Currently I am taking an introductory class in conformal field theory and ran into some confusion. Everywhere I run into Laurent expansions of several important quantities, and (mostly) all of them boil down to the following:

Suppose $\phi (z)$ is a primary field with conformal dimension $h$, then it may be Laurent expanded as:

$$\phi(z) = \sum_{n \in \mathbb{Z}} c_n z^{-n-h}$$

With $c_n$ the Laurent modes (constants). For a particular reference of this, see for example di Francesco's book eqn. (6.7) or Blumenhagen eqn. (2.42) in which it is claimed (without motivation beforehand):

$$T(z) = \sum_{n \in \mathbb{Z}} z^{-n-2} L_n$$

Is there any mathematical or physical reason why we include these conformal weights of the objects in the Laurent expansion (why not simply $\sum_{n \in \mathbb{Z}} z^{n} c_n$) ? Or is it simply convenient in calculations?

Bonus (related) question: in Blumenhagen's book (page 13) the following expression is stated:

$$z'=z + \epsilon(z) = z + \sum_{n \in \mathbb{Z}}\epsilon_n (-z^{n+1})$$

Where $\epsilon(z)$ is assumed to be meromorphic and $\epsilon_n$ constant. Is this particular expansion (so with $z^{n+1}$ instead of $z^n$) similarly done simply for convenience?

  • $\begingroup$ It seems to me that this is a purely mathematical question although it might have an application in physics (as does most mathematics). It might be better on Mathematics SE. $\endgroup$ – sammy gerbil Mar 9 '17 at 0:10
  • $\begingroup$ How is it a mathematics question when it is asking about conformal weights? $\endgroup$ – bolbteppa Mar 9 '17 at 0:11
  • $\begingroup$ When you map between the plane and the cylinder, you are transforming your metric and so using 5.22 of DiFrancesco you pick up the extra $h$ terms, c.f. Ketov Ch. 1, 1.43. $\endgroup$ – bolbteppa Mar 9 '17 at 0:32
  • $\begingroup$ @bolbteppa Ah I see! Thanks for your reply, this indeed solves my question! $\endgroup$ – Wwhite Mar 9 '17 at 14:41

Let me use the book by di Francesco et al. Including the conformal weight in the power of $z$ in the Laurent expansion, as in equation (6.8), allows you to have the nice property (6.10), namely $\phi_{-m} = \phi^{\dagger}_m$. For the energy momentum tensor, it gives $L_{-m} = L^{\dagger}_m$.

It is just a nice convention.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.