# Intuition Behind Semidirect Product Decomposition of Poincare Group

I want to consider the Poincare group $\text{ISO}(1,3)$ as the symmetry group of four-dimensional Minkowski spacetime $\mathbb{R}^{1,3}$. Within the Poincare group, there is a normal subgroup isomorphic to $\mathbb{R}^{1,3}$ consisting of translations. The Lorentz transformations, which here I will call $\mathcal{L}$ also constitute a subgroup of $\text{ISO}(1,3)$, and we hence, we get the semidirect decomposition

$$\text{ISO}(1,3) \cong \mathbb{R}^{1,3} \rtimes_{\varphi} \mathcal{L},$$

where $\varphi: \mathcal{L} \to \text{Aut}(\mathbb{R}^{1,3})$ is action by conjugation. I believe the group operation on the semidirect product is given as follows,

$$(n_{1}, k_{1}) \cdot (n_{2}, k_{2}) = (n_{1} + k_{1} n_{2} k_{1}^{-1}, k_{1}\cdot k_{2}).$$

Please correct me if I'm wrong about anything above, but I've used "$+$" as the group operation on the translations component and "$\cdot$" for composing Lorents transformations.

My question quite simply is: what is the physical interpretation in terms of elementary ideas of the "twisting" taking place in the semidirect product? More specifically, can someone help me appreciate why $n_{1} + k_{1} n_{2} k_{1}^{-1}$ arises instead of $n_{1} + n_{2}$?

• Consider two consecutive Poincaré transformations of the form $x\to kx+n$. – AccidentalFourierTransform Oct 19 '17 at 20:58
• This is simplest to appreciate in the Euclidean case. The first transformation not only translates but also rotates the original vector. Thus, when the second transformation is done, it acts on this already rotated (and translated) vector. It's basically the same argument for Poincare. – ZeroTheHero Oct 19 '17 at 21:02
• @ZeroTheHero that should be an answer ;-P – AccidentalFourierTransform Oct 19 '17 at 21:06
• @AccidentalFourierTransform right... I converted to an answer after all. – ZeroTheHero Oct 19 '17 at 21:12

In the case of the Euclidean group in the plane $E(2)$ (just because the algebra is easy), a general element is of the form $$(R(\theta),x,y)=(1,x,y)\cdot (R(\theta),0,0)$$ where $(R(\theta),0,0)$ is a pure rotation in the place and $(1,x,y)$ is a pure translation takes a vector $(a,b)$ to $(a+x,b+y)$. Thus, \begin{align} (R(\theta_1),x_1,y_1)\cdot(R(\theta_2),x_2,y_2) =(1,x_1,y_1)\cdot (R(\theta_1),0,0)\cdot (1,x_1,y_1)\cdot (R(\theta_1),0,0) \end{align} Acting on the vector $(a,b)$, this gives \begin{align} (R(\theta_1),x_1,y_1)\cdot(R(\theta_2),x_2,y_2)(a,b)&= (1,x_1,y_1)\cdot (R(\theta_1),0,0)\cdot (1,x_2,y_2)(a \cos\theta_2 ,b\cos\theta )\, ,\\ &= (1,x_1,y_1)\cdot (R(\theta_1),0,0)\cdot (a \cos\theta_2+x_2 ,b\cos\theta_2+y_2 )\tag{1} \end{align} This makes it clear that a $(R(\theta_1),x_1,y_1)$ sees the vector $(a \cos\theta_2+x_2 ,b\cos\theta_2+y_2 )$, which is a translation of the rotated original vector.
We can continue from (1) to get \begin{align} &(1,x_1,y_1)\cdot (R(\theta_1),0,0)\cdot (a \cos\theta_2+x_2 ,b\cos\theta_2+y_2 )\\ &\quad = (1,x_1,y_1)\cdot ( (a \cos\theta_2+x_2)\cos\theta_1, (b\cos\theta_2+y_2)\sin\theta_1) \end{align} clearly again showing that the effect of the second rotation is related to the rotated and translated initial vector. The final net translation of the original vector $(a,b)$ includes the effect of the first rotation through the factors $x_2\cos\theta_1$ and $y_2\sin\theta_1$.