1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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. $(A_1,b_1)(A_2,b_2)=(A_1A_2,b_1+b_2)$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.