Talk:Lorentz group

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Old Discussion Archived

This article was changed beyond recognition when I greatly expanded it in July 2005 (and I still have more to do). Unfortunately, in the middle of this process, a serious misunderstanding arose which led to a bit of a kerfluffle. Fortunately, this has been mediated, and I don't think the "flap" is very edifying, so I have archived it at Talk:Lorentz group/Archive.

Please add any comments/suggestins on present article below. I've put in a lot of work on this, so I hope all good Wikipedias will ask me for a response before making major changes. TIA---CH (talk) 04:47, 6 August 2005 (UTC)

Left/Right groups?

What is meant exactly by the left and right groups in the double covering SU(2) → SO(3)? -- Fropuff 06:02, 14 October 2005 (UTC)

You still haven't answered my question CH. This needs to be clarified. At any rate the section on topology needs to be corrected: SO+(1,3) is a trivial bundle over H3; there is no "twist" (any fiber bundle over a contractible base is trivial). -- Fropuff 18:44, 23 January 2006 (UTC)
Sorry, I guess I didn't see your comment before. Yeah, this could be clearer! I meant left and right side of the arrow, not left and right multiplication. Feel free to untwist! A reference specifically discussing the topology of the restricted Lorentz group would be useful. I don't have Hall's book in front of me but there might be something useful in there.---CH 16:20, 24 January 2006 (UTC)
Ahh! Thank you. I couldn't figure out what in the world you meant by that. -- Fropuff 16:30, 24 January 2006 (UTC)

Topology

The article does not state which topology is used for the Lorentz group but mentions connection properties which depend on the topology. I assume it is topology induced by the operator norm which is the same topology that results when the Group is viewed as a finite dimensional vector space? The Infidel 10:41, 21 January 2006 (UTC)

The topology of O(3,1) is the subspace topology inherited from GL(4,R), which itself can be viewed as an open subset in R16 (with the Euclidean topology). The Lorentz group is not a vector space (but it sits inside one), so I'm not sure what you mean by that last statement. -- Fropuff 16:38, 21 January 2006 (UTC)
Thank you. The last statement should be "as (topological) subspace of a finite dimensional vector space". I think we should mention the topology briefly in the article. Any objections? The Infidel 20:17, 22 January 2006 (UTC)
Hi, Infidel, I don't understand what you found confusing about the existing description. Why did you think I might be refering to functional analysis in the context of a finite dimensional real Lie group? Have you seen the book by Hatcher which I mention in the references?
If I understand what you found confusing, I can try to improve this bit myself. I'd like to try to keep the style and emphasis as internally consistent as possible, which probably means that wherever possible I should make any neccessary changes myself.---CH 01:07, 23 January 2006 (UTC)
I have no special background of Lie groups, so I just wondered on which topology the unconnectedness is based. And shortly my mind went astray considering open-compact topology and the like ... The Infidel 18:32, 23 January 2006 (UTC)

Students beware

Just wanted to provide notice that I am only responsible (in part) for the last version I edited; see User:Hillman/Archive. I emphatically do not vouch for anything you might see in more recent versions. I hope for the best, but unfortunately relativity theory attracts many cranks, and at least some future versions of this article are likely to have been vandalized or to contain slanted information, misinformation, or disinformation.

Good luck to all students in your search for information, regardless!---CH 01:50, 1 July 2006 (UTC)

Lie algebra

"its Lie algebra is reducible and can be decomposed to two copies of the Lie algebra of SL(2,R)"

Really? sl(2,R) has rank 1 (because it contains so(2)), so sl(2,R)^2 has rank 2, but so(3,1) has rank 1 (since sl(2,C) has rank 1). What's going on here? Adam1729 07:18, 2 October 2007 (UTC)

No further comment so far? Strictly speaking, the statement is false. As the author has shown before, so(3,1) = sl(2,C) and sl(2,C) = sl(2,R)⊕sl(2,R) as vector spaces but not as Lie algebras. I think however that the author actually refers to the complexified Lie algebra, which is so(4,C) and hence coincides with the complexification of so(4), so the algebra decomposition so(4) = su(2)⊕su(2) alluded to in the article induces so(4,C) = sl(2,C)⊕sl(2,C).

Conjugacy classes

There are infinitely many conjugacy classes. Every rotation of order m is in a different conjugacy class from every rotation of order n. I suspect that when the article says "conjugacy class" it really means "topological closure of conjugacy class". Is this right? Adam1729 07:24, 2 October 2007 (UTC)

Is this the right matrix?

In the text I see this:

$\left[ \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right]$

Should it be this:

$\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right]$

I admit to being way over my head to even try to read this article, for the former matrix looks quite odd to me, while the latter looks commonplace.  Randall Bart   Talk  04:39, 17 December 2009 (UTC)

The first matrix is the rotation generator (infinitesimal excursion from the identity), iJz, which I now labelled. The "commonplace" one you are adducing is a full finite rotation by π/2, namely exp(iπ Jz/2). Try power series expansion of the exponential. This highlights the article's deficiency in not identifying the geometrical entities with the conventional operators employed in quantum mechanics and particle physics, at the very least. Cuzkatzimhut (talk) 19:49, 29 November 2013 (UTC)

Is the Lorentz group a). an abelian group, b). non-abelian group, or c). Sometimes one and sometimes the other? The only thing that I say even connected to these clasifications is a mention of an abelian subgroup here and there. 98.81.17.64 (talk) 21:46, 4 August 2010 (UTC)

I've tweaked the lead to reflect this classification (non-abelian), which is relevant. The article structure needs work. — Quondum 13:30, 15 August 2012 (UTC)

Notation of connected components and covering groups

In Lorentz group#Covering groups it is said that Spin(1,3) is the double cover of SO(1,3), and the double cover of $SO^+(1,3)$, i. e. $SL(2,\mathbb C)$, is called $Spin^+(1,3)$. In Spin group#Indefinite signature, the double cover of $SO^+(1,3) = SO^+(3,1)$ (which is denoted by $SO_0(3,1)$ there) is called $Spin(3,1)$. The notations should be unified in Wikipedia: it should b decided whether $SO^+(p,q)$ or $SO_0(p,q)$ should be used for the connected component of identity of $SO(p,q)$ and whether $Spin(p,q)$ or $Spin^+(p,q)$ should be used to denote the double cover of the identity component. Jaan Vajakas (talk) 12:14, 15 August 2012 (UTC)

The Lie algebra of the Lorentz group

The Lie algebra of SO+(1;3), so(1;3), cannot be decomposed into two copies of sl(2;R). It can't be decomposed at all, because it is simple. What is true is that its complexification, so(1;3)C can be decomposed into sl(2;C) ⊕ sl(2;C). YohanN7 (talk) 20:33, 3 October 2012 (UTC)

Removed this: "Since the Lorentz group is SO+(1,3), its Lie algebra is reducible and can be decomposed to two copies of the Lie algebra of SL(2,R), as will be shown explicitly below (this is the Minkowski space analog of the SO(4) → SU(2) × SU(2) decomposition in a Euclidean space). In particle physics, a state that is invariant under one of these copies of SL(2,R) is said to have chirality, and is either left-handed or right-handed, according to which copy of SL(2,R) it is invariant under."
It wasn't used (and much less proved because it's false) anyway. YohanN7 (talk) 19:01, 8 March 2013 (UTC)

Style problems

The article reads too much like bad lecture notes. It's full of writing in the second person and leaving the reader in suspense ("We will later find such and such..."), working through calculations without telling us where they are going. ("Consider the following..." followed by piles of unmotivated calculations. "But then consider the following!" (jarring and upsetting twist which makes the subject look difficult)) Hopefully I'll be able to make some improvements so it's not necessary to plough through the article line-by-line to understand what it is talking about. Count Truthstein (talk) 17:34, 8 March 2013 (UTC)

map from SL(2,C) to SO+(1,3)

In the section on "Relation to the Möbius group" there is the sentence "SL(2,C) ... is isomorphic to a subset of the Lorentz group". This statement indicates that the map from SL(2,C) to SO+(1,3) is injective. But if I have got this right, this isn't true. The map is surjective, as the next paragraph explains, and the kernel of the map is ±I. - Subh83 (talk | contribs) 19:14, 31 May 2013 (UTC)

"Lorentz transformations are examples of linear transformations"

The article claims that LTs are examples of linear transformations, but isn't half the Lorentz group antilinear, i.e. precisely *not* linear transformations? Matthew.daniels (talk) 15:02, 11 September 2013 (UTC)

Time reversal is a linear operation in the context of the Lorentz group. When one considers the action of time reversal in quantum mechanics, it does become an anti-linear operator on the corresponding Hilbert space, but that is particular to quantum mechanics. --Mark viking (talk) 18:08, 11 September 2013 (UTC)
I have updated Representation theory of the Lorentz group with a new section, Time-reversal, and tried to explain the reasons for the antilinearity. YohanN7 (talk) 09:39, 18 December 2013 (UTC)

Lorentz_group#Parabolic

The formulas are correct only if the matrix is from SL(2,R), i.e. if alpha is real. Otherwise they don't give the right results.--Café Bene (talk) 04:20, 21 November 2013 (UTC) Corrected it.--Café Bene (talk) 04:39, 21 November 2013 (UTC)