0
$\begingroup$

Reading this paper (Handout VIII for the course SYMMETRIES IN PHYSICS, Michael Flohr, Subgroups and Unified Theories), I had my ideas confused: it is said that we can take any $SU(N)$ with its related Dynkin diagram, separate the last dot from the whole diagram and obtain two diagrams made of $N-2$ dots and $1$ dot respectively. So my questions. Am I right thinking that:

  1. I can associate an $SU(2)$ algebra to that single dot.
  2. As a consequence I can write that $su(3) \supset su(2)\oplus su(2)$. Note that $su(3)$ Dynkin diagram has two dots, then two $su(2)$ algebras.
  3. Why instead that single dot is always associated with an $U(1)$ algebra? For example $SU(5)\rightarrow SU(3)\times SU(2) \times U(1)$.
$\endgroup$
3
  • 1
    $\begingroup$ I think you may be conflicting extended and unextended Dykin diagrams here. A single dot is the unextended diagram of SU(2), but the extended diagram has two dots (and four edges, if memory serves me right). The extended diagram of SU(N) has N dots, forming a circle. Dynkin's method of maximal subalgebras involves both extended and unextended diagrams. You may want to read about this method in a more specific source, a book on simple Lie algebras. $\endgroup$ Commented Jan 11, 2021 at 18:42
  • $\begingroup$ Ok, thanks for the clarification. Do you think the paper i linked talks about extended Dynkin diagrams? $\endgroup$ Commented Jan 11, 2021 at 18:52
  • $\begingroup$ Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Commented Jan 11, 2021 at 18:54

1 Answer 1

0
$\begingroup$

Studying more in detail the subject I figured out the problem. Considering $SU(3)$, it is true that we can find two subgroups (both of them are $SU(2)$) that are contained in it. This of course doesn't mean that $SU(3)=SU(2)\otimes SU(2)$, the two subgroups may overlap. $U(1)$ came out from that dot that we left out as it only needs a diagonal generator and no additionals creation-annihilation operators that may be already included in some other subgroup.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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