The first Gauss-Codazzi equation for a metric hypersurface $(S, h_{ab})$ of $(M, g_{ab})$ is: $$\mathcal{R}^{a}_{\ bcd} = -2 \pi^{a}_{\ [c}\pi_{d]b} + h^a_{\ m}h^n_{\ b}h^p_{\ c}h^r_{\ d}R^m_{\ \ npr}$$ Here:
- $R^m_{\ \ npr}$ is the Riemann tensor on $M$;
- $h_{ab}$ is the induced metric on $S$;
- $\mathcal{R}^a_{\ bcd}$ is the Riemann tensor on $S$;
- $\pi_{ab}$ is the second fundamental form on $S$.
I want to derive the following $(*)$: $$\mathcal{R}_{bc} = \pi\ \pi_{bc}-\pi_{ab} \ \pi^a_{\ c}+h^n_{\ b}h^p_{\ c}R_{np}-R_{mbcr}\xi^m\xi^r$$ Here, $\pi = \pi^a_{\ a}$ and $\xi^a$ is a unit vector normal to $S$.
After $a,d$ contraction, I get $\mathcal{R}_{bc} = \pi\ \pi_{bc}-\pi_{ab} \ \pi^a_{\ c}+h^r_{\ m}h^n_{\ b}h^p_{\ c}R^m_{\ \ npr}$. My textbook (Malament 2012, p. 111) suggests substituting $(g^r_{\ m} - \xi^r\xi_m)$ for $h^r_{\ m}$, which gets me the fourth term of $(*)$ only insofar as $h^n_{\ b}h^p_{\ c}R^m_{\ \ npr} = R^m_{\ \ bcr}$, i.e. iff $R^a_{\ bcd}$ is tangent to $S$ in $b$ and $c$.
I'm unable to prove, however, that this is the case, and I'm not even sure that it is. Could anyone help me out with the proof, or perhaps point out if I've made some mistake along the way?
OK, I've got it. To get the fourth term of $(*)$ one does not need to show that $R^a_{\ bcd}$ is tangent to $S$ in $b$ and $c$; it's enough to show that $R_{abcd}\xi^a\xi^d$ is tangent to $S$ in $b$ and $c$.
Using various symmetries of $R_{abcd}$, one can show that:
$$(R_{abcd} + R_{acdb} + R_{adbc})\xi^a\xi^d\xi^b = {\bf 0}$$
To show that $R_{abcd}\xi^a\xi^d$ is tangent to $S$ in $b$, note that:
$$R_{acdb}\xi^a\xi^d\xi^b = R_{dbac}\xi^a\xi^d\xi^b = R_{dabc}\xi^b\xi^d\xi^a = -R_{adbc}\xi^b\xi^d\xi^a$$
So: $(R_{abcd}\xi^a\xi^d)\xi^b = {\bf 0}$.
To show that $R_{abcd}\xi^a\xi^d$ is tangent to $S$ in $c$, note that:
$$R_{adbc}\xi^a\xi^d\xi^b = R_{abdc}\xi^a\xi^b\xi^d = - R_{abcd}\xi^a\xi^b\xi^d$$
So, we have: ${\bf 0} = R_{acdb}\xi^a\xi^d\xi^b = R_{abdc}\xi^a\xi^d\xi^c = -R_{abcd}\xi^a\xi^d\xi^c$; thus: $(R_{abcd}\xi^a\xi^d)\xi^c = {\bf 0}$.