# Infinitesimal transformation of differential forms

The infinitesimal differential forms $dx_1$ and $dx_2$ span a two-dimensional vector space (cotangent space). The transformation $f$ acts on $(x_1,x_2)$ like a coordinate transformation. Thus,

$$dx_i \to du_i = \frac{\partial u_i}{\partial x_j}dx_j,\qquad i = 1,2.$$

Let's look at the infinitesimal transformation of $x_i$, i.e. $$u_i(x) = x_i +\epsilon_i(x).$$

Show that if $f$ is a conformal transformation:

$$\omega(x)\delta_{ij} = -\frac{\partial \epsilon_j}{\partial x^i}-\frac{\partial \epsilon_i}{\partial x^j},$$

where $\omega(x) \in \mathbb{R}$ is a scale factor.

Hint: First, show that $$\delta_{ij}du^idu^j=\left(1+\omega(x)\right)\delta_{kl}dx^kdx^l.$$

I've already tried different things such as substituting the $du$'s in the expression given as a hint. All of this basically led to nothing, so it would be great if somebody could help me find the solution. How should I start?

Maybe I should mention that this is an exercise from a physics workbook, so this question has to be answered using the information given above.

## 1 Answer

You already have all the elements that you need. You just need to equate two expressions for the metric after a conformal transformation. Lets see it.

• First, its change under any transformation $x\mapsto u(x)$ can be calculated as \begin{align} \delta_{ij}dx_idx_j\mapsto& \,\delta_{ij}du_idu_j = \delta_{ij}\frac{\partial u_i}{\partial x_k} \frac{\partial u_j}{\partial x_l}dx_kdx_l = \left(\delta_{ik}+\frac{\partial \epsilon_i}{\partial x_k}\right) \left(\delta_{jl}+\frac{\partial \epsilon_j}{\partial x_l}\right) \delta_{ij}dx_kdx_l\\ =& \left(\delta_i^k\delta_j^l+ \delta_{jl}\frac{\partial \epsilon_i}{\partial x_k}+ \delta_{ik}\frac{\partial \epsilon_j}{\partial x_l}+ O\left(\left(\frac{\partial\epsilon}{\partial x} \right)^2\right)\right)\delta_{ij}dx_kdx_l \\ =& \left(\delta_{kl}+\frac{\partial\epsilon_l}{\partial x_k} +\frac{\partial\epsilon_k}{\partial x_l}+ O\left(\left(\frac{\partial\epsilon}{\partial x}\right)^2\right)\right) dx_kdx_l \end{align}

• On the other side, if it is a conformal transformation it should satisfy that after the transformation the metric is equal to $$\exp(-\omega(x))\delta_{ij}dx_idx_j=\left(1- \omega(x)+O\left(\omega(x)^2\right)\right)\delta_{ij}dx_idx_j.$$

Now, neglecting quadratic or higher powers of $\omega$ and the derivatives of $\epsilon$ one gets:

$$(1+\omega)\delta_{ij}dx_idx_j=\left(\delta_{kl}- \frac{\partial \epsilon_l}{\partial x_k} -\frac{\partial \epsilon_k}{\partial x_l}\right)dx_kdx_l$$

And because the coefficient of each $dx_idx_j$ has to be equal on both sides: $$\omega\delta_{ij}= -\frac{\partial \epsilon_j}{\partial x_i} -\frac{\partial \epsilon_i}{\partial x_j}$$

• Thanks, this makes it clearer. But still, how do I get the desired result? I'm still missing the $\delta_{ij}$ in your result. Dec 20, 2016 at 15:10
• You are right! I was sloppy there. I have edited the answer to fix it Dec 20, 2016 at 16:35