# Complex coordinates in CFT

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 $\mathbb R^2$? (my apologies for the long post)

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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.

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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. –  joshphysics Jan 23 '13 at 2:02
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 $\tilde{L}_m$. –  Matthew Jan 23 '13 at 2:08
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? –  joshphysics Jan 23 '13 at 2:16
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. –  Matthew Jan 23 '13 at 2:24
Excellent! Nothing better than dragging other grad students into your own confused state :) –  joshphysics Jan 23 '13 at 2:36
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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).

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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. –  joshphysics Jan 24 '13 at 18:20
sorry, what i had said previously was not entirely correct. –  yca Jan 24 '13 at 19:13