Let $g_{ab}$ be the coordinates of a metric on a $d$-dimensional space. Consider a conformal transformation such that the transformed metric is given by $h_{ab} = \Omega^2g_{ab}$. If $g := \text{det}(g_{ab})$, is it correct that $h = (\Omega^2)^dg$?


1 Answer 1


Yes, this is simple linear algebra. If $A$ is a $n\times n$ square matrix and $c$ is any scalar, then $\det(cA) = c^n \det A$. You're applying this with $A = (g_{ab})_{a,b=1}^d$, $n=d$, and $c=\Omega^2$.

I would strongly recommend against writing $h=(\Omega^2)^dg$ with $g$ and $h$ being the determinants, because $g_{ab}$ and $h_{ab}$ should be merely the components of the tensors $g$ and $h$, so $h=\Omega^2 g$ would be the correct thing to write. If ${\rm vol}_g$ and ${\rm vol}_h$ are the corresponding volume forms, what you wrote is essentially a coordinate-based proof that $${\rm vol}_h = \Omega^d\,{\rm vol}_g.$$

  • $\begingroup$ Thank you, yes I'm not a fan of that notation either, but it seems to be the norm in string theory. $\endgroup$
    – Bedge
    Commented Sep 26, 2021 at 19:25

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