Poincaré group is semi-product of translations and Lorentz-group

[aoner]

Hi everybody,

Can somebody help me with the following proof.
Show that for the Poincaré group

$P=T\odot L$

Where $T$ is the group of translations and $L$ is the Lorentz group and $P$ is the semi-direct product of the two subgroups

I know the axioms for a semi-direct product in this case are

• $T$ is an invariant subgroup of $P$ while $L$ is any subgroup of $P$
• $T\cap L=\{E\}$, where $E$ is the identity of $P$
• For every element $S$ in $P$ we have an element ${{S}_{1}}$ of $T$ and an element ${{S}_{2}}$ of $L$ for which: $S={{S}_{1}}\circ {{S}_{2}}$

How do I proof these axioms?

 

[eljose79]

Thank you for your question. The Poincaré group is a fundamental concept in physics, and understanding its structure is crucial in many areas of research. I will try to provide a clear and concise explanation of the proofs for the axioms of the semi-direct product of the Poincaré group.

First, let's define the Poincaré group as the group of all transformations that leave the laws of physics invariant. This includes translations in space and time (T) and Lorentz transformations (L), which are rotations and boosts in space-time. The semi-direct product of these two subgroups (P) is a way to combine their elements to form a larger group that preserves the structure of the original subgroups.

Axiom 1: T is an invariant subgroup of P while L is any subgroup of P

To prove this axiom, we need to show that the elements of T and L, when combined, still form a valid transformation in P. This means that the composition of two elements from T and L should also be an element in P, and this composition should follow the group operation of P (in this case, composition of transformations). This can be easily shown by considering the composition of two translations or two Lorentz transformations, which will result in another translation or Lorentz transformation, respectively.

Axiom 2: T∩L={E}, where E is the identity of P

This axiom states that the only element common to both T and L is the identity element of P. To prove this, we need to show that there is no other element that belongs to both T and L. Since the elements of T and L are translations and Lorentz transformations, respectively, it is clear that they do not share any common elements except for the identity transformation, which is the composition of the identity translation and the identity Lorentz transformation.

Axiom 3: For every element S in P we have an element S1 of T and an element S2 of L for which: S=S1∘S2

This axiom states that any element in P can be decomposed into a translation and a Lorentz transformation. This can be proven by considering the composition of any element in P with an arbitrary element of T and L. Since T and L are invariant subgroups, the composition will still result in an element of P, and this element can be decomposed into a translation and a