Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
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]: en.wikipedia.org/wiki/Gauss%E2%80%93Codazzi_equations –  Paul Dec 13 '11 at 9:24
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.