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  SU(2) subgroups of SU(4)?

+ 3 like - 0 dislike
3485 views

The Wikipedia article on Gell-Mann matrices states that there are 3 independent SU(2) subgroups of SU(3). One of them, for example, is given by the generators $\{ \lambda_1, \lambda_2, \lambda_3 \}$, which satisfy the commutation relation of the $\mathfrak{su}(2)$ algebra.

How can I found similar subgroups of SU(4) such that their combination satisfy a commutation relation of the form $[t_a, t_b] = \epsilon_{abc} t_c$ as well?

So far I am aware of three such ways - for example the matrices A, B and B, where $B= i( t_2 + t_{14}) $, $C= i(t_5 - t_{12})$ and $D= i (t_7 + t_{10})$ and $t_i$ are the 4x4 generators of SU(4), obey the above commutation relation.

Are there any more independent ways?

This post imported from StackExchange Mathematics at 2014-11-07 11:08 (UTC), posted by SE-user itsqualtime
asked Nov 6, 2014 in Mathematics by itsqualtime (15 points) [ no revision ]
Related: mathoverflow.net/q/118484 mathoverflow.net/a/65530 Do you have a physical motivation for looking at $\mathrm{SU}(4)$? If not, I think this would be better suited at math.SE.

This post imported from StackExchange Mathematics at 2014-11-07 11:08 (UTC), posted by SE-user ACuriousMind
Yes, we are looking at higher dimensional gauge theories (color electrodynamics). I was actually aware of that question and I have checked all the suggested references - unfortunately most of them are a bit too involved and general (in the case of SU(N), I might have to go that route), but I was hoping this question had already been addressed within the context of SU(4) gauge theories.

This post imported from StackExchange Mathematics at 2014-11-07 11:08 (UTC), posted by SE-user itsqualtime
While there are many pretty results about $\mathrm{SU}(N)$ gauge theories, I know not about such subgroup results. If you were searching for the maximal torus or a similarly "special" subgroup, there are methods, I think, but I can't see anything special about $\mathrm{SU}(2) \subset \mathrm{SU}(4)$. Is there?

This post imported from StackExchange Mathematics at 2014-11-07 11:08 (UTC), posted by SE-user ACuriousMind

2 Answers

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The Dynkin diagram of $SU(4)$ has 3 nodes, which means that it carries three elements in it's Cartan subalgebra. Consider those as three possible choices for $J_z$. Each of those can be attached with a pair of raising/lowering operators $J_{\pm}$ -- to create one $SU(2)$ algebra each.

I presume that argument can be translated to a statement about groups, but I don't think I'm equipped to do that.

This post imported from StackExchange Mathematics at 2014-11-07 11:08 (UTC), posted by SE-user Siva
answered Nov 6, 2014 by Siva (720 points) [ no revision ]
+ 3 like - 0 dislike

You can split the 15 generators of SU(4) into 5 groups of SU(2) generators, below are two ways to do that. The table used below is a product table, the generators in the bottom right square are the products of the first row with the first column.Each of the 5 groups is an anti-commuting triplet. Each of them can be used as a base of SU(2).

This uses the 6 generators of SO(4) below,  SO(4) has two anti-commuting triples while the triples commute between each other.

\begin{equation}SO(4)~\cong~S_i^3~\times~S_j^3\end{equation}

The SO(4) matrices are given below. Note that the colored 3x3 sub matrices are the SO(3) rotation matrices.

Both  SO(4) triples are also alternative bases for quaternians.

Where $*$ is complex conjugation. Further, to get the x,y,z correspondence to the $SO(3)$ generators as mentioned above we have re-associated the x,y,z coordinates with the Pauli matrices as follows: $\sigma^x\!=\!\sigma^3,~\sigma^y\!=\!-\sigma^2,~\sigma^z\!=\!\sigma^1$

Note that this representation is much cleaner as some horrible ones that extend SU(3) to SU(4) 

The full table written out gives:

http://thephysicsquest.blogspot.com/

answered Oct 16, 2018 by Hans de Vries (90 points) [ no revision ]

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