6
$\begingroup$

What is the branching rule for the subgroup $SU(p,q-1)\subset SU(p,q)$, i.e., the structure of the restriction of irreducible, finite-dimensional representations of $SU(p,q)$ to $SU(p,q-1)$? I would appreciate any reference and comments.

$\endgroup$
4
  • 4
    $\begingroup$ Answers completely different according as you mean finite-dimensional or unitary representations (two essentially disjoint classes). $\endgroup$ Nov 16, 2017 at 18:34
  • $\begingroup$ @FrancoisZiegler Thank you fit your answer. I am looking at the finite-dimensional case and am also aware of the Book ba Mckay and Patera. However, this book treats only a limited number of the above groups (rank $\leq 8$). So does this mean that the above branching rule for general p,q is not understood? $\endgroup$
    – Cyrus
    Nov 17, 2017 at 23:28
  • 1
    $\begingroup$ In, this question, the groups $SU(p,q-1),SU(P,q)$ can be replaced by their complexifications, namely $SL_{n-1},SL_n$ with the smaller group viewed as the matrices in the top left hand corner of the bigger group $\endgroup$ Nov 18, 2017 at 1:47
  • 1
    $\begingroup$ Regarding the unitary case, there is an old paper of T. Kobayashi, providing an explicit formula for the restriction $U(p,q)\downarrow U(p,s)\times U(q-s)$, see: projecteuclid.org/download/pdf_1/euclid.pja/1195511349 $\endgroup$ Nov 18, 2017 at 2:39

1 Answer 1

6
$\begingroup$

As was indicated above your question is the equivalent to the branching rules for $sl_{n-1} \to sl_n$. This is a well-known branching rule and it is given by the following formula: If the representation of $sl_n$ is given by the highest weight $(\lambda_1 \geq \cdots \geq\lambda_n)$ then it decomposes upon restricting to $sl_{n-1}$ to the direct sum of irreducible representations with highest weights $(\lambda^{\prime}_1 \geq \cdots \geq\lambda^{\prime}_{n-1})$ satisfying $\lambda_{i+1} \leq \lambda^{\prime}_i\leq \lambda_{i}$. This is equivalent to removing some boxes from the corresponding Young diagramm.

$\endgroup$

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.