Understanding Lorentz Groups and some key subgroups

[JD_PM]

TL;DR

I'd like to gain more insight on the Lorentz Group, its most important subgroups and must-know examples in Physics. At the moment I only know a handful: the orthogonal group $O(3)$, the orthogonal group $SO(3)$ and the proper orthochronous Lorentz group $SO^{\uparrow}(1,3)$

This thread is motivated by samalkhaiat's comment here

That is neither continuous nor connected Lorentz transformation. It is a discrete space-time reversal $(x^{0} , x^{i}) \to (-x^{0} , -x^{i})$. Space reflection $(x^{0} , x^{i}) \to (x^{0} , -x^{i})$; time reversal $(x^{0} , x^{i}) \to (-x^{0} , x^{i})$ and space-time reversal form disjoint subsets and are not continuously connected to the identity. In English, $x \to – x$ does not belong to the proper orthochronous Lorentz group $SO^{\uparrow}(1,3)$. By the “Lorentz group”, we always mean the real semi-simple Lie group $SO^{\uparrow}(1,3)$.

I know that the Lorentz Group is formed by all matrices that satisfy

$$\eta = \Lambda^{T} \eta \Lambda \tag{1.1}$$

Which is equivalent to

$$\eta_{\mu\nu}\Lambda^{\mu}{}_{\rho}\Lambda^{\nu}{}_{\sigma} = \eta_{\rho \sigma} \tag{1.2}$$

If we add no more restrictions and define the group under the inner product we end up with the group $O(1,3)$.

I also know that all orthogonal matrices must satisfy $1_3 =R^T 1_3 R$. They include not only rotational matrices but also space and time reversals, also known as parity transformations (i.e. $(x^{0} , x^{i}) \to (x^{0} , -x^{i})$) and $(x^{0} , x^{i}) \to (-x^{0} , x^{i})$ respectively)

I have some questions:

1) Is the Lorentz group always defined under the inner product? The only examples I know are; for instance: the orthogonal group $O(3)$, the orthogonal group $SO(3)$ and the proper orthochronous Lorentz group $SO^{\uparrow}(1,3)$

2) I've read that $O(1,3)$ has one positive and one negative eigenvalue of its defining symmetric matrix. I do not see why, how could we prove it?

3)

$x \to – x$ does not belong to the proper orthochronous Lorentz group $SO^{\uparrow}(1,3)$. By the “Lorentz group”, we always mean the real semi-simple Lie group $SO^{\uparrow}(1,3)$.

If I am not mistaken, this is because $SO^{\uparrow}(1,3)$ requires matrices to fulfil not only $1_3 =R^T 1_3 R$ condition but also that its determinant must be 1. The later condition is not fulfilled by matrices representing parity transformations (spatial, time and space-time reversals), but how can I prove this?Sources:

SpaceTime & Geometry by Carroll, pages 12,13,14

vanhees71 QFT manuscript, section 3.1.
Any help is appreciated.

Thank you

 

[PeroK]

If I am not mistaken, this is because $SO^{\uparrow}(1,3)$ requires matrices to fulfil not only $1_3 =R^T 1_3 R$ condition but also that its determinant must be 1. The later condition is not fulfilled by matrices representing parity transformations (spatial, time and space-time reversals), but how can I prove this?

Just an observation that the determinant is a continuous mapping from the set of all matrices to the set of real numbers. You cannot smoothly map a matrix with determinant $1$ to a matrix with determinant $-1$, without going through all other values (intermediate value theorem).

And, in fact, the existence of a continuous function from a set into a two-point set $\{-1, 1 \}$ is the definition of a disconnected set. Hence $O(3)$ comprises two disconnected subsets.

 

[JD_PM]

Hi PeroK

Just an observation that the determinant is a continuous mapping from the set of all matrices to the set of real numbers.

Please let me write down the mathematical definition of the determinant for future copy-paste reference

$$f:\Re^{n \times n} \rightarrow \Re : A = \Big(
\begin{pmatrix}
R_1 \\
. \\
. \\
R_n
\end{pmatrix}\Big) \mapsto f(A)$$

You cannot smoothly map a matrix with determinant $1$ to a matrix with determinant $-1$, without going through all other values (intermediate value theorem).

And, in fact, the existence of a continuous function from a set into a two-point set $\{-1, 1 \}$ is the definition of a disconnected set. Hence $O(3)$ comprises two disconnected subsets.

Oh so space reflections $(x^{0} , x^{i}) \to (x^{0} , -x^{i})$ and time reversals $(x^{0} , x^{i}) \to (-x^{0} , x^{i})$ are two disconnected subsets of $O(3)$? Then, do disconnected and disjoint mean the same?

 

[PeroK]

Then, do disconnected and disjoint mean the same?

Disjoint refers to any two sets having no members in common. Disconnected-ness is a property of a topological space, which roughly means that it is not in one piece. For example, the real numbers can be expressed as the union of two disjoint sets.

$\mathbb{R} = (-\infty, 0] \cup (0, +\infty)$

But, $\mathbb{R}$ is connected (with the usual topology).

A topological space, $S$, is disconnecetd iff there exists a continuous function from $S$ onto a two-point set.

For example, the set $S = (-\infty, 0) \cup (0, +\infty)$ is disconnected. You can see this as the function $f: S \rightarrow \{0, 1\}$:

$f(x) = 0 \ (x < 0)$, and $f(x) = 1 \ (x > 0)$ is continuous (despite what some people may say!).

PS $O(3)$ is disconnected. Using the properties of the determinant gives you the easiest proof.