SU(N) tensor products for model building

A useful tensor-product answer identifies the group, every highest weight, and every multiplicity. Dimension labels provide a readable shorthand, but they can hide a conjugation or identify several different irreducible representations with the same number. The SU(N) workbench keeps Dynkin labels beside dimensions throughout the calculation.

All representations below are finite-dimensional continuous complex representations of compact $SU(N)$. “Real” or “quaternionic” describes the type of a complex irrep; it does not change the coefficient field of the decomposition. The method reference specifies the Young-diagram convention, exact algorithm, and checks.

Start with labels, then use dimensions

The defining representation has labels $(1,0,\ldots,0)$. Its conjugate is $(0,\ldots,0,1)$. More generally,

$$\overline{(a_1,a_2,\ldots,a_{N-1})} =(a_{N-1},\ldots,a_2,a_1).$$

The fundamental weight $\omega_k$ gives $\Lambda^k\mathbb C^N$, of dimension $\binom Nk$. The corresponding Young diagram is a column of height $k$. The adjoint has labels $(1,0,\ldots,0,1)$ for $N\geq3$ and dimension $N^2-1$; for $SU(2)$ its single label is $2$. These identifications follow from the standard type-A highest weights and determinant-one restriction; see SageMath, Weyl-character conventions.

The public calculation supports $SU(2)$ through $SU(6)$. Each input label is at most $12$, and the sum of the canonical Young-diagram sizes must be at most $16$. For labels $a$, that size is $\sum_i i a_i$. Thus an antifundamental costs $N-1$ boxes despite having dimension $N$. This convention makes the input bound transparent; it is not a physical distinction between a representation and its conjugate.

SU(5): the square of the 10

Take $\mathbf{10}=\Lambda^2\mathbf5$, with labels $(0,1,0,0)$. Its square is

$$(0,1,0,0)\otimes(0,1,0,0) =(0,0,0,1)\oplus(1,0,1,0)\oplus(0,2,0,0).$$

Output labels	Dimension	Multiplicity	Center charge modulo 5
$(0,0,0,1)$	$5$	$1$	$4$
$(1,0,1,0)$	$45$	$1$	$4$
$(0,2,0,0)$	$50$	$1$	$4$

The first summand is $\overline{\mathbf5}$. The other two are specified by their labels so that a bar convention in a different table cannot change the meaning. Dimensions give $100=5+45+50$. The polynomial diagrams are $(1,1,1,1)$, $(2,1,1)$, and $(2,2)$, all containing four boxes. The dimension pattern agrees with Feger–Kephart, LieART, Table A.38.

Calculate this product.

There is a direct construction of the first channel. For two alternating tensors $A^{ij}$ and $B^{kl}$,

$$C_m=\epsilon_{mijkl}A^{ij}B^{kl}$$

transforms in $\overline{\mathbf5}$. Exchanging $A$ and $B$ exchanges two pairs of indices and gives a plus sign. This identifies the channel as symmetric under exchanging the two representation factors, without constructing every Clebsch–Gordan coefficient. Contracting $C_m$ with a fundamental $H^m$ gives an invariant in $\mathbf{10}\otimes\mathbf{10}\otimes\mathbf5$. Whether a particular field operator survives also depends on its Lorentz, flavor, and statistics assignments.

Changing one input changes the channels. For example,

$$(0,1,0,0)\otimes(0,0,0,1) =(1,0,0,0)\oplus(0,1,0,1), \qquad 10\cdot5=5+45.$$

The dimension-$45$ output here is conjugate to the one in the previous product. Keeping only “45” would conceal this difference. Open $\mathbf{10}\otimes\overline{\mathbf5}$.

SU(5): two adjoints in the adjoint square

For the adjoint $\mathbf{24}=(1,0,0,1)$, the answer is

$$\mathbf{24}\otimes\mathbf{24} =\mathbf1\oplus2\mathbf{24}\oplus\mathbf{75} \oplus\mathbf{126}\oplus\overline{\mathbf{126}} \oplus\mathbf{200}.$$

Here the two dimension-$126$ terms mean the conjugate labels $(0,1,0,2)$ and $(2,0,1,0)$; the $75$ and $200$ have labels $(0,1,1,0)$ and $(2,0,0,2)$. Their placement agrees with LieART, Table A.38, and

$$576=1+2\cdot24+75+126+126+200.$$

The coefficient two is the dimension of an adjoint multiplicity space. There are two independent equivariant maps from the product to an adjoint; the decomposition does not supply a preferred normalized basis for them.

In the traceless-matrix realization one can recognize two channels directly:

$$(A,B)\longmapsto[A,B], \qquad (A,B)\longmapsto AB+BA-\frac25\operatorname{tr}(AB)\mathbf1_5.$$

Both transform by conjugation; the first is alternating and the second symmetric under exchange. They explain why an adjoint occurs in each exchange sector. They do not constitute a complete choice of intertwiners for every summand. Calculate the adjoint square.

SU(4) and SU(6): a singlet does not determine reality

Two useful exterior-power examples are

$$\begin{aligned} SU(4):\quad (0,1,0)\otimes(0,1,0) &=(0,0,0)\oplus(1,0,1)\oplus(0,2,0),\\ 6\cdot6&=1+15+20,\\[3pt] SU(6):\quad (0,0,1,0,0)\otimes(0,0,1,0,0) &=(0,0,0,0,0)\oplus(1,0,0,0,1)\\ &\quad\oplus(0,0,2,0,0)\oplus(0,1,0,1,0),\\ 20\cdot20&=1+35+175+189. \end{aligned}$$

These dimension decompositions appear in LieART, Tables A.37 and A.39. Both input representations are self-dual and both squares contain one singlet. Nevertheless, the $SU(4)$ $\mathbf6$ is of real type, whereas the $SU(6)$ $\mathbf{20}=\Lambda^3\mathbf6$ is quaternionic.

The difference can be seen through the invariant pairing

$$\Lambda^k\mathbb C^{2k}\times\Lambda^k\mathbb C^{2k} \longrightarrow\Lambda^{2k}\mathbb C^{2k}\cong\mathbb C, \qquad (u,v)\longmapsto u\wedge v.$$

Its exchange sign is $(-1)^{k^2}=(-1)^k$. The nondegenerate pairing is symmetric for $k=2$ and alternating for $k=3$. For an irreducible unitary representation, these cases give real and quaternionic type, respectively; see Adams, §5 and Lemma 5.3. The singlet count alone cannot distinguish the two.

Calculate $\mathbf6\otimes\mathbf6$ · Calculate $\mathbf{20}\otimes\mathbf{20}$

Global form and physical operators

The center charge of labels $a$ is $t(a)=\sum_i i a_i\pmod N$. For a divisor $d$ of $N$, a representation descends to $SU(N)/\mathbb Z_d$ exactly when $t(a)=0\pmod d$, where $\mathbb Z_d$ is the unique central subgroup of order $d$. Indeed, its generator is $z^{N/d}$ and acts by $e^{2\pi i t(a)/d}$. This is the type-A instance of the weight-lattice description in Conrad’s center notes.

A product may have neutral total charge even when its individual factors do not descend. Thus $\mathbf N\otimes\overline{\mathbf N}$ defines a representation of $PSU(N)$, while either factor separately requires the simply connected cover. In a product gauge group, an abelian charge can also compensate a nontrivial center action through a diagonal quotient. Record the actual global group before interpreting a field list.

For irreducible $R,S$, Schur’s lemma gives

$$\dim\operatorname{Hom}_{SU(N)}(\mathbf1,R\otimes S) =\dim\operatorname{Hom}_{SU(N)}(R^*,S) =\begin{cases}1,&S\cong R^*,\\0,&\text{otherwise}.\end{cases}$$

For three or more factors, intermediate multiplicities matter. Repeatedly decompose products and retain those multiplicities at every stage. This counts invariant tensors in the unrestricted tensor product. An operator built from identical fields can occupy a smaller space: commuting scalar factors select symmetric powers, while Grassmann components select alternating powers of the full component space. Spinor, flavor, derivative, and gauge indices must be included before imposing exchange symmetry.

After decomposing the parent representation, the charged block-branching workbench can restrict a selected summand along a specified block embedding and retain its abelian charge. Restriction requires the embedding and charge normalization; a bare subgroup name is insufficient.

For research use, save the Dynkin labels, multiplicities, global-group choice, and exported convention and algorithm versions. The JSON also retains determinant columns and exact dimension checks. These make a calculation inspectable; normalized invariant tensors, a full operator basis, and independent expert review remain separate tasks.