Undergrad Understanding Lorentz Groups and some key subgroups

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The discussion centers on the properties of the Lorentz group, specifically the proper orthochronous Lorentz group SO^{\uparrow}(1,3) and its relationship with transformations like space and time reversals. It emphasizes that transformations such as x \to -x do not belong to SO^{\uparrow}(1,3) due to the requirement that matrices must have a determinant of 1, which parity transformations do not satisfy. The thread also clarifies the distinction between disconnected and disjoint sets, noting that while O(3) comprises two disconnected subsets, they are not the same as being disjoint. The continuity of the determinant function is highlighted as a key reason for the disconnected nature of certain transformation groups. Overall, the conversation deepens the understanding of the mathematical structure underlying Lorentz transformations.
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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 [itex]SO^{\uparrow}(1,3)[/itex]
This thread is motivated by samalkhaiat's comment here

samalkhaiat said:
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)
samalkhaiat said:
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 :biggrin:
 
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JD_PM said:
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.
 
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Hi PeroK

PeroK said:
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

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

PeroK said:
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?
 
JD_PM said:
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.
 
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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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