# 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 $\displaystyle {\nabla _\beta }{n_\alpha } = - {K_{\alpha \beta }} - {a_\alpha }{n_\beta }.$

We now calculate $a$ in terms of the lapse function $N$. Thanks to $n=N\nabla t$, we find that $\displaystyle\begin{gathered} {a_\alpha } = {n^\mu }{\nabla _\mu }{n_\alpha } \hfill \\ \quad\,= - {n^\mu }{\nabla _\mu }(N{\nabla _\alpha }t) \hfill \\ \quad\,= - {n^\mu }{\nabla _\mu }N{\nabla _\alpha }t - N{n^\mu }\underbrace {{\nabla _\mu }({\nabla _\alpha }t)}_{{\nabla _\alpha }{\nabla _\mu }t} \hfill \\ \quad\,= \frac{1}{N}{n_\alpha }{n^\mu }{\nabla _\mu }N - N{n^\mu }{\nabla _\alpha }( - \frac{1}{N}{n_\mu }) \hfill \\ \quad\,= \frac{1}{N}{n_\alpha }{n^\mu }{\nabla _\mu }N - \underbrace {{n^\mu }{\nabla _\alpha }{n_\mu }}_0 + N{n^\mu }{n_\mu }{\nabla _\alpha }(\frac{1}{N}) \hfill \\ \quad\,= \frac{1}{N}{n_\alpha }{n^\mu }{\nabla _\mu }N - \frac{1}{N}\underbrace {{n^\mu }{n_\mu }}_{ - 1}{\nabla _\alpha }N \hfill \\ \quad\,= \frac{1}{N}({\nabla _\alpha }N + {n_\alpha }{n^\mu }{\nabla _\mu }N) \hfill \\ \quad\,= \frac{1}{N}\nabla _\alpha ^MN \hfill \\ \quad\,= \nabla _\alpha ^M\ln N. \hfill \\ \end{gathered}$

Therefore, we can write $\displaystyle {\nabla _\beta }{n_\alpha } = - {K_{\alpha \beta }} - \nabla _\alpha ^M \ln N {n_\beta }.$

Using this, we can calculate ${\gamma _{\alpha \mu }}{n^\sigma }\gamma _\beta ^\nu R_{\rho \nu \sigma }^\mu {n^\rho }$ as follows. $\displaystyle\begin{gathered} {\gamma _{\alpha \mu }}{n^\sigma }\gamma _\beta ^\nu R_{\rho \nu \sigma }^\mu {n^\rho } = {\gamma _{\alpha \mu }}{n^\sigma }\gamma _\beta ^\nu ( - {\nabla _\nu }K_\sigma ^\mu - {\nabla _\nu }[{({\nabla ^M})^\mu }\ln N{n_\sigma }] + {\nabla _\sigma }K_\nu ^\mu + {\nabla _\sigma }[{({\nabla ^M})^\mu }\ln N{n_\nu }]) \hfill \\ \qquad\qquad\qquad= {\gamma _{\alpha \mu }}\gamma _\beta ^\nu ( - \underbrace {{n^\sigma }{\nabla _\nu }K_\sigma ^\mu }_{{n^\sigma }K_\sigma ^\mu = 0} - \underbrace {{n^\sigma }{n_\sigma }}_{ - 1}{\nabla _\nu }[{({\nabla ^M})^\mu }\ln N] - [{({\nabla ^M})^\mu }\ln N]\underbrace {{n^\sigma }{\nabla _\nu }{n_\sigma }}_0 \hfill \\ \qquad\qquad\qquad\qquad+ {n^\sigma }{\nabla _\sigma }K_\nu ^\mu + {n^\sigma }\underbrace {{n_\nu }}_{\gamma _\beta ^\nu {n_\nu } = 0}{\nabla _\sigma }[{({\nabla ^M})^\mu }\ln N] + [{({\nabla ^M})^\mu }\ln N]\underbrace {{n^\sigma }{\nabla _\sigma }{n_\nu }}_{\nabla _\nu ^M\ln N}) \hfill \\ \qquad\qquad\qquad= {\gamma _{\alpha \mu }}\gamma _\beta ^\nu (K_\sigma ^\mu {\nabla _\nu }{n^\sigma } + {\nabla _\nu }[{({\nabla ^M})^\mu }\ln N] + {n^\sigma }{\nabla _\sigma }K_\nu ^\mu + [{({\nabla ^M})^\mu }\ln N][{({\nabla ^M})_\nu }\ln N]) \hfill \\ \qquad\qquad\qquad= - {K_{\alpha \sigma }}K_\beta ^\sigma + {({\nabla ^M})_\beta }[{({\nabla ^M})_\alpha }\ln N] + \gamma _\alpha ^\mu \gamma _\beta ^\nu {n^\sigma }{\nabla _\sigma }{K_{\mu \nu }} + [{({\nabla ^M})_\alpha }\ln N][{({\nabla ^M})_\beta }\ln N] \hfill \\ \qquad\qquad\qquad= - {K_{\alpha \sigma }}K_\beta ^\sigma + \frac{1}{N}{({\nabla ^M})_\beta }[{({\nabla ^M})_\alpha }N] + \gamma _\alpha ^\mu \gamma _\beta ^\nu {n^\sigma }{\nabla _\sigma }{K_{\mu \nu }} \hfill .\end{gathered}$

For any normal vector $m$, say $m=Nn$, we obtain the following Lie derivative $\displaystyle {\mathcal L_m}{K_{\alpha \beta }} = {m^\mu }{\nabla _\mu }{K_{\alpha \beta }} + {K_{\mu \beta }}{\nabla _\alpha }{m^\mu } + {K_{\alpha \mu }}{\nabla _\beta }{m^\mu }.$

Since $\nabla m = \nabla (Nn) = N\nabla n + n \otimes \nabla N$, we find that $\displaystyle {\nabla _\beta }{m^\alpha } = - NK_\beta ^\alpha - {({\nabla ^M})^\alpha }N{n_\beta } + {n^\alpha }{\nabla _\beta }N.$

Hence, $\displaystyle\begin{gathered} {\mathcal L_m}{K_{\alpha \beta }} = {m^\mu }{\nabla _\mu }{K_{\alpha \beta }} - 2N{K_{\alpha \mu }}K_\beta ^\mu - {K_{\mu \beta }}{({\nabla ^M})^\mu }N{n_\alpha } - {K_{\alpha \mu }}{({\nabla ^M})^\alpha }N{n_\beta } \hfill \\ \qquad\qquad+ \underbrace {{K_{\mu \beta }}{n^\mu }}_0{\nabla _\alpha }N + \underbrace {{K_{\alpha \mu }}{n^\alpha }}_0{\nabla _\beta }N. \hfill \\ \qquad\quad\,= {Nn^\mu }{\nabla _\mu }{K_{\alpha \beta }} - 2N{K_{\alpha \mu }}K_\beta ^\mu - {K_{\mu \beta }}{({\nabla ^M})^\mu }N{n_\alpha } - {K_{\alpha \mu }}{({\nabla ^M})^\alpha }N{n_\beta }. \hfill \\ \end{gathered}$

We now project the above equation onto $M$ and thanks to the fact that $\mathcal L_mK$ tangents to $M$ because $K$ does, we arrive at $\displaystyle {\mathcal L_m}{K_{\alpha \beta }} = N\gamma _\alpha ^\mu \gamma _\beta ^\nu {n^\sigma }{\nabla _\sigma }{K_{\mu \nu }} - 2N{K_{\alpha \mu }}K_\beta ^\mu ,$

which yields $\displaystyle\gamma _\alpha ^\mu \gamma _\beta ^\nu {n^\sigma }{\nabla _\sigma }{K_{\mu \nu }} = \frac{1}{N}{\mathcal L_m}{K_{\alpha \beta }} + 2{K_{\alpha \mu }}K_\beta ^\mu.$

Thus, we have proved that $\displaystyle {\gamma _{\alpha \mu }}{n^\sigma }\gamma _\beta ^\nu R_{\rho \nu \sigma }^\mu {n^\rho } = \frac{1}{N}{\mathcal L_m}{K_{\alpha \beta }} + \frac{1}{N}{({\nabla ^M})_\beta }{({\nabla ^M})_\alpha }N + {K_{\alpha \mu }}K_\beta ^\mu.$

This is the so-called Ricci equation. Note that we can write $\displaystyle {({\nabla ^M})_\beta }{({\nabla ^M})_\alpha } = {({\nabla ^M})_\alpha }{({\nabla ^M})_\beta }$

since $\nabla^M$ has no torsion.

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