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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 Lorentz 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 $n_{1} + n_{2}$?

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    $\begingroup$ Consider two consecutive Poincaré transformations of the form $x\to kx+n$. $\endgroup$ Oct 19, 2017 at 20:58
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    $\begingroup$ 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. $\endgroup$ Oct 19, 2017 at 21:02
  • $\begingroup$ @ZeroTheHero that should be an answer ;-P $\endgroup$ Oct 19, 2017 at 21:06
  • $\begingroup$ @AccidentalFourierTransform right... I converted to an answer after all. $\endgroup$ Oct 19, 2017 at 21:12

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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.

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 $y_2\sin\theta_1$.

It's basically the same argument for Poincare, except that the algebra is more complicated.

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  • $\begingroup$ Thanks a lot! At first, I was for some reason thinking this twisting phenomena was related to the elementary principles of relativity and then I just recalled even in the classical setting you have a semidirect decomposition of the Euclidean group, as you say. But this clears up why exactly this arises. Thanks again. $\endgroup$
    – Benighted
    Oct 19, 2017 at 22:39
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    $\begingroup$ @StephenPietromonaco yes this kind of structure holds for any semi-direct product. $\endgroup$ Oct 19, 2017 at 23:17
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Well, using your definition (never mind what any of it means for now): several conclusions can be drawn. First, let $1$ denote the identity of the $k$ group, and $(k,k') ↦ kk'$ the product operation for that group. Second, let $0$ denote the identity of the $n$ group, and $(n,n') ↦ n + n'$ its product operation, which is written as a sum because (apparently) we're assuming the $n$ group to be Abelian. We're also assuming that the $k$ group acts on both sides of the $n$ group as a conjugate representation, and that this action is denoted algebraically in the obvious way; so we have the properties that $$(k'k)n(k'k)^{-1} = k'(knk^{-1}){k'}^{-1}, \hspace 1em 1n1^{-1} = n,$$ as well as linearity, $$k0k^{-1} = 0, \hspace 1em k(n + n')k^{-1} = knk^{-1} + kn'k^{-1}.$$

First, we note that from the definition you posed, the following are properties of the combined group: $$ (n, 1)·(0, k) = (n, k), \hspace 1em (0, k)·(n, 1) = (knk^{-1}, 1)·(0, k),\\ (n, 1)·(n', 1) = (n + n', 1), \hspace 1em (0, k)·(0, k') = (0, kk'). $$ Note, also that the conjugate action actually becomes a conjugate action, now: $$(knk^{-1},1) = (0, k)·(n, 1)·{(0, k)}^{-1}.$$

Next, we can equivalently treat the combined group as generated from the disjoint union of the $n$ and $k$ groups by punning each $n$ as $(n, 1)$, and each $k$ as $(0, k)$, thereby allowing us to rewrite the fundamental properties above as: $$ n·k = (n, k), \hspace 1em k·n = knk^{-1}·k, \hspace 1em n·n' = n + n', \hspace 1em k·k' = kk', \hspace 1em knk^{-1} = k·n·k^{-1}. $$ So, you can see - when rendered this way in algebraic form - that the twisting condition is just a form of associativity, in disguise, with the conjugate action now actually being written as such.

Taking these, instead, as the fundamental relations and applying them to your product, we get $$\begin{align} (n, k)·(n', k') &= n·k·n'·k' \\ &= n·k·n'·k^{-1}·k·k' \\ &= n·kn'k^{-1}·k·k' \\ &= (n + kn'k^{-1})·kk' \\ &= (n + kn'k^{-1}, kk'). \end{align}$$

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