\item Using the coordinate transformations~$\mathbf{e}_{a'}=\mathbf{e}_a M^a{}_{a'}$,~$\tilde{\omega}^{a'}= M^{a'}{}_a \tilde{\omega}^a$ and the inverse matrices~$M^{a'}{}_b =(M^{b}{}_{a'})^{-1}$, prove that the components of the vectors,~$\mathbf{A}= A^a \mathbf{e}_a$, dual-vectors,~$\mathbf{\tilde{B}}= B_b \tilde{\omega}^b$, and mixed tensors,~$\mathbf{T}= T^a{}_b \mathbf{e}_a \tilde{\omega}^b$, are transformed as~$A^{a'}= M^{a'}{}_a A^a$,~$B_{a'}= B_a M^a{}_{a'}$ and~$T^{a'}{}_{b'}= M^{a'}{}_a T^a{}_b M^b{}_{b'}$, respectively.
(4 marks) \\\\
\textit{Proof.}~$\mathbf{e}_{a'}=\mathbf{e}_a M^a{}_{a'}$ and~$M^{a'}{}_b =(M^{b}{}_{a'})^{-1}$ give~$\mathbf{e}_a = M^{a'}{}_a \mathbf{e}_{a'}$. Using this result, we get~$\mathbf{A}= A^a \mathbf{e}_a = A^a M^{a'}{}_a \mathbf{e}_{a'}= M^{a'}{}_a A^a \mathbf{e}_{a'}$. As we also have~$\mathbf{A}= A^{a'}\mathbf{e}_{a'}$, we get the relations. \\
The inverse relation also gives~$\tilde{\omega}^a = M^a{}_{a'}\tilde{\omega}^{a'}$. This gives~$\mathbf{\tilde{B}}= B_a \tilde{\omega}^a = B_a M^a{}_{a'}\tilde{\omega}^{a'}$. Since~$\mathbf{\tilde{B}}= B_{a'}\tilde{\omega}^{a'}$, we get the relations. \\
The above relations also give~$\mathbf{T}= T^a{}_b \mathbf{e}_a \tilde{\omega}^b = T^a{}_b M^{a'}{}_a \mathbf{e}_{a'} M^b{}_{b'}\tilde{\omega}^{b'}= M^{a'}{}_a T^a{}_b M^b{}_{b'}\mathbf{e}_{a'}\tilde{\omega}^{b'}$. Since we also have~$\mathbf{T}= T^{a'}{}_{b'}\mathbf{e}_{a'}\tilde{\omega}^{b'}$, we get the relations.~$\blacksquare$
\item Using the Bianchi identities,~$\nabla_e {}^{(4)} R_{abcd}+\nabla_d {}^{(4)} R_{abec}+\nabla_c {}^{(4)} R_{abde}=0$, prove that~$\nabla^a G_{ab}=0$, where~$G_{ab}={}^{(4)} R_{ab}-\frac{1}{2} g_{ab}{}^{(4)} R$.
(2 marks) \\\\
\textit{Proof.} The Bianchi identity gives~$g^{ac} g^{bd}(\nabla_e {}^{(4)} R_{abcd}+\nabla_d {}^{(4)} R_{abec}+\nabla_c {}^{(4)} R_{abde})=0$, or~$\nabla_e {}^{(4)} R -\nabla^b {}^{(4)} R_{be}-\nabla^a {}^{(4)} R_{ae}=0$. Relabeling the indices, we get,~$\nabla^a ({}^{(4)} R_{ab}-\frac{1}{2} g_{ab}{}^{(4)} R)=0$, or~$\nabla^a G_{ab}=0$.~$\blacksquare$
\item Given the Schwarzschild metric expresed in Schwarzschild coordinates
and the Kerr metric in Boyer-Lindquist coordinates
\begin{align*}
ds^2 = & - \left( 1 - \frac{2Mr}{\Sigma}\right) dt^2 - \frac{4 a M r \sin^2 \theta}{\Sigma}\, dt \, d\phi + \frac{\Sigma}{\Delta} dr^2 + \Sigma d\theta^2 \\
& + \left( r^2 + a^2 + \frac{2 a^2 M r \sin^2 \theta}{\Sigma}\right) \sin^2 \theta d\phi^2 \, ,
\end{align*}
where~$a \equiv J/M$,~$\Delta\equiv r^2-2 M r + a^2$,~$\Sigma\equiv r^2+ a^2\cos^2\theta$,~$M$ and~$J$ are the mass and angular momentum of the black hole, respectively. Derive the curvature invariants~$R_{ab} R^{ab}$ and~$R$ for both of them in the vacuum case. %[\textbf{Hint:} You do not need to derive any of the tensors from the first principles.]
(2 marks) \\\\
\textit{Proof}. Taking the trace of the vacuum Einstein equation gives~$R =0$, which, in turn, gives~$R_{ab}=0$. Therefore, all the curvature invariants asked above are~$0$.~$\blacksquare$
%\item Derive the expressions for the event horizon and outer boundary of the ergosphere of the Kerr metric expressed in Boyer-Lindquist coordinates.
%(2 marks)
\item Exercise 2.1 of Baumgarte and Shapiro. (2 marks) \\
\textit{Proof.} Using the Bianchi identities~$\nabla_b G^{ab}=0$ and the conservation law for the matter field~$\nabla_b T^{ab}=0$, we get~$\nabla_b (G^{ab}-8\pi T^{ab})=0$. Expanding in terms of partial derivatives, we get
Imposing the constraint~$G^{a0}= T^{a0}$ in the above expresions then gives~$\p_i (G^{ai}-8\pi T^{ai})=\Gamma^a_{bc}(G^{bc}-8\pi T^{bc})=\Gamma^c_{bc}(G^{ab}-8\pi T^{ab})=0$, reducing~\eqref{Constraint_evolution} to~$\p_0(G^{a0}-8\pi T^{a0})=0$. Therefore, whenever the constraints are satisfied on the initial slice, and all the dynamical variables satisfy the Einstein field equations, the constraints will be satisfied forever, as their time variation will always be zero.~$\blacksquare$
\item Exercise 2.12-2.14 of Baumgarte and Shapiro. (4 marks) \\\\
Solving~$r^2-2Mr + a^2\cos^2\theta=0$, we get~$r = M +(M^2- a^2\cos^2\theta)^{1/2}$. which is the radial coordinate of the ergosphere. Therefore,~$t^a t_a > 0$ inside the ergosphere and~$< 0$ outside it.
(b) The aymptotic values of various quantities are $\alpha=\sqrt{1-2M/r}$, \\
$\beta_i =(0,0,-2Ma \sin^2\theta/r)$, $\gamma_{ij}=$ diag$((1-2M/r)^{-1}, r^2-2Mr, r^2\sin^2\theta)$ and $\gamma^{ij}=$ diag$((1-2M/r), r^{-2}, (r^2\sin^2\theta)^{-1})$. As here~$\p_t \gamma_{ij}=0$, we get~$K_{ij}= D_{(i}\beta_{j)}/\alpha$. The only nonzero components of~$K_{ij}$ are~$K_{r\phi}$ and $K_{\theta\phi}$. As only~$\Gamma^\phi{}_{r \phi}=2/r$ and~$\Gamma^\phi{}_{\theta\phi}=2\cot\theta$ will contribute to the covariant derivatives of~$\beta_i$, we get~$D_r \beta_\phi=\p_r \beta_\phi-\Gamma^\phi{}_{r \phi}=2Ma \sin^2\theta/r^2+4Ma \sin^2\theta/r^2=6Ma \sin^2\theta/r^2$, and $D_\theta\beta_\phi=\p_\theta\beta_\phi-\Gamma^\phi{}_{\theta\phi}=-4Ma \sin\theta\cos\theta/r +4Ma \sin\theta\cos\theta/r =0$. And since $D_\phi\beta_\theta=-\Gamma^\phi{}_{\theta\phi}\beta_\phi=4Ma \sin\theta\cos\theta/r$ and $D_\phi\beta_r =-\Gamma^\phi{}_{r \phi}\beta_\phi=4Ma \sin^2\theta/r^2$, I am not sure if it is the only nonvanishing component.
