1} \,,\end{aligned}$$ where ${\bf 1}$ is the unit matrix. Taking the derivative of this, we have the relation between the derivative of the deviation matrix and $\stackrel{\odot}J{}^i$, $$\begin{aligned}
D^i\Psi=M_{[1]}\stackrel{\odot}J{}^iM_{[0]}^{-1} \,, 
\label{eq:deriv}\end{aligned}$$ where $D_i$ is a covariant derivative associated with the abstract three-metric $\gamma$. Taking, further, the divergence of the above formula and also the trace of the matrix elements, we have the following divergence identity $$\begin{aligned}
D_i D^i {\rm tr} \Psi 
 = {\rm tr} \left( \stackrel{\odot}J{}^{Ti} M_{[1]}\stackrel{\odot}J{}^iM_{[0]}^{-1} \right) \,,   
\label{id:global-divergence}\end{aligned}$$ where we have also used the conservation equation $D_i J{}^i =0$. Then, integrating this divergence identity over the region $\Sigma=\{(\rho,z)|\rho\ge 0,\ -\infty<z<\infty \}$, we obtain the Mazur identity, $$\begin{aligned}
\int_{\partial \Sigma}\rho \partial_p {\rm tr} \Psi dS^p 
= \int_{\Sigma}\rho \hat h_{pq}{\rm tr} 
 ({\cal M}^{Tp} \: {\cal M}^q)d\rho dz \,, 
\label{eq:id} \end{aligned}$$ where $\hat h_{pq}$ is the two-dimensional flat metric $$\begin{aligned}
\hat h=d\rho^2+dz^2 \,, \end{aligned}$$ and the matrix ${\cal M}$ is defined by $$\begin{aligned}
{\cal M}^p=\hat g_{[0]}^{-1}\: \stackrel{\odot}J{}^{Tp} \: \hat g_{[1]} \,. \end{aligned}$$

Now we note that the right-hand side of the identity, (\[eq:id\]), is non-negative. Therefore, if we impose the boundary conditions at $\partial\Sigma$, under which the left-hand side of Eq.(\[eq:id\]) vanishes, then we must have $\stackrel{\odot}J{}^i=0$. In that case, it 