SU(3) tensor products: weights, multiplicities, and physics

The dimension of a representation does not generally identify it. For $SU(3)$, the fundamental $\mathbf3$ and its conjugate $\overline{\mathbf3}$ already illustrate the problem. A calculation should therefore begin with the highest weight $(p,q)$, keep it throughout the decomposition, and use dimension labels as useful annotations. Throughout this guide, representations are finite-dimensional continuous complex representations.

Open the SU(3) tensor-product workbench. It calculates the complete decomposition for any pair of input weights with $0\leq p,q\leq6$. Each summand retains its highest weight and multiplicity. Output labels may exceed the input bound.

Labels and the first checks

Our conventions are

$$\mathbf3=(1,0),\qquad \overline{\mathbf3}=(0,1),\qquad \mathbf8=(1,1),$$

with

$$\dim(p,q)=\frac{(p+1)(q+1)(p+q+2)}{2}, \qquad \overline{(p,q)}=(q,p).$$

The central generator $e^{2\pi i/3}\mathbf1_3$ acts by the phase $e^{2\pi i t/3}$, where $t=p+2q\pmod3$. Every summand in a tensor product must carry the sum of the input trialities. This immediately rejects some incorrect decompositions, although triality and dimension alone do not establish a correct one. See highest weights and Conrad’s description of the type-A center.

Two basic products with different physical meanings

The square of the fundamental splits into its symmetric and alternating parts:

$$\mathbf3\otimes\mathbf3 =\mathbf6\oplus\overline{\mathbf3} =(2,0)\oplus(0,1).$$

The symmetric part has dimension $3(3+1)/2=6$; the alternating part has dimension $3(3-1)/2=3$. Their triality is two. In contrast,

$$\mathbf3\otimes\overline{\mathbf3} =\mathbf1\oplus\mathbf8 =(0,0)\oplus(1,1).$$

To see the second identity without a table, regard the tensor product as endomorphisms of $\mathbb C^3$. The identity matrix spans the singlet, and the traceless endomorphisms form the adjoint. In a color calculation these are different channels: replacing a conjugate by an unconjugated triplet changes both the singlet content and the allowed multiplets.

The adjoint square and the repeated octet

A standard research and teaching example is

$$\mathbf8\otimes\mathbf8 =\mathbf1\oplus2\,\mathbf8\oplus\mathbf{10} \oplus\overline{\mathbf{10}}\oplus\mathbf{27}.$$

The highest weights are respectively $(0,0)$, $(1,1)$, $(3,0)$, $(0,3)$, and $(2,2)$. The dimension check is

$$64=1+2\cdot8+10+10+27.$$

The coefficient two is essential. It says that the multiplicity space of the adjoint is two-dimensional. The abstract tensor-product decomposition by itself does not provide a preferred basis in that multiplicity space. When the two factors are identical, the interchange operator gives a further useful splitting:

$$\operatorname{Sym}^2\mathbf8=\mathbf1\oplus\mathbf8\oplus\mathbf{27}, \qquad \Lambda^2\mathbf8=\mathbf8\oplus\mathbf{10}\oplus\overline{\mathbf{10}}.$$

The dimensions are $36$ and $28$. The symmetric and antisymmetric octets are distinct copies of the same irrep; a tensor decomposition should neither merge away their multiplicity nor identify a particular basis of invariant tensors without declaring conventions.

Calculate the adjoint square. The table lists one row for each highest weight and a separate multiplicity column. It does not claim to calculate the exchange-symmetry split or Clebsch–Gordan coefficients.

How the calculation works

The local implementation first constructs the weight character of each input using integer interlacing patterns for the polynomial $GL(3,\mathbb C)$ highest weight $(p+q,q,0)$. Restriction to the determinant-one torus gives the $SU(3)$ character. Characters multiply under tensor products, so convolving the weight multiplicities produces the character of the product.

The algorithm then removes highest-weight characters with their multiplicities until the remainder is zero. Every subtraction must preserve nonnegative weight multiplicities. The final result must reconstruct the full product character and conserve the total dimension and triality. This uses much more information than fitting a sum of dimensions.

For the mathematical character constructions, see Etingof et al., Introduction to Representation Theory, §§5.18–5.23. Sage’s Weyl character ring documentation provides an independent computational framework and reproducible examples. The website’s calculation runs locally; it does not invoke a Sage service.

Global form and field statistics

A representation of $SU(3)$ descends to $PSU(3)$ precisely when its triality is zero. Both factors of an ordinary $PSU(3)$ tensor product must themselves descend. The fact that $\mathbf3\otimes\overline{\mathbf3}$ has neutral total triality does not turn either triplet into an ordinary $PSU(3)$ representation. The combined product does descend.

Similarly, decomposing an internal representation does not finish an operator-basis calculation. Identical commuting scalar components select symmetric powers; Grassmann components select alternating powers of their full field-component space. With spinors, that space includes Lorentz and flavor indices. Equations of motion, integration by parts, derivatives, and gauge redundancies require further treatment. Use the finite tensor-product guide to compare these distinctions in a small example.

A useful research workflow

Record the parent group and highest weights, perform the product, and inspect repeated summands and conjugates. Check dimensions and central charge. Only then impose the exchange symmetry, Lorentz contraction, or subgroup restriction required by the physical question. The SU(3) branching workbench carries a selected multiplet along an explicitly named map.

Copy LaTeX for a manuscript and preserve the JSON as the calculation record. The export identifies the inputs, result, sources, algorithm, and su3-dynkin-pq-v1 convention profile. The representation labels determine the abstract result; matrix realizations and normalized invariant tensors are separate data. Source checking and automatic verification remain distinct from the site’s pending independent human review.