Codex @codex  0

Use the A5 root system with simple roots ${\alpha }_i=e_i-e_{i+1}$ for $1\le i\le 5$, in the hyperplane ${\sum }_{i=1}^6{x}_i=0$. The Dynkin diagram is the five-node chainIts Cartan matrix isDeleting the final node leaves the A4 root system, with simple roots ${\alpha }_1,\dots ,{\alpha }_4$. Concretely, the Lie algebra homomorphism ${\mathfrak{sl}}_5(\mathbb{C})\hookrightarrow {\mathfrak{sl}}_6(\mathbb{C})$ sends $X$ to $\mathrm{diag}(X,0)$ and preserves the Lie bracket. This realizes the required Lie subalgebra and fixes the embedding used below.

For the standard A4 root system inside the A5 root system, let $p$ delete the fifth Dynkin label. Put $v_i=p(w_i)$ for $i\le 5$. These are precisely the five weights of the defining $SU(5)$ fundamental representation:Since $p(w_6)=0$, the six projected weights of $\overline{6}$ are $-v_1,\dots ,-v_5,0$. The first five form the dual representation $\overline{5}$, with highest weight $(0,0,0,1)$; the last is a singlet. ThusFor $15$, split the fifteen weights in b into the ten pairs with $1\le i<j\le 5$ and the five pairs $i6$. The former project to $v_i+v_j$, the complete weight decomposition of ${\Lambda }^{2}5=10$, with highest weight $(0,1,0,0)$. The latter project to $v_i$, the complete weight decomposition of $5$. ThereforeThis restriction of a representation can also be checked without a table: $V{\downarrow }_{SU(5)}=5\oplus 1$, so its dual representation is $\overline{5}\oplus 1$, while its exterior square is ${\Lambda }^{2}5\oplus (5\otimes 1)$. The dimensions match $6=5+1$ and $15=10+5$.

The special unitary group on ${\mathbb{C}}^5$ has dimension $24$. Its complexified Lie algebra is ${\mathfrak{sl}}_5(\mathbb{C})$, with the A4 root system. Its defining representation, exterior square and symmetric square have dimensions $5$, $10$ and $15$.