& = r_1 + r_2 - 2 \langle x_1,y_2 \rangle.
        \end{aligned} 
    \end{spreadlines}$$ This product becomes bilinear after homogenization, and the projective coordinates of $\sigma_{12}$ are $$\label{equation:product}
    \big( \underbrace{h_1 h_2}_{h_{12}}: \underbrace{M_1
    M_2}_{M_{12}}:
    \underbrace{M_2^t x_1 + h_1 x_2}_{x_{12}}: \underbrace{h_2 y_1
    + M_1 y_2}_{y_{12}}: \underbrace{h_2 r_1 + h_1 r_2 - 2\langle
    x_1,y_2\rangle}_{r_{12}} \big).$$ (The matrices and vectors appearing in the above coordinates should be replaced by their entries.)

\[proposition:product\] Let $(h_1:M_1:x_1:y_1:r_1),(h_2:M_2:x_2:y_2:r_2) \in X$. Then the above product is defined if at least one of $h_1$, $h_2$ is not equal to zero.

Assume that the product is undefined, which means that all entries are zero. In particular, $h_1 h_2 = 0$. We assume $h_1 = 0$ (the other case $h_2 = 0$ can be treated analogously). Assume, indirectly, that $h_2\ne 0$. Since $\det(M_2) = h_2^3$, it follows that $M_2$ is invertible. Since $M_1 M_2 = 0$, it follows that $M_1=0$. Since $M_2^t x_1 + h_1 x_2=0$, it follows that $x_1 = 0$. Since $h_2 r_1 + h_1 r_2 - 2 \langle x_1,
    y_2 \rangle = 0$, it follows that $r_1 = 0$. Then all coordinates of the first element vanish, a contradiction.

Since the $h$–coordinate of any element $\sigma \in \Phi({\mathrm{SE}_3})$ is always different 