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I have been studying Conformal field theory for the past one week from the books by Blumenhagen and Di Francesco etal.

If I understand correctly, whenever one talks of 'local (infinitesimal) conformal transformations' on the complex plane (for eg. $ z \rightarrow z + \epsilon(z) $ ) being described by holomorphic functions, one implicitly assumes that there is an open subset (say $U$) containing a point of the Complex plane on which the function $\epsilon(z)$ is well-defined and in fact holomorphic. My first question is whether this is the right way to look at local conformal transformations. If not, then, what exactly does one mean by 'local' conformal transformations?

Now, if the answer to the earlier question is yes, then I don't see why both the above books expand $\epsilon(z)$ as a Laurent series about 0 when deriving the Witt algebra. The Laurent series will have its own domain of convergence which may not contain the open set $U$. So can anybody explain why it is justified to assume that $\epsilon$ has a Laurent series about 0 and use it to derive the form of the generators of the Witt algebra? One could in principle choose any point (say $a$) of the open set and choose the neighbourhood $U$ to be sufficiently 'small' so that $\epsilon(z)$ is represented by a Taylor Series around $a$. One will in general get a smaller algebra by this procedure.

And I repeat: if the answer to my first question is no, could anybody then clarify the meaning of local conformal transformations in the first place?

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  • $\begingroup$ If you want to know how is the mathematical rigorous treatment o infinitesimal conformal transformation see the book "Advances in Moduli Theory" by Ueno and Shimizu. $\endgroup$
    – user40276
    Commented May 14, 2015 at 10:04

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Comments to the question (v2):

  1. To be specific, let us assume that the underlying 2D manifold is the Riemann sphere $S^2\cong \mathbb{C}\cup\{\infty\}$.

  2. The group of globally defined conformal transformations (connected to the identity) is the 6-dimensional group $${\rm Conf}_0(p,q)~\cong~SO^+(1,3)~\cong~PSL(2,\mathbb{C})$$ of Moebius transformations.

  3. Mathematically speaking, one should consider the groupoid of locally defined conformal transformations. See also e.g. this Phys.SE post.

  4. The (complex) Witt algebra is the Lie algebra of meromorphic vector fields on the Riemann sphere $S^2$. It is also the complexification of the Lie algebra of vector fields on a circle $S^1$.

  5. From a physical perspective, if we e.g. think in terms of a closed string in the operator formalism, i.e. a simple closed curve $\gamma\subset S^2$ encircling some marked point $a\in S^2$, then by the Riemann mapping theorem, we can choose a coordinate patch so that $\gamma$ is a unit circle in that coordinate system.

  6. A Laurent series about the point $a\in S^2$, that we assume is defined in some annulus $A \supset \gamma$, will contain all local deformations/Fourier modes of the string, and is therefore useful for a physical description.

  7. The point $a$ is typically associated with the infinite past $\tau=-\infty$. By change of coordinates, we may assume $a=0$.

  8. A Taylor series (as opposed to a Laurent series) about the point $a$ will e.g. miss half the Fourier modes of the string.

  9. On the other hand, if the Riemann sphere $S^2$ has $n\geq 2$ marked points $a_i\in S^2$, $i\in\{1, \ldots, n\}$, corresponding to $n$ local operator insertions, then in the $i$'th local coordinate neighbourhood $U_i$ around $a_i$, a Taylor series is typical an adequate physical description.

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  • $\begingroup$ @Qmechanic The group of globally defined conformal transformations should be the smooth vector fields of the circle, whose complexification yields the Witt algebra. The Möbius group $PSL(2,\mathbb{C})$ is the subgroup of the conformal transformations that leaves the vacuum state invariant, isn't it? $\endgroup$
    – gented
    Commented Aug 8, 2016 at 19:21
  • $\begingroup$ The group of globally defined conformal transformations (connected to the identity) on the Riemann sphere is the Moebius group $PSL(2,\mathbb{C})$. $\endgroup$
    – Qmechanic
    Commented Dec 16, 2016 at 12:11
  • $\begingroup$ 1. Thank you for the explanation, but it remains unclear to me whether we are considering meromorphic vectors with poles only at $0$ and $\infty$ or general meromorphic vectors, it seems in both cases we can expand in $z^n\partial_z$. 2. And if we are considering general meromorphic vectors, physically why? $\endgroup$ Commented Dec 13, 2022 at 6:12
  • $\begingroup$ 1. Holomorphic on an annulus. 2. There could e.g. be additional poles due to operator insertions. $\endgroup$
    – Qmechanic
    Commented Dec 13, 2022 at 8:13

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