Important revisions have been made in "the Extra Bit"
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Relevant material in Latex:
Metric
\begin{equation}c^2d\tau^2=c^2dt^2-dx2-dy^2-dz^2 \end{equation} (1)
\begin{equation}c^2=c^2\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dx}{d\tau}\right)^2-\left(\frac{dy}{d\tau}\right)^2-\left(\frac{dz}{d\tau}\right)^2\end{equation}
\begin{equation}c^2=c^2{v_t}^2-{v_x}^2-{v_y}^2-{v_z}^2\end{equation}(2)
We consider two proper velocities on the same manifold
\begin{equation}c^2=c^2{v_{1t}}^2-{v_{1x}}^2-{v_{1y}}^2-{v_{1z}}^2\end{equation}(3.1)
\begin{equation}c^2=c^2{v_{2t}}^2-{v_{2x}}^2-{v_{2y}}^2-{v_{2z}}^2\end{equation}(3.2)
Adding (3.1) and (3.2) we obtain
\begin{equation}2c^2=c^2\left({v_{1t}}^2+{v_{2t}}^2\right)-\left({v_{1x}}^2+{v_{2x}}^2\right)-\left({v_{1y}}^2+{v_{2y}}^2\right)-\left({v_{1z}}^2+{v_{2z}}^2\right)\end{equation}
\begin{equation}2c^2=c^2\left(v_{1t}+v_{2t}\right)^2-\left(v_{1x}+v_{2x}\right)^2-\left(v_{1y}+v_{2y}\right)^2-\left(v_{1z}+v_{2z}\right)^2-2v_1\dot v_2\end{equation}
\begin{equation}2c^2+2v_1\dot v_2=c^2\left(v_{1t}+v_{2t}\right)^2-\left(v_{1x}+v_{2x}\right)^2-\left(v_{1y}+v_{2y}\right)^2-left(v_{1z}+v_{2z}\right)^2\end{equation}
Since v1.v2>=c^2 we have
\begin{equation}c^2\left({v_{1t}}+{v_{2t}}\right)^2-\left({v_{1x}}+{v_{2 x}}\right)^2-\left({v_{1y}}+{v_{1y}}\right)^2-\left({v_{1z}}+{v_{1z}}\right)^2\ge 4c^2\end{equation}(4)
\begin{equation}\left(v_1+v2)\dot (v_1+v_2)\right) \ge 4c^2\end{equation}(5)
If $v_1+v_2$ is a proper velocity then
\begin{equation}c^2=c^2\left(v_{1t}+v_{2t})\right)^2-\left(v_{1x}+v_{2x})\right)^2-\left(v_{1y}+v_{2y})\right)^2-\left(v_{1z}+v_{2z})\right)^2\end{equation} (6)
\begin{equation}c^2=c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2+ c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2+2v_1.v_2\end{equation}
\begin{equation}c^2=c^2+c^2+2v_1.v_2\end{equation}(7)
Therefore
\begin{equation}v_1.v_2\le -½ c^2\end{equation}(8)
which is not true since
\begin{equation}v.v=c^2\end{equation}
Therefore $$v_1+v_2$$ is not a four vector if $$v_1$$ and $$v_2$$ are four vectors
Again if $$v_1-v_2$$ is a four vector then
\begin{equation}c^2=c^2\left(v_{1t}-v_{2t})\right)^2-\left(v_{1x}-v_{2x})\right)^2-\left(v_{1y}-v_{2y})\right)^2-\left(v_{1z}-v_{2z})\right)^2\end{equation} (9)
\begin{equation}c^2=c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2+ c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2-2v_1.v_2\end{equation}
\begin{equation}c^2=c^2+c^2-2v_1.v_2\end{equation}
\begin{equation} ½ c^2=v_1.v_2\end{equation} (10)
But the above formula is not a valid one. Given two infinitesimally close four velocities their difference is not a four velocity. Therefore the manifold has to be a perforated one. The manifold indeed is a mesh of worldlines and each world line is a train of proper velocity four vectors as tangents. A particle moves along a timelike path and therefore each point on it has a four velocity as a tangent representing the motion. The manifold is discrete and that presents difficulty an impossibility to be precise with procedure like differentiation.