from zero, then Proposition \[proposition:product\] ensures that left and right multiplication by $\sigma$, which a priori are maps from $\Phi({\mathrm{SE}_3})$ to itself, extend to linear maps $X \longrightarrow X$.

\[remark:action\] We specialize Equation to the cases of left and right multiplication by translations or rotations along the origin. We fix $\sigma \in X$ and we suppose that it has coordinates $\sigma = (h: M: x: y:
r)$.

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-   Given a vector $s \in {\mathbb{R}}^3$, the translation by $s$ gives the following element $\sigma' \in \Phi({\mathrm{SE}_3})$: $$\sigma' = \big( 1: \mathrm{id}: -s: s: \langle s, s \rangle \big).$$ Then left multiplication by $\sigma'$ provides $$\sigma' {\thinspace}\sigma \; = \; \big( h: M: -M^t s + x: hs + y: h {\thinspace}\langle s, s \rangle + r + 2 \langle s, y \rangle \big),$$ while right multiplication by $\sigma'$ provides $$\sigma {\thinspace}\sigma' \; = \; \big( h: M: x - hs: y + Ms: r + h {\thinspace}\langle
                s, s \rangle - 2 \langle x, s \rangle \big).$$

-   Given an orthogonal matrix $M' \in {\mathrm{SO}_3}$, the rotation around the origin by $M'$ gives the following element $\sigma' \in \Phi({\mathrm{SE}_3})$: $$\sigma' \; = \; (1: M: 0: 0: 0).$$ Then left 