Suppose the metric $g$ with respect to coordinate $X^{a}=(x^{i},y^{m})$ has the form $g=h_{ij}(y)dx^{i}dx^{j}+k_{mn}(y)dy^{m}dy^{n}$, and $R_{x^{i}}^{x^{i}}=R_{x^{1}}^{x^{1}}+....+R_{x^{n}}^{x^{n}}=0$, then the square root of det of $h_{ij}$ is harmonic in the metric $k_{mn}$, ie,
\[D^{m}D_{m}\sqrt{|h|}=0\]
where $D$ is the connection of $k_{mn}$.
Motivation:
in the page 166 Wald says: the equation (7.1.20) yields (7.1.21),
here the two equations are,
\[0=R_{t}^{t}+R_{\phi}^{\phi}=(\nabla_{a}t)R_{b}^{a}\xi^{b}+(\nabla_{a}\phi)R_{b}^{a}\psi^{b}\]
and,
\[D^{a}D_{a}\rho=0\]
where the metric ansatz is,
\[ds^{2}=-V(\rho,z)(dt-w(\rho,z)d\phi)^{2}+V^{-1}\rho^{2}d\phi^{2}+\Omega(\rho,z)^{2}(d\rho^{2}+\Lambda(\rho,z)dz^{2})\]
it can be checked by some direct calculation and the two are equal to $\frac{\sqrt{\Lambda}_{,z}}{\rho\Omega^{2}\sqrt{\Lambda}}$, up to a factor, and certainly it holds not accidentally, which motivites the foregoing results.
Question:
Can it be derived more geometrically using something like foliation?
And what is the intuitive picture?
Here is a proof I got. It's direct, but with less intuition.
Proof:
There are n Killing vector fields $(A_{i})^{a}$ wrt $x^{i}$ for $g$ is independent of $x^{i}$, which leads to, (notice: no summation here, and $x^{i}$ is a scalar function)
\[R_{x^{i}}^{x^{i}}=\nabla_{a}x^{i}\cdot\nabla_{c}\nabla^{c}(A_{i})^{a}\]
for a antisymmetric (2,0) tensor, the covariant derivative is,
\[\nabla_{c}A^{ca}=\frac{1}{\sqrt{|g|}}\frac{\partial(A^{ca}\sqrt{|g|})}{\partial X^{c}}\]
applying to (\ref{a}),
\[\nabla_{c}\nabla^{c}(A_{i})^{a}=\frac{1}{\sqrt{|g|}}\frac{\partial(\nabla^{m}(A_{i})^{a}\sqrt{|g|})}{\partial y^{m}}\]
we need to calculate $\nabla^{m}(A_{i})^{a}$,
\[\nabla^{m}(A_{i})^{a}=g^{mb}\Gamma_{\ bd}^{a}(A_{i})^{d}=k^{mn}\Gamma_{\ nx^{i}}^{a}\]
and,
\[\Gamma_{\ nx^{i}}^{a}=g^{ac}(g_{cn,x^{i}}+g_{cx^{i},n}-g_{nx^{i},c})=g^{ac}g_{cx^{i},n}=h^{ai}h_{ix^{i},n}\]
(notice: the non-vanishing components of the connection are $\Gamma_{\ xy}^{x},\ \Gamma_{\ xx}^{y},\ \Gamma_{\ yy}^{y}$.)
substitute it into (\ref{eq:b}),
\[\nabla_{c}\nabla^{c}(A_{i})^{a}=\frac{1}{\sqrt{|g|}}\frac{\partial(k^{mn}h^{ai}h_{ix^{i},n}\sqrt{|g|})}{\partial y^{m}}\]
$\nabla_{a}x^{i}\cdot$ is to pick out the $x^{i}$ term of the index $a$, thus summing up (\ref{eq:c}) over $x^{i}-$s, the vanishment of partial trace of Ricci tensor is equivalent to,
\[\frac{1}{\sqrt{|g|}}\frac{\partial(k^{mn}h^{ji}h_{ij,n}\sqrt{|g|})}{\partial y^{m}}=0\]
while for $D^{m}D_{m}\sqrt{|h|}=\frac{1}{\sqrt{|k|}}\frac{\partial}{\partial y^{m}}(k^{mn}\sqrt{|k|}\frac{\partial}{\partial y^{n}}(\sqrt{|h|}))$, the last ingrediant left is,
\[\frac{\partial}{\partial y^{m}}h=hh^{ij}\frac{\partial}{\partial y^{m}}h_{ij}\]
Q&A (4871)
