The Setup: Let's say we want to study a Euclidean $\mathrm{CFT}_2$ on $\mathbb R^2$ with coordinates $\sigma^1$ and $\sigma^2$ and metric

$ds^2 = (d\sigma^1)^2+(d\sigma^2)^2$.

It seems to me that in the usual discussion (e.g. di Francesco, Ginsparg, Polchinski), one proceeds to consider an analytic continuation of the CFT to $\mathbb C^2$ with coordinates $z^1, z^2$ and complex metric

$ds^2 = (dz^1)^2+(dz^2)^2$

and then, one performs the coordinate transformation $z = z^1+iz^2$ and $\bar z = z^1-iz^2$. In this way the coordinates $z$ and $\bar z$ can be considered "independent" because they are coordinates on a complex two-dimensional manifold. Also, in these coordinates the metric becomes

$ds^2 = dz\,d\bar z$

and it becomes clear that conformal mappings consist of mappings: $(z, \bar z)\to (f(z), g(\bar z))$.

My confusion is this: Since our original theory was on $\mathbb R^2$, books say that when we do calculations, we should consider the physical theory as living on the copy of $\mathbb R^2$ embedded in $\mathbb C^2$ given by the condition $\bar z = z^*$. But consider the mapping $(z, \bar z)\to (z^2, \bar z)$. This is a conformal mapping on $\mathbb C^2$, but it does not map the surface $\bar z = z^*$ to itself; for example the point $(z, \bar z)=(2,2)$ gets mapped to the point $(z^2, \bar z) =(4,2)$ and $2$ is clearly not equal to $4^*$. In particular, it seems to me that analytic continuation to a CFT on $\mathbb C^2$ enlarges the set of mappings one can have, so what relevance does it really have to the original CFT on $\mathbb R^2$?


2 Answers 2


The definition of a conformal mapping in this situation is one that takes $(z,\overline{z})\to (f(z),f^*(z))$, where $f(z)$ is holomorphic. So the example you gave isn't actually conformal.

To be concrete, let's take a free boson. A conformal transformation acts as

\begin{align*} \delta \phi&=-\epsilon v\partial \phi-\epsilon v^*\overline{\partial }\phi\\ &\approx \phi(z,\overline{z})-\phi(z+\epsilon v,\overline{z}+\epsilon v^*), \end{align*}

where $v$ is holomorphic. So its clear that the conformal transformation acts on $z$ and $\overline{z}$ as $(z,\overline{z})\mapsto (f(z),f^*(z))$.

The confusing thing, which I think you're referring to, is that any vector $v^a$ on $\mathbf{C}^2$ such that $v^z$ is holomorphic and $v^{\overline{z}}$ is antiholomorphic satisfies $\mathcal{L}_v \delta_{ab}\propto \delta_{ab}$, and is therefore a conformal transformation. However, we know that $v^z$ and $v^{\overline{z}}$ are complex conjugates since the theory is really living on $\mathbf{R}^2$, so we should only consider such $v$'s for CFTs.

  • $\begingroup$ When I was originally learning CFT, that's what I thought as well, but 1) isn't it true that the mapping $(z, \bar z)\to (z^2, \bar z)$ leads to $2dz d\bar z$ which is conformal on $\mathbb C^2$ but just not on the surface $\bar z = z^*$? and 2) isn't it precisely the fact that one can perform independent transformations of $z$ and $\bar z$ that gives independent copies of the conformal algebra after complexification? Thanks for the help Matthew. $\endgroup$ Jan 23, 2013 at 2:02
  • 1
    $\begingroup$ Sorry, I think we had overlapping edits... my last paragraph should answer your #1. For #2, it's true that $L_m$ and $\tilde{L}_m$ are independent, but in the end $\delta X$ only involves a certain linear combination of $L_m$ and $\tilde{L}_m$. $\endgroup$
    – Matthew
    Jan 23, 2013 at 2:08
  • $\begingroup$ No prob. Ok so what I gather from this is that when we want to perform "physically permissible" conformal mappings or whatever one would call such things, we restrict ourselves to performing conformal mappings on $\mathbb C^2$ of the form $(z, \bar z)\to (f(z), f^*(z))$. Is this true? If so, why does it seem like people do independent things to $z$ and $\bar z$ in physics books/papers regarding CFTs on $\mathbb R^2$? What physics could one gather from such beasts? Perhaps I'm just imagining things? $\endgroup$ Jan 23, 2013 at 2:16
  • $\begingroup$ Maybe you're thinking about the fact that if you had a chiral boson like $\phi(z)$, then it transforms as $\phi'(f(z))=\phi(z)$, so it doesn't see the antiholomorphic side. There are also these warped CFTs, as considered by Hofman+Strominger, where the scaling only acts on the left-movers. By the way, you just confused everyone working in the cubicles around me at a grad school that will remain unnamed. $\endgroup$
    – Matthew
    Jan 23, 2013 at 2:24
  • 2
    $\begingroup$ Excellent! Nothing better than dragging other grad students into your own confused state :) $\endgroup$ Jan 23, 2013 at 2:36

I think you took the wrong steps. This is my take: we first identify $\mathbb{R}^2 \cong \mathbb{C}$ the natural way. Then here we take leap to $\mathbb{C}^2$ just for computational convenience. This is because in the world of complex functions sometimes it is easier to talk not of $z \in \mathbb{C}$ alone but of $z$ and $\bar z$ $\in \mathbb{C}$ at the same time as independent variables. This because if we write $z=x+iy$, we can have functions such as $f_1=x^2+y^2=z\bar z$ and also functions such as $f_2=x^2-y^2+i2xy = z^2$. So as you can see some functions over $\mathbb{C}$ can be written as functions of $z$ alone (holomorphic functions), others as functions of $(z,\bar z)$, and others with $\bar z$ as their only argument. So if we treat $z$ and $\bar z$ as different arguments, we naturally end up on $\mathbb{C}^2$.

About your example: You are still working with a 2-dim'l CFT. The map you defined is not a holomorphic map over $\mathbb{C}$ so by now you have strayed off the path of transformations of $\mathbb{C}$ ( or $\mathbb{S}^1$. if you consider global transformations).

  • $\begingroup$ I agree with the math in the first paragraph; if we extend a function $f$ on $\mathbb C$ to a function $\hat f$ on $\mathbb C^2$ in the natural way, then if the extension satisfies $\partial \hat f/\partial \bar z= 0$, the original function $f$ is holomorphic. I disagree however that the second two metrics I wrote down are not metrics on $\mathbb C^2$ if that's what you're saying. Moreover, the map I wrote down is $\mathbb C^2\to \mathbb C^2$, so I agree it's not holomorphic in the sense you describe; but my point was more in the spirit captured by Matt's answer. $\endgroup$ Jan 24, 2013 at 18:20
  • $\begingroup$ sorry, what i had said previously was not entirely correct. $\endgroup$
    – yca
    Jan 24, 2013 at 19:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.