# Ngô Quốc Anh

## June 5, 2011

### Variation of the determinant

Filed under: Riemannian geometry — Ngô Quốc Anh @ 6:05

Today, we shall prove the following identity

$\displaystyle\boxed{\delta \det(g) = \delta \det (g_{\mu\nu}) = \det(g )g^{\mu\nu} \delta g_{\mu\nu} }.$

First, by the Jacobi formula, we know that for any matrix $A$

$\displaystyle d\mbox{det} (A) = \mbox{tr} (\mbox{adj}(A) \, dA)$

where $d$ is a differential of $A$. Since

$\displaystyle A^{-1}=\frac{1}{\det (A)}\mbox{adj}(A),$

we can rewrite the Jacobi formula as follows

$\displaystyle d\mbox{det} (A) = \det (A)\mbox{tr} (A^{-1}dA).$

We now make use this rule with $A$ replaced by metric $g$ and $d$ replaced by $\delta$. Obviously,

$\displaystyle \delta \det(g) =\det(g)\mbox{tr}(g^{-1}\delta g)= \det(g )g^{\mu\nu} \delta g_{\mu\nu}$

where we have used the fact that $g^{-1}=g^{\mu\nu}$ and that

$\displaystyle \mbox{tr}(AB)=\sum_{i,j}A_{ij}B_{ij}.$