Is The Oscillator Formalism Complete?

We have undoubtedly constructed, in an earlier post, unitary irreducible representations of $ \text{SU}(d) $, but we might ask ourselves: are these all the unitary irreducible representations of $ \text{SU}(d) $? That is, is the above construction exhaustive?

To answer this question, it is helpful to quickly recall what differentiates the Lie group $ \text{SU}(2) $ from $ \text{SU}(d) $ for $ d \geq 3 $. Let us begin by observing that the definition of $ \text{SU}(d) $ itself suggests a representation: that of $ d $-dimensional complex column vector, which transforms under the action of an element $ U \in \text{SU}(d) $ as \begin{equation} U : x ^{i} \mapsto U ^{i} {} _ {j} x ^{j} \ . \end{equation} Another representation is obtained by the corresponding row vectors: \begin{equation} \label{eq:row-vector-transformation} U : y _ {i} \mapsto y _ {j} \overline{U} ^{j} {} _ {i} \ , \end{equation} where the bar denotes complex conjugation. By unitarity \begin{equation} \label{eq:unitarity} \overline{U} ^{j} {} _ {i} U ^{i} {} _ {k} = \delta ^{j} {} _ {k} \ , \end{equation} the singlet $ x _ {i}y ^{i} $ is invariant, since \begin{equation} U: x _ {i}y ^{i} \mapsto x _ {j} \cdot \overline{U} ^{j} {} _ {i} U ^{i} {} _ {k} \cdot y ^{k} = x _ {j} \cdot \delta ^{j} {} _ {k} \cdot y ^{k} = x _ {i} y ^{i} \ . \end{equation} Products of these primitive upper- and lower-indexed objects clearly carry other representations of $ \text{SU}(d) $, but they are not in general reducible, and one must then construct invariant combinations that do not mix under the action of elements of the unitary group, and discard those combinations that can be obtained from tensors with smaller rank. Once this is done, one is (ostensibly) left with new irreducible representations.

Let’s recall how this is done for the case of $ d=2 $. Define the antisymmetric $ 2 $-tensor $ \epsilon _ {ij} $ such that $ \epsilon _ {12} = 1 $. For \emph{any} $ 2 \times 2 $ complex matrix $ U $, it is a fact that \begin{equation} \label{eq:epsilon-invariance-1} U ^{i} {} _ {k} U ^{j} {} _ {\ell } \epsilon _ {ij} = \operatorname{det} U \times \epsilon _ {k \ell } \ . \end{equation} In our case, $ \operatorname{det} U = 1 $ since our $ U \in \text{SU}(2) $. The form of the identity in \eqref{eq:epsilon-invariance-1}, although true, is a little unsatisfactory as objects with lower indices ought to transfer analogously to \eqref{eq:row-vector-transformation}. Multiplying \eqref{eq:epsilon-invariance-1} by $ \overline{U} ^{k} {} _ {m} \overline{U} ^{\ell } {} _ {n} $ and using \eqref{eq:unitarity}, we get \begin{equation} \label{eq:epsilon-invariance-2} \epsilon _ {k \ell }\overline{U} ^{k} {} _ {m} \overline{U} ^{\ell } {} _ {n} = \epsilon _ {mn} \ , \end{equation} which shows us that the $ \epsilon $-tensor is an invariant tensor of $ \text{SU}(2) $.

The nice thing about this antisymmetric tensor is that it can be used to raise and lower indices: $ x _ {i} = \epsilon _ {ij} x ^{j} $. This is quite special to $ \text{SU}(2) $, and is the statement that the representations $ \mathbf{2} $ and $ \mathbf{\overline{2} } $ (viz. the fundamental and antifundamental) are isomorphic. More generally, this means we can safely consider (say) tensors with lower indices only.

Next, consider a tensor $ T _ {ij} $ that can be split into symmetric and antisymmetric components $ T _ {[ij]} $ and $ T _ {(ij)} $, and note that the antisymmetric part can be written as $ T _ {ij} = \epsilon _ {ij} \phi $, which in turn expresses the decomposition $ \mathbf{2} \otimes \mathbf{2} = \mathbf{3} \oplus \mathbf{1} $ so in this case only a symmetric tensor and a scalar remain. More generally, for a tensor of arbitrary rank, this trick can be used repeatedly \begin{equation} T _ {\cdots [i| \cdots |j] \cdots } = \epsilon _ {ij} S _ {\cdots } \ , \end{equation} to show that irreducible representations correspond to the set of completely symmetric tensors. A rank-$ n $ tensor that is completely symmetric would have dimension $ n+1 $, which is precisely what we found via the Jordan-Schwinger construction.

Note, however, that for the case of $ \text{SU}(3) $, the invariant tensor $ \epsilon _ {ijk} $ has three indices, and therefore cannot be used to raise and lower indices, and so we must consider tensors with both upper and lower indices of the form $ T ^{i _ {1} \cdots i _ {n}} {} _ {j _ {1} \cdots j _ {m}} $. However, we may yet express the antisymmetric parts of tensors in terms of tensors of lower rank, just as we did for the case of $ \text{SU}(2) $, as follows: \begin{equation} T ^{\cdots [i| \cdots |j] \cdots } {} _ {\cdots } = \epsilon ^{ijk} S ^{\cdots } {} _ {k \cdots } \ . \end{equation} This teaches us that for the case of $ \text{SU}(3) $ we can restrict ourselves to the study of tensors that are completely symmetric in both upper and lower indices. Further, since traces of such tensors are themselves representations, we need only consider traceless representations, i.e. ones where every contraction of upper and lower indices vanishes. The space of traceless symmetric tensors with $ n $ upper and $ m $ lower indices will be denoted $ (n,m) $.

What is the dimension of $ (n,m) $? The answer to this question follows from the following observation: the trace map sends \begin{equation} \mathbf{3} ^{\odot n} \otimes \mathbf{\overline{3} } ^{\odot m} \rightarrow\mathbf{3} ^{\odot n-1} \otimes \mathbf{\overline{3} } ^{\odot m-1} \ , \end{equation} where $ \left( \bullet \right) ^{\odot n} $ denotes the $ n ^{\text{th}} $ symmetric tensor product of $ \bullet $. Further, since the kernel of this map is precisely the irreducible representation $ (n,m) $, it must be that

\[\begin{align} \operatorname{dim}(n,m) &= \operatorname{dim}\left(\mathbf{3} ^{\odot n} \otimes \mathbf{\overline{3} } ^{\odot m}\right) - \operatorname{dim}\left(\mathbf{3} ^{\odot n-1} \otimes \mathbf{\overline{3} } ^{\odot m-1}\right) \ , \\ &= \binom{n+2}{2} \binom{m+2}{2} - \binom{n+1}{2} \binom{n+1}{2} \ , \\ &= \frac{1}{2} \left( n+1 \right) \left( m+1 \right) \left( n+m+2 \right) \ . \label{eq:oscillator-su3-dimension} \end{align}\]

The second line follows from a simple exercise in combinatorics. On comparing the dimension formula above with \eqref{eq:oscillator-su3-dimension}, we see that the oscillator formalism in the previous section only constructed the representations $ \left( N,0 \right) $.

We conclude from this discussion that the oscillator formalism only constructs symmetric representations. For the case of $ \text{SU}(2) $, this exhausts all representations, but for $ \text{SU}(d) $ with $ d \geq 3 $, there are representations that our construction has yet to reproduce. Concretely, for the case of $ \text{SU}(3) $ we do not yet know how to construct representations of the form $ (n,m) $ for $ m \neq 0 $. It is natural, then, to ask if some modification of our construction can generate these missing representations.