Talk:Irreducible representation

Irrelevancies/dubious content
Let's settle the these matters in this section.


 * Why is the transformation of the spinor fields according to the representation of the Lorentz group irrelevant, and why is it irrelevant/dubious that the literature uses Lorentz group representations to construct relativistic wave equations? The paper by T. Jaroszewicz and P.S Kurzepa discusses this, and so does the book by E. Abers (chapter 12). Well, it may be erroneous to say the equations are "derived from" the irreps, but one way or another they are used...


 * Why is the section irreducible representation irrelevant when it's (supposed to be) about the irreps of the Lorentz group?

More posts will probably follow... M&and;Ŝc2ħεИτlk 13:46, 8 July 2013 (UTC)


 * If I remember correctly, the group of spatial rotations SO(3) has one irreducible representation for spin 1/2. And there are two ways to extend it to a representation of the Lorentz group which differ in how they handle boosts. The four-vector representation is a tensor product of these two. JRSpriggs (talk) 14:08, 8 July 2013 (UTC)
 * No, you do not. In short:
 * There are infinite families of irreps;
 * There are two complex spin-½ irreps of SU(2) (or projective representations of SO(3)), they differ only in reversing of the complex structure. When you extend SO(3) to SO(3,1) then yes, they also differ in how they handle boosts, but these are distinct irreps over complex numbers already in 3-d.
 * Recall representation theory of SU(2) and representation theory of the Lorentz group. Incnis Mrsi (talk) 17:30, 8 July 2013 (UTC)
 * The article should not put such gross emphasis on the Lorentz group and classification of its irreps. Real irreps certainly are used for description of spin, but they do not constitute a comprehensive theory of spin: it is a quantum-mechanical phenomenon and is too complex to be compressed to this article. Incnis Mrsi (talk) 17:30, 8 July 2013 (UTC)


 * JRSpriggs: Yes, the correspondence between SO(3) and SU(2) does seem familiar and is given in many books, but how tensor products fit in is still confusing for me. It'll absorb in someday. (I was being presumptuous in the lead by writing that reps could be reduced into tensor products of irreps and no further, which as user:Mathsci and Incnis Mrsi indicated here that this isn't the case...).


 * Incnis Mrsi: Well, yes this article shouldn't dwell too much on just the Lorentz group but as you can see there are plenty of empty sections on other groups to fill in, so it wouldn't just be the Lorentz group. This article needs to indicate how and what the irreps of the Lorentz group are with some synopsis of their importance. M&and;Ŝc2ħεИτlk 12:16, 9 July 2013 (UTC)


 * Agreed and no longer an issue, the empty sections have been deleted. M&and;Ŝc2ħεИτlk 17:19, 10 November 2013 (UTC)

Irreducible and indecomposable
User:R.e.b. and IP User:96.60.16.101 have done a good job at correcting the article after I tried to change from a redirect to an article and failed. However, R.e.b. introduced the concept of "indecomposable" as different to "irreducible". Maybe the term is confused in places (not just me), maybe the terminology is just used differently. Note R.e.b. actually wrote indecomposable without adding citations to support this terminology, and exactly where I added references which do in fact use the term "irreducible representation".

To quote User:YohanN7 from Talk:Symmetry in quantum mechanics:


 * "Indecomposable" refers (in Halls book) to the Lie algebra itself, while irreducible (universally) refers to reps. Technicalities about the terminology: Suppose that there is an invariant subspace U of a rep. Then the orthogonal complement of U may or may not be an invariant subspace itself. In the former case, the rep is reducible (to U and its complement), but not in the latter case. If the space breaks up completely (by repeatedly finding invariant subspaces and their invariant complements), then the rep is completely reducible. If every rep of a group/algebra breaks up this way, it has the complete reducibility property. Semisimple Lie algebras all have the complete reducibility property.


 * I have quickly glanced the new article, and I suggest that the term indecomposable should be avoided because it isn't standard, and apparently has multiple meanings. YohanN7 (talk) 14:46, 10 November 2013 (UTC)"

When possible, if others could add more references on the usage of the terminology, it would definitely help. M&and;Ŝc2ħεИτlk 17:16, 10 November 2013 (UTC)

Properties of irreducible representation
http://mathworld.wolfram.com/IrreducibleRepresentation.html lists several interesting properties of irreducible representations. I think those properties might deserve a mention on this page, seeing as this page currently does not contain much information beyond the basic definitions. 167.220.232.21 (talk) 08:55, 26 April 2014 (UTC)

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Must mention that k > 1 for decomposability BEFORE making a false statement
The (sub-)section Decomposable and Indecomposable representations contains this passage:

-- The representation can be decomposed into a direct sum of k matrices:


 * $$ D'(a) = P^{-1} D(a) P = \begin{pmatrix}

D^{(1)}(a) & 0 & \cdots & 0 \\ 0 & D^{(2)}(a) & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & D^{(k)}(a) \\ \end{pmatrix} = D^{(1)}(a) \oplus D^{(2)}(a) \oplus \cdots \oplus D^{(k)}(a) $$

so $D(a)$ is decomposable, ... --

Maybe I missed someplace where it says that k > 1 for decomposability.

But it is ridiculous to completely miss the point by claiming decomposability simply because a displayed equation shows more than one summand.

The condition that k > 1 must be mentioned before making a statement that takes that as an assumption.

As it stands now, this subsection mentions k = 1 only for indecomposability and only in the very last sentence.2600:1700:E1C0:F340:FD35:462B:BD8B:E7F1 (talk) 02:22, 16 November 2018 (UTC)