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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:

  1. 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?

  2. 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?

  3. 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.

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  • $\begingroup$ I believe this question of mine on Math.SE, and the great answer by Jack Lee (author of Introduction to Smooth Manifolds), may be useful to you: math.stackexchange.com/q/2501573 $\endgroup$
    – Gold
    Apr 13, 2022 at 23:30

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In fact this has been clearly stated in the follow paragraph of Ginsparg's note.

Since two independent algebras naturally arise, it is frequently useful to regard z and z as independent coordinates. (More formally, we would say that since the action of the conformal group in two dimensions factorizes into independent actions on z and z, Green functions of a 2d conformal field theory may be continued to a larger domain in which z and z are treated as independent variables.) In terms of the original coordinates (x1, x2) ∈ $R^2$, this amounts to taking instead (x1, x2) ∈ C2, and then the transformation to z, z coordinates is just a change of variables. In C2, the surface defined by z = z∗ is the ‘real’ surface on which we recover (x, y) ∈ R2. This procedure allows the algebra A ⊕ $\overline{A}$ to act naturally on C2, and the ‘physical’ condition z = z∗ is left to be imposed at our convenience. The real surface specified by this condition is preserved by the subalgebra of A⊕$\overline{A}$ generated by ℓn+ℓn and i(ℓn−ℓn). In the sections that follow, we shall frequently use the independence of the algebras A and A to justify ignoring anti-holomorphic dependence for simplicity, then reconstruct it afterwards by adding terms with bars where appropriate.

(sorry that I do not type tex command but just copy the above context from pdf)

for example consider the generator for rotation operation, denoted as $l$ . Points in the total space is $(z,\overline{z})$, acting $\exp(i\epsilon l)$ on it just let it goes to $(e^{i\epsilon}z,\overline{z})$ which only rotating $z$ but do not rotate $\overline{z}$. To actually rotate $(x,y)$ you should in fact act $l+\overline{l}$ which both changing $z$ and $\overline{z}$. However, since it's obvious that the change of $\overline{z}$ could be related to those of $z$ when we restrict the total space on the "physical" plane, so we usually only talk about how $z$ changes.

In a word, $\overline{l}$ has nothing to do with the anti-holomorphic transformation. It's how $\overline{z}$ changes under holomorphic transformation.

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  • $\begingroup$ "Since two independent algebras naturally arise, it is frequently useful to regard z and z as independent coordinates. (More formally, we would say that since the action of the conformal group in two dimensions factorizes into independent actions on z and z, Green functions of a 2d conformal field theory may be continued to a larger domain in which z and z are treated as independent variables.)" -- This is what I don't understand. To me it seems that only one algebra arises from infinitesimal 2d conformal transformations. $\endgroup$ Apr 14, 2022 at 15:26
  • $\begingroup$ Specifically, he says that the action of the conformal group "factorizes" into independent actions on z and zbar. I am assuming by factorizes he means that the group has product structure GxG', and that an element of the group (g, h) acts on (z, zbar) as (g(z), h(zbar)). But again, I fail to see how this is the case, since once you know how z changes you know zbar changes, so the group actions cannot be independent. $\endgroup$ Apr 14, 2022 at 15:32

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