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If $M$ is a s-R(semi-Riemannian) submanifold of a s-R manifold $\overline{M}$ the function $R^{\perp}:\mathfrak{X}(M)\times\mathfrak{X}(M)\times\mathfrak{X}(M)^\perp\rightarrow\mathfrak{X}(M)^\perp$ given by

$$R_{VW}^\perp X=D_{[V,W]}^\perp X-[D_V^\perp,D_W^\perp]X $$

is called the normal curvature tensor of $M$. The question is, how can I prove that it satisfies the Ricci equation:

$$\langle R_{VW}^\perp X,Y\rangle=\langle \bar{R}_{VW} X,Y\rangle+\langle\widetilde{II}(V,X),\widetilde{II}(W,Y)\rangle-\langle\widetilde{II}(V,Y),\widetilde{II}(W,X)\rangle,$$

where $X,Y\in\mathfrak{X}^\perp(M)$?

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This is [Gauss-Codazzi equation][1] in the Riemannian case. Have you ever try to look at the proof of the Riemannian case to see if everything goes through for the semi-Riemannian case? [1]: – Paul Dec 13 '11 at 9:24

Write $D$ for the covariant derivative on $\overline M$ and decompose $D$ in to tangential and normal components, i.e. $$D = D^T + D^{\perp}$$ where $D^T_XY$ is the projection of $D_XY$ onto $TM$ and $D^{\perp}_XY$ is the projection of $D_XY$ onto $\nu(M)$, the normal bundle of $M$ in $\overline M$. That is if $n$ is a normal vector to $M$, write $D^{\perp}_XY = g(D_XY,n)n$. Now write down your definition of curvature $$R_{VW}X = D_{[V,W]}X - [D_V,D_W]X$$ and make the substitution $D = D^T + D^{\perp}$. When the dust clears (and if you know the definition of $\widetilde{II}$) you will get the Gauss and Codazzi equations that you desire. I will also remark that this is standard in any textbook on the subject- see for example Barret O'Neil's book Semi-Riemannian Geometry.

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