I am reading Ginsparg's notes on 2D-CFT, and I am deeply confused about why Ginsparg states after (1.8) that the conformal algebra for 2d Euclidean space consists of two copies of the Witt algebra.
My understanding (primarily from Martin Schottenloher's book) is that a conformal transformation between two Riemannian manifolds $(M, g)$ and $(M', g')$ is a diffeomorphism $f:M\to M'$ such
$$ f^*(g') = \Omega^2 g $$ where $f^*(g')$ is the pullback of $g'$ and $\Omega$ is a positive smooth function on $M$. In the case where $M=M'=\mathbb{R}^2$ and $g$ and $g'$ are the standard Euclidean metric, the condition that a map
$$f:\mathbb{R}^2\to\mathbb{R}^2 \\ (x,y)\mapsto (u(x,y), v(x,y))$$
is conformal becomes
$$ (\partial_x u)^2+(\partial_x v)^2 = (\partial_y u)^2+(\partial_y v)^2 = \Omega(x,y)^2 \neq 0, $$ $$ \partial_x u\, \partial_y u + \partial_x v \,\partial_y v = 0. $$ Identifying $\mathbb{R}^2$ with $\mathbb{C}$, the above conditions are equivalent to the map $f:\mathbb{C}\to\mathbb{C}$ defined by $x+iy \mapsto u(x,y)+iv(x,y)$ being either holomorphic or anti-holomorphic (along with some local invertibility condition).
I am comfortable with all of this so far. Now, following the approach in section 1.1 of Ginsparg, we determine the conformal algebra by looking at infinitesimal transformations. In other words, we consider a one-parameter family of conformal transformations $f_\epsilon:\mathbb{C}\to\mathbb{C}$ such that $f_0$ is the identity mapping on $\mathbb{C}$. In the case $f_\epsilon$ is holomorphic, we can we expand it as
$$ f_\epsilon(z) = z+\epsilon h(z)+\mathcal{O}(\epsilon^2) $$
with $h$ holomorphic. We may then expand $h$ in a Laurent series in $z$, and identify $l_n=-z^{n+1} \partial_z$ as the vector fields which generate infinitesimal holomorphic transformations.
Here comes where I get lost. If we try to do the same thing in the case where $f_\epsilon$ is anti-holomorphic in an attempt to find the generators of anti-holomorphic transformations, we run into the issue that $f_0$ is the identity map, and the identity map is not anti-holomorphic. In other words, it does not make sense to discuss infinitesimal anti-holomorphic transformations because anti-holomorphic transformations are disconnected from the identity. Indeed, the answer to this post also acknowledges that 2d infinitesimal conformal transformations consist only holomorphic maps. The fact that anti-holomorphic maps are disconnected from the identity makes intuitive sense as well. An anti-holomorphic map is holomorphic map composed with complex conjugation, which acts as a reflection in the complex plane, so the case is similar to the two disconnected components of the group $O(n)$.
Ginsparg however includes the vector fields $-\bar{z}^{n+1} \partial_\bar{z}$ in the generators for the 2d conformal algebra, arguing that they "act" on $\bar{z}$ rather on $z$. I have spent much time (without success) trying to interpret his argument in a way that seems consistent with viewing conformal algebra as generating infinitesimal conformal transformations.
So the main questions I have are:
Am I correct in viewing the conformal algebra associated with a space as the generators of infinitesimal conformal transformations? I understand that there are compactification and locality details to consider, but is this view at least correct in spirit?
If such a viewpoint is correct, why are the generators $-\bar{z}^{n+1} \partial_\bar{z}$ included in the 2d conformal algebra, and what infinitesimal transformations do they generate on the plane? That is, if $f_\lambda:\mathbb{C}\to\mathbb{C}$ is the one-parameter family generated by $-\bar{z}^{n+1} \partial_\bar{z}$, and I give you an arbitrary complex number $z=x+iy$, what is $f_\lambda(z)$? Is $f_\lambda$ holomorphic or anti-holomorphic?
If such a viewpoint is incorrect, what is the correct viewpoint and how does it encapsulate the cases Ginsparg covers in section 1.1, which seem consistent with viewing the conformal algebra as generators of infinitesimal conformal transformations?
I have been agonizing over this for a while, and all the literature I can find is extremely terse on this point.