# 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

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

Using this, one gets

$\displaystyle\delta \sqrt {\det (g)} = - \frac{1}{{2\sqrt {\det (g)} }}\delta g = \frac{1}{2}\sqrt {\det (g)} ({g^{\mu \nu }}\delta {g_{\mu \nu }}) = - \frac{1}{2}\sqrt {\det (g)} ({g_{\mu \nu }}\delta {g^{\mu \nu }}).$

A simple application of this is that if we denote by $S$ the Einstein-Hilbert action, in the following form,

$\displaystyle S=\int_M Rdv_g =\int_M R \sqrt{\det (g)} dx$

where $R$ denotes the scalar curvature, then if we require $\delta S=0$, then this is about to say that

$\displaystyle\sqrt {\det (g)} \delta R + R\delta \sqrt {\det (g)} = 0,$

i.e.,

$\displaystyle\sqrt {\det (g)} {R_{\mu \nu }}\delta {g^{\mu \nu }} - \frac{1}{2}R\sqrt {\det (g)} {g_{\mu \nu }}\delta {g^{\mu \nu }} = 0.$

In other words,

$\displaystyle\sqrt {\det (g)} \delta {g^{\mu \nu }}\left( {{R_{\mu \nu }} - \frac{1}{2}{g_{\mu \nu }}R} \right) = 0,$

i.e.,

$\displaystyle R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R =0$

which is nothing but the Einstein equation in vacuum.

Source: Wiki.