How can the algebra of infinitesimal conformal transformations be infinite dimensional (in 2D)?

Question

In Blumenhagen's book "Introduction to Conformal Field Theory", I found the statement

The algebra of infinitesimal conformal transformations in an Euclidean 2-dimensional space is infinite dimensional.

He concludes this after finding that the generators of the infinitesimal conformal transformations (i.e. a basis for its Lie algebra) are $l_n=-z^{n+1}\partial$ and $\bar l_n= -\bar z^{n+1}\bar\partial$ for $n\in\mathbb N$.
My problem with this is: shouldn't the elements of the Lie algebra be linear combinations (with real coefficients) of the basis elements $\partial,\bar\partial$ of $T_eG$? (Instead of a roduct of the derivatives with polynomials). Furthermore (and related to the previous point), if our Lie group is 2-dimensional (as a manifold), that implies that its Lie algebra is 2-dimensinal (as vector space), so definitely not infinitely dimensional. What is going on here?

Answer 1

You are confusing 2 different things:

A d-dimensional Lie group has a d-dimensional algebra which is the tangent space at the identity.

Here, there is an infinite dimensional group acting on a 2 dimensional space. There is a representation of the algebra as vector fields (derivations) on this space, given by the formulas you stated. (infinitesimal conformal transformations)

The generators should depend on the point z, as each conformal transformations act differently around each point - for example, the map $z$ to $z^2$ acts infinitisimaly as $z+\delta z$ to $z+2z\delta z$. The dependence on the point is easiliy understood - for example under this map, the more far z is away from the origin, it is stretched more.