# Understanding Lorentz Groups and some key subgroups

• I

## Summary:

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 $SO^{\uparrow}(1,3)$

## Main Question or Discussion Point

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

etotheipi

PeroK
Homework Helper
Gold Member
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
Homework Helper
Gold Member
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.

JD_PM