group-theorylie-algebraspecial-relativity
It can be shown easily, by introducing new generators from the usual ones that we can think of the Lie algebra of the Lorentz group as being built up by two copies of the $SU(2)$ Lie algebra:
$$ \mathfrak{so}(3,1) \cong \mathfrak{su}(2) \oplus \mathfrak{su}(2) $$
The Poincare group is a semidirect product of the translations and the Lorentz group.
Is there a similar relation for the Lie algebra of the Poincare group?
Preliminary remarks.
As Danu writes in his comment, the physics of the other four generators has to do with spacetime translations, one for each spatial direction, and one for time. But how do we see this explicitly in the math behind the somewhat odd-looking presentation of the Poincare group and its Lie algebra that Hall discusses.
First, recall that any $d+1$-dimensional Lorentz transformation is a Linear transformation on $\mathbb R^{d+1}$, so it can be representation by multiplication by a $(d+1)\times(d+1)$ matrix $\Lambda$.
Second, and most crucially, recall that translations of $\mathbb R^{d+1}$ are not linear transformations; there is no way to write spacetime translation as multiplication by a matrix.
However, here's the really cool thing. If we embed a copy of $(d+1)$-dimensional spacetime into the vector space $\mathbb R^{d+2}$, namely into a space with one higher dimension, then we can implement translations as linear transformations. Here's how it works.
The main construction.
For each $x\in \mathbb R^{d+1}$, we associate an element of $\mathbb R^{d+2}$ as follows: \begin{align} x \mapsto \begin{pmatrix} x \\ 1 \\ \end{pmatrix} \end{align} Now, for each Lorentz transformation $\Lambda\in \mathrm O(d,1)$, and for each spacetime translation characterized by a vector $a\in\mathbb R^{d+1}$, we form the matrix \begin{align} \begin{pmatrix} \Lambda & a \\ 0 & 1 \\ \end{pmatrix} \tag{$\star$} \end{align} and we notice what this matrix does to the embedded copies of points in $\mathbb R^{d+1}$; \begin{align} \begin{pmatrix} \Lambda & a \\ 0 & 1 \\ \end{pmatrix} \begin{pmatrix} x \\ 1 \\ \end{pmatrix} = \begin{pmatrix} \Lambda x+a \\ 1 \\ \end{pmatrix} \end{align} Whoah! That's really cool! What has happened here is that when we augment the dimension of spacetime by one with the embedding give above, and when we correspondingly embed Lorentz transformations and translations appropriately into square matrices of dimension $d+2$, then we actually do get a way of representing both Lorentz transformations and translations as linear transformations on $\mathbb R^{d+2}$ that act in precisely the correct way on the copy of Minkowski space embedded in $\mathbb R^{d+2}$!
In other words, the Poincare group in $d+1$ dimensions can be thought of as the set of all $(d+2)\times(d+2)$ matrices of the form $(\star)$ where $\Lambda\in \mathrm O(d,1)$ and $a\in\mathbb R^{d+1}$.
What about the Lie algebra?
A natural question then arises: what do the Lie algebra elements look like as matrices when we represent the Lie group elements this way? Well, I quick standard computation will show you that the Lie algebra of the Poincare group can, in this representation, be regarded as all matrices of the form \begin{align} \begin{pmatrix} X & \epsilon \\ 0 & 0 \\ \end{pmatrix} \tag{$\star$} \end{align} where $X\in\mathfrak{so}(d,1)$ and $\epsilon\in \mathbb R^{d+1}$, precisely as Hall indicates. But from the remarks above, we see clearly that the parameter $\epsilon$ precisely corresponds to the generators that span the subspace of the Poincare algebra that yield spacetime translations.
Unfortunately you can't just define the Poincare group to be that, because in the standard treatment it is defined a little bit differently. What you defined is actually the Lorentz group. The Poincare group contains translations as well.
The Lorentz group $O(1,3)$ is the group of all $\Lambda \in GL(4, \mathbb{R})$ such that
$$\Lambda^T \eta \Lambda = \eta,$$
with $\eta = \operatorname{diag}(-1,1,1,1)$. This can be seen as the group of all "changes of orthonormal frames in spacetime".
Remember that one orthonormal reference frame is a set of vectors $\{e_\mu\}$ such that $g(e_\mu,e_\nu)=\eta_{\mu\nu}$. In that sense, given two such frames, the change of frame that takes components in one of them to components in the other is given by these elements.
With this, the proper Lorentz group is $SO(1,3)$ which basically means that you pick all elements of $O(1,3)$ with determinant $+1$. The orthocronous part just means that if $\Lambda \in O(1,3)$ has $\Lambda^0_0 > 0$ then it preserves the sense of time-like vectors.
Some authors seems to include "by default" the orthocronous requirement in the group $SO(1,3)$ (see for example Analysis, Manifolds and Physics by Choquet-Bruhat, vol. 1, page 290). Others leave it separately, so that you end up with a group $SO^+(1,3)$, but this is a question of convention.
Now the Poincare group $P(1,3)$ (which I don't know any standard notation for) is the group of all Lorentz transformations together with all spacetime translations. In other words, we have:
$$P(1,3)= \{(a,\Lambda) : a\in \mathbb{R}^4, \Lambda \in SO(1,3)\}$$
together with the multiplication defined by
$$(a_1,\Lambda_1)\cdot (a_2,\Lambda_2)=(a_1+\Lambda_1 a_2, \Lambda_1\Lambda_2)$$
Think like that: while elements of $SO(1,3)$ relates orthonormal frames with coincident origins, elements of $P(1,3)$ also allows for the shift of origin as well.
The action of $SO(1,3)$ in the Minkowski vector space $\mathbb{R}^{1,3}$ (not to be confused with flat spacetime - this is actually the "model" for spacetime's tangent spaces, which just happens to be possible to identify with spacetime itself in the flat case) is given by usual matrix multiplication, i.e., given $ \Lambda \in SO(1,3)$ you have:
$$\Lambda \cdot v = \Lambda v, \quad \forall v\in \mathbb{R}^{1,3}.$$
On the other hand, the action of $P(1,3)$ on the Minkowski vector space $\mathbb{R}^{1,3}$ is characterized by the fact that given $(a,\Lambda)\in P(1,3)$ you have:
$$(a,\Lambda)\cdot v = a + \Lambda v, \quad \forall v\in \mathbb{R}^{1,3}.$$
I don't know if it helps, but people like to compare this to the case in $\mathbb{R}^3$ where you have the rotation group $SO(3)$ and the Euclidean group $E(3)$ comprising rotations in $SO(3)$ with translations in $\mathbb{R}^3$ thus forming the group of rigid motions. This could be seen as the analogous construction in spacetime.
EDIT: Regarding the semi-direct product construction mentioned in comments, recall that given groups $N,H$ with $\varphi : H\to \operatorname{Aut}(N)$ a homomorphism into the group of automorphisms of $N$, we can build the semi-direct product as the set $N\times H$ with the product:
$$(a,b)\cdot (c,d)=(a\varphi(b)(c), bd)$$
the resulting group is denoted $N\rtimes H$. In the particular case it is clear that we have this construction with $N = \mathbb{R}^{1,3}$, $H = SO(1,3)$ and $\varphi : SO(1,3)\to \operatorname{Aut}(\mathbb{R}^{1,3})$ given by $\varphi(\Lambda)(v) = \Lambda v$. Thus
$$P(1,3)=\mathbb{R}^{1,3}\rtimes SO(1,3)$$
Best Answer
Well, as it is noted in the first comment, it is not true that the Lorentz group algebra is isomorphic to the vector space sum of two su(2) algebras, but the complexification of the Lorentz algebra is isomorphic to the vector space sum of two copies of the sl(2) algebra seen as a vector space over $\mathbb C$. Mathematically:
$$ \mathfrak{lor}^{\mathbb C} (1,3) \simeq \mathfrak{sl}_\mathbb{C} (2,\mathbb C) \oplus \mathfrak{sl}_\mathbb{C} (2,\mathbb C) $$
Moving to the Poincaré group, there is no corresponding relation, because the Poincaré algebra, unlike the Lorentz one, is not semisimple, it has a non-trivial abelian subalgebra, namely, the algebra of 4-translations. So any Cartan or Iwasawa decomposition of the algebra does not exist. One can still have a Levi decomposition as in the answer by Qmechanic below.