Special Euclidean group:

$\rm SE(3)=SO(3)\rtimes \mathbb{R}^3$

How to explain this expression of $\rm SE(3)$, about rigid body workspace.

  • $\begingroup$ It's the usual decomposition "linear part + translation" which works for the affine group too. $\endgroup$ – user228113 Dec 26 '17 at 14:25
  • $\begingroup$ so, why we need define the semidirect product ? $\endgroup$ – Ben Dec 26 '17 at 15:11
  • $\begingroup$ We need to define the semidirect product in order to explain the relation between $SE(3)$, $SO(3)$, and $\mathbb{R}^3$. $\endgroup$ – Lee Mosher Dec 26 '17 at 16:08

Let us represent the map $x\mapsto Ax+b$ by the pair $(A,b)$. Then composing the maps corresponding to $(A_1,b_1)$ with the one corresponding to $(A_2,b_2)$, you get the map corresponding to $(A_1A_2,b_1+A_1b_2)$. This is the way how multiplication in a semi-direct product works, whereas in a direct product, you would simple have component-wise multiplication, i.e. $(A_1,b_1)(A_2,b_2)=(A_1A_2,b_1+b_2)$.

  • $\begingroup$ very clearly, thx $\endgroup$ – Ben Dec 27 '17 at 9:17
  • $\begingroup$ So the $b$ is invariant subgroup, right? $\endgroup$ – Ben Dec 27 '17 at 9:21
  • $\begingroup$ Yes, the elements of the form $(I,b)$ form a normal subgroup isomorphic to $\mathbb R^3$. $\endgroup$ – Andreas Cap Dec 27 '17 at 10:04

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