Where does the relative velocity composition law equation come from? I'm trying to understand this paper (equation $2.8$ specifically):
Bini, D., Carini, P., & Jantzen, R. T. ($1995$). Relative observer kinematics in general relativity. Classical and Quantum Gravity

one wishes to express the relative velocity $\nu(U,u)$ of $U$ with respect to the first
observer in terms of the one $\nu(U,u')$ with respect to the second observer, and the relative
velocity $\nu(u',,u)$ or $\nu(u,u')$ between the two observers.

where $u$ is a $4$-velocity, $u'$ is another $4$-velocity, the relative velocity $ν(u',u)$ (of $u'$ with respect to $u$), the gamma factor is $\gamma(u',u) = u \cdot u'$, $P(u)$ is the projection orthogonal to $u$ in each tangent space (spatial projection with respect to $u$) and the relative projection operator is given by $P(u,u') = P(u)P(u')$
How does one derive or cross verify the above equation? In particular:
$$ \nu(U,u') = - [ u \cdot \nu(U,u') + P(u,u') \nu(U,u')]  $$
My Attempt
From $(2.1c)$ and $(2.1d)$ we have:
$$\frac{U}{\gamma (U,u')} = \gamma(u',u) (u + \nu(u',u)) + \nu(U,u') $$
From a variation of $2.2$ we know:
$$ \frac{P(u')}{\gamma(u',U)} U = \nu(U,u')$$
a
Applying the projector operator $P(u')$:
$$ \nu(U,u') = P(u') \gamma(u',u) (u + \nu(u',u)) + \nu(U,u') $$
which is nothing like their expression.
 A: From my answer ( https://physics.stackexchange.com/a/723077/148184 ) to your earlier question ( Understanding where equation $2.3$ comes from? ) here are some possibly useful relations:

*

*(essentially 2.2) $P_{(u)}{}^a{}_b \stackrel{\tiny (-+++)}{\equiv} (\delta^a{}_b + u^a u_b)$ implies $$\delta^a{}_b \stackrel{\tiny (-+++)}{\equiv} ( - u^a u_b) +P_{(u)}{}^a{}_b  $$

*(2.1c) $u'=u'_{\| u}+u'_{\perp u}= \gamma[ u + V_{(u',u)} ]$ implies that
$$V_{(u',{\color{red}u})} \perp {\color{red}u} \qquad\mbox{note the second argument is ${\color{red}u}$}$$

*Thus, $$P_{({\color{red}u})}V_{(u',{\color{red}u})}=V_{(u',{\color{red}u})}$$
and similarly
$$P_{(u')}V_{(u,u')}=V_{(u,u')}$$
but
$$P_{(u)}V_{(u,u')}=-\gamma V_{(u',u)}$$
and similarly
$$P_{(u')}V_{(u',u)}=-\gamma V_{(u,u')}.$$
So, following the top of p. 2553,
to get $V_{(U,u')}$, either

*

*use the $U$-equation in terms of $u'$ (2.1b), solve for $V_{(U,u')}$.
to get $$V_{(U,u')}=\frac{1}{\gamma} U -u'$$
but this doesn't seem to make progress toward his equations.

*or, instead, use the projection property
$$V_{(U,u')}=P_{(u')}V_{(U,u')}$$
which we write in index-notation as
$$V_{(U,u')}^a=P_{(u')}{}^a{}_bV_{(U,u')}^b$$
then use $\delta^a{}_b \stackrel{\tiny (-+++)}{\equiv} ( - u^a u_b) +P_{(u)}
{}^a{}_b  $ and the projection property
\begin{align}
V_{(U,u')}^a 
&=\delta^a{}_c P_{(u')}{}^c{}_b V_{(U,u')}^b \\
&={\color{red}{(-u^a{}u_c +P_{(u)}{}^a{}_c)}} P_{(u')}{}^c{}_b V_{(U,u')}^b\\
&=-u^a{}u_c P_{(u')}{}^c{}_b V_{(U,u')}^b  +P_{(u)}{}^a{}_c P_{(u')}{}^c{}_b 
V_{(U,u')}^b\\
&=-u^a{}u_c  V_{(U,u')}^c \qquad\ +P_{(u)}{}^a{}_c P_{(u')}{}^c{}_b 
V_{(U,u')}^b\\
&=-u^a{}[u_c  V_{(U,u')}^c] \!\! \qquad + \quad P_{(u,u')}{}^a{}_b 
V_{(U,u')}^b
\end{align}
to get his first line.
It was the use of $P_{(u,u')}$ in the first line that led me in this direction!!
Next, using (2.1d)
\begin{align}
&=-u^a{}[u_c  \qquad\ \qquad V_{(U,u')}^c]  +P_{(u,u')}{}^a{}_b 
V_{(U,u')}^b\\
&=
-u^a{}[ \color{red}{\gamma (u'_c+V_{(u,u')}{}_c)}  V_{(U,u')}^c]  +P_{(u,u')}{}^a{}_b 
V_{(U,u')}^b \qquad\mbox{gets his second line.}
\end{align}
Finally, by distributing
\begin{align}
&=
-u^a{}[ \color{red}{\gamma u'_c}V_{(U,u')}^c+ \color{red}{\gamma V_{(u,u')}{}_c}  V_{(U,u')}^c)]  +P_{(u,u')}{}^a{}_b 
V_{(U,u')}^b\\
&=
-u^a{}[ {\qquad 0\quad}+ \color{red}{\gamma V_{(u,u')}{}_c}  V_{(U,u')}^c)]  +P_{(u,u')}{}^a{}_b 
V_{(U,u')}^b
\end{align}
we get his third line, Eq. (2.8).
