4
$\begingroup$

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?

$\endgroup$
5
  • 2
    $\begingroup$ 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. $\endgroup$ – ACuriousMind Nov 6 '14 at 19:32
  • 1
    $\begingroup$ 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. $\endgroup$ – itsqualtime Nov 6 '14 at 20:41
  • $\begingroup$ 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? $\endgroup$ – ACuriousMind Nov 6 '14 at 20:45
  • $\begingroup$ I guess it is only "special" in the sense that since there is a n x n matrix representation of SU(2) for any n, it becomes possible to obtain isotropic theories out of any SU(n) gauge by mapping SU(n) subalgebras into SU(2) (at least we've been able to do that up to SU(4)). $\endgroup$ – itsqualtime Nov 7 '14 at 20:09
  • $\begingroup$ There's the generalized Gell-Mann Matrices: math.stackexchange.com/q/4100876/202425 $\endgroup$ – Nike Dattani Apr 13 at 19:07
2
$\begingroup$

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.

EDIT:

BTW, a simple way to see this is the following: Consider the generators on the diagonal (you'll have $N-1$ independent ones, due to the tracelessness condition) as the $J_z$. Then, the corresponding $J_\pm$ are those generators that are one step off-diagonal above/below, and there are exactly $N-1$ of them.

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

$\endgroup$
13
  • $\begingroup$ I'll look into Dynkin diagram, thanks. Do you think this could be generalized to arbitrary SU(N) groups? $\endgroup$ – itsqualtime Nov 7 '14 at 20:01
  • $\begingroup$ Yes, an arbitrary $SU(N)$ group has $N-1$ such nodes, and hence as many independent $SU(2)$s sitting inside. In some sense, that is an essential part of the approach to classifying lie algebras. Since I know that you're a physicist, take a look here:phyweb.lbl.gov/~rncahn/www/liealgebras/book.html $\endgroup$ – Siva Nov 8 '14 at 2:03
  • $\begingroup$ Them what about SU(3)? I get three independent subgroups (see the Gell-Mann matrices article), yet only 2 nodes. Or are they not quite related? Anyway, I'll look more into it, thanks for the ref. above. $\endgroup$ – itsqualtime Nov 8 '14 at 4:26
  • $\begingroup$ IMHO: If I understand that link correctly, it is extending my statement by a little bit. Eg: think of the meson octet (representation of $SU(3)$. The two $J_3$ generators I mentioned are the $\hat{i}$ and $-\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}$ on that diagram. In the language of the Wiki article, (for reasons I do not understand) it also includes the third possible generator $-\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j}$ as "independent" (even though those three are no linearly independent). To me, the sense in which those $SU(3)$ groups are "independent" depends on context. $\endgroup$ – Siva Nov 8 '14 at 6:23
  • 1
    $\begingroup$ Yes, that is precisely what I meant. (Those notes are quite comprehensive!) BTW, I think that in my hurry I made a typo in my sketch (I think the bottom right corner of my PDF must contain $\lambda_8$ instead of $\lambda_3$ and a missing $\pm$). That gives you the "second" SU(2). Note that the way I worded my answer, the three SU(2)s are not linearly independent. $\endgroup$ – Siva Jan 4 '15 at 1:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.