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

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?

## 1 Answer

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.

• Thanks for your answer ! I see your point, but for a single generator $X(z)$ there might be more than one Laurent expansion (one for each annulus centered at $0$ with some inner and outer radius, for instance if one has singularities at each $w_n = ni$) right? So when defining this algebra we should consider the restriction $X|_A :A\to TA\subset T\mathbb{C}$ to each annulus to be a distinct local infinitesimal conformal transformation, instead of considering $X(z)$ as a single object defined in the whole $\mathbb{C}\setminus S$ where $S$ are its singularities?
– Gold
Jan 7, 2020 at 23:19
• @user1620696 On every annulus centered at $z=0$ and by $z\mapsto 1/z$ on every annulus centered at $z = \infty$, these things generate meromorphic vector fields on it. You cannot necessarily get a global $\mathbb{C}\setminus S$ out of a single Laurent series, but locally, every point of $\mathbb{C}\setminus S$ is part of such an annulus, and you could patch together the whole thing by considering the Laurent series on the different annuli covering it. But since we're considering infinitesimal/local transformations anyway, that doesn't play any relevant role here. Jan 7, 2020 at 23:37
• If we demand invariance/proper transformation under all these series, we are demanding invariance under the local flow of all these meromorphic vector fields. Jan 7, 2020 at 23:38