Ginsparg Applied Conformal Field Theory

[physicus]

Ginsparg "Applied Conformal Field Theory"

I have some questions concerning the first two chapters of Ginsparg's text on CFT, which can be found here

1. In equation (2.1) primary fields of conformal weight $(h,\overline{h})$ are introduced as fields that transform the following way:
$\Phi(z,\overline{z}) \to \left(\frac{\partial f}{\partial z}\right)^h \left(\frac{\partial \overline{f}}{\partial \overline{z}}\right)^\overline{h}\Phi(f(z), \overline{f}(\overline{z}))$
Then, $\Phi(z,\overline{z})dz^h d\overline{z}^\overline{h}$ is supposed to be invariant. I don't understand that statement. $ds^2$ transforms the following way: $ds^2 \to \left(\frac{\partial f}{\partial z}\right) \left(\frac{\partial \overline{f}}{\partial \overline{z}}\right)ds^2$
Therefore, I suppose $dz \to \left(\frac{\partial f}{\partial z}\right)dz, d\overline{z} \to \left(\frac{\partial \overline{f}}{\partial \overline{z}}\right)d\overline{z}$
That leads to $\Phi(z,\overline{z})dz^h d\overline{z}^\overline{h} \to \left(\frac{\partial f}{\partial z}\right)^{2h} \left(\frac{\partial \overline{f}}{\partial \overline{z}}\right)^{2\overline{h}}\Phi(f(z), \overline{f}(\overline{z}))dz^h d\overline{z}^\overline{h}$
I don't see why this transformation behaviour should be called invariant.

2. In the section on two dimensional CFT Ginsparg compactifies the spatial coordinate in oder to eliminate infrared divergences. Is there an easy way to understand why this compactification should lead to the elimination of infrared divergences? Why does this not further constrain the generalitiy of the theory?

3. After the compactification of the spatial coordinate Ginsparg maps the resulting cylinder to the complex plane. There we can use the known tools of complex analysis to proceed, in particular complex line integration. He introduces the radial ordering operator
$\begin{equation}R(A(z)B(w))=<br /> \left\{<br /> \begin{aligned}<br /> A(z)B(w) & \quad |z|>|w|\\<br /> B(w)A(z) & \quad |z|<|w|<br /> \end{aligned}<br /> \right.<br /> \end{equation}$
Then, Ginsparg claims, the equal-time commutator of a local operator $A$ with the spatial integral of an operator $B$ becomes the contour integral of the radially ordered product:
$\left[\int dx B, A\right]_{e.t.} \to \oint dz \: R(B(z)A(w))$
I don't see why this is the case.

I am happy about answers to any of the three questions.

 

[fzero]

That leads to $\Phi(z,\overline{z})dz^h d\overline{z}^\overline{h} \to \left(\frac{\partial f}{\partial z}\right)^{2h} \left(\frac{\partial \overline{f}}{\partial \overline{z}}\right)^{2\overline{h}}\Phi(f(z), \overline{f}(\overline{z}))dz^h d\overline{z}^\overline{h}$
I don't see why this transformation behaviour should be called invariant.

You can show that

$$\Phi(f(z), \overline{f}(\overline{z}))df^h d\overline{f}^\overline{h} = \left(\frac{\partial z}{\partial f}\right)^{h}\left(\frac{\partial \overline{z}}{\partial \overline{f}}\right)^{\overline{h}}\Phi(z, \overline{z}) \left(\frac{\partial f}{\partial z}\right)^{h} dz^h \left(\frac{\partial \overline{f}}{\partial \overline{z}}\right)^{\overline{h}} d\overline{z}^\overline{h} = \Phi(z, \overline{z}) dz^h d\overline{z}^\overline{h}.$$

This is the usual definition of invariance.

2. In the section on two dimensional CFT Ginsparg compactifies the spatial coordinate in oder to eliminate infrared divergences. Is there an easy way to understand why this compactification should lead to the elimination of infrared divergences? Why does this not further constrain the generalitiy of the theory?

Infrared divergences would arise from long distances in the spatial coordinate, but we cut these off by setting a finite radius. This is like putting a free particle in a finite box so that we get normalizable wavefunctions, or even just putting a gas in a finite box in stat mech. Alternatively, we could demand that our physical fields have "compact support," which roughly means that all functions are nonzero over a finite area and vanish at $\pm$ spatial infinity. The generality of results is not constrained if you can show that there are no divergences in observables when you take the radius to infinity (the large volume limit).

3. After the compactification of the spatial coordinate Ginsparg maps the resulting cylinder to the complex plane. There we can use the known tools of complex analysis to proceed, in particular complex line integration. He introduces the radial ordering operator
$\begin{equation}R(A(z)B(w))=<br /> \left\{<br /> \begin{aligned}<br /> A(z)B(w) & \quad |z|>|w|\\<br /> B(w)A(z) & \quad |z|<|w|<br /> \end{aligned}<br /> \right.<br /> \end{equation}$
Then, Ginsparg claims, the equal-time commutator of a local operator $A$ with the spatial integral of an operator $B$ becomes the contour integral of the radially ordered product:
$\left[\int dx B, A\right]_{e.t.} \to \oint dz \: R(B(z)A(w))$
I don't see why this is the case.

The $\int dx \rightarrow \int d\theta$ after mapping to the complex plane is explained above (2.7). The issue of radial ordering is explained in the paragraphs below (2.8), where he explains that convergence of observables is tied to time ordering. So we're really computing an "equal-time" commutation relation by slightly deforming one of the times to $t\pm \delta t$ and then letting $\delta t$ go to zero. Perhaps you can make more sense of it by mapping fig. 2 from the plane to the cylinder and seeing that the arc around $w$ corresponds to this time deformation.

 

[physicus]

Hi, thank you for your answers. They were very helpful. However, I still have some problems concerning my first question.
In other literature I often find the following transformation law for primary fields:
$\Phi(z,\overline{z}) \to \Phi'(z',\overline{z}')=\left(\frac{\partial f}{\partial z}\right)^{-h}\left(\frac{\partial \overline{f}}{\partial \overline{z}}\right)^{-\overline{h}}\Phi(z,\overline{z})$
If I use that definition I can follow your argument.

But how does that agree with
$\Phi(z,\overline{z}) \to \left(\frac{\partial f}{\partial z}\right)^h \left(\frac{\partial \overline{f}}{\partial \overline{z}}\right)^\overline{h}\Phi(f(z), \overline{f}(\overline{z}))$
Is the right hand side $\Phi'(z,\overline{z})$ or $\Phi'(f(z),\overline{f}(\overline{z}))$?

 

[fzero]

Hi, thank you for your answers. They were very helpful. However, I still have some problems concerning my first question.
In other literature I often find the following transformation law for primary fields:
$\Phi(z,\overline{z}) \to \Phi'(z',\overline{z}')=\left(\frac{\partial f}{\partial z}\right)^{-h}\left(\frac{\partial \overline{f}}{\partial \overline{z}}\right)^{-\overline{h}}\Phi(z,\overline{z})$
If I use that definition I can follow your argument.

But how does that agree with
$\Phi(z,\overline{z}) \to \left(\frac{\partial f}{\partial z}\right)^h \left(\frac{\partial \overline{f}}{\partial \overline{z}}\right)^\overline{h}\Phi(f(z), \overline{f}(\overline{z}))$
Is the right hand side $\Phi'(z,\overline{z})$ or $\Phi'(f(z),\overline{f}(\overline{z}))$?

I would say that

$$\Phi'(f(z),\overline{f}(\overline{z})) = \left(\frac{\partial f}{\partial z}\right)^h \left(\frac{\partial \overline{f}}{\partial \overline{z}}\right)^\overline{h}\Phi(f(z), \overline{f}(\overline{z})).$$

Ginsparg states things backwards, but it should be clear that he's using the requirement that $\Phi(z,\bar{z}) dz^h d\bar{z}^\bar{h}$ be invariant in order to define how the scalar field transforms.

 

[physicus]

Ok, thanks a lot for your help!