Breaking down SU(N) representation into smaller groups

[diegzumillo]

Hi all

I have a shallow understanding of group theory but until now it was sufficient. I'm trying to generalize a problem, it's a Lagrangian with SU(N) symmetry but I changed some basic quantity that makes calculations hard by using a general SU(N) representation basis. Hopefully the details of the problem are not important though, as I just want to rewrite these matrices in a more useful way. Say I have SU(4), it has 15 generators, right, so it looks plausible to replace this basis with 5 sets of SU(2) matrices instead. That would simplify my calculations! But looking at the actual matrices of SU(4) it's hard to see how to break them into SU(2) sets.

This seems to me like basic stuff I don't know. Anyone cares to nudge me in the right direction?

 

[fresh_42]

Hi all

I have a shallow understanding of group theory but until now it was sufficient. I'm trying to generalize a problem, it's a Lagrangian with SU(N) symmetry but I changed some basic quantity that makes calculations hard by using a general SU(N) representation basis. Hopefully the details of the problem are not important though, as I just want to rewrite these matrices in a more useful way. Say I have SU(4), it has 15 generators, right, so it looks plausible to replace this basis with 5 sets of SU(2) matrices instead. That would simplify my calculations! But looking at the actual matrices of SU(4) it's hard to see how to break them into SU(2) sets.

This seems to me like basic stuff I don't know. Anyone cares to nudge me in the right direction?

Of course it all depends on what you want to achieve, but there is no natural such way. $\mathfrak{su}(n)$ are simple Lie algebras, and representations can only be split by a decomposition into ideals, diagonal matrices if the product is direct, upper triangular matrices if the product is semidirect. As $\mathfrak{su}(n)$ has no ideals and the group only a central discrete group of unities, there is no natural way to do so. You can only change the basis and therewith the matrix representations of the generators. However, the representations of the groups / Lie algebras are all known, at least in principle, which reduces your choices further.

In short: If you restrict yourself to embedded generators of another group, the representations of the larger group will lead you out of them, i.e. the matrices are still full, i.e. no zero blocks.

 

[Hans de Vries]

Here are the two ways that you can split the 15 generators of SU(4) into 5 groups of SU(2) generators:
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:

 

[diegzumillo]

Thanks! I appreciate the help. This is an old topic, at the end I went with a very anticlimactic SU(n1)xSU(n2)x some extra matrices (with n=n1+n2). It made sense in the problem at hand.