Conformal Field Theory

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1. Mar 19, 2015

Ace10

Hi all,

my question is rather a simple one and regards conformal transformations. On "Applied CFT" by P.Ginsparg, http://arxiv.org/pdf/hep-th/9108028.pdf , on page 10, gives the transformation rule of a quasi primary field and relates the exponent of 1.12 to the one of 1.10. My first question is how can I obtain 1.10 and secondly, how the first equation of 1.11 is related to the one of 1.12..

I know that under dilatations: x'→λx , but how can I write this for a field? It has to do with the Jacobian 1.10? Is this somehow related to the volume element? (I see the determinant of the metric in the denominator and I think that is related to the volume element but I'm not sure..)

Thank you very much in advance for your help.

2. Mar 21, 2015

samalkhaiat

The defining relation of the conformal group $C ( 1 , n - 1 )$ is given by $$\bar{g}_{a b} ( x ) = \frac{\partial \bar{x}^{c}}{\partial x^{a}} \frac{\partial \bar{x}^{d}}{\partial x^{b}} \ \eta_{c d} = S( x ) \ \eta_{a b} . \ \ \ (1)$$ Taking the determinants and assuming even-dimensional space-time with signature $(1 , n - 1)$, we find $$( - \bar{g} ) = | \frac{\partial \bar{x}}{\partial x} |^{2} = S^{n} ,$$ or $$\frac{1}{\sqrt{- g}} = S^{- \frac{n}{2}} = | \frac{\partial x}{\partial \bar{x}} | .$$ From this, we obtain $$\frac{1}{\sqrt{S( x )}} = | \frac{\partial x}{\partial \bar{x}} |^{\frac{1}{n}} . \ \ \ \ \ \ \ \ (2)$$ In order to understand how the fields transform, you really need to study the representation theory of the conformal algebra. You can find more details in
However, we can do it loosely in here. Let us rewrite (1) in the form $$\left( \frac{1}{\sqrt{S}} \frac{\partial \bar{x}^{c}}{\partial x^{a}} \right) \left( \frac{1}{\sqrt{S}} \frac{\partial \bar{x}^{d}}{\partial x^{b}} \right) \eta_{c d} = \eta_{a b} .$$ Therefore, it is clear that the matrix $$\Lambda ( x ) \equiv \frac{1}{\sqrt{S}} \frac{\partial \bar{x}}{\partial x} ,$$ is an element of the Lorentz group $SO(1,n-1)$. Moreover, this $\Lambda (x)$ forms a linear representation of the conformal group. This is because both $\frac{ \partial \bar{x}^{a}}{\partial x^{b}}$ and $\sqrt{S}=| \frac{\partial \bar{x}}{\partial x}|^{\frac{1}{n}}$ are themselves linear representations. This allows us to extend any linear representation of the Poincare group to the full conformal group. Therefore, given the finite-dimensional (matrix) representation $\Lambda \rightarrow D(\Lambda) , \ \forall \Lambda \in SO(1,n-1)$, the conformal transformation $x \rightarrow \bar{x}$ can be represented by $$\mathcal{C}(\frac{\partial \bar{x}}{\partial x})= \left( \sqrt{S(x)} \right)^{ - \Delta} \times D( \Lambda(x)) = \left( \sqrt{S(x)} \right)^{ - \Delta} \times D( \frac{1}{\sqrt{S(x)}}\ \frac{\partial \bar{x}}{\partial x} ) ,$$ where $\Delta$ is a real number (the scaling dimension) if $D(\Lambda)$ is irreducible (Schur's lemma), otherwise a matrix satisfying $[\Delta,D(\Lambda)]=0$. Indeed, all finite-dimensional representations of $C(1,3)$ are completely specified by the finite-dimentional irreducible representations $(j_{1},j_{2})$ of the Lorentz group $SO(1,3)$ and those of the non-compact group of pure dilatations $SO(1,1)$ labelled by the scaling dimension $\mathcal{R}(\sqrt{S(x)}) = \left(\sqrt{S} \right)^{- \Delta}$. For example, if $V^{a}(x)$ is a field transforming in the vector representation of the Lorentz group: $\tilde{V}^{a}(\tilde{x}) = \Lambda^{a}{}_{b} \ V^{c}(x) ; \ x^{a}\rightarrow \tilde{x}^{a} = \Lambda^{a}{}_{c} \ x^{c}$, then, under a conformal transformation $x \rightarrow \bar{x}$, we have $$\bar{V}^{a} ( \bar{x} ) = \mathcal{C}^{a}{}_{c} ( \frac{\partial \bar{x}}{\partial c} ) \ V^{c} ( x ) , \ \ \ \ (3)$$ where $$\mathcal{C}^{a}_{c} ( \partial \bar{x} / \partial x ) = \left( \sqrt{S} \right)^{- \Delta} \ D^{a}{}_{c}( \frac{1}{\sqrt{S(x)}} \ \frac{\partial\bar{x}}{\partial x} ) = \left( \sqrt{S} \right)^{- \Delta - 1} \frac{\partial \bar{x}^{a}}{\partial x^{c}} . \ \ \ (4)$$ So, under pure dilatations; $\bar{x}^{a} = e^{-\alpha}\ x^{a}$, we have $$\frac{\partial \bar{x}^{a}}{\partial x^{c}} = e^{- \alpha} \delta^{a}_{c} , \ \ \Rightarrow \ | \frac{\partial \bar{x}}{\partial x} | = e^{- n \alpha} , \ \ \ (5)$$ and, therefore $$\left( \sqrt{S} \right)^{- \Delta - 1} = | \frac{\partial \bar{x}}{\partial x} |^{\frac{- \Delta - 1}{n}} = e^{\alpha \Delta + \alpha} . \ \ \ (6)$$ Putting eq’s (4), (5) and (6) in equation (3), we find $$\bar{V}^{a}( \bar{x} ) = e^{\alpha \Delta} \ V^{a} ( x ) \equiv | \frac{\partial x}{\partial \bar{x}} |^{\frac{\Delta}{n}} V^{a} ( x ) .$$

Sam

3. Mar 21, 2015

Ace10

Wow Sam, thank you very much for your detailed reply! I realised I was treating the representations the wrong way! Thank you very much!

By the way, just to ask, if we have an odd dimensional spacetime, say:
g=diag{...-1... , ...+1...} (n-times -1 with n odd),
then we use absolute value at the deteminant under the squareroot, right? So one can say that in general case we follow the notation with the absolute value?

4. Mar 23, 2015

samalkhaiat

For $(p,q)$ signature in $n = |p| + |q|$ dimensions, you have $$\det (g_{ab}) = (-1)^{|q|} |g| = (-1)^{n} (-1)^{- |p|} |g| .$$ So
$$\det(g_{ab}) = (-1)^{|p|} |g| , \ \ \mbox{for} \ \ n = 2k ,$$ $$\det(g_{ab}) = (-1)^{|p| + 1} |g| , \ \ \mbox{for} \ \ n = 2k + 1 .$$

5. Mar 26, 2015

Ace10

Right, thank you very much!