sakes I reach there fast ;) – Aneesh Mar 15 '17 at 18:31
• @Antoine It's actually much more subtle. Conformal normal ordering commutes with derivatives when the Ricci scalar and its derivatives vanish, which is not the case on general Riemann surfaces (the obstruction being the Euler number). More precisely, the two commute if curvature is stored in global information (in the transition functions) rather than locally, but they do not commute on curved surfaces if Ricci curvature is encoded locally. – Wakabaloola Oct 1 '19 at 7:16