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


Create a free website or blog at