\chapter{Curvature of space}
\pagebreak[4]
\section{p82 - Exercise}
\begin{tcolorbox}
Explain why the surfaces of an ordinary cylinder and an ordinary cone are to be regarded as "flat" in the sense of our definition.
\end{tcolorbox}
The reason is because those surfaces can be "unwrapped" like the figure below shows. 
\begin{figure}[h]
%\includegraphics[scale=.5]{Conemapping.jpg}

\input{D:/MathLatex/images/fig_p82_21_a.tex}
\caption{Unwrapping of a cone}
\end{figure}\\
For the cone, we can for each point $\ P $ on the cone, lying on a distance $\ h $ from the apex and making an angle $\theta $,  associate on a plane,  a  point $ P^* $ lying at the same  distance $ h $ from the apex, taken as origin for the coordinate system, and making an angle $\phi =  \theta \sin{\alpha} $ with $ \alpha $ the angle of the cone. This pair of coordinates are polar coordinates with $ r \in (-\infty, + \infty)$ and $\theta \in [0, 2k\pi)$. 
The same reasoning can be applied to a cylinder which is a cone with the apex at $\infty$. In that case the coordinate system becomes a Cartesian coordinate system. \\
As a continuous mapping exist from polar to orthogonal Cartesian coordinates both  coordinate system can be written under the required form (3.101) and so can be called "flat".

$$\blacklozenge$$
\newpage

\section{p83 - Exercise}
\begin{tcolorbox}
What are the values of $R^s_{.rmn}$ in an Euclidean plane, the coordinates being rectangular Cartesians? Deduce the values of the components of this tensor for polar coordinates from its tensor character, or else by direct calculation.
\end{tcolorbox}
See 2.64 page 139 (exercise 18).
$$\blacklozenge$$
\newpage

\section{p86 - Exercise}
\begin{tcolorbox}
Show that in a $V_2$ all the components of the covariant curvature tensor either vanish or are expressible in terms of $R_{1212}$.
\end{tcolorbox}
We have (3.115) and (3.116)
\begin{align}
\left \{ \begin{array}{l}
\ R_{rsmn} =  - R_{srmn}\\
\ R_{rsmn} =  - R_{rsnm}\\
\ R_{rsmn} =   R_{mnrs}\\
\ R_{rsmn} + R_{rmns}+R_{rnsm}=0
\end{array} \right.
\end{align}
It is clear from the two first identities that  in pairs $(rs)$ and $(mn)$ both indices have to be different when the tensor is not $0$. So we only have to consider $R_{1212}$,  $R_{1221}$, $R_{2112}$ and $R_{2121}$.\\
The two first  identities gives us:
\begin{align}
 R_{1221}= -R_{1212}\\
 R_{2112}= -R_{1212}\\
 R_{2121}= -R_{2112} = R_{1212}
\end{align}
The third identity does give us any additional information.
The fourth identity gives us only trivial statements:
\begin{align}
 R_{1212} + \underbrace{R_{1122}}_{=0}+\underbrace{R_{1221}}_{=- R_{1212} } = 0\\
 \underbrace{R_{1221}}_{=- R_{1212}} + R_{1212}+\underbrace{R_{1122}}_{=0 } = 0\\
 \underbrace{R_{2112}}_{=- R_{1212}} + \underbrace{R_{2121}}_{=R_{1212}}+\underbrace{R_{2211}}_{=0} = 0\\
 \underbrace{R_{2121}}_{= R_{1212}} + \underbrace{R_{2211}}_{=0}+\underbrace{R_{2112}}_{= -R_{1212}} = 0
\end{align}
$\textbf{Conclusion:}$\\
We get the identities (2), (3) and (4) in function of $R_{1212}$ and all vanish if $R_{1212} = 0$
$$\blacklozenge$$
\newpage

\section{p86-87 - clarification}
\begin{tcolorbox}
\textit{The number of independent components of the covariant curvature tensor in a space of N dimensions is} $$\frac{1}{12}N^2\left(N^2-1\right)$$
\end{tcolorbox}
We have (3.115) and (3.116)
\begin{align}
\left \{ \begin{array}{l}
R_{rsmn} =  - R_{srmn}\\
R_{rsmn} =  - R_{rsnm}\\
R_{rsmn} =   R_{mnrs}\\
R_{rsmn} + R_{rmns}+R_{rnsm}=0
\end{array}\right.
\end{align}
It is clear from the two first identities that  in the tuple $(rs)$ and $(mn)$ both indices have to be different when the component is not $0$. So we only have to consider the component with the pair of tuples $(r,s)$ and $m,n)$ with $r \neq s$ and $m \neq n$.
For the tuple $(r,s)$ we have $N$ possibilities to draw an index for $r$ but for $s$ only $N-1$ indices remain as $r \neq s$. So for the tuple $(r,s)$ we get $N(N-1)$ possibilities. But note by the first identity $R_{rsmn} =  - R_{srmn}$ that we only have to consider the half of this quantity as once we have chosen a tuple $(r,s)$ we also know  the component for the tuple $(s,r)$. So the total number of possibilities we have for  $(r,s)$ is $M = \half N(N-1)$. The same yields for the tuple $(mn)$. So, we get in total $M^2$ possibilities according to the two first identities.\\
The third identity $R_{rsmn} =   R_{mnrs}$ puts an extra constraint on the number of possibilities as we have to subtract from $M^2$ the number of possibilities covered by this third identity. Note that, once we have chosen a tuple $(rs)$ we have to exclude the tuple $(m,n) =  (r,s)$ as the identity $R_{rsrs} =   R_{rsrs}$ becomes trivial.. So for the first tuple we have $M$ possibilities, but once chosen, only $M-1$ remain for the second tuple. So we get $M(M-1)$ possibilities. But, again we only have to take half of these possibilities as the identities $R_{rsmn} =   R_{mnrs}$ and $ R_{mnrs} = R_{rsmn}$ are equivalent.\\
So the total number of possibilities reduces to $$ M^2 - \half M(M-1) \quad \text{with} \quad M=\half N(N-1) $$
What about the fourth identity $$R_{rsmn} + R_{rmns}+R_{rnsm}=0$$
First we note that this identity implies that all indices are different as it becomes trivial in the other cases. This is a consequence of the first 3 identities. Indeed, we know already that
\begin{align}
\left \{ \begin{array}{l}
\ r \neq s\\
\ m \neq n\\
\ (r,s) \neq (m,n)\\
\end{array} \right.
\end{align}
Let's consider the following cases
\begin{align}
\left \{ \begin{array}{llll}
\ r = m&\rightarrow m\neq s \  m\neq n \ r\neq n&\rightarrow & R_{rsrn} + \underbrace{R_{rrns}}_{=0}+\underbrace{R_{rnsr}}_{= -R_{rnrs}= -R_{rsrn}}=0\\
\ r = n&\rightarrow n\neq s \  m\neq n \ r\neq s&\rightarrow & R_{rsmr} + \underbrace{R_{rmrs}}_{= -R_{mrrs}= -R_{rsmr}}+\underbrace{R_{rrsm}}_{=0}=0\\
\ s = m&\rightarrow m\neq r \  n\neq s \ r\neq s&\rightarrow & R_{rssn} + \underbrace{R_{rsns}}_{= -R_{rssn}}+\underbrace{R_{rnss}}_{=0}=0\\
\ s = n&\rightarrow r\neq s \  m\neq s \ m\neq n&\rightarrow & R_{rsms} + \underbrace{R_{rmss}}_{=0}+\underbrace{R_{rssm}}_{= -R_{rsms}}=0
\end{array} \right.
\end{align}
So indeed, once two indices are equal, the fourth identity becomes trivial and does not put extra constraints to the number of possibilities. For the tuple $(r,s,m,n)$ we have $N$ possibilities to draw an index for $r$, for $s$ only $N-1$, for $m$ only $N-2$ and for $n$ only $N-3$ indices remain as $r \neq s\neq m\neq n$. The maximum number of constraint generated by the fourth identity is thus $$N(N-1)(N-2)(N-3)$$
But here again double counts occur. Indeed the fourth identity is true for the $6$ tuples $$(rsmn),(rmsn) ,(rmns),(rsnm),(rnsm),(rnms)$$ as first entry    in the identity. The same reasoning is valid for the tuples $(n...), \ (s...) \ (m...) $. \\So in total we get $6\times4 = 24$ equivalent identities and the number of constraints generated by the fourth identity reduces to  $$\frac{1}{24}N(N-1)(N-2)(N-3)$$
Note that this number of constraints vanish for $N \leq 3$.\\
Putting it all together the number of independent components of $R_{rsmn}$ becomes
\begin{align}
\mho &= M^2 - \half M(M-1)-\frac{1}{24}N(N-1)(N-2)(N-3)\\
&= \half M(M+1)-\frac{1}{24}N(N-1)(N-2)(N-3)\\
&= \frac{1}{8} N(N-1) \left(N(N-1)+2\right)-\frac{1}{24}N(N-1)(N-2)(N-3)\\
&= \frac{N}{24} \left( 3N^2-6N^2+9N-6-N^3+3N^2+3N^2-9N-2N+6 \right)\\
&= \frac{1}{12}N^2 \left( N^2-1\right)
\end{align}
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\newpage

\section{p87 - Exercise}
\begin{tcolorbox}
Using the fact that the absolute derivative of the fundamental tensor vanishes, prove that 3.107 may be written $$ \frac{\delta^2 T_r}{\delta u \delta v} - \frac{\delta^2 T_r}{\delta v \delta u} = R_{rpmn}T^p\partial_u x^m \partial_v x^n $$
\end{tcolorbox}
By 2.519 and 2.619 we have
\begin{align}
\fdv{T_r}{u} &= \fdv{(a_{rk}T^k)}{u} 
 = \underbrace{\fdv{(a_{rk})}{u}}_{=0}T^k+a_{rk}\fdv{(T^k)}{u}\\
 &= a_{rk}T^k_{|n} \partial_u x^n\\
 \Rightarrow\quad \frac{\delta^2 T_r}{\delta u \delta v} &= \frac{\delta (a_{rk}T^k_{|n} \partial_u x^n)}{\delta v}\\
  & = \underbrace{\frac{\delta (a_{rk})}{\delta v}}_{=0}T^k_{|n} \partial_u x^n+a_{rk}\frac{\delta (T^k_{|n} )}{\delta v}\partial_u x^n+a_{rk}T^k_{|n} \delta \frac{(\partial_u x^n)}{\delta v}\\
   &= a_{rk}\underbrace{\frac{\delta (T^k_{|n} )}{\delta v}}_{=T^k_{|nm}\partial_v x^m} \partial_u x^n+a_{rk}T^k_{|n} \underbrace{\delta \frac{(\partial_u x^n)}{\delta v}}_{=\left(\partial_u x^n \right)_{|m}\partial_v x^m}\\
   &= a_{rk}T^k_{|nm}\partial_v x^m \partial_u x^n+a_{rk}T^k_{|n} \underbrace{\left(\partial_u x^n \right)_{|m}}_{ = \partial_m \left(\partial_u x^n \right) + \Gamma^n_{pm}\partial_u x^p}\partial_v x^m\\
   &= a_{rk}T^k_{|nm}\partial_v x^m \partial_u x^n+a_{rk}T^k_{|n} \left( \underbrace{\partial_m \left(\partial_u x^n \right)}_{=\partial_u \left( \delta_m^n \right)=0} + \Gamma^n_{pm}\partial_u x^p \right)\partial_v x^m\\
   &= a_{rk}T^k_{|nm}\partial_v x^m \partial_u x^n+a_{rk}T^k_{|n} \Gamma^n_{pm}\partial_u x^p \partial_v x^m
\end{align}
Hence we have 
\begin{align}
\frac{\delta^2 T_r}{\delta v \delta u} &=a_{rk}T^k_{|nm}\partial_u x^m \partial_v x^n+a_{rk}T^k_{|n} \Gamma^n_{pm}\partial_v x^p \partial_u x^m\\
&=a_{rk}T^k_{|mn}\partial_u x^n \partial_v x^m+a_{rk}T^k_{|n} \Gamma^n_{mp}\partial_v x^m \partial_u x^p\\
\Rightarrow\quad \frac{\delta^2 T_r}{\delta u \delta v} - \frac{\delta^2 T_r}{\delta v \delta u} &= a_{rk}T^k_{|nm}\partial_v x^m \partial_u x^n - a_{rk}T^k_{|mn}\partial_u x^n \partial_v x^m\\
&= \left(a_{rk}T^k_{|nm} - a_{rk}T^k_{|mn}\right)\partial_u x^n \partial_v x^m\\
&= \left(\underbrace{T_{r|nm} - T_{r|mn}}_{= -R_{rpmn}T^p}\right)\partial_u x^n \partial_v x^m\\
&= -\underbrace{R_{rpmn}}_{=-R_{rpnm}} T^p\partial_u x^n \partial_v x^m = 
R_{rpmn} T^p\partial_u x^m \partial_v x^n
\end{align}
$$\blacklozenge$$
\newpage

\section{p91 - Exercise}
\begin{tcolorbox}
Would the study of geodesic deviation enable us to distinguish between a plane and a right circular cylinder?
\end{tcolorbox}
For geodesic lines we have
\begin{align}
\dv[2]{x^r}{u} + \Gamma^r_{mn}\dv{x^m}{u}\dv{x^n}{u} = 0\\
\end{align}
with the fundamental tensor for a cylinder (see exercise page 27)
\begin{align} 
(a_{mn}) = \begin{pmatrix}
 1& 0 \\
0 & r^2 \\
\end{pmatrix}
\end{align}
As no element of this tensor is a function of the coordinates, it is clear that all Christoffel symbols vanish and the geodesic curve are solutions of the simple system of $2^{nd}$ order differential equations
\begin{align}
\dv[2]{x^r}{u}  = 0
\end{align}
Hence
\begin{align}
x^r  &= \kappa u + \mu\\
\text{or}\quad x^r &= \kappa^r u
\end{align}
by choosing the origin of the coordinates system with the initial condition position of the point.
Choosing polar coordinates $\phi, z$ as coordinates, the distance one walks when following a geodesic is given by
\begin{align}
 s= \int_{u_0}^{u_1}\sqrt{(r^2(d\phi)^2 + (dz)^2)}
\end{align}
and if we take $u=\phi$ as independent a parameter, by (6) we get 
\begin{align}
 s-s_0 &= \int_{\psi_0}^{\psi_1}\sqrt{(r^2 + \kappa^2)}d\psi \equiv g(\psi - \psi_0)\\
 \text{or}\quad s&= \int_{\psi_0}^{\psi_1}\sqrt{(r^2 + \kappa^2)}d\psi \equiv g\psi
\end{align}
by introducing transformed coordinates $s^{'}= s-s_0$ and $\psi^{'} =\psi - \psi_0$.\\
and thus we get 
\begin{align}
 z = m s^{'}
\end{align}
\begin{figure}[H]
%\centering
\begin{minipage}[t]{.4\textwidth}
%\centering
\vspace{0pt}
%\includegraphics[scale=.5]{p85_ex1.png}
\input{D:/MathLatex/images/fig_p91_153_a.tex}
\end{minipage}\hfill
\caption{Geodesics on a cylinder}
\label{fig:fig_p91_153_a}
\end{figure}
The above figure illustrates what an observer living in the manifold $\Omega$ sees when walking along geodesics on the cylinder.  He only can measure the distance $s^{'}$ and the displacement along $z$ and by (10) can only draw a chart like the one seen on the right of the cylinder. A "flatlander" living in the mapped manifold $M(\Omega)$ would see the same chart when walking along geodesics in his plane.\\\\
\textbf{Conclusion:} No, studying the geodesic deviation on a right circular cylinder does not enable us to say on which surface we are.
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\newpage

\section{p93 - Exercise}
\begin{tcolorbox}
For rectangular cartesians in Euclidean 3-space, show that the general solution of 3.311 is $\eta^r = A^r s+B^r $, where $A^, \ B^r$ are constants. verify this by elementary geometry. 
\end{tcolorbox}
We have equation 3.111
\begin{align}
\fdv[2]{\eta^r}{s}+ R^r_{.smn} p^s \eta^m p^n=0
\end{align}
From exercise on page 83 we know that $R^r_{.smn}=0$ in an Euclidean space. Also, in such spaces, the Christoffels symbols vanish and equation (1) reduces to $ \dv[2]{\eta}{s} = 0$. And so, $$\eta^r = A^r s+B^r $$

This is also easily deducted from a geometrical point of view. In an Euclidean space, the geodesics are straight lines. For an infinitesimal change in the geodesic family parameter $v$, we can assume that a vector, going perpendicular from 1 point from one geodesic with parameter $v$  to another infinitesimal close  geodesic with parameter $v+dv$,  will also be perpendicular on this geodesic. This situation is depicted in fig. ~\ref{fig:fig_p93_16}(a). We conclude that $\overrightarrow{AA^{'}} \ \parallel \ \overrightarrow{PP^{'}}$. This can also be deducted from Thales theorem (see fig.~\ref{fig:fig_p93_16} (b)). 
\begin{figure}[H]%
    \centering
    \subfloat[]{\input{D:/MathLatex/images/fig_p93_16_a.tex}}
	\qquad
    \subfloat[]{\input{D:/MathLatex/images/fig_p93_16_b.tex}}
\caption{Geometrical deduction of the geodesical deviation equation in an Euclidean space.}
\label{fig:fig_p93_16}
\end{figure}
Hence, we than can say that,
$$ \frac{\left|AA^{'}\right|}{u_0}=\frac{\left|PP^{'}\right|}{u}  $$
or, $$ \eta^r = A^r u + B^r $$ as the reference point $A$ can be chosen arbitrarily on the line $AP$.
$$\blacklozenge$$
\newpage


\section{p96 - Clarification}
\begin{tcolorbox}
... But under parallel propagation along a geodesic, a vector makes a constant angle with the geodesic; following the vector round the small quadrilateral, it is easy to see that the angle through which the vector has turned on completion of the circuit is $E$, the excess of the angle-sum of four right angles...
\end{tcolorbox}
\begin{figure}[h]
\input{D:/MathLatex/images/fig_p96_3415_a.tex}
\caption{Parallel transportation along a  closed path}
\label{fig:fig_p96_3415_a}
\end{figure}
Consider 4 geodesics $\gamma_{1}, \gamma_{2},\gamma_{3},\gamma_{4}$ close to each other so that they form a small quadrilateral. At each intersection they form an angle $\Psi_{i\rightarrow i+1}$. A vector $\overrightarrow{u_{0}}$ is transported parallely along the path starting at the intersection $S$ of $\gamma_{1},\gamma_{4}$ and ends as vector $\overrightarrow{u_{t}}$ at the same point $S$. In general $\overrightarrow{u_{0}}\ne \overrightarrow{u_{t}}$ and will differ by a small angle $\delta\theta$. Let's investigate the relationship between $\delta\theta$ and the $\Psi_{i\rightarrow i+1}$.\\
\begin{figure}[h]
\input{D:/MathLatex/images/fig_p96_3415_b.tex}
\caption{Relationship between parallel transportation along a closed path and the excess of the angle-sum over four right angles of a quadilateral.}
\label{fig:fig_p96_3415_b}
\end{figure}\\
Let $\xi_{i}$ and $\xi^{o}_{i} \ (i=1,2,3,4)$ be respectively, tangent unit vectors to the  geodesics at the beginning and at the end of the intersections of the geodesics. Be $\widehat{\xi}_{i}$ and $\widehat{\xi}^{o}_{i} \ (i=1,2,3,4)$ the angles of these vectors relative to an arbitrary reference vector and be $\widehat{\tau}_{i}$ and $\widehat{\tau}^{o}_{i} \ (i=0,2,3,4)$ the angles of the transported vector (relative to this arbitrary reference vector) at the intersections of the geodesics.
Then,
\begin{align}
\left \{ \begin{array}{ll}
\widehat{\tau}_{0} = \widehat{\xi}_{1}+\theta_{1}&\\
\widehat{\tau}_{1} = \widehat{\xi}^{o}_{1}+\theta_{1}&\widehat{\tau}_{1} = \widehat{\xi}_{2}+\theta_{2}\\
\widehat{\tau}_{2} = \widehat{\xi}^{o}_{2}+\theta_{1}&\widehat{\tau}_{2} = \widehat{\xi}_{3}+\theta_{3}\\
\widehat{\tau}_{3} = \widehat{\xi}^{o}_{3}+\theta_{3}&\widehat{\tau}_{3} = \widehat{\xi}_{4}+\theta_{4}\\
\widehat{\tau}_{4} = \widehat{\xi}^{o}_{4}+\theta_{4}&\\
\end{array} \right.
\end{align}
We have also
\begin{align}
\left \{ \begin{array}{l}
\widehat{\xi}_{2}  - \widehat{\xi}^{o}_{1} = \Psi_{1 \rightarrow 2}\\
\widehat{\xi}_{3}  - \widehat{\xi}^{o}_{2} = \Psi_{2 \rightarrow 3}\\
\widehat{\xi}_{4}  - \widehat{\xi}^{o}_{3} = \Psi_{3 \rightarrow 4}\\
\widehat{\xi}_{1}  - \widehat{\xi}^{o}_{4} = \Psi_{4 \rightarrow 1}\\
\widehat{\tau}_{4}-\widehat{\tau}_{0} = \delta \theta\\ 
\end{array} \right.
\end{align}
Combining (1) and (2)
\begin{align}
\left \{ \begin{array}{l}
\Psi_{1 \rightarrow 2} = \theta_{1}-\theta_{2}\\
\Psi_{2 \rightarrow 3}= \theta_{2}-\theta_{3}\\
\Psi_{3 \rightarrow 4}= \theta_{3}-\theta_{4}\\
\Psi_{4 \rightarrow 1}= \theta_{4}-\theta_{1}- \delta \theta\\
\end{array} \right.
\end{align}
and so 
\begin{align}
\delta \theta = -(\Psi_{1 \rightarrow 2}+\Psi_{2 \rightarrow 3}+\Psi_{3 \rightarrow 4}+\Psi_{4 \rightarrow 1})
\end{align}\\
Note that these relationships are valid on the (curved) $V_{2}$ manifold. In order to go further we map the quadrilateral, on the manifold, on it's tangent plane (see fig.~\ref{fig:fig_p96_3415_d} (a) hereunder) - supposing that the quadrilateral is infinitesimally small and that we can find a conformal map from $\gamma$ to $T(\gamma)$.
\begin{figure}[h]%
    \centering
    \subfloat[]{\input{D:/MathLatex/images/fig_p96_3415_c.tex}}
	\qquad
    \subfloat[]{\input{D:/MathLatex/images/fig_p96_3415_d.tex}}
\caption{Relationship between parallel transportation along a closed path and the excess of the angle-sum over four right angles of a quadrilateral.}
\label{fig:fig_p96_3415_d}
\end{figure}\\
\\ Let's look at the point $p_2$ on $\partial \Omega$. We have $\nu_2 = \epsilon^{-}+\Psi_{1 \rightarrow 2} +\epsilon^{+}= \Psi_{1 \rightarrow 2}+\epsilon$. In general,
\begin{align}
\sum_{i=1}^{4} \nu_i &= 2\pi \\
\underbrace{\sum_{i=1}^{4} \Psi_{i \rightarrow i+1}}_{= - \delta \theta} +\sum_{i=1}^{4} \epsilon_{i} &= \sum_{i=1}^{4} \frac{\pi}{2} \\
\Rightarrow \quad - \delta \theta &= \sum_{i=1}^{4}(\frac{\pi}{2} - \epsilon _{i})
\end{align}
Calling $\frac{\pi}{2} - \epsilon _{i}$ the excess, we get the assertion made.

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\newpage

\section{p98 - Clarification}
\begin{tcolorbox}
... it is easy to see that the expansion takes the form $$\mathbf{3.425.}\spatie  \eta=\theta\left(s-\frac{1}{6}\epsilon K s^3 + \dots\right)$$
\end{tcolorbox}
Expanding $\eta$ in a power series gives
\begin{align}
\eta &= \underbrace{\left.\eta\right|_0}_{=0} + \underbrace{\left.\dv{\eta}{s}\right|_0 }_{=\theta}s -\half\underbrace{\left.\dv[2]{\eta}{s}\right|_0}_{=0}s^2 + \frac{1}{6}\left.\dv[3]{\eta}{s}\right|_0s^3 + \dots \\ 
\dv[2]{(1)}{s} \quad \Rightarrow\quad \dv[2]{\eta}{s} &=\left.\dv[3]{\eta}{s}\right|_0s+ \dots\\
\text{for } \lim_{ s \to 0} \ \text{ we have } \eta \approx \theta s \quad \text{so (2)} \quad \Rightarrow\quad \dv[2]{\eta}{s} &=\left.\dv[3]{\eta}{s}\right|_0\frac{\eta}{\theta}+ \dots\\
\lim_{ s \to 0}\quad \Rightarrow \quad \left.\dv[3]{\eta}{s}\right|_0 &= \theta\underbrace{\lim_{ s \to 0}\frac{1}{\eta}\dv[2]{\eta}{s}}_{= -\epsilon K}\\
\Rightarrow \quad \eta&=\theta\left(s-\frac{1}{6}\epsilon K s^3 + \dots\right)
\end{align}
$$\blacklozenge$$
\newpage

\section{p102 - Clarification}
\begin{tcolorbox}
Still using the same notation and Fig. 7, we have at B the three vectors $(T^r)_1,\ (T^r)_2,\ Y_r$ ...  it follows that $$\textbf{3.516.}\quad \quad (\Delta T^r)_A(Y_r)_{A'2}= - (\Delta T^r)_B(Y_r)_B$$
\end{tcolorbox}
First we note that for an invariant "propagated parallely" along a curve we have $$\fdv{(T^r Y_r)}{u} = \fdv{T^r} {u}Y_r+T^r\fdv{Y_r}{u}=0$$
\begin{figure}[h]
\input{D:/MathLatex/images/fig_p102_3516_a.tex}
\caption{Parallel transportation along a  closed path}
\label{fig:fig_p102_3516_a}
\end{figure}
In fig. ~\ref{fig:fig_p102_3516_a} we use the following convention:\\
- a one black arrowed line means a forward propagation from $A$ to $B$ along $C_1$\\
- a double black arrowed line means a forward propagation from $A$ to $B$ along $C_2$\\
- a double open arrowed line means a backward propagation from $B$ to $A$ along $C_2$\\\\
In order to find the angular displacement of the vector $(T^r)_0$ when propagated parallely from $A$ to $B$ and back to $A$ along different paths, we follow  the dash dotted line in Fig. 1.7. Following that path, we end with the vector $(T^r)_0+ (\Delta T^r)_{A})$ in $A$ and have the vector $(T^r)_1$ as intermediate forward propagation from $A$ to $B$ along $C_1$, that vector being transported backwards to $A$ along $C_2$.\\\\
Note that for a vector the forward and backward propagation along the same curve is a null operation:$$(T^r)_0 \underset{\underset{C_2}{A\rightarrow B}}{\rightarrow} (T^r)_2 \underset{\underset{C_2}{B\rightarrow A}}{\rightarrow} (T^r)_0$$
Also, we have in Fig.1.7.
$$(T^r)_0 \underset{\underset{C_1}{A\rightarrow B}}{\rightarrow} (T^r)_1 \underset{\underset{C_2}{B\rightarrow A}}{\rightarrow} (T^r)_0+ (\Delta T^r)_{A}$$
$$(Y_r)_{B} \underset{\underset{C_2}{B\rightarrow A}}{\rightarrow} (Y_r)_{A,2}$$
At $A$ we form the following invariants
\begin{align}
\left \{ \begin{array}{l}
 ((T^r)_0+ (\Delta T^r)_{A})\ (Y_r)_{A,2}\\
 (T^r)_0 \ (Y_r)_{A,2}
\end{array} \right.
\end{align}\\
and at $B$
\begin{align}
\left \{ \begin{array}{l}
 (T^r)_1\ (Y_r)_{B}\\
 (T^r)_2 \ (Y_r)_{B} = ((T^r)_1+(\Delta T^r)_B )\ (Y_r)_{B}
\end{array} \right.
\end{align}\\

Due to the null effect of  parallel propagation on invariants, we  get
\begin{align}
 (T^r)_1\ (Y_r)_{B} &= ((T^r)_0+ (\Delta T^r)_{A})\ (Y_r)_{A,2}\\
 ((T^r)_1+(\Delta T^r)_B )\ (Y_r)_{B}&=(T^r)_0 \ (Y_r)_{A,2}\\
 \text{(3)-(4)}\quad \Rightarrow \quad -(\Delta T^r)_B )\ (Y_r)_{B}&=(\Delta T^r)_{A})\ (Y_r)_{A,2}
\end{align}\\
$$\blacklozenge$$
\newpage


\section{p105 - Clarification}
\begin{tcolorbox}
\begin{align*}
\textbf{3.521.} \spatie \dv{I}{v} &= \int_{u_1}^{u_2}\partial_v\left(T_n\partial_u x^n\right)du\\
&= \int_{u_1}^{u_2}\fdv{T_n}{v}\partial_u x^n du + \int_{u_1}^{u_2}T_n\fdv{\left(\partial_u x^n\right)}{v} du. 
\end{align*}
Now $\fdv{T_n}{v}=0$, since $T_r$ is propagated along $\textit{all}$ curves in $V_n$.
\end{tcolorbox}
To better understand this last statement recall that from 3.515, we have
\begin{align*}
 \left( \Delta T^r \right)_B \left( Y_r \right)_B &= \int \int Y_rR^r_{.pmn}T^p\partial _u x^m \partial _v x^n du dv\\
 &= 0 \quad \text{as} \spatie R^r_{.pmn}=0
\end{align*}
As $\left( Y_r \right)_B$ is arbitrary we have $\left( \Delta T^r \right)_B =0$. Consider fig. ~\ref{fig:fig_p105_3521_a} below.\\
\begin{figure}[H]
\center
\input{D:/MathLatex/images/fig_p105_3521_a.tex}
\caption{Parallel transportation along a path in a space with zero curvature tensor}
\label{fig:fig_p105_3521_a}
\end{figure}
Consider the path $A\rightarrow P \rightarrow P^{'}$, $ P$ being situated at the parametric coordinates $(u, v)$ and $P^{'}$ at $(u, v + dv)$. For this path we have also $\left( \Delta T^r \right)_{P,P^{'}} =0$. So $(T^r)_{u,v} = (T^r)_A$ and $(T^r)_{u,v+dv} = (T^r)_{u,v}$ and thus $\fdv{T_r}{v}=0$.
$$\blacklozenge$$
\newpage

\section{p108 - Exercise 1}
\begin{tcolorbox}
Taking polar coordinates on a sphere of radius a, calculate the curvature tensor, the Ricci tensor, and the curvature invariant.
\end{tcolorbox}
We have 
\begin{align}
 \Phi = a^2d\theta^2 + a^2\sin^2 \theta d\phi^2
\end{align}
We only have to calculate $R_{1212}$ (see exercise page 86).
\begin{align}
 R_{\theta\phi\theta\phi}&= \partial_{\theta}\underbrace{[\phi \phi,\theta]}_{= -a^2\sin\theta\cos\theta} -\partial_{\phi}\underbrace{[\phi \theta,\theta]}_{=0} + \underbrace{\Gamma^{\theta}_{\phi\theta}}_{=0}[\theta\phi,\theta]+ \underbrace{\Gamma^{\phi}_{\phi\theta}[\theta\phi,\phi]}_{=a^2\cos^2\theta} - \underbrace{\Gamma^{\theta}_{\phi\phi}}_{=0}[\theta\theta,\theta]- \underbrace{\Gamma^{\phi}_{\phi\phi}}_{=0}[\theta\theta,\phi]\\
 &= a^2\sin^2\theta\\
 \text{3.208. : }& \quad \frac{R_{11}}{a_{11}}=\frac{R_{22}}{a_{22}}=-\frac{R_{\theta\phi\theta\phi}}{det(a_{mn})}\quad\Rightarrow \quad \left \{ \begin{array}{l}
 R_{11} =  -1\\
 R_{12} = 0\\
 R_{22} =  -\sin^2\theta
\end{array} \right.\\
\text{3.210. : }& \quad R=-\frac{2}{det(a_{mn})}R_{\theta\phi\theta\phi}\quad\Rightarrow \quad R= -\frac{2}{a^2}
\end{align}
$$\blacklozenge$$
\newpage

\section{p108 - Exercise 2}
\begin{tcolorbox}
Take as manifold $V_2$ the surface of an ordinary right circular cone, and consider one of the circular sections. A vector in $V_2$ is propagated parallely round this circle. Show that its direction is changed on completion of the circuit. can you reconcile this result with the fact that $V_2$ is flat?
\end{tcolorbox}
\begin{figure}[h]
\center
\input{D:/MathLatex/images/fig_p108_Ex2_a.tex}
\caption{Intrinsic coordinates on  a cone}
\label{fig:fig_p108_Ex2_a}
\end{figure}
We take as coordinate system $\left(u,\theta \right)$, embedded in the manifold, $u$ being the distance of the generator of the considered point to the apex of the cone and $\theta$ the angle with a arbitrary vector laying in a plane perpendicular to the axis of the cone.\\
It is not hard to see that the fundamental form for this manifold is $$ \Phi = du^2+ \underbrace{k}_{= \ \sin^2 \alpha}u^2 d\theta^2 $$ 
We have 
\begin{align}
 (a_{mn})= \begin{pmatrix}
 1&0  \\
 0& k u^2 \\
\end{pmatrix}\quad
 (a^{mn})= \begin{pmatrix}
 1&0  \\
 0& \frac{1}{k u^2} \\
\end{pmatrix}\\
 \begin{pmatrix}
 \left[ mn,u \right] \\
 \left[ mn,\theta \right] \\
\end{pmatrix}=\begin{pmatrix}
 0&0&-k u \\
 k u&0&0 \\
\end{pmatrix}\\
 \begin{pmatrix}
 \Gamma^u_{mn} \\
 \Gamma^{\theta}_{mn} \\
\end{pmatrix}=\begin{pmatrix}
 0&0&-k u \\
 0&\frac{1}{u}&0 \\
\end{pmatrix}
\end{align}

Let's calculate the curvature tensor. From the exercise on page 86 we know that for a $V_2$ all components of the curvature tensor can be expressed in terms of $R_{1212}$. We have
\begin{align}
R_{u\theta u\theta} &= \left \{ \begin{array}{l}
\frac{1}{2}\left(\partial_{\theta u}a_{u \theta}+\partial_{u \theta}a_{\theta u}-\partial_{\theta \theta}a_{uu}-\partial_{u u}a_{\theta \theta}  \right) \\
 + a^{pq}\left([u \theta,p][\theta u,q] -[u u,p][\theta \theta,q]  \right)
\end{array} \right.
&= \left \{ \begin{array}{l}
-k \\
 + a^{uu}\left(\underbrace{[u \theta,u][\theta u,u] -[u u,u][\theta \theta,u]}_{=0}  \right)\\
 + \underbrace{a^{u \theta}}_{=0}\left([u \theta,u][\theta u,\theta] -[u u,u][\theta \theta,\theta]  \right)\\
 + \underbrace{a^{\theta u}}_{=0}\left([u \theta,\theta][\theta u,u] -[u u,\theta][\theta \theta,u]  \right)\\
 + \underbrace{a^{\theta \theta}}_{=\frac{1}{ku^2}}\left(\underbrace{[u \theta,\theta][\theta u,\theta]}_{= k^2u^2} -\underbrace{[u u,\theta][\theta \theta,\theta]}_{=0}  \right)\\
\end{array} \right.
\end{align}
So indeed all components of the curvature tensor vanish and hence $V_2$ is flat.

Let's now calculate the parallel transportation of a vector $T^r$ along a circle somewhere on the cone. Taking $\theta$ as the parameter of the curve, the equation of the curve is $(u=u_0,\ \theta) \quad \theta \in \ \left[0, 2\pi \right)$. We have for parallel transportation along that curve $\fdv{T^r}{\theta}=0$ and get
\begin{align}
 & \left \{ \begin{array}{l}
\dv{T^u}{\theta} + \Gamma^u_{\theta \theta}T^{\theta}\dv{\theta}{\theta} =0\\\\
\dv{T^{\theta}}{\theta} + \Gamma^{\theta}_{u \theta}T^{u}\dv{\theta}{\theta}+ \Gamma^{\theta}_{\theta u}T^{\theta}\underbrace{\dv{u}{\theta}}_{=0} =0\quad \left(\dv{u}{\theta} = 0 \quad \text{as} \quad u= C^{st} \right)
\end{array} \right.\\
\Rightarrow \quad
 & \left \{ \begin{array}{l}
\dot{T^u} -ku_0 T^{\theta}=0\\\\
\dot{T}^{\theta} + \frac{1}{u_0}T^{u}=0
\end{array} \right.\\
\Rightarrow \quad
 & \left \{ \begin{array}{l}
\frac{\ddot{T}^u}{T^{u}} = -k\\\\
\dot{T}^{\theta} =- \frac{T^{u}}{u_0}
\end{array} \right.
\end{align}
From the first equation in (11) we deduce that a solution can be of the form\\ $T^u = p^{'}\left( e^{(a \theta + b^{'})}+ e^{-(a \theta + b^{'})}\right)$. Substituting in (12) we see that $a^2 = -k \rightarrow a = \pm i\sqrt{k}$. Replacing $b^{'}$ by $ib$ the solution for the system of differential equations (12) becomes
\begin{align}
T^u  &= p^{'}\left( e^{i(\sqrt{k} \theta + b)}+ e^{-i(\sqrt{k} \theta + b)}\right)\\
 &= p\cos{\left(\sqrt{k} \theta + b \right)}\\
 \Leftrightarrow \quad &=   C_1\sin{\sqrt{k}\theta}+ C_2\cos{\sqrt{k}\theta}\\
 \text{(15) in (12) gives:}\quad & \left \{ \begin{array}{l}
T^u = C_1\sin{\sqrt{k}\theta}+ C_2\cos{\sqrt{k}\theta}\\\\
T^{\theta}  =  \frac{1}{\sqrt{k}u_0}\left(C_1\cos{\sqrt{k}\theta}- C_2\sin{\sqrt{k}\theta}\right)\\
\end{array} \right.\\
\text{with} \quad & \left \{ \begin{array}{l}
 C_1=\sqrt{k}u_0\left.T^{\theta}\right|_{\theta=0} \\\\
 C_2=\left.T^{u}\right|_{\theta=0}\\
\end{array} \right.\\
\Rightarrow \quad & \left \{ \begin{array}{l}
T^u = \sqrt{k} u_0 T^{\theta}_{0}\sin{\sqrt{k}\theta}+ T^{u}_{0}\cos{\sqrt{k}\theta}\\\\
T^{\theta}  =   T^{\theta}_{0}\cos{\sqrt{k}\theta}- \frac{T^{u}_{0}}{\sqrt{k}u_0} \sin{\sqrt{k}\theta}\\
\end{array}\right.
\end{align}\\
Let's now compute the angle $\phi$ between the starting vector $T^r_{0}$ and the vector $T^r$ parallely transported over an angle $\theta$ on the circle. We have (see $\textbf{(2.301.)}$ and $\textbf{(2.312.)}$):
\begin{align}
\left \{ \begin{array}{l}
 \left| T^r_0 \right|^2 =  \left(T^u_0\right)^2 + k u^2_0\left(T^{\theta}_0\right)^2\\\\
 \left| T^r \right|^2 =  \left(T^u_0\right)^2 + k u^2_0\left(T^{\theta}_0\right)^2\\\\
 \cos{\phi} =  \frac{T^u_0 T^u+k u^2_0T^{\theta}_0 T^{\theta} }{\left(T^u_0\right)^2 + k u^2_0\left(T^{\theta}_0\right)^2}
\end{array} \right.
\end{align}
The last equation in (19) becomes
\begin{align}
\cos{\phi} &= \cos{\sqrt{k}\theta} \\
\Rightarrow \quad \phi &= \sqrt{k}\theta + 2m\pi\quad\quad (m= 0,\pm 1,\pm 2, \dots)
\end{align}
So for $\theta = 2\pi$ the angle between the starting vector and the transported vector is not $2m\pi$. 
\begin{figure}[H]%
    \centering
    \subfloat[]{\input{D:/MathLatex/images/fig_p108_Ex2_b1.tex}}
	\qquad
    \subfloat[]{\input{D:/MathLatex/images/fig_p108_Ex2_b2.tex}}
\caption{Relationship between parallel transportation along a circle of a cone and unwrapping a cone .}
\label{fig:fig_p108_Ex2_b2}
\end{figure}
Fig. ~\ref{fig:fig_p108_Ex2_b2} illustrates the analogy between 
(a): the $\parallel$ transportation of a vector $P$ along a circular curve $\gamma$ on the cone over an angle $\theta$ giving a vector $P_t$ making an angle $\phi = \sqrt{k}\theta$ with the starting vector, and \\
(b): the result of "unwrapping" a cone. Be two vectors $\overrightarrow{OP}$ and $\overrightarrow{OP_t}$, $P$ and $P_t$ being two points placed at a distance $r\theta$ along a circle  with radius $r$. Placing the vector $\overrightarrow{OP}$ in the $XY$-plane and unwrapping the cone over an angle $\theta$ will map the vector $\overrightarrow{OP_t}$ in the $XY$-plane to a vector $\overrightarrow{OP_t^{*}}$ making an angle $\phi = \sqrt{k}\theta$ with the vector $\overrightarrow{OP}$.\\
The transported vector will only coincide with the initial vector for $\sqrt{k}=\sin{\alpha}  = \frac{1}{n}\quad (n= 1, 2, \dots)$. Fig. ~\ref{fig:fig_p108_Ex2_b3} illustrates this in the $X-Y$-plane,  for n=3. Only after a transportation over three periods, will the transported vector coincide with the initial vector. The dotted area represented the unwrapped cone.
\begin{figure}[H]%
    \centering
{\input{D:/MathLatex/images/fig_p108_Ex2_b3.tex}}
\caption{Parallel transportation along a circle of a cone with $\sin{\alpha}  = \frac{1}{3}$ .}
\label{fig:fig_p108_Ex2_b3}
\end{figure}
$$\blacklozenge$$
\newpage

\section{p109 - Exercise 3 and 4}
$\mathbf{Exercise 3.}$\\
\begin{tcolorbox}
Consider the equations $$ \left( R_{mn} -\theta a_{mn}\right)X^n = 0$$ where $R_{mn}$ is the Ricci tensor in a $V_N \ (N>2)$, $\theta$ an invariant, and $X^n$ a vector. Show that, if these equations are to be consistent, $\theta$ must have one of a certain set of $N$ values, and that the vectors $X^n$ corresponding to different values of $\theta$ are perpendicular to one another. (The directions of these vectors are called the $\mathit{Ricci \ principal \ directions}$).
\end{tcolorbox}

\begin{align}
\left( R_{mn} -\theta a_{mn}\right)X^n = 0\\
 (1)\times (a^{mp}) \quad \Rightarrow \quad a^{mp} R_{mn}X^n  -\theta \underbrace{a^{mp}a_{mn}}_{=\delta^p_n}X^n  =0\\
 \Rightarrow \quad a^{mp} R_{mn}X^n  -\theta X^p  =0\\
\end{align}
Define $T^{p}_{n} =  a^{mp} R_{mn} $, then (3) can be written in matrix form with $\mathbf{T}\equiv (T^{p}_{n})$ , $\mathbf{X}\equiv (X^p)$ and $\mathbf{I} \equiv (\delta^i_j)$
\begin{align}
\left(\mathbf{T}-\theta \mathbf{I}\right)\mathbf{X} =0
\end{align}
This is an eigenvector equation with $\mathbf{T}$ being Hermitian i.e. $\mathbf{T}^{\dag} =\mathbf{T}$ . Indeed, obviously $\mathbf{\overline{T}} = \textbf{T}$ and  
\begin{align}
\mathbf{T}^T &=  \left(\mathbf{AR}\right)^T\\
 &= \mathbf{R}^T \mathbf{A}^T\\
 \Leftrightarrow \left(T^{j}_{i}\right)  &= \left(R_{kj}\right)^T\left(a^{ik}\right)^T\\
 &= \left(R_{jk}\right)\left(a^{ki}\right)
\end{align}         
as both $ R_{jk}, \ a^{ki}$ are symmetric we  have
\begin{align}
\left(T^{j}_{i}\right) &= \left(R_{kj}\right)\left(a^{ik}\right)\\
 &= \left(T^{i}_{j}\right)\\
 \Rightarrow \spatie \mathbf{T}^{\dag} &=\mathbf{T}
\end{align} 
This means that the $N$ roots of $det\left(\mathbf{T}-\theta \mathbf{I_n}\right)=0$, which is a necessary condition to have equation (5) consistent, are real. Hence $\theta$ will take $N$ values, being the eigenvalues of the transformation matrix $\textbf{T}$. If all eigenvalues have multiplicity one, then  the N eigenvectors in (5) corresponding to the $N$ eigenvalues will be  orthogonal to each other. But, can eigenvalues with algebraic multiplicity $ m > 1$ occur? The answer is yes. Let's rewrite $P(\theta) = det\left(\mathbf{T}-\theta \mathbf{I_n}\right)$ as $\theta^{N} + q_i \theta^{n-i}  \quad (i= N-1,N-2,\dots, 1)$ with $q_i$ functions of the Ricci tensor components. The condition for eigenvalues with algebraic multiplicity $ m > 1$  to occur is that the determinant of the Sylvester matrix of the two following polynomial should be zero.
\begin{align}
\left \{ \begin{array}{l}
 P(\theta) = \theta^{N} + q_i \theta^{n-i}  \quad (i= N-1,N-2,\dots, 1)\\
 \dv{P(\theta)}{\theta} = N \theta^{N-1} + (n-i)q_i \theta^{n-i-1}  \quad (i= N-1,N-2,\dots, 1)
\end{array} \right.
\end{align}
The associated Sylvester matrix with these two polynomials will be of the form $$ S \left( P(\theta),\dv{P(\theta)}{\theta}\right) = \left( \begin{array}{cccccccc}
 1&q_{N-1}&\dots&q_1&0&0&\dots&0\\
 0&1&q_{N-1}&\dots&q_1&0&\dots&0\\
 \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\
 0&\dots&0&1&q_{N-1}&\dots &q_2&q_1\\
 N&(N-1)q_{N-1}&\dots&q_1&0&0&\dots&0\\
 0&N&(N-1)q_{N-1}&\dots&q_1&0&\dots&0\\
 \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\
 0&\dots&0&N&(N-1)q_{N-1}&\dots &2q_2&q_1\\
\end{array} \right)
$$
If the determinant of this matrix is not zero, then there will be no algebraic multiplicity. In the other case, one has to check whether in the eigenspace, related to the eigenvalues with algebraic multiplicity $ m > 1$, $m$ linear independent eigenvectors can be found.\\\\ 
$\mathbf{Exercise \ 4.}$\\
\begin{tcolorbox}
What becomes of the Ricci principal directions (see above) if $N=2$?
\end{tcolorbox}
From $\mathbf{3.208.}$ we have
\begin{align}
\frac{R_{11}}{a_{11}} &= \frac{R_{12}}{a_{12}}=\frac{R_{22}}{a_{22}}=-\frac{R_{1212}}{a}\\
\Rightarrow\quad & \left \{ \begin{array}{l}
 T^{1}_{1} = a^{11}R_{11} +  a^{12}R_{21}\\
 T^{1}_{2} = a^{11}R_{12} +  a^{12}R_{22}\\
 T^{2}_{2} = a^{21}R_{12} +  a^{22}R_{22}\\
\end{array} \right.\\
\text{put } K=-\frac{R_{1212}}{a} \quad \Rightarrow\quad & \left \{ \begin{array}{l}
 T^{1}_{1} = K a^{11}a_{11} +  K a^{12}a_{12}\\
 T^{1}_{2} = Ka^{11}a_{12} +  Ka^{12}a_{22}\\
 T^{2}_{2} = Ka^{21}a_{12} +  Ka^{22}a_{22}\\
\end{array} \right.\\
\Rightarrow\quad & \left \{ \begin{array}{l}
 T^{1}_{1} = K\delta^1_1 = K \\
 T^{1}_{2} = K\delta^1_2 = 0\\
 T^{2}_{2} = K\delta^2_2 = K\\
\end{array} \right.
\end{align}
hence the characteristic equation $det\left(\mathbf{T}-\theta \mathbf{I_n} \right)=0$ becomes $ (K-\theta)^2=0$ . So only one value of $\theta$ exists as $\theta = K$.  Equation (5) becomes $\mathbf{0}\mathbf{X}=0$. So we can chose any pair of linear independent vectors as eigenvectors and can make them perpendicular to one another.
$$\blacklozenge$$
\newpage

\section{p109 - Exercise 5}
\begin{tcolorbox}
Suppose that two spaces $V_N$, $V^{'}_N$ have metric tensors $a_{mn}, \ a^{'}_{mn}$ such that $\ a^{'}_{mn}=k\ a_{mn}  $, where $k$ is a constant. Write down the relations between the curvature tensors, the Ricci tensors, and the curvature invariants of the two spaces.
\end{tcolorbox}
We have 
\begin{align}
 ds^2 &= a_{mn}dx^{m} dx^{n}\\
 ds^{'2}&= a^{'}_{mn} dx^{'m}dx^{'n}
\end{align}
But let's be careful: there is no reason to assume that  $dx^m = dx^{'m}$. Let's embed the two spaces in a space $V_{N+1}$. If an observer in that space sees two displacements  $ds^2$ and $ds^{'2}$ which for him have the same magnitude, we have
\begin{figure}[H]
\centering
\begin{minipage}[t]{.4\textwidth}
%\centering
\vspace{0pt}
%\includegraphics[scale=.5]{p85_ex1.png}
\input{D:/MathLatex/images/fig_p109_Ex5_a.tex}
\end{minipage}\hfill
\caption{Embedded $V_N$ spaces}
\label{fig:p109_Ex5_a}
\end{figure}
\begin{align}
 ds^{'2} &= ds^2\\
 \Rightarrow \spatie a^{'}_{mn} dx^{'m}dx^{'n} &= a_{mn}dx^{m} dx^{n}\\
 \Rightarrow \spatie k a_{mn} dx^{'m}dx^{'n} &= a_{mn}dx^{m} dx^{n}\\
 \Rightarrow \spatie dx^{'m} &= \frac{1}{\sqrt{k}}dx^{m}
\end{align}
We have also
\begin{align}
 a^{'}_{mk}a^{'kn} &= \delta^n_m\\
 \Rightarrow \spatie ka_{mp}a^{'pn} &= \delta^n_m\\
 \Rightarrow \spatie a^{'pn} &= \frac{1}{k}a^{pn}
\end{align}
And get the following relations
\begin{align}
\left \{ \begin{array}{l}
 [mn,r]^{'} = \sqrt{k}^3[mn,r]\\\\
 \Gamma^{'r}_{.mn} =\sqrt{k}\Gamma^{r}_{.mn}
 \end {array}\right.
 \end{align}
 \begin{align}
 \Rightarrow \spatie R^{'s}_{\ .rmn} &= \frac{\partial \Gamma^{'s}_{\ .rn}}{\partial x^{'m}} + \dots\\
 \Rightarrow \spatie R^{'s}_{\ .rmn} &= \frac{\sqrt{k}\partial \Gamma^{s}_{.rn}}{\partial \left(\frac{x^{m}}{\sqrt{k}}\right)} + \dots\\
 \Rightarrow \spatie R^{'s}_{\ .rmn} &= k R^{s}_{.rmn}\\
 \times \ a^{'ks}\spatie \Rightarrow \quad a^{'ks}R^{'s}_{\ .rmn} &= k a^{'ks}R^{s}_{.rmn}\\
 \Rightarrow \quad R^{'}_{krmn} &= k \frac{1}{k}a^{ks}R^{s}_{.rmn}\\
 \Rightarrow \quad R^{'}_{krmn} &=R_{krmn}\\
 R_{rm} = R^n_{.rmn} \spatie \Rightarrow \quad R^{'}_{\ mn} &= k R_{rmn}\\
 R = a^{mn}R_{mn} \spatie \Rightarrow \quad R^{'} &= a^{'mn}R^{'}_{mn}\\
 \Rightarrow \spatie  R^{'} &= \frac{1}{k}a^{mn}k R_{mn}\\
 \Rightarrow \spatie  R^{'} &=  R
\end{align}\\\\
$\mathbf{Summary}$\\
 \begin{align}
R^{'s}_{\ .rmn} &= k R^{s}_{.rmn}\\
R^{'}_{krmn} &=R_{krmn}\\
R^{'}_{ mn} &= k R_{rmn}\\
R^{'} &=  R
\end{align}
$$\blacklozenge$$
\newpage





\section{p109 - Exercise 6}
\begin{tcolorbox}
For an orthogonal coordinates system in a $V_2$ we have $$ ds^2=a_{11}\left( dx^1\right)^2+a_{22}\left( dx^2\right)^2$$
Show that 
$$\frac{1}{a}R_{1212}= -\half\frac{1}{\sqrt{a}}\left[\partial_1\left(\frac{1}{\sqrt{a}}\partial_1 a_{22}\right)+\partial_2\left(\frac{1}{\sqrt{a}}\partial_2 a_{11}\right)\right]$$
\end{tcolorbox}
We have 
\begin{align}
\left(a_{mn}\right)= \begin{pmatrix}
 a_{11}& 0 \\
 0& a_{22} \\
\end{pmatrix}\quad \left(a^{mn}\right)= \frac{1}{a}\begin{pmatrix}
 a_{22}& 0 \\
 0& a_{11} \\
\end{pmatrix}\quad
 a= a_{11}a_{22}
\end{align}
We have also \\
\begin{align}
R &= -\frac{2}{a}R_{1212}\\
R =  a^{mn}R_{mn}\quad\Rightarrow\quad R &= a^{11}R_{11}+ a^{22}R_{22}
\end{align}
Looking at the pattern generated by equations $(2)$ and $(3)$ suggests that using these equations could lead to the proposed equation. Let's have a try ...
\begin{align}
& \left \{ \begin{array}{ll}
\Gamma^1_{11} = \half\frac{a_{22}}{a}\partial_1 a_{11}&\Gamma^1_{22} =- \half\frac{a_{22}}{a}\partial_1 a_{22}\\\\
\Gamma^2_{11} =- \half\frac{a_{11}}{a}\partial_2 a_{11}&\Gamma^2_{22} = \half\frac{a_{11}}{a}\partial_2 a_{22}\\\\
\Gamma^1_{12} = \half\frac{a_{22}}{a}\partial_2 a_{11}&\Gamma^2_{12} = \half\frac{a_{11}}{a}\partial_1 a_{22}
\end {array}\right. \\
\text{3.205.}\quad\Rightarrow\quad & R_{rm} = \half\partial_{rm} \log a - \half \Gamma^p_{rm}\partial_p \log a-\partial_n\Gamma^n_{rm} + \Gamma^p_{rn}\Gamma^n_{pm}\\
\Rightarrow\quad &\left \{ \begin{array}{l}
 R_{11} =  \half\partial_{11} \log a - \half \Gamma^1_{11}\partial_1 \log a- \half \Gamma^2_{11}\partial_2 \log a \\
 -\partial_1\Gamma^1_{11}-\partial_2\Gamma^2_{11} + 
 \\ \Gamma^1_{11}\Gamma^1_{11}+ \Gamma^1_{12}\Gamma^2_{11}+ \Gamma^2_{11}\Gamma^1_{21}+ \Gamma^2_{12}\Gamma^2_{21}\\\\
  R_{22} =  \half\partial_{22} \log a - \half \Gamma^1_{22}\partial_1 \log a- \half \Gamma^2_{22}\partial_2 \log a \\
 -\partial_1\Gamma^1_{22}-\partial_2\Gamma^2_{22} + 
 \\ \Gamma^1_{21}\Gamma^1_{12}+ \Gamma^1_{22}\Gamma^2_{12}+ \Gamma^2_{21}\Gamma^1_{22}+ \Gamma^2_{22}\Gamma^2_{22}\\\\
 \end {array}\right.
 \end{align}
 %**************************************************
 \begin{align}
 \Rightarrow\quad &\left \{ \begin{array}{l}
 R_{11} =  \half\partial_{11} \log a - \half \half\frac{a_{22}}{a}\partial_1 a_{11}\partial_1 \log a- \half (- \half\frac{a_{11}}{a}\partial_2 a_{11})\partial_2 \log a \\
 -\partial_1(\half\frac{a_{22}}{a}\partial_1 a_{11})-\partial_2(- \half\frac{a_{11}}{a}\partial_2 a_{11}) + 
 \\ \half\frac{a_{22}}{a}\partial_1 a_{11}\half\frac{a_{22}}{a}\partial_1 a_{11}+ \half\frac{a_{22}}{a}\partial_2 a_{11}(- \half\frac{a_{11}}{a}\partial_2 a_{11})+ \\
 (- \half\frac{a_{11}}{a}\partial_2 a_{11})\half\frac{a_{22}}{a}\partial_2 a_{11}+ \half\frac{a_{11}}{a}\partial_1 a_{22}\half\frac{a_{11}}{a}\partial_1 a_{22}\\\\
  R_{22} =  \half\partial_{22} \log a - \half (- \half\frac{a_{22}}{a}\partial_1 a_{22})\partial_1 \log a- \half \half\frac{a_{11}}{a}\partial_2 a_{22}\partial_2 \log a \\
 -\partial_1(- \half\frac{a_{22}}{a}\partial_1 a_{22})-\partial_2(\half\frac{a_{11}}{a}\partial_2 a_{22}) + 
 \\ \half\frac{a_{22}}{a}\partial_2 a_{11}\half\frac{a_{22}}{a}\partial_2 a_{11}+ (- \half\frac{a_{22}}{a}\partial_1 a_{22})\half\frac{a_{11}}{a}\partial_1 a_{22}+ \\
 \half\frac{a_{11}}{a}\partial_1 a_{22}(- \half\frac{a_{22}}{a}\partial_1 a_{22})+ \half\frac{a_{11}}{a}\partial_2 a_{22}\half\frac{a_{11}}{a}\partial_2 a_{22}\\\\
 \end {array}\right.
 \end{align}
 Simplifying the notational burden by  replacing $a_{11}$ by $\gamma$ and $a_{22}$ by $\eta$:
  \begin{align}
 \Rightarrow\quad &\left \{ \begin{array}{l}
 R_{11} =  \half\partial_{11} \log a - \half \half\frac{1}{\gamma}\partial_1 \gamma\partial_1 \log a+ \half \half\frac{1}{\eta}\partial_2 \gamma\partial_2 \log a \\
 -\half\partial_1(\frac{1}{\gamma}\partial_1 \gamma)+\half\partial_2(\frac{1}{\eta}\partial_2 \gamma)
 \\ + \half\half\frac{1}{\gamma}\frac{1}{\gamma}\partial_1 \gamma\partial_1 \gamma- \half\half\frac{1}{\gamma}\frac{1}{\eta}\partial_2 \gamma \partial_2 \gamma \\
 - \half\half\frac{1}{\gamma}\frac{1}{\eta}\partial_2 \gamma\partial_2 \gamma+ \half\half\frac{1}{\eta}\frac{1}{\eta}\partial_1 \eta\partial_1 \eta\\\\
  R_{22} =  \half\partial_{22} \log a + \half\half\frac{1}{\gamma}\partial_1 \eta\partial_1 \log a- \half \half\frac{1}{\eta}\partial_2 \eta\partial_2 \log a \\
 +\half\partial_1(\frac{1}{\gamma}\partial_1 \eta)-\half\partial_2(\frac{1}{\eta}\partial_2 \eta)  
 \\ + \half\half\frac{1}{\gamma}\frac{1}{\gamma}\partial_2 \gamma\partial_2 \gamma- \half\half\frac{1}{\gamma}\frac{1}{\eta}\partial_1 \eta\partial_1 \eta \\
 - \half\half\frac{1}{\gamma}\frac{1}{\eta}\partial_1 \eta \partial_1 \eta+ \half\half\frac{1}{\eta}\frac{1}{\eta}\partial_2 \eta\partial_2 \eta\\\\
 \end {array}\right.
 \end{align}
 %****************************************************
 Noting that $\partial_{ii}\log a = \partial_{i}\left(\frac{1}{a_{11}}\partial_{i} a_{11}\right) + \partial_{i}\left(\frac{1}{a_{22}}\partial_{i} a_{22}\right)$ and $\partial_{i}\log a = \frac{1}{a_{11}}\partial_{i} a_{11} + \frac{1}{a_{22}}\partial_{i} a_{22}\quad (i=1,2)$, we get:\\
 \begin{align}
 %****************************
  \left| \begin{array}{c}
 2R_{11} = \\\\
  \underbrace{\partial_{1}\left(\frac{1}{\gamma}\partial_{1} \gamma\right)}_{*}+\partial_{1}\left(\frac{1}{\eta}\partial_{1} \eta\right) \\\\
  - \underbrace{\half \frac{1}{\gamma}\frac{1}{\gamma}(\partial_1 \gamma )^2}_{-} - \half \frac{1}{\gamma}\frac{1}{\eta}\partial_1 \gamma \partial_{1} \eta \\\\
   + \half\frac{1}{\gamma}\frac{1}{\eta}(\partial_2 \gamma)^2 +\half\frac{1}{\eta}\frac{1}{\eta}\partial_2 \gamma \partial_{2} \eta \\\\
 -\underbrace{\partial_1(\frac{1}{\gamma}\partial_1 \gamma)}_{*}+\partial_2( \frac{1}{\eta}\partial_2 \gamma)  
 \\\\
  + \underbrace{\half\frac{1}{\gamma}\frac{1}{\gamma}(\partial_1 \gamma)^2}_{-}- \underbrace{\half \frac{1}{\gamma}\frac{1}{\eta}(\partial_2 \gamma )^2}_{+} \\\\
 - \underbrace{\half \frac{1}{\gamma}\frac{1}{\eta}(\partial_2 \gamma)^2}_{+}+ \half\frac{1}{\eta}\frac{1}{\eta}(\partial_1 \eta)^2\\\\
 \end {array}\quad
 \right.
 \left | \begin{array}{c}
  2R_{22} = \\\\
    \partial_{2}\left(\frac{1}{\gamma}\partial_{2} \gamma\right)+\underbrace{\partial_{2}\left(\frac{1}{\eta}\partial_{2} \eta\right)}_{*} \\\\
   + \half\frac{1}{\gamma}\frac{1}{\gamma}\partial_{1} \gamma\partial_1 \eta  + \half\frac{1}{\gamma}\frac{1}{\eta}(\partial_1 \eta )^2 \\\\
   - \half \frac{1}{\gamma}\frac{1}{\eta}\partial_{2} \gamma\partial_2 \eta  - \underbrace{\half \frac{1}{\eta}\frac{1}{\eta}(\partial_2 \eta)^2 }_{-} \\\\
 +\partial_1(\frac{1}{\gamma}\partial_1 \eta)-\underbrace{\partial_2( \frac{1}{\eta}\partial_2 \eta)}_{*}  
 \\\\
  + \half\frac{1}{\gamma}\frac{1}{\gamma}(\partial_2 \gamma)^2- \underbrace{\half\frac{1}{\gamma}\frac{1}{\eta}(\partial_1 \eta )^2}_{+}\\\\
 - \underbrace{\half\frac{1}{\gamma}\frac{1}{\eta}(\partial_1 \eta)^2}_{+}+ \underbrace{\half \frac{1}{\eta}\frac{1}{\eta}(\partial_2 \eta)^2}_{-}\\\\
 \end {array}\right|
\end{align}
 \begin{align}
 %****************************
  \Rightarrow\quad &\left| \begin{array}{c}
  2R_{11} = \\\\
 \partial_{1}\left(\frac{1}{\eta}\partial_{1} \eta\right)+ \partial_2( \frac{1}{\eta}\partial_2 \gamma)\\\\
  + \half\frac{1}{\eta}\frac{1}{\eta}(\partial_1 \eta)^2 - \half\frac{1}{\gamma}\frac{1}{\eta}(\partial_2 \gamma)^2\\\\ - \half \frac{1}{\gamma}\frac{1}{\eta}\partial_1 \gamma \partial_{1} \eta  +\half\frac{1}{\eta}\frac{1}{\eta}\partial_2 \gamma \partial_{2} \eta \\
\end{array}\right.\quad
\left|\begin{array}{c}
  \spatie 2R_{22} = \\\\
    \partial_1(\frac{1}{\gamma}\partial_1 \eta) +\partial_{2}\left(\frac{1}{\gamma}\partial_{2} \gamma\right)\\\\
    - \half\frac{1}{\gamma}\frac{1}{\eta}(\partial_1 \eta )^2 + \half\frac{1}{\gamma}\frac{1}{\gamma}(\partial_2 \gamma)^2\\\\
  + \half\frac{1}{\gamma}\frac{1}{\gamma}\partial_{1} \gamma\partial_1 \eta  - \half \frac{1}{\gamma}\frac{1}{\eta}\partial_{2} \gamma\partial_2 \eta  \\
 \end {array}\right|
 \end{align}\\\\ 
  %****************************
  
  Be $R = \frac{1}{\gamma}R_{11}+ \frac{1}{\eta}R_{22}$, all first order derivatives vanish and we get,\\\\
  \begin{align}
  \frac{1}{\gamma}R_{11}+ \frac{1}{\eta}R_{22} = & 
 \half \left[ \frac{1}{\eta}\partial_1(\frac{1}{\gamma}\partial_1 \eta)+ \frac{1}{\gamma} \partial_{1}(\frac{1}{\eta}\partial_{1} \eta)\right]
    + \half  \left[\frac{1}{\gamma}\partial_{2}(\frac{1}{\gamma}\partial_{2} \gamma)+ \frac{1}{\eta}\partial_2( \frac{1}{\eta}\partial_2 \gamma) \right] 
\end{align}\\
We further simplify this expression. Considering the symmetry of $(11)$ we only explicit the calculations for the first terms in $\partial_1$.
\begin{align}
 \frac{1}{\eta}\partial_1(\frac{1}{\gamma}\partial_1 \eta)+ \frac{1}{\gamma} \partial_{1}(\frac{1}{\eta}\partial_{1} \eta)&=\frac{1}{\eta}\partial_1(\frac{1}{\sqrt{\gamma}}\frac{1}{\sqrt{\gamma}}\frac{\sqrt{\eta}}{\sqrt{\eta}}\partial_1 \eta)+ \frac{1}{\gamma} \partial_{1}(\frac{1}{\sqrt{\eta}}\frac{1}{\sqrt{\eta}}\frac{\sqrt{\gamma}}{\sqrt{\gamma}}\partial_{1} \eta)\\
 &=\frac{1}{\eta}\partial_1\left[\left(\frac{\eta}{\gamma}\right)^{\half}\frac{1}{\sqrt{a}}\partial_1 \eta\right]+ \frac{1}{\gamma} \partial_{1}\left[\left(\frac{\eta}{\gamma}\right)^{-\half}\frac{1}{\sqrt{a}}\partial_{1} \eta\right]\\
 &=\left \{ \begin{array}{l}
\underbrace{\frac{1}{\eta}\left(\frac{\eta}{\gamma}\right)^{\half}}_{= \frac{1}{\sqrt{a}}}\partial_1\left[\frac{1}{\sqrt{a}}\partial_1 \eta\right]+ \underbrace{\frac{1}{\gamma} \left(\frac{\eta}{\gamma}\right)^{-\half}}_{= \frac{1}{\sqrt{a}}}\partial_{1}\left[\frac{1}{\sqrt{a}}\partial_{1} \eta\right]\\\\
+\frac{1}{\sqrt{a}}\partial_1 \eta\underbrace{\left[\frac{1}{\eta}\partial_1\left(\frac{\eta}{\gamma}\right)^{\half}+ \frac{1}{\gamma} \partial_{1}\left(\frac{\eta}{\gamma}\right)^{-\half}\right]}_{=0}
 \end {array}\right.\\
  &= 2\frac{1}{\sqrt{a}}\partial_1\left[\frac{1}{\sqrt{a}}\partial_1 a_{22}\right]\\
  \Rightarrow \quad  \half \left[ \frac{1}{\eta}\partial_1(\frac{1}{\gamma}\partial_1 \eta) + \frac{1}{\gamma} \partial_{1}(\frac{1}{\eta}\partial_{1} \eta)\right] &=  \frac{1}{\sqrt{a}}\partial_1\left[\frac{1}{\sqrt{a}}\partial_1 a_{22}\right]  
\end{align}
Using $(16)$ and the same calculations for the terms in $\partial_2$ and using $(2)$ and $(3)$ we  get $$\frac{1}{a}R_{1212}= -\half\frac{1}{\sqrt{a}}\left[\partial_1\left(\frac{1}{\sqrt{a}}\partial_1 a_{22}\right)+\partial_2\left(\frac{1}{\sqrt{a}}\partial_2 a_{11}\right)\right]$$
$$\blacklozenge$$
\newpage

\section{p109 - Exercise 7}
\begin{tcolorbox}
Suppose that in a $V_3$ the metric is :$$ ds^2= (h_1dx^1)^2+(h_2dx^2)^2+(h_3dx^3)^2$$ whre $h_1, h_2, h_3$ are functions of the three coordinates. Calculate the curvature  tensor in terms of the $h^{'}s$ and their derivatives. Check your result by nothing that the curvature tensor will vanish if $h_1$ is a function of $x^1$ only, $h_2$ a function of $x^2$ only, and $h_3$  a function of $x^3$ only.
\end{tcolorbox}
From $\mathbf{3.115.}$ and $\mathbf{3.115.}$ we get for the non vanishing components of the covariant curvature tensor($6$ independent components to calculate):
\begin{align}
R_{1212} =\left\{ \begin{array}{c}
- R_{1221} \\
- R_{2112} \\
 R_{2121} \\
\end{array}\right.  \quad R_{2323} =\left\{ \begin{array}{c}
- R_{2332} \\
- R_{3223} \\
 R_{3232} \\
\end{array}\right.  \quad R_{1313} =\left\{ \begin{array}{c}
- R_{1331} \\
- R_{3113} \\
 R_{3131} \\
\end{array}\right.  \\
R_{1213} =\left\{ \begin{array}{c}
- R_{1231} \\
 R_{1312} \\
 -R_{1321} \\
- R_{2113} \\
 R_{2131} \\
 -R_{3112} \\
 R_{3121} \\
\end{array}\right.  \quad
R_{1223} =\left\{ \begin{array}{c}
- R_{1232} \\
- R_{2123} \\
 R_{2132} \\
 R_{2312} \\
 -R_{2321} \\
 R_{3212} \\
 -R_{3221} \\
\end{array}\right. \quad
R_{1323} =\left\{ \begin{array}{c}
- R_{1332} \\
 R_{2313} \\
 -R_{2331} \\
- R_{3123} \\
 R_{3132} \\
 -R_{3213} \\
 R_{3231} \\
\end{array}\right.
\end{align}
The metrix tensors:
\begin{align}
(a_{mn})  = \begin{pmatrix}
 h^2_1& 0&0 \\
 0& h^2_2&0 \\
 0& 0&h^2_3
\end{pmatrix}\quad (a_{mn})  = \begin{pmatrix}
 \frac{1}{h^2_1}& 0&0 \\
 0& \frac{1}{h^2_2}&0 \\
 0& 0&\frac{1}{h^2_3}
\end{pmatrix}
\end{align}
The Christoffel symbols:
\begin{align}
\begin{array}{lll}
\ [11,1]= h_{1} \partial_{1} h_{1} & [11,2]=- h_{1} \partial_{2} h_{1} & [11,3]= -h_{1} \partial_{3} h_{1}\\
\ [12,1]= h_{1} \partial_{2} h_{1} & [12,2]= h_{2} \partial_{1} h_{2} & [12,3]= 0\\
\ [22,1]= -h_{2} \partial_{1} h_{2} & [22,2]=h_{2} \partial_{2} h_{2} & [22,3]= -h_{2} \partial_{3} h_{2}\\
\  [23,1]= 0 & [23,2]= h_{2} \partial_{3} h_{2} & [23,3]= h_{3} \partial_{2} h_{3}\\
\ [33,1]= -h_{3} \partial_{1} h_{3} & [33,2]=- h_{3} \partial_{2} h_{3} & [33,3]= -h_{3} \partial_{3} h_{3}\\
\ [31,1]= h_{1} \partial_{3} h_{1} & [31,2]=0 & [31,3]= h_{3} \partial_{1} h_{3}\\
\end{array}
\end{align}
\begin{align}
\begin{array}{lll}
\Gamma^1_{11}=\frac{1}{h_{1}}\partial_{1}{h_{1}} &\Gamma^2_{11}=-\frac{h_{1}}{h_{2}^2}\partial_{2}{h_{1}}  & \Gamma^3_{11}= -\frac{h_{1}}{h_{3}^2}\partial_{3}{h_{1}}   \\
\Gamma^1_{12}= \frac{1}{h_{1}}\partial_{2}{h_{1}} &\Gamma^2_{12}=\frac{1}{h_{2}}\partial_{1}{h_{2}} & \Gamma^3_{12}=0\\
\Gamma^1_{22}= -\frac{h_{2}}{h_{1}^2}\partial_{1}{h_{2}} &\Gamma^2_{22}=\frac{1}{h_{2}}\partial_{2}{h_{2}} & \Gamma^3_{22}=-\frac{h_{2}}{h_{3}^2}\partial_{3}{h_{2}} \\
\Gamma^1_{23}=0 &\Gamma^2_{23}=\frac{1}{h_{2}}\partial_{3}{h_{2}} & \Gamma^3_{23}= \frac{1}{h_{3}}\partial_{2}{h_{3}}\\
\Gamma^1_{33}= -\frac{h_{3}}{h_{1}^2}\partial_{1}{h_{3}}&\Gamma^2_{33}=-\frac{h_{3}}{h_{2}^2}\partial_{2}{h_{3}} & \Gamma^3_{33}= \frac{1}{h_{3}}\partial_{3}{h_{3}}\\
\Gamma^1_{31}=\frac{1}{h_{1}}\partial_{3}{h_{1}} &\Gamma^2_{31}=0 & \Gamma^3_{31}=\frac{1}{h_{3}}\partial_{1}{h_{3}} \\
\end{array}
\end{align}
We use $3.113.$ $$R_{rsmn}= \partial_m[sn,r] -\partial_n[sm,r]+\Gamma^p_{sm}[rn,p]-\Gamma^p_{sn}[rm,p]$$\\

Note that we only have to perform the full calculation for two curvature tensors e.g. $R_{1212}$ and $R_{1213}$  as the others can be retrieved by using adequate indices renaming and use of the identities $3.115.$
\begin{align}
R_{1212}&=
-h_2\partial_{11}^2(h_2)-h_1\partial_{22}^2(h_1)
+\frac{h_2}{h_1}\partial_1 h_1\partial_1 h_2+\frac{h_1}{h_2}\partial_2 h_1\partial_2 h_2-\frac{h_1 h_2}{h_3^2}\partial_3 h_1\partial_3 h_2\\
R_{2323}&=
-h_3\partial_{22}^2(h_3)-h_2\partial_{33}^2(h_2)
+\frac{h_3}{h_2}\partial_2 h_2\partial_2 h_3+\frac{h_2}{h_3}\partial_3 h_2\partial_3 h_3-\frac{h_2 h_3}{h_1^2}\partial_1 h_2\partial_1 h_3\\
R_{1313}&=
-h_3\partial_{11}^2(h_3)-h_1\partial_{33}^2(h_1)
+\frac{h_3}{h_1}\partial_1 h_1\partial_1 h_3+\frac{h_1}{h_3}\partial_3 h_1\partial_3 h_3-\frac{h_1 h_3}{h_2^2}\partial_2 h_1\partial_2 h_3
\end{align}
\begin{align}
R_{1213}&=-h_1\partial_{32}^2(h_1)+\frac{h_1}{h_3}\partial_2 h_3\partial_3 h_1+\frac{h_1}{h_2}\partial_2 h_1\partial_3 h_2\\
R_{1223}&=h_2\partial_{31}^2(h_2)-\frac{h_2}{h_1}\partial_1 h_2\partial_3 h_1-\frac{h_2}{h_3}\partial_3 h_2\partial_1 h_3\\
R_{1323}&=
-h_3\partial_{21}^2(h_3)+\frac{h_3}{h_1}\partial_1 h_3\partial_3 h_1+\frac{h_3}{h_2}\partial_2 h_3\partial_1 h_2
\end{align}
And, indeed, all curvature tensors vanish when the $h_i$ are only a function of the indices' dimension.
$$\blacklozenge$$
\newpage


\section{p109 - Exercise 8}
\begin{tcolorbox}
In relativity we encounter the metric form $$ \Phi= e^{\alpha} + e^{x^1}\left[ \left( dx^2 \right)^2 + \sin^2 x^2 \left(dx^3 \right)^2 \right] - e^{\gamma}\left(dx^4\right)^2$$ where $\alpha$ and $\gamma$ are functions of $x^1$ and $x^4$ only.\\
Show that the complete set of non-zero components of the Einstein tensor (see equation (3.214)) for the form given above are as follows
\begin{align*}
G^1_{.1} &= e^{-\alpha}\left( -\kwart -\half\gamma_1\right) + e^{-x^1}\\
G^2_{.2} &= e^{\alpha} \left(-\kwart - \half \gamma_{11} - \kwart \gamma_1^2 -\kwart\gamma_1+\kwart \alpha _1 +\kwart \alpha _1 \gamma_1 \right) \\
&+e^{\gamma} \left(\half \alpha_{44} + \kwart\alpha_4^2 -\kwart\alpha_4\gamma_4\right)\\
G^3_{.3} &= G^2_{.2} \\
G^4_{.4} &= e^{-\alpha}\left( -\frac{3}{4} -\half\alpha_1\right) + e^{-x^1}\\
 e^{\alpha} G^4_{.1} &= -e^{\gamma} G^4_{.1} = -\half \alpha_4
\end{align*}
The subscript on $\alpha$ and $\gamma$ indicate partial derivatives with respect to $x^1$ and $x^4$.
\end{tcolorbox}
We have 
\begin{align}
\left(a_{mn}\right)= \begin{pmatrix}
 e^{\alpha}& 0&0&0 \\
 0& e^{x^1} &0&0\\
 0& 0 &e^{x^1}\sin^2 x^2&0\\
 0& 0 &0&-e^{\gamma}\\
\end{pmatrix}\quad 
\left(a^{mn}\right)= \begin{pmatrix}
 {e^{-\alpha}}& 0&0&0 \\
 0& {e^{-x^1}} &0&0\\
 0& 0 &\frac{e^{-x^1}}{\sin^2 x^2}&0\\
 0& 0 &0&-{e^{-\gamma}}\\
\end{pmatrix}
\end{align}
And will use the following definitions:
\begin{align}
 G^n_{.t} &= R^n_{.t} - \half \delta^n_t R\\
 R^n_{.t} &= a^{nk}R_{kt}\\
 R_{kt} &= a^{sn}R_{sktn}\\
 R &= a^{kt}R_{kt}
\end{align}
Considering that the non-diagonal components of $a_{mn}$ vanish and as $R_{sktn}=0$ when $s=k$ or $t=n$, we can write :
\begin{align}
\begin{pmatrix}
 R_{11}\\
 R_{22}\\
 R_{33}\\
 R_{44}\\
\end{pmatrix}&=
\begin{pmatrix}
 0& R_{2112} & R_{3113} & R_{4114} \\
 R_{1221}& 0 &  R_{3223}& R_{4224} \\
 R_{1331}&R_{2332}  & 0 & R_{4334} \\
 R_{1441}& R_{2442} & R_{3443}& 0 \\
\end{pmatrix}\begin{pmatrix}
 a^{11}\\
 a^{22}\\
 a^{33}\\
 a^{44}\\
\end{pmatrix}
\end{align}
\begin{align}
\begin{pmatrix}
 R_{12}\\
 R_{13}\\
 R_{14}\\
 R_{23}\\
 R_{24}\\
 R_{34}\\
\end{pmatrix}&=
\begin{pmatrix}
 0& 0& R_{3123} & R_{4124}  \\
 0& R_{2132} &  0& R_{4134} \\
 0&R_{2142}  &R_{3143} & 0 \\
 R_{1231}& 0 &0& R_{4234} \\
 R_{1241}& 0 &R_{3243}& 0 \\
 R_{1341}& R_{2342} &0& 0 \\
\end{pmatrix}\begin{pmatrix}
 a^{11}\\
 a^{22}\\
 a^{33}\\
 a^{44}\\
\end{pmatrix}\\
\end{align}
The Christoffel symbols of the first kind are:
\begin{align}
\begin{array}{llll}
 \ [11,1]=\half\alpha _1 e^{\alpha}& \ [11,2]=0& \ [11,3]= 0& \ [11,4]=-\half\alpha _4 e^{\alpha} \\
 \ [12,1]=0& \ [12,2]=\half e^{x^1}& \ [12,3]= 0& \ [12,4]=0 \\
 \ [13,1]=0& \ [13,2]=0& \ [13,3]= \half e^{x^1}\sin^2 x^2& \ [13,4]=0\\
 \ [14,1]=\half\alpha _4 e^{\alpha}& \ [14,2]=0& \ [14,3]= 0& \ [14,4]=-\half\gamma _1 e^{\gamma} \\
 \ [22,1]=-\half e^{x^1}& \ [22,2]=0& \ [22,3]= 0& \ [22,4]=0\\
 \ [23,1]=0& \ [23,2]=0& \ [23,3]= \half e^{x^1}\sin 2x^2  & \ [23,4]=0\\
 \ [24,1]=0& \ [24,2]=0& \ [24,3]= 0& \ [24,4]=0 \\
 \ [33,1]=-\half e^{x^1}sin^2 x^2& \ [33,2]=-\half e^{x^1}\sin 2x^2 & \ [33,3]= 0& \ [33,4]=0\\
 \ [34,1]=0& \ [34,2]=0& \ [34,3]= 0& \ [34,4]=0 \\
 \ [44,1]=\half \gamma _1 e^{\gamma}& \ [44,2]=0& \ [44,3]= 0& \ [44,4]=-\half\gamma _4 e^{\gamma} \\
\end{array}
\end{align}
We use $3.114.$ and considering that $a_{mn}= a^{mn}= 0$ for $m\ne n$: $$R_{rsmn}= \left\{\begin{array}{l}
\half\left(\partial^2_{sm}a_{rn}+\partial^2_{rn}a_{sm}\right)\\
+ \frac{1}{e^{\alpha}}\left( [rn,1] [sm,1] -[rm,1] [sn,1]  \right)\\
+ \frac{1}{e^{x^1}}\left([rn,2] [sm,2] -[rm,2][sn,2] \right)\\
+ \frac{1}{e^{x^1}\sin^2 x^2}\left( [rn,3][sm,3]-[rm,3][sn,3]  \right)\\
-\frac{1}{e^{\gamma}}\left( [rn,4][sm,4] -[rm,4][sn,4]  \right)
\end{array}\right.$$\\


%************************************************************
Giving:
\begin{align}
R_{2112}&= \left\{\begin{array}{l}
\half\left(\partial^2_{11} a_{22}+\partial^2_{22} a_{11}\right)\\\\
+ \frac{1}{e^{\alpha}}\left( [22,1] [11,1] -[21,1] [12,1]  \right)\\\\
+ \frac{1}{e^{x^1}}\left( [22,2]  [11,2] -[21,2] [12,2]  \right)\\\\
+ \frac{1}{e^{x^1}\sin^2 x^2}\left( [22,3] [11,3] -[21,3] [12,3]   \right)\\\\
-\frac{1}{e^{\gamma}}\left( [22,4]  [11,4] -[21,4] [12,4]   \right)
\end {array}\right.\\
\end{align}
\begin{align}
R_{3113}&= \left\{\begin{array}{l}
\half\left(\partial^2_{11} a_{33}+\partial^2_{33} a_{11}\right)\\\\
+ \frac{1}{e^{\alpha}}\left( [33,1] [11,1] -[31,1] [13,1]  \right)\\\\
+ \frac{1}{e^{x^1}}\left( [33,2] [11,2] -[31,2] [13,2]   \right)\\\\
+ \frac{1}{e^{x^1}\sin^2 x^2}\left( [33,3] [11,3] -[31,3] [13,3]  \right)\\\\
-\frac{1}{e^{\gamma}}\left( [33,4]  [11,4] -[31,4] [13,4]   \right)
\end {array}\right.\\
\end{align}
\begin{align}
R_{4114}&= \left\{\begin{array}{l}
\half\left(\partial^2_{11} a_{44}+\partial^2_{44} a_{11}\right)\\\\
+ \frac{1}{e^{\alpha}}\left( [44,1] [11,1] -[41,1] [14,1]  \right)\\\\
+ \frac{1}{e^{x^1}}\left( [44,2]  [11,2]  -[41,2] [14,2]  \right)\\\\
+ \frac{1}{e^{x^1}\sin^2 x^2}\left( [44,3]  [11,3] -[41,3] [14,3]  \right)\\\\
-\frac{1}{e^{\gamma}}\left( [44,4] [11,4] -[41,4] [14,4]  \right)
\end {array}\right.\\
\end{align}
\begin{align}
R_{3223}&= \left\{\begin{array}{l}
\half\left(\partial^2_{22} a_{33}+\partial^2_{33} a_{22}\right)\\\\
+ \frac{1}{e^{\alpha}}\left( [33,1] [22,1] -[32,1] [23,1]  \right)\\\\
+ \frac{1}{e^{x^1}}\left( [33,2] [22,2] -[32,2] [23,2]   \right)\\\\
+ \frac{1}{e^{x^1}\sin^2 x^2}\left( [33,3]  [22,3] -[32,3] [23,3]  \right)\\\\
-\frac{1}{e^{\gamma}}\left( [33,4]  [22,4] -[32,4] [23,4]   \right)
\end {array}\right.\\
\end{align}
\begin{align}
R_{4224}&= \left\{\begin{array}{l}
\half\left(\partial^2_{22} a_{44}+\partial^2_{44} a_{22}\right)\\\\
+ \frac{1}{e^{\alpha}}\left( [44,1] [22,1] -[42,1] [24,1]  \right)\\\\
+ \frac{1}{e^{x^1}}\left( [44,2] [22,2] -[42,2] [24,2]  \right)\\\\
+ \frac{1}{e^{x^1}\sin^2 x^2}\left( [44,3] [22,3] -[42,3] [24,3]  \right)\\\\
-\frac{1}{e^{\gamma}}\left( [44,4] [22,4] -[42,4] [24,4]  \right)
\end {array}\right.\\
\end{align}
\begin{align}
R_{4334}&= \left\{\begin{array}{l}
\half\left(\partial^2_{33} a_{44}+\partial^2_{44} a_{33}\right)\\\\
+ \frac{1}{e^{\alpha}}\left( [44,1] [33,1] -[43,1] [34,1]  \right)\\\\
+ \frac{1}{e^{x^1}}\left( [44,2] [33,2] -[43,2] [34,2]  \right)\\\\
+ \frac{1}{e^{x^1}\sin^2 x^2}\left( [44,3] [33,3] -[43,3] [34,3]  \right)\\\\
-\frac{1}{e^{\gamma}}\left( [44,4] [33,4] -[43,4] [34,4]  \right)
\end {array}\right.\\
\end{align}
\begin{align}
R_{3123}&= \left\{\begin{array}{l}
\half\left(\partial^2_{12} a_{33}\right)\\\\
+ \frac{1}{e^{\alpha}}\left( [33,1] [12,1] -[32,1] [13,1]  \right)\\\\
+ \frac{1}{e^{x^1}}\left( [33,2] [12,2] -[32,2] [13,2]  \right)\\\\
+ \frac{1}{e^{x^1}\sin^2 x^2}\left( [33,3] [12,3] -[32,3] [13,3]  \right)\\\\
-\frac{1}{e^{\gamma}}\left( [33,4] [12,4] -[32,4] [13,4]  \right)
\end {array}\right.\\
\end{align}
\begin{align}
R_{4124}&= \left\{\begin{array}{l}
\half\left(\partial^2_{12} a_{44}\right)\\\\
+ \frac{1}{e^{\alpha}}\left( [44,1] [12,1] -[42,1] [14,1]  \right)\\\\
+ \frac{1}{e^{x^1}}\left( [44,2] [12,2] -[42,2] [14,2]  \right)\\\\
+ \frac{1}{e^{x^1}\sin^2 x^2}\left( [44,3] [12,3] -[42,3] [14,3]  \right)\\\\
-\frac{1}{e^{\gamma}}\left( [44,4] [12,4] -[42,4] [14,4]  \right)
\end {array}\right.\\
\end{align}
\begin{align}
R_{2132}&= \left\{\begin{array}{l}
\half\left(\partial^2_{13} a_{22}\right)\\\\
+ \frac{1}{e^{\alpha}}\left( [22,1] [13,1] -[23,1] [12,1]  \right)\\\\
+ \frac{1}{e^{x^1}}\left( [22,2] [13,2] -[23,2] [12,2]  \right)\\\\
+ \frac{1}{e^{x^1}\sin^2 x^2}\left( [22,3] [13,3] -[23,3] [12,3]  \right)\\\\
-\frac{1}{e^{\gamma}}\left( [22,4] [13,4] -[23,4] [12,4]  \right)
\end {array}\right.\\
\end{align}
\begin{align}
R_{4134}&= \left\{\begin{array}{l}
\half\left(\partial^2_{13} a_{44}\right)\\\\
+ \frac{1}{e^{\alpha}}\left( [44,1] [13,1] -[43,1] [14,1]  \right)\\\\
+ \frac{1}{e^{x^1}}\left( [44,2] [13,2] -[43,2] [14,2]  \right)\\\\
+ \frac{1}{e^{x^1}\sin^2 x^2}\left( [44,3] [13,3] -[43,3] [14,3]  \right)\\\\
-\frac{1}{e^{\gamma}}\left( [44,4] [13,4] -[43,4] [14,4]  \right)
\end {array}\right.\\
\end{align}
\begin{align}
R_{2142}&= \left\{\begin{array}{l}
\half\left(\partial^2_{14} a_{22}\right)\\\\
+ \frac{1}{e^{\alpha}}\left( [22,1] [14,1] -[24,1] [12,1]  \right)\\\\
+ \frac{1}{e^{x^1}}\left( [22,2] [14,2] -[24,2] [12,2]  \right)\\\\
+ \frac{1}{e^{x^1}\sin^2 x^2}\left( [22,3] [14,3] -[24,3] [12,3]  \right)\\\\
-\frac{1}{e^{\gamma}}\left( [22,4] [14,4] -[24,4] [12,4]  \right)
\end {array}\right.\\
\end{align}
\begin{align}
R_{3143}&= \left\{\begin{array}{l}
\half\left(\partial^2_{14} a_{33}\right)\\\\
+ \frac{1}{e^{\alpha}}\left( [33,1] [14,1] -[34,1] [13,1]  \right)\\\\
+ \frac{1}{e^{x^1}}\left( [33,2] [14,2] -[34,2] [13,2]  \right)\\\\
+ \frac{1}{e^{x^1}\sin^2 x^2}\left( [33,3] [14,3] -[34,3] [13,3]  \right)\\\\
-\frac{1}{e^{\gamma}}\left( [33,4] [14,4] -[34,4] [13,4]  \right)
\end {array}\right.\\
\end{align}
\begin{align}
R_{1231}&= \left\{\begin{array}{l}
\half\left(\partial^2_{23} a_{11}\right)\\\\
+ \frac{1}{e^{\alpha}}\left( [11,1] [23,1] -[13,1] [21,1]  \right)\\\\
+ \frac{1}{e^{x^1}}\left( [11,2] [23,2] -[13,2] [21,2]  \right)\\\\
+ \frac{1}{e^{x^1}\sin^2 x^2}\left( [11,3] [23,3] -[13,3] [21,3]  \right)\\\\
-\frac{1}{e^{\gamma}}\left( [11,4] [23,4] -[13,4] [21,4]  \right)
\end {array}\right.\\
\end{align}
\begin{align}
R_{4234}&= \left\{\begin{array}{l}
\half\left(\partial^2_{23} a_{44}\right)\\\\
+ \frac{1}{e^{\alpha}}\left( [44,1] [23,1] -[43,1] [24,1]  \right)\\\\
+ \frac{1}{e^{x^1}}\left( [44,2] [23,2] -[43,2] [24,2]  \right)\\\\
+ \frac{1}{e^{x^1}\sin^2 x^2}\left( [44,3] [23,3] -[43,3] [24,3]  \right)\\\\
-\frac{1}{e^{\gamma}}\left( [44,4] [23,4] -[43,4] [24,4]  \right)
\end {array}\right.\\
\end{align}
\begin{align}
R_{1241}&= \left\{\begin{array}{l}
\half\left(\partial^2_{24} a_{11}\right)\\\\
+ \frac{1}{e^{\alpha}}\left( [11,1] [24,1] -[14,1] [21,1]  \right)\\\\
+ \frac{1}{e^{x^1}}\left( [11,2] [24,2] -[14,2] [21,2]  \right)\\\\
+ \frac{1}{e^{x^1}\sin^2 x^2}\left( [11,3] [24,3] -[14,3] [21,3]  \right)\\\\
-\frac{1}{e^{\gamma}}\left( [11,4] [24,4] -[14,4] [21,4]  \right)
\end {array}\right.\\
\end{align}
\begin{align}
R_{3243}&= \left\{\begin{array}{l}
\half\left(\partial^2_{24} a_{33}\right)\\\\
+ \frac{1}{e^{\alpha}}\left( [33,1] [24,1] -[34,1] [23,1]  \right)\\\\
+ \frac{1}{e^{x^1}}\left( [33,2] [24,2] -[34,2] [23,2]  \right)\\\\
+ \frac{1}{e^{x^1}\sin^2 x^2}\left( [33,3] [24,3] -[34,3] [23,3]  \right)\\\\
-\frac{1}{e^{\gamma}}\left( [33,4] [24,4] -[34,4] [23,4]  \right)
\end {array}\right.\\
\end{align}
\begin{align}
R_{1341}&= \left\{\begin{array}{l}
\half\left(\partial^2_{34} a_{11}\right)\\\\
+ \frac{1}{e^{\alpha}}\left( [11,1] [34,1] -[14,1] [31,1]  \right)\\\\
+ \frac{1}{e^{x^1}}\left( [11,2] [34,2] -[14,2] [31,2]  \right)\\\\
+ \frac{1}{e^{x^1}\sin^2 x^2}\left( [11,3] [34,3] -[14,3] [31,3]  \right)\\\\
-\frac{1}{e^{\gamma}}\left( [11,4] [34,4] -[14,4] [31,4]  \right)
\end {array}\right.\\
\end{align}
\begin{align}
R_{2342}&= \left\{\begin{array}{l}
\half\left(\partial^2_{34} a_{22}\right)\\\\
+ \frac{1}{e^{\alpha}}\left( [22,1] [34,1] -[24,1] [32,1]  \right)\\\\
+ \frac{1}{e^{x^1}}\left( [22,2] [34,2] -[24,2] [32,2]  \right)\\\\
+ \frac{1}{e^{x^1}\sin^2 x^2}\left( [22,3] [34,3] -[24,3] [32,3]  \right)\\\\
-\frac{1}{e^{\gamma}}\left( [22,4] [34,4] -[24,4] [32,4]  \right)
\end {array}\right.\\
\end{align}
%++++++++++++++++++++++++++++++++++++++++++++++++++
Considering that $[mn,q]= 0$ for $m\ne n \ne q \ne m$ and replacing the remaining Christoffels symbols:
\begin{align}
R_{2112}&= \left\{\begin{array}{l}
\half\left( \partial^2_{11} a_{22} + \partial^2_{22} a_{11} \right)\\\\
-\kwart\alpha _1 e^{x^1} 
\end {array}\right.\\
\end{align}
\begin{align}
R_{3113}&= \left\{\begin{array}{l}
\half\left( \partial^2_{11} a_{33} + \partial^2_{33} a_{11} \right)\\\\
 -\kwart\left(1+\alpha _1 \right) e^{x^1}sin^2 x^2\\
\end {array}\right.\\
\end{align}
\begin{align}
R_{4114}&= \left\{\begin{array}{l}
\half\left( \partial^2_{11} a_{44} + \partial^2_{44} a_{11} \right)\\\\
+ \kwart\left( \alpha _1\gamma _1 e^{\gamma}   - \alpha_4^2 e^{\alpha}\right)
+\kwart\left(  \alpha _4\gamma _4  e^{\alpha}   - \gamma _1^2 e^{\gamma} \right)
\end {array}\right.\\
\end{align}
\begin{align}
R_{4114}&= \left\{\begin{array}{l}
\half\left( \partial^2_{11} a_{44} + \partial^2_{44} a_{11} \right)\\\\
+ \kwart\left( \alpha _1  \gamma _1 e^{\gamma}-\alpha _4 ^2  e^{\alpha}    \right)
-\kwart\left(\alpha _4\gamma _4 e^{\alpha}- \gamma _1^2 e^{\gamma}   \right)
\end {array}\right.\\
\end{align}
\begin{align}
R_{3223}&= \left\{\begin{array}{l}
\half\left( \partial^2_{22} a_{33} + \partial^2_{33} a_{22} \right)\\\\
+ \kwart\frac{e^{x^1}}{e^{2\alpha}}   sin^2 x^2 - e^{x^1} \cos^2x^2 
\end {array}\right.\\
\end{align}
\begin{align}
R_{4224}&= \left\{\begin{array}{l}
\half\left( \partial^2_{22} a_{44} + \partial^2_{44} a_{22} \right)\\\\
-\kwart \gamma _1  \frac{e^{\gamma}e^{x^1}}{e^{\alpha}}\\
\end {array}\right.\\
\end{align}
\begin{align}
R_{4334}&= \left\{\begin{array}{l}
\half\left( \partial^2_{33} a_{44} + \partial^2_{44} a_{33} \right)\\\\
-\kwart \gamma _1  \frac{e^{\gamma}e^{x^1}}{e^{\alpha}}\sin^2 x^2 \\
\end {array}\right.\\
\end{align}
\begin{align}
R_{3123}&= 
\half\left( \partial^2_{12} a_{33}\right)=0\\
R_{4124}&= 
\half\left( \partial^2_{12} a_{44}  \right)=0\\
R_{2132}&= 
\half\left( \partial^2_{13} a_{22}  \right)=0\\
R_{4134}&= 
\half\left( \partial^2_{13} a_{44} \right)=0\\
R_{2142}&= 
 e^{-\alpha}\left([22,1][14,1]\right)\\
R_{3143}&= 
e^{-\alpha}\left([33,1][14,1]\right)\\
R_{1231}&= 
\half\left( \partial^2_{23} a_{11} \right)=0\\
R_{4234}&= 
\half\left( \partial^2_{23} a_{44}  \right)=0\\
R_{1241}&= 
\half\left( \partial^2_{24} a_{11} \right)=0\\
R_{3243}&= 
\half\left( \partial^2_{24} a_{33}  \right)=0\\
R_{1341}&= 
\half\left( \partial^2_{34} a_{11} \right)=0\\
R_{2342}&= 
\half\left( \partial^2_{34} a_{22} \right)=0
\end{align}

Giving:
\begin{align}
R_{2112}&= \kwart \left(1-\alpha _1 \right) e^{x^1} \\
R_{3113}&=  \kwart\left(1-\alpha _1 \right) e^{x^1}\sin^2 x^2 \\
R_{4114}&= \left\{ \begin{array}{l}
\half e^{\alpha} \left(\alpha_{44}  + \half\alpha_{4}^2 
- \half \alpha _4\gamma _4  \right) \\ -\half e^{\gamma} \left(\gamma_{11} +\half\gamma_{1}^2 
-  \half \alpha _1\gamma _1  
\right)
\end{array}\right.
\\
R_{3223}&=
 \left(\kwart\frac{e^{x^1}}{e^{\alpha}}-1 \right)e^{x^1}\sin^2 x^2\\
R_{4224}&=
-\kwart \gamma _1  \frac{e^{\gamma}e^{x^1}}{e^{\alpha}}\\
R_{4334}&=
-\kwart \gamma _1  \frac{e^{\gamma}e^{x^1}}{e^{\alpha}}sin^2 x^2 \\
R_{2142}&= -\kwart\alpha_4 e^{x^1}\\
R_{3143}&= -\kwart\alpha_4 e^{x^1}\sin^2 x^2
\end{align}

As all other curvature components vanish, we get
\begin{align}
\begin{pmatrix}
 R_{11}\\\\
 R_{22}\\\\
 R_{33}\\\\
 R_{44}\\\\
\end{pmatrix}&=P
\begin{pmatrix}
 {e^{-\alpha}}\\\\
 {e^{-x^1}}\\\\
 \frac{e^{-x^1}}{\sin^2 x^2}\\\\
 -{e^{-\gamma}}\\\\
\end{pmatrix}
\end{align}
With
\begin{align}
 P&=
\begin{pmatrix}
 0&\kwart \left(1-\alpha _1 \right) e^{x^1} & \kwart\left(1-\alpha _1 \right) e^{x^1}\sin^2 x^2 & \left. \begin{array}{l}
\half e^{\alpha} \left(\alpha_{44}  + \half\alpha_{4}^2 
- \half \alpha _4\gamma _4  \right) \\ -\half e^{\gamma} \left(\gamma_{11} +\half\gamma_{1}^2 
-  \half \alpha _1\gamma _1  
\right)
\end{array}\right. \\\\
  .& 0 &  \left(\kwart\frac{e^{x^1}}{e^{\alpha}}-1 \right)e^{x^1}\sin^2 x^2& -\kwart \gamma _1  \frac{e^{\gamma}e^{x^1}}{e^{\alpha}} \\\\
 . & . & 0 & -\kwart \gamma _1  \frac{e^{\gamma}e^{x^1}}{e^{\alpha}}sin^2 x^2 \\\\
 . &  . &. & 0 \\
\end{pmatrix}
\end{align}
and
\begin{align}
\begin{pmatrix}
 R_{12}\\
 R_{13}\\
 R_{14}\\
 R_{23}\\
 R_{24}\\
 R_{34}\\
\end{pmatrix}&=
\begin{pmatrix}
 0& 0& 0 & 0  \\
 0& 0 &  0& 0 \\
 0&-\kwart\alpha_4 e^{x^1}  &-\kwart\alpha_4 e^{x^1}\sin^2 x^2 & 0 \\
 0& 0 &0& 0 \\
 0& 0 &0& 0 \\
 0& 0 &0& 0 \\
\end{pmatrix}\begin{pmatrix}
 {e^{-\alpha}}\\\\
 {e^{-x^1}}\\\\
 \frac{e^{-x^1}}{\sin^2 x^2}\\\\
 -{e^{-\gamma}}\\\\
\end{pmatrix}
\end{align}
Finally  we can compute the Einstein tensor:
\begin{align}
 R &= a^{11}R_{11}+a^{22}R_{22}+a^{33}R_{33}+a^{44}R_{44}\\
 &= \left\{\begin{array}{l}
 e^{-\alpha}\left(\gamma_{11} + \half \gamma_1^2 -\half\alpha_1 \gamma_1\right)\\\\
 e^{-\gamma}\left(\alpha_{44} + \half \alpha_4^2 -\half\alpha_4 \gamma_4\right)\\\\
 +e^{-\alpha}\left(\frac{3}{2}+\gamma_1-\alpha_1\right)\\\\
 - 2e^{-x^1}
 \end{array}\right.
\end{align}
\begin{align}
G^1_{.1} &= e^{-\alpha}R_{11} - \half R\\
 &= e^{-\alpha}\left(-\kwart - \half \gamma_1\right)+e^{-x^1}
\end{align}
\begin{align}
G^2_{.2} &= e^{-x^1}R_{22} - \half R\\
 &= \left\{\begin{array}{l}e^{-\alpha}\left(-\kwart - \half \gamma_{11} - \kwart \gamma_{1}^2 -\kwart \gamma_{1} + \kwart\alpha_1\gamma_1 + \kwart \alpha_1\right)\\\\
 +e^{-\gamma}\left( \half \alpha_{44} +\kwart \alpha_{4}^2 - \kwart\alpha_4\gamma_4 \right)
 \end{array}\right.
\end{align}
\begin{align}
G^3_{.3} &= \frac{e^{-x^1}}{\sin^2 x^2}R_{33} - \half R\\
 &= \left\{\begin{array}{l}e^{-\alpha}\left(-\kwart - \half \gamma_{11} - \kwart \gamma_{1}^2 -\kwart \gamma_{1} + \kwart\alpha_1\gamma_1 + \kwart \alpha_1\right)\\\\
 +e^{-\gamma}\left( \half \alpha_{44} +\kwart \alpha_{4}^2 - \kwart\alpha_4\gamma_4 \right)
 \end{array}\right.
\end{align}
\begin{align}
G^4_{.4} &= -e^{-\gamma}R_{44} - \half R\\
 &= e^{-\alpha}\left(-\frac{3}{4} + \half \alpha_1\right)+e^{-x^1}
\end{align}

\begin{align}
G^1_{.4} &= e^{-\alpha}R_{14}\\
 &= -\half e^{-\alpha} \alpha_4
\end{align}
\begin{align}
G^4_{.1} &=- e^{-\gamma}R_{14}\\
 &= \half e^{-\gamma} \alpha_4
\end{align}
$$\blacklozenge$$
\newpage

\section{p110 - Exercise 9}
\begin{tcolorbox}
If we change the metric tensor  from $a_{mn}$ to $a_{mn}+b_{mn}$  where $b_{mn}$ is small, calculate the principal parts of the increment in the components of the curvature tensor.
\end{tcolorbox}
Let's start with one form of the curvature tensor
\begin{align}
R_{rsmn} &= \left\{\begin{array}{l}\half\left(\partial^2_{sm} a_{rn}+\partial^2_{rn} a_{sm}-\partial^2_{sn} a_{rm}-\partial^2_{rm} a_{sn}\right) \\
+ a^{pq}\left([rn,p][sm,q] - [rm,p][sn,q]\right)\end{array}\right.
\end{align}
Be $R_{rsmn}^{'}$ and $a^{'}_{rn}$ the components of the tensors after changing the metric form to $a_{mn}+b_{mn}$.\\
Then
\begin{align}
R_{rsmn}^{'} &= \left\{\begin{array}{l} \half\left(\partial^2_{sm} a^{'}_{rn}+\partial^2_{rn} a^{'}_{sm}-\partial^2_{sn} a^{'}_{rm}-\partial^2_{rm} a^{'}_{sn}\right) \\
+ a^{'pq}\left([rn,p]^{'}[sm,q]^{'} - [rm,p]^{'}[sn,q]^{'}\right)\end{array}\right.
\end{align}
At the point where we want to calculate the increment of the curvature tensor, we can choose Riemannian coordinates related to the metric $a^{'}_{mn}$. Then, the Christoffel symbols vanish at that point as origin  and (2) becomes
\begin{align}
R_{rsmn}^{'} &= \half\left(\partial^2_{sm} a^{'}_{rn}+\partial^2_{rn} a^{'}_{sm}-\partial^2_{sn} a^{'}_{rm}-\partial^2_{rm} a^{'}_{sn}\right) \\
 &= \left\{\begin{array}{l}\half\left(\partial^2_{sm} a^{}_{rn}+\partial^2_{rn} a^{}_{sm}-\partial^2_{sn} a^{}_{rm}-\partial^2_{rm} a^{}_{sn}\right)\\\\+\half\left(\partial^2_{sm} b^{}_{rn}+\partial^2_{rn} b^{}_{sm}-\partial^2_{sn} b^{}_{rm}-\partial^2_{rm} b^{}_{sn}\right) \end{array}\right.\\
  &= \left\{\begin{array}{l}R_{rsmn}-a^{pq}\left([rn,p][sm,q] - [rm,p][sn,q]\right)\\\\+\half\left(\partial^2_{sm} b^{}_{rn}+\partial^2_{rn} b^{}_{sm}-\partial^2_{sn} b^{}_{rm}-\partial^2_{rm} b^{}_{sn}\right) \end{array}\right.\\
  \Rightarrow \quad \Delta R_{rsmn}&= \left\{\begin{array}{l}-a^{pq}\left([rn,p][sm,q] - [rm,p][sn,q]\right)\\\\
  +\half\left(\partial^2_{sm} b^{}_{rn}+\partial^2_{rn} b^{}_{sm}-\partial^2_{sn} b^{}_{rm}-\partial^2_{rm} b^{}_{sn}\right) \end{array}\right.
\end{align}
As $[rn,p]^{'}=0$ we have $[rn,p]^{"}=-[rn,p]$ (the suffix $"$ referring to $b_{mn}$).
So we have for $[rn,s]^{"}$ and $[ms,r]^{"}$
\begin{align}
\partial_r b_{sn} + \partial_n b_{rs}-\partial_s b_{rn} &= -\partial_r a_{sn} - \partial_n a_{rs}+\partial_s a_{rn} \\
\partial_m b_{rs} + \partial_s b_{rm}-\partial_r b_{sm} &= -\partial_m a_{rs} - \partial_s a_{rm}+\partial_r a_{sm} \\
\partial_m (7) \quad\Rightarrow\quad \partial^2_{rm} b_{sn} + \partial^2_{mn} b_{rs}-\partial^2_{sm} b_{rn} &= -\partial^2_{rm} a_{sn} - \partial^2_{mn} a_{rs}+\partial^2_{sm} a_{rn}\\
\partial_n (8) \quad\Rightarrow\quad \partial^2_{mn} b_{rs} + \partial^2_{sn} b_{rm}-\partial^2_{rn} b_{sm} &= -\partial^2_{mn} a_{rs} - \partial^2_{sn} a_{rm}+\partial^2_{rn} a_{sm}
\end{align}
Combining $(9)$ and $(10)$ in $(6)$, we get
\begin{align}
\Delta R_{rsmn}&= \left\{\begin{array}{l}-a^{pq}\left([rn,p][sm,q] - [rm,p][sn,q]\right)\\\\
  +\half\left(\partial^2_{rm} a_{sn} + \partial^2_{mn} a_{rs}-\partial^2_{sm} a_{rn}\right)\\\\ 
+\half\left(\partial^2_{mn} a_{rs} + \partial^2_{sn} a_{rm}-\partial^2_{rn} a_{sm}\right)\\\\  
+\half\left( \partial^2_{mn} b_{rs}+\partial^2_{mn} b_{rs}\right)\\\\
  \end{array}\right.\\
  &= \left\{\begin{array}{l}a^{pq}\left( [rm,p][sn,q]-[rn,p][sm,q]\right)\\\\
  +\half\left(\partial^2_{rm} a_{sn} + \partial^2_{sn} a_{rm} -\partial^2_{sm} a_{rn}-\partial^2_{rn} a_{sm}\right)\\\\  
  +\half\left(\partial^2_{mn} a_{rs}+\partial^2_{mn} a_{rs}\right)\\\\
+\half\left( \partial^2_{mn} b_{rs}+\partial^2_{mn} b_{rs}\right)\\\\
  \end{array}\right.\\
  &= \left\{\begin{array}{l}R_{rsnm}\\\\  
  +\half\left(\partial^2_{mn} a_{rs}+\partial^2_{mn} a_{rs}\right)\\\\
+\half\left( \partial^2_{mn} b_{rs}+\partial^2_{mn} b_{rs}\right)\\\\
  \end{array}\right.\\
    &= -R_{rsmn}
  +\partial^2_{mn} a_{rs}+\partial^2_{mn} b_{rs}
\end{align}
Can I simplify further ??
$$\blacklozenge$$
\newpage


\section{p110 - Exercise 10}
\begin{tcolorbox}
If we use normal coordinates in a Riemannian $V_n$, the metric form is as in equation $2.630$. For this coordinate system, express the curvature tensor, the Ricci tensor, and the curvature invariant in terms of the corresponding quantities for the $(N-1)-\text{space}$ $x^N = C^{st}$ and certain additional terms. Check these additional terms by nothing that they must have tensor character with respect to transformations of the coordinates $x^1, x^2, \dots, x^{N-1}$.
\end{tcolorbox}
$$R_{rsmn} = \partial_m[sn,r]-\partial_n[sm,r] + \Gamma^p_{sm}[rn,p] -\Gamma^p_{sn}[rm,p]$$
We indicate by $R^{'}_{\dots}$  the quantity generated by the previous definition, but restricted to the subspace $V_{N-1}$. Calculating $R_{\alpha\beta\gamma\delta}$ restricted to the subspace $V_{N-1}$ gives:
\begin{align}
 R_{\alpha\beta\gamma\delta}&= \underbrace{\partial_{\gamma}[\beta \delta, \alpha]-\partial_{\delta}[\beta \gamma, \alpha] + \Gamma^{\nu}_{\beta \gamma}[\alpha \delta, \nu] -\Gamma^{\nu}_{\beta \delta}[\alpha \gamma, \nu]}_{= R^{'}_{\alpha\beta\gamma\delta}}+ \Gamma^{N}_{\beta \gamma}[\alpha \delta,N] -\Gamma^{N}_{\beta \delta}[\alpha \gamma,N]\\
 &= R^{'}_{\alpha\beta\gamma\delta}+ \underbrace{\Gamma^{N}_{\beta \gamma}}_{ -\half\frac{1}{a_{NN}}\partial_N a_{\beta\gamma}}\underbrace{[\alpha \delta,N]}_{  -\half \partial_N a_{\alpha\delta}} -\underbrace{\Gamma^{N}_{\beta \delta}}_{-\half\frac{1}{a_{NN}}\partial_N a_{\beta\delta}  }\underbrace{[\alpha \gamma,N]}_{ -\half \partial_N a_{\alpha\gamma}}\quad \text{(see page 66/67)}\\
\Rightarrow\spatie R_{\alpha\beta\gamma\delta}&= R^{'}_{\alpha\beta\gamma\delta}+ \kwart\frac{1}{a_{NN}}\left(\partial_N a_{\beta\gamma} \partial_N a_{\alpha\delta}-\partial_N a_{\beta\delta} \partial_N a_{\alpha\gamma}\right)
\end{align}
We calculate $R_{\alpha\beta}$:
\begin{align}
R_{\alpha\beta} &= a^{sn} R_{s\alpha\beta n}\\
 &= \underbrace{a^{\nu\mu} R_{\nu\alpha\beta \mu}}_{R^{'}_{\alpha\beta}} + \underbrace{a^{N \mu}}_{=0} R_{N \alpha\beta \mu}+\underbrace{a^{\nu N}}_{=0} R_{\nu\alpha\beta N}+\underbrace{a^{NN}}_{=\frac{1}{a_{NN}}} R_{N \alpha\beta N}\\
\text{(5)}\Rightarrow \spatie R_{\alpha\beta} &=R^{'}_{\alpha\beta} + \frac{1}{a_{NN}} R_{N\alpha\beta N}\\
R_{N \alpha\beta N}&=\left\{\begin{array}{l}\partial_{\beta}[\alpha N,N]-\partial_{N}[\alpha\beta,N]\\\\ + \Gamma^{\nu}_{\alpha\beta}[NN,\nu] -\Gamma^{\nu}_{\alpha N}[N \beta,\nu]\\\\
+ \Gamma^{N}_{\alpha\beta}[NN,N] -\Gamma^{N}_{\alpha N}[N \beta,N]
\end{array}\right.\\
&=\left\{\begin{array}{l}R^{'}_{N \alpha\beta N}\\\\
+ \underbrace{\Gamma^{N}_{\alpha\beta}}_{-\half\frac{1}{a_{NN}}\partial_N a_{\alpha\beta}}\underbrace{[NN,N]}_{\half\frac{1}{a_{NN}}\partial_N a_{NN}} -\underbrace{\Gamma^{N}_{\alpha N}}_{\half\frac{1}{a_{NN}}\partial_{\alpha} a_{NN}}\underbrace{[N \beta,N]}_{\half\partial_{\beta} a_{NN}}\end{array}\right.\\
&=R^{'}_{N \alpha\beta N}
 -\kwart\frac{1}{a_{NN}}\left( \frac{1}{a_{NN}}\partial_N a_{\alpha\beta}\partial_N a_{NN}+\partial_{\alpha} a_{NN}\partial_{\beta} a_{NN}\right)
\end{align}
\begin{align}
 \text{(6) and (9)}\Rightarrow \spatie R_{\alpha\beta} &=\left\{\begin{array}{l}R^{'}_{\alpha\beta} + \frac{1}{a_{NN}}R^{'}_{N \alpha\beta N} \\\\ -\kwart\frac{1}{a_{NN}^2}\left( \frac{1}{a_{NN}}\partial_N a_{\alpha\beta}\partial_N a_{NN}+\partial_{\alpha} a_{NN}\partial_{\beta} a_{NN}\right)\end{array}\right.
\end{align}
And the invariant curvature:
\begin{align}
R &= a^{mn} R_{mn}\\
&= \underbrace{a^{\mu\nu} R_{\mu\nu}}_{= R^{'}}+\underbrace{a^{N\nu}}_{=0} R_{N\nu}+\underbrace{a^{\mu N}}_{=0} R_{\mu N}+a^{NN} R_{NN}\\
&= R^{'}+a^{NN} R_{NN}\\
R_{NN}&= a_{sn}R_{sNNn}\\
&= \underbrace{a_{\mu\nu}R_{\mu N N\nu}}_{=R^{'}_{NN}}+\underbrace{a_{N\nu}R_{N N N\nu}}_{=0}+\underbrace{a_{\mu N}R_{\mu N NN}}_{=0}+\underbrace{a_{NN}R_{N N NN}}_{=0}\\
\text{(13) and (15)}\Rightarrow \spatie R &= R^{'}+\frac{1}{a_{NN}} R^{'}_{NN}
\end{align}
We now check the tensor character of the residuals in $(3$), $(10)$ and $(16)$
Let's define the following coordinate transformations 
$a=a(\alpha,\beta,\gamma,\dots), \ b=b(\alpha,\beta,\gamma,\dots),...$ and $N = N$.
Hence,
\begin{align}
a_{\alpha\beta}= a_{ab}\partial_\alpha a \partial_\beta b
\end{align}
Now,
$$i)\spatie \mathbf{R_{\alpha\beta\gamma\delta}= R^{'}_{\alpha\beta\gamma\delta}+ \kwart\frac{1}{a_{NN}}\underbrace{\left(\partial_N a_{\beta\gamma} \partial_N a_{\alpha\delta}-\partial_N a_{\beta\delta} \partial_N a_{\alpha\gamma}\right)}_{T_{\alpha\beta\gamma\delta}=tensor?}}$$
\begin{align}
T_{\alpha\beta\gamma\delta}&=\left\{\begin{array}{l}\partial_N\left( a_{ab}\partial_\beta a \partial_\gamma b\right)\partial_N\left( a_{cd}\partial_\alpha c \partial_\delta d\right)\\\\
-\partial_N \left(a_{ab}\partial_\beta a \partial_\delta b \right)\partial_N\left(a_{cd}\partial_\alpha c \partial_\gamma d\right)
\end{array}\right.\\
&=\left\{\begin{array}{l}
\partial_N a_{ab}\partial_N a_{cd}\partial_\beta a \partial_\gamma b\partial_\alpha c \partial_\delta d\\\\
-\partial_N a_{ab}\partial_N a_{cd}\partial_\beta a \partial_\delta b \partial_\alpha c \partial_\gamma d\\\\
+\text{other terms}
\end{array}\right.\\
&=\left\{\begin{array}{l}
\left(\underbrace{\partial_N a_{cb}\partial_N a_{ad}
-\partial_N a_{bd}\partial_N a_{ac}}_{\equiv T_{abcd}}\right)\partial_\alpha a\partial_\beta b \partial_\gamma g \partial_\delta d\\\\
+\text{other terms}
\end{array}\right.
\end{align}
So we see that $T_{abcd}$ transforms to $T_{\alpha\beta\gamma\delta}$ as a tensor provided that the other terms in $(20)$ vanish. This is the case as the other terms will all have elements containing factors like $\partial2_{N\nu} x$ which are zero as the $x$ are no explicit function of $N$.
$$ii)\spatie \mathbf{R_{\alpha\beta} =\left\{\begin{array}{l}R^{'}_{\alpha\beta} + \frac{1}{a_{NN}}R^{'}_{N \alpha\beta N} \\\\ -\kwart\frac{1}{a_{NN}^2}\left(\underbrace{ \frac{1}{a_{NN}}\partial_N a_{\alpha\beta}\partial_N a_{NN}+\partial_{\alpha} a_{NN}\partial_{\beta} a_{NN}}_{\equiv T_{\alpha\beta} = tensor?}\right)\end{array}\right.}$$
\begin{align}
T_{\alpha\beta} &= \frac{1}{a_{NN}}\partial_N \left(a_{ab}\partial_\alpha a \partial_\beta b\right)\partial_N a_{NN}+\partial_{a} a_{NN}\partial_{b} a_{NN}\partial_{\alpha}a\partial_{\beta} b\\
&= \frac{1}{a_{NN}}\partial_N a_{ab}\partial_\alpha a \partial_\beta b\partial_N a_{NN}+\partial_{a} a_{NN}\partial_{b} a_{NN}\partial_{\alpha}a\partial_{\beta} b +\text{ other terms}\\
&= \left(\underbrace{\frac{1}{a_{NN}}\partial_N a_{ab}\partial_N a_{NN}+\partial_{a} a_{NN}\partial_{b} a_{NN}}_{\equiv T_{ab}}\right)\partial_{\alpha} a\partial_{\beta} b +\text{ other terms}
\end{align}
Again the other terms vanish - see previous curvature tensor - and so $T_{ab}$ transforms to $T_{\alpha\beta}$ as a tensor.
$$iii)\spatie \mathbf{R = R^{'}+\frac{1}{a_{NN}} R^{'}_{NN}}$$
The last term is an invariant and by definition  at tensor.
$$\blacklozenge$$
\newpage



\section{p110 - Exercise 11}
\begin{tcolorbox}
Prove that $$T^{mn}_{\ \ \ \ |mn} =T^{mn}_{\ \ \ \ |nm}$$ whet $T^{mn}$ is not necessarily symmetric.
\end{tcolorbox}
\begin{align}
T^{mn}_{\ \ \ \ |s} &= \partial_s T^{mn} + \Gamma^m_{ks}T^{kn} + \Gamma^n_{ks}T^{mk}\\
T^{mn}_{\ \ \ \ |st} &= (\partial_s T^{mn})_{|t} + (\Gamma^m_{ks})_{|t}T^{kn} + (\Gamma^n_{ks})_{|t}T^{mk}+ \Gamma^m_{ks}(T^{kn})_{|t} + \Gamma^n_{ks}(T^{mk})_{|t})
\end{align}
We choose Riemannian coordinates and take as origin the point where we want to check the asked identity. At that point the Christoffel symbols vanish and $(2)$ becomes
\begin{align}
T^{mn}_{\ \ \ \ |st} &= (\partial_s T^{mn})_{|t} + (\Gamma^m_{ks})_{|t}T^{kn} + (\Gamma^n_{ks})_{|t}T^{mk}\\
&= \left\{\begin{array}{l}\partial^2_{st}T^{mn}+\quad \text{terms in}\quad \Gamma\text{'s (=0)}\\\\ + T^{kn} (\Gamma^m_{ks})_{|t}\\\\ + T^{mk} (\Gamma^n_{ks})_{|t}\end{array}\right.\\
&= \left\{\begin{array}{l}\partial^2_{st}T^{mn}\\\\ + T^{kn} \left[\underbrace{a^{mp}_{\ \ \ |t}}_{=0}[ks,p] + \half a^{mp} \left( (\partial_{k} a_{sp})_{|t} + (\partial_{s} a_{kp})_{|t} - (\partial_{p} a_{ks})_{|t} \right) \right]\\\\  + T^{mk} \left[\underbrace{a^{np}_{\ \ \ |t}}_{=0}[ks,p] + \half a^{np} \left( (\partial_{k} a_{sp})_{|t} + (\partial_{s} a_{kp})_{|t} - (\partial_{p} a_{ks})_{|t} \right) \right]\end{array}\right.\\
&= \left\{\begin{array}{l}\partial^2_{st}T^{mn}\\\\ +\half a^{mp} \left( \partial^2_{kt} a_{sp} + \partial^2_{st} a_{kp} - \partial^2_{pt} a_{ks} \right) T^{kn}  \\\\  + \half   a^{np} \left( \partial^2_{kt} a_{sp} + \partial^2_{st} a_{kp} - \partial^2_{pt} a_{ks}\right)T^{mk} \end{array}\right.
\end{align}
In the same way we get
\begin{align}
T^{mn}_{\ \ \ \ |ts} &= \left\{\begin{array}{l}\partial^2_{st}T^{mn}\\\\ +\half a^{mp} \left( \partial^2_{ks} a_{tp} + \partial^2_{st} a_{kp} - \partial^2_{ps} a_{kt} \right) T^{kn}  \\\\  + \half   a^{np} \left( \partial^2_{ks} a_{tp} + \partial^2_{st} a_{kp} - \partial^2_{ps} a_{kt}\right)T^{mk} \end{array}\right.
\end{align}
Hence 
\begin{align}
2\left(T^{mn}_{\ \ \ \ |st}-T^{mn}_{\ \ \ \ |ts}\right) &= \left\{\begin{array}{l} \ a^{mp} \left( \partial^2_{kt} a_{sp} + \partial^2_{st} a_{kp} - \partial^2_{pt} a_{ks} - \partial^2_{ks} a_{tp} - \partial^2_{st} a_{kp} + \partial^2_{ps} a_{kt}  \right) T^{kn}  \\\\  + a^{np} \left( \partial^2_{kt} a_{sp} + \partial^2_{st} a_{kp} - \partial^2_{pt} a_{ks}- \partial^2_{ks} a_{tp} - \partial^2_{st} a_{kp} + \partial^2_{ps} a_{kt}\right)T^{mk} \end{array}\right.\\
&= \left\{\begin{array}{l} \ a^{mp} \left( \partial^2_{kt} a_{sp} - \partial^2_{pt} a_{ks} - \partial^2_{ks} a_{tp} + \partial^2_{ps} a_{kt}  \right) T^{kn}  \\\\  + a^{np} \left( \partial^2_{kt} a_{sp}  - \partial^2_{pt} a_{ks}- \partial^2_{ks} a_{tp} + \partial^2_{ps} a_{kt}\right)T^{mk} \end{array}\right.
\end{align}
Putting $s=m$ and $t=n$:
\begin{align}
2\left(T^{mn}_{\ \ \ \ |mn}-T^{mn}_{\ \ \ \ |nm}\right) 
&= \left\{\begin{array}{l} \ a^{mp} \left( \partial^2_{kn} a_{mp} - \partial^2_{pn} a_{km} - \partial^2_{km} a_{np} + \partial^2_{pm} a_{kn}  \right) T^{kn}  \\\\  + a^{np} \left( \partial^2_{kn} a_{mp}  - \partial^2_{pn} a_{km}- \partial^2_{km} a_{np} + \partial^2_{pm} a_{kn}\right)T^{mk} \end{array}\right.
\end{align}
In the last term of $(10)$ we can swap the indices $m$ and $n$ giving
\begin{align}
2\left(T^{mn}_{\ \ \ \ |mn}-T^{mn}_{\ \ \ \ |nm}\right) 
&= \left\{\begin{array}{l} \ a^{mp} \left( \partial^2_{kn} a_{mp} - \partial^2_{pn} a_{km} - \partial^2_{km} a_{np} + \partial^2_{pm} a_{kn}  \right) T^{kn}  \\\\  + a^{mp} \left( \partial^2_{km} a_{np}  - \partial^2_{pm} a_{kn}- \partial^2_{kn} a_{mp} + \partial^2_{pn} a_{km}\right)T^{nk} \end{array}\right.
\end{align}
In the second term of $(11)$ we may again swap the indices $n$ and $k$ giving
\begin{align}
2\left(T^{mn}_{\ \ \ \ |mn}-T^{mn}_{\ \ \ \ |nm}\right) 
&= \left\{\begin{array}{l} \ a^{mp} \left( \cancel{\partial^2_{kn} a_{mp}} - \partial^2_{pn} a_{km} - \partial^2_{km} a_{np} + \bcancel{\partial^2_{pm} a_{kn} } \right) T^{kn}  \\\\  + a^{mp} \left( \partial^2_{nm} a_{kp}  - \bcancel{\partial^2_{pm} a_{kn}}- \cancel{\partial^2_{kn} a_{mp}} + \partial^2_{pk} a_{nm}\right)T^{kn} \end{array}\right.\\\\
&=  \left( \underbrace{a^{mp}\partial^2_{nm} a_{kp}}_{\text{swap m and p}}  + \underbrace{a^{mp}\partial^2_{pk} a_{nm}}_{\text{swap m and p}} - a^{mp}\partial^2_{pn} a_{km} -a^{mp} \partial^2_{km} a_{np} \right) T^{kn}\\
&=  \left( a^{mp}\partial^2_{np} a_{km}  + a^{mp}\partial^2_{mk} a_{np} - a^{mp}\partial^2_{pn} a_{km} -a^{mp} \partial^2_{km} a_{np} \right) T^{kn}\\
&=0
\end{align}
$$\blacklozenge$$
\newpage

\section{p110 - Exercise 12a}
\begin{tcolorbox}
Prove that the quantities $$G^{mn}+ \frac{1}{2a}\frac{\delta^2}{\delta x^r \delta x^s}\left[ a \left( a^{mn}a^{rs}-a^{mr}a^{ns} \right)\right]$$ can be expressed in terms of the metric tensor and its first derivatives.
\end{tcolorbox}
Let's define 
\begin{align}
K^{mn} &\equiv \frac{1}{2a}\frac{\delta^2}{\delta x^r \delta x^s}\left[ a \left( a^{mn}a^{rs}-a^{mr}a^{ns} \right)\right]\\
T^{mn} &\equiv G^{mn}+ K^{mn}
\end{align}
The strategy to proof this, is to separate in both terms of the expression, the parts that can be expressed in $a_{ij}, a^{ij}, \partial_k a_{ij}, \partial_k a^{ij}$ and the parts of higher order differentiation.
We begin with the second term.\\
As the covariant derivatives of the metric tensor vanish, we get
\begin{align}
K^{mn} &=\frac{1}{2a}\left[ a \left( a^{mn}a^{rs}-a^{mr}a^{ns} \right)\right] _{|rs}\\
&= \frac{1}{2a}\left[ a_{|r} \left( a^{mn}a^{rs}-a^{mr}a^{ns} \right)+ a \underbrace{\left( a^{mn}a^{rs}-a^{mr}a^{ns} \right)_{|r}}_{=0}\right] _{|s}\\
&= \frac{1}{2a}\left[ a_{|r} \left( a^{mn}a^{rs}-a^{mr}a^{ns} \right)\right] _{|s}\\
&= \frac{1}{2a}\left[ \underbrace{a_{|rs}}_{=\partial^2_{rs} a} \left( a^{mn}a^{rs}-a^{mr}a^{ns} \right)+a_{|r} \underbrace{\left( a^{mn}a^{rs}-a^{mr}a^{ns} \right)_{|s}}_{=0} \right]
\end{align}
Considering, 
\begin{align}
\partial^2_{rs} \ln a&= \partial_s \left(\frac{1}{a}\partial_r a\right)\\
&= \frac{1}{a}\partial^2_{rs}a-\frac{1}{a^2}\partial_r a\partial_s a\\
\Rightarrow\spatie \frac{1}{a}\partial^2_{rs}a &= \partial^2_{rs} \ln a+ \frac{1}{a^2}\partial_r a\partial_s a
\end{align}
\begin{align}
\Rightarrow \quad K^{mn} &= \half\left( a^{mn}a^{rs}-a^{mr}a^{ns} \right)\partial^2_{rs}\ln a+\mathcal{E}^{mn} 
\end{align}
with $\mathcal{E}^{mn}$ being a function in  $a_{ij}, a^{ij}, \partial_k a_{ij}, \partial_k a^{ij}$ only.\\
Note that $\mathcal{E}^{mn}$ is a symmetrical object in $m,n$. Indeed,
\begin{align}
\mathcal{E}^{mn} &= \frac{1}{2a^2}\left( a^{mn}a^{rs}-a^{mr}a^{ns} \right)\partial_r a\partial_s a\\
\Rightarrow\spatie \mathcal{E}^{nm} &= \frac{1}{2a^2}\left( a^{nm}a^{rs}-a^{nr}a^{ms} \right)\partial_r a\partial_s a\\
&= \frac{1}{2a^2}\left( a^{mn}a^{sr}-a^{ns}a^{mr} \right)\partial_s a\partial_r a\\
&= \mathcal{E}^{mn}
\end{align}
But by $(2.541.)$ : $\partial_i \ln a = 2\Gamma^t_{it}$
hence,
\begin{align}
\partial_{rs} \ln a &= 2\partial_{s}\Gamma^t_{rt} = 2\partial_{r}\Gamma^t_{st}\\
\Rightarrow\quad K^{mn} &= \left( a^{mn}a^{rs}-a^{mr}a^{ns} \right)\partial_{s}\Gamma^t_{rt}+\mathcal{E}^{mn}\\
\end{align}
Considering $(10)$ and  swapping dummy indices we get the following expressions for $K^{mn}$
\begin{align}
K^{mn} &= \left( a^{mn}a^{rs}-a^{mr}a^{ns} \right)\partial_{s}\Gamma^t_{rt}+\mathcal{E}^{mn}\\
K^{mn} &= \left( a^{mn}a^{rs}-a^{mr}a^{ns} \right)\partial_{r}\Gamma^t_{st}+\mathcal{E}^{mn}\\
K^{mn} &= \left( a^{mn}a^{rs}-a^{ms}a^{nr} \right)\partial_{s}\Gamma^t_{rt}+\mathcal{E}^{mn}\\
K^{mn} &= \left( a^{mn}a^{rs}-a^{mr}a^{ns} \right)\partial_{r}\Gamma^t_{st}+\mathcal{E}^{mn}\\
\end{align}
We now rewrite $G^{mn}$. \\
We have:
\begin{align}
G^{mn} &= a^{nk}G^m_{. \ k}\\
G^m_{. \ k} &= R^m_{. \ k}-\half\delta^m_k R\\
R^m_{. \ k} &= a^{mp} R_{pk}\\
R &= a^{pk}R_{pk}
\end{align}
And by $3.203.$
\begin{align}
R_{pk} = \partial_k \Gamma^t_{pt} - \partial_t \Gamma^t_{pk} + \mathcal{F}^{pk}
\end{align}
With $\mathcal{F}^{pk}$ a function in  $a_{ij}, a^{ij}, \partial_k a_{ij}, \partial_k a^{ij}$ only. Note that $\mathcal{F}^{pk}$ is a symmetrical object in $p,k$ as $R_{pk}$ is a symmetrical tensor in $p,k$.
Hence by $(15)$ to $(19)$:
\begin{align}
G^{mn} &= a^{nr} R^m_{. \ r}-\half a^{nr}\delta^m_k R \\
&=  a^{nr} a^{ms} R_{pk}-\half a^{nm}a^{rs}R_{rs} \\
&=  \left( a^{nr} a^{ms} -\half a^{nm}a^{rs}\right)R_{rs} \\
&=  \left( a^{nr} a^{ms} -\half a^{nm}a^{rs}\right)\left(\partial_r \Gamma^t_{st} - \partial_t \Gamma^t_{rs} + \mathcal{F}^{rs}\right) \\
&=  \left( a^{nr} a^{ms} -\half a^{nm}a^{rs}\right)\left(\partial_r \Gamma^t_{st} - \partial_t \Gamma^t_{rs} \right)+ \mathcal{H}^{mn}\\
&= a^{nr} a^{ms}\partial_r \Gamma^t_{st} -\half a^{nm}a^{rs}\partial_r \Gamma^t_{st} -  a^{nr} a^{ms}\partial_t \Gamma^t_{rs} +\half a^{nm}a^{rs}\partial_t \Gamma^t_{rs}+ \mathcal{H}^{mn}
\end{align}
with $\mathcal{H}^{mn}$ a function in  $a_{ij}, a^{ij}, \partial_k a_{ij}, \partial_k a^{ij}$ only.\\
Note that $\mathcal{H}^{mn}$ is a symmetrical object in $m,n$ . Indeed,
\begin{align}
\mathcal{H}^{mn} &= \left( a^{nr} a^{ms} -\half a^{nm}a^{rs}\right)\mathcal{F}^{rs}\\
\Rightarrow\spatie \mathcal{H}^{nm} &= \left( a^{mr} a^{ns} -\half a^{mn}a^{rs}\right)\mathcal{F}^{rs}\\
&= \left( a^{ms} a^{nr} -\half a^{mn}a^{rs}\right)\mathcal{F}^{rs}\\
&= \mathcal{H}^{mn}
\end{align}
 
Putting  $(10)$ and $(2)$ together with $\mathcal{L}^{mn}=\mathcal{E}^{mn}+ \mathcal{H}^{mn}$ we get,
\begin{align}
T^{mn} &= Q^{mn}+\mathcal{L}^{mn}
\end{align}
with $\mathcal{L}^{mn}$ a symmetrical object in $m,n$ and depending only on terms in $a_{ij}, a^{ij}, \partial_k a_{ij}, \partial_k a^{ij}$  and  
\begin{align}
 Q^{mn} &\equiv \left\{ \begin{array}{l}
\cancel{a^{nr} a^{ms}\partial_r \Gamma^t_{st}} -\underbrace{\half a^{nm}a^{rs}\partial_r \Gamma^t_{st}}_{*} \\\\ 
-  a^{nr} a^{ms}\partial_t \Gamma^t_{rs} +\half a^{nm}a^{rs}\partial_t \Gamma^t_{rs}\\\\
+ \underbrace{a^{mn}a^{rs}\partial_{s}\Gamma^t_{rt}}_{*}-\cancel{a^{mr}a^{ns}\partial_{s}\Gamma^t_{rt}}\\\\
\end{array} \right. \\
&= \half a^{mn}a^{rs}\left(\partial_{s}\Gamma^t_{rt}+\partial_t \Gamma^t_{rs}\right)-  a^{nr} a^{ms}\partial_t \Gamma^t_{rs}
\end{align}
$Q^{mn} $ a symmetrical object in $m,n$


$$\blacklozenge$$
\newpage

