This is planned as a Q&A session, hopefully it serves people who seek a mathematical foundation to (relatively) known results in standard textbooks on QM or QFT.

Question: What is the topology of the full Poincaré group in 1+3 spacetime dimensions? What about the restricted Poincaré group?


For a generic 4-vector $x^\mu$ in the (1+3) Minkowski spacetime of metric tensor $\eta = \text{diag} (+---)$, a Poincaré transformation $P_4$ acts on $x^\mu$ as $$x'^\mu = P_4 x^\mu = \Lambda^{\mu}_{~\nu} x^{\nu} + a^\mu$$

Here The Topology of the Lorentz group in 1+3 dimensions it is proven that the set of Lorentz transformations $\{\Lambda\}$, by mapping it into $\mathbb R^{16}$ as a vector space and topologizing it with the supremum norm, becomes a topological Lie group. The polar decomposition theorem for the Lorentz group (a glimpse on it one can find here: https://arxiv.org/abs/math-ph/0211047) creates the topological decomposition (i.e. as a manifold, not as a group direct product, which it isn't):

$$(\mathcal L) = \text O (1,3) = \text O (3) \times \mathbb R^3 $$

This gives us a hint on how to look for the topology of the full Poincaré group. First ensure it is a topological group by a proper embedding(*) into $\text{GL}(n,\mathbb R)$. This is done by creating the generic 5x5 matrix

$$ P_5 = \left(\begin{array} \ \Lambda_{4\times 4} \ a_{4\times 1} \\ \bf{0} _{1\times 4} \ 1 \end{array}\right)\in \text{GL}(5,\mathbb R), \Lambda\in\text O(1,3), a\in\text T_4$$ which acts on a generic 5x1 matrix $ A = \left(\begin{array} \ x^\mu \\ 1 \end{array}\right)$.

Endowing this set of matrices with the same supremum norm, it becomes a topological space. One can show that the set of all $P_5$'s becomes a group under matrix multiplication (prove it!) and under the norm it becomes a topological group (the norm ensures the continuity of group operations). A Lie group actually, as it a closed subgroup of $\text{GL}(5,\mathbb R)$. Then we can easily establish the group isomorphism between the set of all Poincaré transformations $P_4$ and the set of all $P_5$ (prove it!).

Therefore, the topology of the Poincaré group in 1+3 dimensions is the topology of the group of matrices $P_5$. This group is immediately seen to be connected, locally connected and non-compact (hence by Schreier's theorem, it admits a simply connected non-compact universal covering group). Moreover, we have the matrix decomposition:

$$ P_5 = \left(\begin{array} \ 1 \ 0 \ 0 \ 0 \ a^0 \\ 0 \ 1 \ 0 \ 0 \ a^1 \\ 0 \ 0 \ 1 \ 0 \ a^2 \\ 0 \ 0 \ 0 \ 1 \ a^3 \\ 0 \ 0\ 0 \ 0 \ 1 \end{array}\right)\left(\begin{array} \ \Lambda_{4\times 4} \ 0_{1\times 4} \\ 0_{4\times 1} \ 1 \end{array}\right) $$

which entails the topological decomposition (homeomorphism):

$$ P_5 = \mathcal P (1,3) = \mathbb R^4 \times \text O (1,3) = \mathbb R^4 \times \text O (3) \times \mathbb R^3. $$

The latter is the topological structure of the set of all Poincaré transformations. Specializing on restricted Lorentz transformations, we have

$$ \left(P_5\right)_{+}^{\uparrow} = \mathcal P_{+}^{\uparrow} (1,3) = \mathbb R^4 \times \text{SO} (3) \times \mathbb R^3. $$

By this topological decomposition we obtain that the restricted Poincaré group is double-connected (its fundamental group has two elements, $\pm \bf{I}_{2\times 2}$), hence its universal covering group $\widetilde{\mathcal{P}_{+}^{\uparrow}} $ is a double cover.

(*) In the comments below, Rod questions if the word "embedding" is properly used (from a diff. geom. point of view) or perhaps "immersion" would have been the right word. Here we are dealing with matrix groups which are very particular Lie groups, i.e. analytic submanifolds of $\mathbb R^n$ for various values of "n". The finesse which distinguishes an immersion (a local mapping) from embedding (global mapping) doesn't exist, because any two topologies on $\mathbb R^n$ are equivalent.

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    $\begingroup$ I haven't had time to read this in detail (I'm not the fastest understander in the West by any means), but some comments are in order: firstly (somewhat pedantically) all Lie groups are topological groups, so "topological Lie group" is a tautology. Do you mean to say that the embedding process bestows a new topology (i.e. that inherited as relative to the topology of $\mathbb{R}^{16}$) on it that is not the Lie group topology? But this would seem to tell against the word "embedding"; I haven't read the paper in detail, but if it is indeed an embedding as opposed to an immersion, then .... $\endgroup$ – WetSavannaAnimal Oct 30 '17 at 0:40
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    $\begingroup$ the Lie group is closed in the superset's topology and the Lie group and relative topologies are the same. Secondly, I've added quite a few words to clear up a mistake I made on reading this the first time: please check that the sense is what you mean. Great idea BTW: I like this kind of Q&A for people to share their efforts to understand a topic; it's a bit like Fermi's famous group paper reading sessions, which my own PhD supervisor emulated. $\endgroup$ – WetSavannaAnimal Oct 30 '17 at 0:42
  • $\begingroup$ The supremum norm is equivalent to other norms one can place on $\text{Mat}_{n\times n} (\mathbb R)$, so the topology of the submanifold can be safely taken to be the subspace (inherited) one. We don't have Klein bottles or Moebius strips here, so everything seems straightforward. $\endgroup$ – DanielC Oct 30 '17 at 9:16
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    $\begingroup$ What I am getting at: is the inherited topology is the same as the Lie group topology, i.e. that generated by a base of sets of the form $\exp(\mathbf{h})$ where $\mathbf{h}\subseteq \mathfrak{p}(1,\,3)$ is a neighborhood in the Lie algebra? This happens whenever the Lie group is closed in $\text{Mat}_{n\times n} (\mathbb R)$. It doesn't happen, for example, for an irrational slope subgroup of the Abelian torus group. This is the kind of thing I mean - one has to be careful with inherited topologies. $\endgroup$ – WetSavannaAnimal Oct 30 '17 at 9:24
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    $\begingroup$ The proof is well known, as long as one can prove the Lie group is closed in the superset's topology. Now I guess that's fairly clear thinking about this some more; it would take a little time to put my hand on / come up with a proof. $\endgroup$ – WetSavannaAnimal Oct 30 '17 at 10:20

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