# Ngô Quốc Anh

## December 25, 2012

### The Ricci equation

Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 2:58

Let us consider the Ricci identity defining the Riemann tensor $R$ as measuring the lack of commutation of two successive covariant derivatives with respect to the connection $\nabla$ associated with $V$‘s metric $g$ $\nabla_\nu\nabla_\sigma w^\mu- \nabla_\sigma\nabla_\nu w^\mu=R_{\rho\nu\sigma}^\mu w^\rho.$

By using $w$ as the normal vector $n$, we find that $\nabla_\nu\nabla_\sigma n^\mu- \nabla_\sigma\nabla_\nu n^\mu=R_{\rho\nu\sigma}^\mu n^\rho.$

We now project this tensor twice onto $M$, here the indices $\mu$ and $\nu$, and once along $n$, here the index $\sigma$. Using the previous topic, we find that $\displaystyle {\gamma _{\alpha \mu }}{n^\sigma }\gamma _\beta ^\nu ({\nabla _\nu }{\nabla _\sigma }{n^\mu } - {\nabla _\sigma }{\nabla _\nu }{n^\mu }) = {\gamma _{\alpha \mu }}{n^\sigma }\gamma _\beta ^\nu R_{\rho \nu \sigma }^\mu {n^\rho }.$

We now denote by $a$ the following vector $\nabla_n n$. It is clear that $a$ is a tangent vector.

This is clear since $\displaystyle n\cdot a = n \cdot \nabla_n n = \frac{1}{2}\nabla_n (n \cdot n)=0.$

By extending the second fundamental $K$ on $M$ given by $K(u,v)=- u \cdot \nabla_v n$ onto $V$, we find that $K(u,v)=- u \cdot \nabla_v n - (a \cdot u)(n \cdot v)$ for all vectors $u, v \in T_p(M)$.

Again, this is clear since $\displaystyle \begin{gathered} K(u,v)\mathop = \limits^{\rm def} - (u + (n \cdot u)n) \cdot {\nabla _{v + (n \cdot v)n}}n \hfill \\ \qquad\quad\,\,= - (u + (n \cdot u)n) \cdot ({\nabla _v}n + (n \cdot v){\nabla _n}n) \hfill\\ \qquad\quad\,\,= - u \cdot {\nabla _v}n - (n \cdot v)u \cdot \underbrace {{\nabla _n}n}_a - (n \cdot u)\underbrace {n \cdot {\nabla _v}n}_0 - (n \cdot u)(n \cdot v)\underbrace {n \cdot {\nabla _n}n}_0 \hfill \\ \qquad\quad\,\,= - u \cdot {\nabla _v}n - (a \cdot u)(n \cdot v). \hfill \\ \end{gathered}$

Thus, in components, we have