How does the metric change under scale transformations? I was reading Zee's book in Group Theory, chapter VIII.2 (The Conformal Algebra) and I'm stuck in something probably easy:
In p.516, he states that under a scale transformation, $x'^{\mu} = \lambda x^{\mu}$, the metric changes as:
$g'_{\rho\sigma}(x') = g'_{\rho\sigma}(\lambda x) = \lambda^2g_{\mu\nu}(x)$
However, this result seems wrong to me. First, the indices are wrong and using the general formula for transforming the metric I get:
$g'_{\rho\sigma}(x') = g_{\mu\nu}(x)\frac{\partial x^{\mu}}{\partial x'^{\rho}}\frac{\partial x^{\nu}}{\partial x'^{\sigma}} = \frac{g_{\mu\nu}(x)}{\lambda^2}\frac{\partial x^{\mu}}{\partial x^{\rho}}\frac{\partial x^{\nu}}{\partial x^{\sigma}} = \frac{g_{\mu\nu}(x)}{\lambda^2} \delta^{\mu}_{\rho}\delta^{\nu}_{\sigma} = \frac{g_{\rho\sigma}(x)}{\lambda^2}$
Is something wrong in my logic?  
 A: I think he means something more mundane i.e $g_{\mu \nu} dx^{\mu}dx^{\nu} \rightarrow g_{\mu \nu} d (\lambda x^{\mu})d( \lambda x^{\nu}) = g_{\mu \nu} \lambda^2 dx^{\mu}dx^{\nu} $. Then we can define new components $g_{\mu \nu}' =g_{\mu \nu} \lambda^2 $. I think the confusion comes from the fact that we are not actually talking about a coordinate transformation (at least not in the usual sense). Consider the following: I have basis set $\{e_i\}$ and $\{e_j\}$. We can write a vector in each as follows $v=v^ie_i$ or $v=v^je_j$. There is a linear transformation $A_{ij}e_j = e_i$. This means $v = v^iA_{ij}e_j$ and therefore switching from the $\{e_i\}$ basis to $\{e_j\}$ requires multiplication of $v$ by $A^T$ or orthogonal transformations specifically $A^{-1}$.Something like this is what you have in mind with the equations you wrote.
I think what Zee has in mind is the following. In a specific coordinate system $\{e_a\}$, we can write the components of the covariant tensor as $g_{ab}=e_a \cdot e_b$. But in this coordinate system perform a scaling transformation. What do the matrix elements of the matrix tensor look like? Well, $e_a \rightarrow \lambda e_a$ so $g_{ab}' = \lambda e_a \cdot \lambda e_b = \lambda^2 e_a \cdot e_b = \lambda^2 g_{ab}$
A: There is nothing wrong with your logic.
I believe that you have found a typographical error in Zee's book.
Congratulations!
