= Solution
Write $[r_1,r_2,\ldots]$ for a <Young diagram> with row lengths $r_i$, and depict an empty box by $\square$. Appending the box of the <fundamental representation> to another single box gives a row or a column:
$$
\square\otimes\square
=(\square\,\square)\ \oplus\ \begin{array}{l}\square\\[-4pt]\square\end{array},
\qquad [1]\otimes[1]=[2]\oplus[1,1].
$$
The corresponding <symmetric tensors> and <antisymmetric tensors> have dimensions $4\cdot5/2=10$ and $4\cdot3/2=6$, so
$$
\boxed{\mathbf4\otimes\mathbf4=\mathbf{10}_{\mathrm{s}}\oplus\mathbf6_{\mathrm{a}}.}
$$
For the <conjugate representation>, the invariant volume tensor identifies $\bar{\mathbf4}\cong\Lambda^3\mathbf4$, represented by a column of three boxes. Adding one box gives
$$
[1]\otimes[1,1,1]=[2,1,1]\oplus[1,1,1,1],
\qquad
\square\otimes\begin{array}{l}\square\\[-4pt]\square\\[-4pt]\square\end{array}
=\begin{array}{l}\square\,\square\\[-4pt]\square\\[-4pt]\square\end{array}
\ \oplus\ \begin{array}{l}\square\\[-4pt]\square\\[-4pt]\square\\[-4pt]\square\end{array}.
$$
A column of four boxes is the <determinant representation>, which is trivial on $SU(4)$. The remaining <Young diagram> has <Dynkin labels> $(1,0,1)$ and dimension fifteen. For example the <hook-content formula> gives
$$
\dim[2,1,1]=\frac{4\cdot5\cdot3\cdot2}{4\cdot1\cdot2\cdot1}=15.
$$
Equivalently, a mixed <tensor> $M^i{}_j$ transforms by $M\mapsto UMU^\dagger$ and splits into its <trace> and <traceless matrix> parts. Therefore
$$
\boxed{\mathbf4\otimes\bar{\mathbf4}=\mathbf1\oplus\mathbf{15},}
$$
where $\mathbf{15}$ is the <Adjoint representation>.
Back to article page
