MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I read in at least one paper and in the wiki below

that the 56 symmetric irrep of SU(6) breaks down into 10^{3/2} + 8^{1/2} irreps of SU(3)xSU(2). Here the first is 40 dimensional (10 of SU(3) x 4 of SU(2)) and the second is 16 dimensional (8 of SU(3) x 2 of SU(2)).

The problem is that this is simply not true. I checked the branching rules and this just doesn't show up. Can someone please double check if the above decomposition is correct or not.


share|cite|improve this question
It's always sensible to approach Wikipedia entries with some caution, but I'm not clear about what "I checked the branching rules" involved on your part. (It might help to mention your own source for branching rules.) Computations of this sort don't usually have intuitive outcomes and can be tricky, but Robert Bryant seems to have checked carefully enough. – Jim Humphreys Dec 27 '12 at 23:58
I use GAP (groups algorithms and programming) with sla package for quick branching rules (it works with algebras...) I also have a collection of code I've written over the years that is somewhat mature and reliable...but it's always good to check these calculations from different angles... – Y Macdisi Dec 28 '12 at 0:15
up vote 10 down vote accepted

You appear to have made a mistake in your calculation of the branching rules. The answer given in the wiki is correct, but it seems that you are using the 'wrong' subgroup of $\mathrm{SU}(6)$. Perhaps you are using the subgroup isomorphic to $\mathrm{SU}(2)\times\mathrm{SU}(3)$ under which the fundamental $\mathrm{SU}(6)$-representation $\mathbb{C}^6$ breaks up as $\mathbb{C}\oplus\mathbb{C}^2\oplus\mathbb{C}^3$.

In the wiki page you cite, the author is using the subgroup under which the fundamental $\mathrm{SU}(6)$-representation is still irreducible, but is a tensor product of $V$, the $2$-dimensional representation of $\mathrm{SU}(2)$, and $W$, the $3$-dimensional representation of $\mathrm{SU}(3)$. Then the statement (which is correct) is that, as $\bigl(\mathrm{SU}(2)\times\mathrm{SU}(3)\bigr)$-representations, one has $$ \mathsf{S}^3(V\otimes W)\simeq \bigl(\mathsf{S}^3(V)\otimes \mathsf{S}^3(W)\bigr) \oplus \bigl(V\otimes (W\otimes W^*)_0\bigr). $$ These are the two irreducible subspaces that you see in the answer.

This is a special case of the general formula for $\mathsf{S}^3(V\otimes W)$ as a sum of tensor products of representations of $\mathrm{SL}(V)$ and $\mathrm{SL}(W)$. [One term is missing because $\Lambda^3(V)=0$, and I have used other identifications that hold because the ranks of the two factor groups are so small.]

share|cite|improve this answer
In a pure $SU(3) \otimes SU(2)$ setting, you're right : $S^3(3 \otimes 2) \cong (10 \otimes 4) \oplus (8 \otimes 2)$ or in terms of dimensions $S^3(6)=40+16$; the 6,40,and 16 are all $SU(3) \otimes SU(2)$ irreps. – Y Macdisi Dec 28 '12 at 0:06
@Y Macdisi: Let $V$ be the $2$-dimensional rep of $\mathrm{SU}(2)$ and $W$ be the $3$-dimensional rep of $\mathrm{SU}(3)$. Then $U = V\otimes W$ has dimension $6$ and is irreducible as a rep of the product $G = \mathrm{SU}(2)\times \mathrm{SU}(3)$. Since $G$ is compact and since the top exterior power of $U = V\otimes W$ is clearly trivial as a $G$-rep, there is some $G$-invariant, definite Hermitian form on $U$. Since $G$ lies in the stabilizer of this Hermitian form and complex volume form, it follows that $G$ lies in a copy of $\mathrm{SU}(6)$ acting on $U$, and this action must be irred. – Robert Bryant Dec 28 '12 at 0:45
@Y Macdisi: (continued) Thus, you can see that it is indeed possible that $\mathrm{SU}(6)$ contains a copy of $G$ that acts irreducibly on $U\simeq \mathbb{C}^6$. Moreover, under $G$, the vector space $\mathsf{S}(U)$ does indeed break up into a $40$-dimensional and a $16$-dimesional representation of $G$, each of which is irreducible. – Robert Bryant Dec 28 '12 at 0:55
@Y Macdisi: Yes, in fact, it is not hard to show that there are exactly two conjugacy classes of subgroups of $\mathrm{SU}(6)$ that are isomorphic to $\mathrm{SU}(3)\times\mathrm{SU}(2)$, and they are represented by the two that we have been discussing. This kind of thing happens more generally: Typically, when $p$ and $q$ are bigger than $1$, there will be more than one conjugacy class of subgroups of $\mathrm{SU}(pq)$ that are isomorphic to $\mathrm{SU}(p) \times\mathrm{SU}(q)$ or a finite quotient (if $p$ and $q$ are not relatively prime) of that product. – Robert Bryant Dec 28 '12 at 20:50
@Y Macdisi: You are right about the third case at the Lie algebra level, but the third case does not give an embedding of $\mathrm{SU}(3)\times\mathrm{SU}(2)$ into $\mathrm{SU}(6)$ because it's not injective on the $\mathrm{SU}(2)$ factor, rather, this case produces a copy of $\mathrm{SU}(3)\times\mathrm{SO}(3)$ in $\mathrm{SU}(6)$. – Robert Bryant Dec 28 '12 at 22:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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