What's the meaning of semidirect product?

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.

• It's the usual decomposition "linear part + translation" which works for the affine group too. – user228113 Dec 26 '17 at 14:25
• so, why we need define the semidirect product ? – Ben Dec 26 '17 at 15:11
• We need to define the semidirect product in order to explain the relation between $SE(3)$, $SO(3)$, and $\mathbb{R}^3$. – 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. $(A_1,b_1)(A_2,b_2)=(A_1A_2,b_1+b_2)$.
• So the $b$ is invariant subgroup, right? – Ben Dec 27 '17 at 9:21
• Yes, the elements of the form $(I,b)$ form a normal subgroup isomorphic to $\mathbb R^3$. – Andreas Cap Dec 27 '17 at 10:04