# Generators of conformal isometries on two-dimensional manifolds

I'm studying Conformal Field Theory via Paul Ginsparg's lecture notes and have a doubt regarding the Witt algebra.

Let $$(M,g)$$ be a smooth manifold with a metric tensor $$g$$ which we shall assume to have Euclidean signature for simplicity. A conformal isometry is a diffeomorphism $$\phi :M\to M$$ such that $$\phi_\ast g = \Omega^2 g$$ for some conformal factor.

We wish to understand the generators of such transformations, i.e., vector fields $$X\in \Gamma(TM)$$ whose flow are conformal isometries. This demands that the Lie derivative of $$g$$ be $$\mathfrak{L}_Xg=\alpha g.$$

Now, in components this is

$$\nabla_\mu X_\nu+\nabla_\nu X_\mu=\alpha g_{\mu\nu}.$$

Now suppose that $$D = 2$$ and $$g_{\mu\nu}=\delta_{\mu\nu}$$. The resulting equations are the Cauchy-Riemann equations:

$$\partial_1 X_2 = -\partial_2 X_1,\quad \partial_1 X_1 = \partial_2 X_2. \tag{1}$$

This I can understand. The problem is that one usually takes a big leap from here and says:

1. That $$\phi : M\to M$$ must be a holomorphic or anti-holomorphic function; If we introduce complex coordinates on $$M$$, $$(z,\bar{z}) : U\subset M\to \mathbb{C}^2$$ this means $$\phi$$ is either a function only of $$z$$ or only of $$\bar{z}$$.

2. That the algebra of the generators of conformal isometries is generated by $$\ell_n=-z^{n-1}\partial_z,\quad \bar{\ell}_n=-\bar{z}^{n-1}\partial_\bar{z}.$$

Now I fail to see how do we get from Eq. (1) to the two conclusions. All I know is that if the function $$f : \mathbb{C}\to \mathbb{C}$$ written $$f(x,y) = u(x,y)+i v(x,y)$$ has $$u,v$$ satisfying the Cauchy-Riemann equations, then $$f$$ is holomorphic.

It is not the case here. What satisfies the Cauchy Riemann equations are the components of vector fields generators of conformal isometries, not the maps $$\phi : M\to M$$ themselves.

Also, I fail to see why $$\ell_n,\bar{\ell}_n$$ are the generators.

So how do we get from the vector fields satisfying the Cauchy Riemann equations to the conclusion that the finite diffeomorphisms are holomorphic/anti-holomorphic and that such vector fields are spanned by the $$\ell_n,\bar{\ell}_n$$?

What's written is indeed not quite right. (A mathematician would never write what he did.) First of all, there is a confusion between the group and pseudogroup. Let $$M$$ be a Riemann surface (equivalently, an oriented Riemannian surface with Riemannian metric considered up to multiplication by a positive scalar function.) The group $$G$$ of conformal transformations of $$M$$, denoted $$Conf(M)$$, consists of holomorphic and antiholomorphic automorphisms of $$M$$. More precisely, $$Conf(M)$$ contains an index 2 subgroup consisting to holomorphic automorphisms. One can put a "reasonable" topology on this group (say, topology of local uniform convergence) which makes $$G=Conf(M)$$ a topological group; this group turns out to be a (finite dimensional) Lie group. Then only holomorphic automorphisms can be close to the identity in this group, so considering antiholomorphic vector fields as elements of the Lie algebra of $$G$$ is utterly wrong. The Lie algebra consists of holomorphic vector fields. There are some issues here in the case of noncompact surfaces since not every holomorphic vector field can be integrated in a 1-parameter family of conformal automorphisms. However, if $$M$$ is compact, this is not a problem.
However, what Ginsparg wants to consider is something more complicated (and he does not, or cannot, tell you the difference). He wants to consider a pseudogroup of conformal transformations. A formal definition would take awhile but it suffices to say at this point that this pseudogroup consists of locally defined conformal automorphisms, i.e. biholomorphic maps between open subsets of $$M$$. Composition of these maps is not always possible (it is might be defined on an empty subset). It is a bit better to consider the subgroup of the above pseudogroup consisting of conformal transformations fixing a point $$p\in M$$ and defined on some (open) neighborhoods of $$p$$: $$g: U\to V, p\in U\cap V, g(p)=p,$$ where $$U, V$$ are neighborhoods of $$p$$. (A mathematician would use the word "germs" here: One identifies maps which agree on a neighborhood of $$p$$. This is not entirely vacuous since neighborhoods can be disconnected.) This becomes an infinite dimensional Lie group (once equipped with a suitable topology). The Lie algebra of this group consists of (germs at $$p$$) holomorphic vectors fields on (neighborhoods of $$p$$) vanishing at $$p$$. The holomorphic (local) transformations in this group again form a neighborhood of the identity, hence, if you are looking only at the Lie algebra, you should consider only holomorphic vector fields. (And disregard the antiholomorphic ones!). Now, since all the discussion is local, it suffices to consider open subsets of the complex plane where $$p=0$$.
Then vector fields as above are all of the form $$f(z)\partial_{z},$$ where $$f(0)=0$$ and $$f$$ is holomorphic at some neighborhood of $$0$$. Now, just forget about $$\partial_z$$ and think about $$f(z)$$. Saying that $$f$$ is holomorphic is equivalent to saying that $$f$$ is complex-analytic at $$0$$, i.e. is represented by its Taylor series: $$f(z)=\sum_{n=1}^\infty a_n z^n.$$ (Again, in some neighborhood of $$0$$.) In other words, the vector fields $$z^n\partial_z, n\ge 1$$, are topological generators of the Lie algebra of holomorphic vector fields at $$0$$ (vanishing at $$0$$). There is an analytical issue hidden here since not every power series has positive radius of convergence. One can either settle for the Lie algebra of formal power series (where you do not care if the series converges) or stick with the Lie algebra of (local) holomorphic vector fields. Both viewpoints are common. Hope it helps.