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The irreps of (complexified) $\mathfrak{su}(n)$ are labelled by highest weight Dynkin labels $(a_1, \ldots, a_{n-1})$, and are often referred to simply by their dimension, e.g. $\mathbf{3}$ to label $(1, 0)$ of $\mathfrak{su}(3)$. Now consider the $(2, 1)$ and $(4, 0)$ irreps of $\mathfrak{su}(3)$: both have dimension 15, but they are not dual (the dual of $(2, 1)$ is $(1, 2)$). Is there a standard way in the physicists notation to distinguish such irreps? If not, then is there ever any (physical) need to do so?

xzd209
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2 Answers2

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? Of course physicists use the standard Dynkin indices, as you confirm by using your Slansky 1981, Table 23, p 92.

Even though your two irreps have the same $$ d(p,q)= \tfrac{1}{2} (p+1)(q+1)(p+q+2) ~~~~\leadsto ~~~ 15, $$ Their quadratic Casimirs entering in the β function of QCD are different, eigenvalues $$ (p^2+q^2+3p+3q+pq)/3 ~~~~\leadsto ~~~ 28/3 \leftrightarrow 16/3, $$ respectively; and the cubic Casimir eigenvalues, rescaled to the respective anomaly coefficients, are $$ (p-q)(3+p+2q)(3+q+2p)/18 ~~~~\leadsto ~~~ 154/9 \leftrightarrow 28/9 , $$ so they'd contribute differently to anomalies.

Cosmas Zachos
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  • Interesting! It is certainly common to ignore Dynkin indices and just to write e.g. $\mathbf{15}$ though. Is there any way to distinguish the two irreps using this notation, or are we simply meant to assume that it refers to the 15-dimensional irrep with the lowest weight (i.e. $(2, 1)$ rather than $(4, 0)$)? – xzd209 Jan 10 '21 at 15:17
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    No, the dimensionality is not a unique label of a representation. I've no idea what might be meant to be assumed. Perhaps that there is some 15 in the Clebsch reduction that will not feature in the discussion again, and the number of states are a correction check in the counting of states... – Cosmas Zachos Jan 10 '21 at 15:25
  • If I recall the convention is that 15 is the totally symmetric one, (4,0), probably because it arises more naturally from multiplying together smaller irreps. – Luke Pritchett Jan 10 '21 at 15:43
  • @LukePritchet Actually, Slansky uses 15 for (2,1) and 15' for (4,0), but who cares? Given ambiguity, any reasonable author would specify exactly what she/he means through Dynkin labels. – Cosmas Zachos Jan 10 '21 at 19:23
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The irreps are really different, as illustrated by branching rules to subgroup and other relevant quantum numbers.

For instance the $(4,0)$ contains angular momenta $L=4,2,0$ but the $(2,1)$ contains $L=3,2,1$. The branch to $\mathfrak{su}(2)$ irreps is also different.

The irrep $(4,0)$ does not have weight multiplicities but the irrep $(2,1)$ has three weights occurring twice. Thus the physical contents are certainly different.

ZeroTheHero
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