Poincaré group as a topology and a Lie group manifold After some thinking on how to make a topology of the Poincaré group Am not convinced on how it turns out to be in my head : i was thinking about making balls $B_r (w) $ but didn't get much of that, because am not used to this way of thinking. (Butt i reeaaally want to know how to do it).
So i skipped the topology question and switched to the manifold one, I thought of making at first a set $$P(1,3)= \{(L,a)|L\in SO (1,3),a\in R^{3,1}  \}$$ and then an operation $$(L_1,a_1)(L_2,a_2)=(L_2L_1,L_2a_1+a_2)$$
and an inverse $$(L,a)^{-1} =(L^{-1},-L^{-1}a)$$
so the idea was to search for an atlas which i didnt know how to construct, and i find it complex to think of the Poincaré group as two groups.
I also think that my definition of the set is ill defined i think it needs the orthogonality equation of the SO (1,3).
i hope you can help me on this one or give me a book or a YT lecture so i can develop this idea in secure.
 A: We can see that as a set the Poincare group $P=SO(1,3)\rtimes \mathbb{R}^{1,3}$ is the cartesian product $SO(1,3)\times \mathbb{R}^4$. As a topological space, $SO(1,3)$ inherits its topology from $GL(4,\mathbb{R})\subset \mathbb{R}^4$ by the subspace topology; that is $\mathcal{O}_{SO(1,3)}=\{ U\cap SO(1,3)| U\in \mathcal{O}_{GL(4)}\}$, where $\mathcal{O}_{GL(4)}$ is defined using the subspace topology inherited from $\mathbb{R}^{4\times 4}$ defined similarly.
We can endow $P$ with a smooth structure by endowing each of its factors $SO(1,3)$ and $\mathbb{R}^4$ with smooth atlases. The closed-subgroup theorem says that if $H\subset G$ is a (topologically) closed subgroup of a Lie group $G$, then $H$ is a Lie group (and hence a smooth manifold). So for our case, it suffices to show that $SO(1,3)$ is a closed subset of $GL(4)$.
Let $(\Lambda_n)$ be a sequence of matrices in $SO(1,3)$ that converge to some matrix $\Lambda\in GL(4)$. For each $n$, $\Lambda_n^T \eta \Lambda_n=\eta$. Since matrix multiplication and the transpose is continuous, $$\lim_{n\to\infty}(\Lambda_n^T\eta \Lambda_n)=(\lim_{n\to\infty} \Lambda_n)^T\eta(\lim_{n\to\infty}\Lambda_n)=\Lambda^T\eta\Lambda$$
Because each of the $\Lambda_n$'s are in $SO(1,3)$, $\Lambda_n^T\eta\Lambda_n=\eta$ for each $n$ and hence $$\lim_{n\to\infty}(\Lambda_n^T\eta \Lambda_n)=\lim_{n\to\infty}\eta=\eta=\Lambda^T\eta\Lambda$$
Hence $\Lambda\in SO(1,3)$. Since any set subset of $\mathbb{R}^n$ which contains all of its limit points is closed, $SO(1,3)$ is a closed subset of $\mathbb{R}^{4\times 4}$ and therefore a closed subset of $GL(4)$. The closed subgroup theorem then guarantees that $SO(1,3)$ is a Lie subgroup of $GL(4)$ and hence a smooth manifold.
