Why allow only one singularity at $z =0$ when defining the local conformal algebra in 2D?

Question

The reference I'm using for this question is the review "Applied CFT" by Paul Ginsparg. In section 1.2 (Conformal algebra in 2 dimensions) he argues that if the metric is the Euclidean one $g_{\mu\nu}=\delta_{\mu\nu}$, then the equations for one infinitesimal conformal diffeomorphism read $$\partial_1\epsilon_1=\partial_2\epsilon_2,\quad\partial_1\epsilon_2=-\partial_2\epsilon_1.$$

He then remarks that these are just the Cauchy-Riemann equations and hence we might as well introduce holomorphic coordinates $(z,\bar{z})$ defined by $z = x^1+ix^2$ and $\bar{z}=x^1-ix^2$ and exchange $\epsilon^1,\epsilon^2$ by $\epsilon(z)$ and $\bar{\epsilon}(\bar{z})$ defined to be $$\epsilon(z)=\epsilon^1+i\epsilon^2,\quad \bar{\epsilon}(\bar{z})=\epsilon^1-i\epsilon ^2.$$

He briefly mentions that two-dimensional conformal transformations are then analytic transformations $$z\mapsto f(z),\quad \bar{z}\mapsto \bar{f}(\bar{z})\tag{1.5}.$$

He then says:

To calculate the commutation relations of the generators of the conformal algebra, i.e. infinitesimal conformal transformations of the form (1.5), we take for basis $$z\mapsto z'=z+\epsilon_n(z),\quad \bar{z}\mapsto \bar{z}'=\bar{z}+\bar{\epsilon}_n(\bar{z}),\quad n\in \mathbb{Z},$$ where $$\epsilon_n(z)=-z^{n+1},\quad \bar{\epsilon}_n(\bar{z})=-\bar{z}^{n+1}\tag{1.7}$$

I understand that the globally defined conformal transformations coincide with the globally defined analytic functions. So it seems we would like to look at the locally defined $f : U\subset \mathbb{C}\to V\subset \mathbb{C}$ (strictly speaking these ones do not form a group but that's not the point of this question).

Now such $f$ is locally defined for either one of two reasons: or because it is a restriction of a globally defined conformal map, or because it can't be further extended. The first situation gives nothing new, the second one does.

So in my understanding, if $f$ cannot be further extended is because it has a singularity somewhere.

It seems to me that Eq. (1.7) allows for the most general generator which has exactly one singularity at $z =0$.

Why one makes this choice? I mean if we are looking at the whole algebra of generators of conformal transformations which can't be globally defined why not allow for singularities of various points $S\subset \mathbb{C}$ only subject to the restriction that $S$ be closed? This seems to be the most general thing that can't be globally defined.

Why take the local conformal algebra as simply the generators which are singular only at zero and analytic everywhere else?

Answer 1

Your claim that (1.7) only considers infinitesimal conformal transformations with singularity at zero is simply wrong:

The most general generator that you get from (1.7) is a linear combination of all the basic generators there, i.e. a Laurent series $\sum a_n z^n$. While the center of a Laurent series may be its only singularity, it need not be - it has an inner and outer convergence radius on which some divergences lie, and that's it.

So this treatment of local infinitesimal conformal transformations does allow for singularities other than at zero.