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The root system $D_5$ consists of the forty roots $\pm e_i\pm e_j$, $1\le i<j\le 5$, in an orthonormal basis. One choice of simple roots is $e_1-e_2,e_2-e_3,e_3-e_4,e_4-e_5,e_4+e_5$. The Dynkin diagram has a fork at node $3$ with ends $4$ and $5$. Its Weyl group permutes coordinates and changes an even number of signs. The associated complex simple Lie algebra is $\mathfrak{so}(10,\mathbb{C})$. Its two half-spin representations have dimension $16$, with highest weights $\frac{1}{2}(e_1+e_2+e_3+e_4\mp e_5)$.

For the D5 root system, take an orthonormal basis $e_1,\dots ,e_5$ and simple roots ${\alpha }_i=e_i-e_{i+1}$ for $i=1,2,3,4$, together with ${\alpha }_5=e_4+e_5$. The Dynkin diagram has single bonds $1-2$, $2-3$, $3-4$, $3-5$, with the fork at node $3$. Thus its Cartan matrix isThe fundamental weight ${\omega }_5=\frac{1}{2}(e_1+e_2+e_3+e_4+e_5)$ has Dynkin labels $(0,0,0,0,1)$. Its pairing with every positive coroot $e_i\pm e_j$, $i<j$, is zero or one. Hence its half-spin representation is a minuscule representation, so its weight system is a single Weyl group orbit with multiplicity one. The Weyl group of the D5 root system permutes coordinates and changes an even number of signs. Consequently its complete weight system isThere are $1+\left(\genfrac{}{}{0ex}{}{5}{2}\right)+\left(\genfrac{}{}{0ex}{}{5}{4}\right)=16$ such weights. Their Dynkin labels are $((s_1-s_2)/2,(s_2-s_3)/2,(s_3-s_4)/2,(s_4-s_5)/2,(s_4+s_5)/2)$, explicitlyHere the Lie algebra is $\mathfrak{so}(10,\mathbb{C})$, or its compact real form when discussing compact groups. Globally the half-spin representation is a representation of $\mathrm{Spin}(10)$, and does not descend to $SO(10)$; the question only specifies the Lie algebra.

Choose the first $SU(2)$ as $S{U}_w(2)$ and embed $S{U}_c(3)$ in the upper-left three-by-three block of $SU(4)$. The fundamental representations then restrict as $4\to 3\oplus 1$ and $\overline{4}\to \overline{3}\oplus 1$. Thus the representation in solution, forgetting hypercharge, restricts toThis contains every required $S{U}_w(2)\times S{U}_c(3)$ representation, with one extra singlet.

For the hypercharge embedding, let $X=\mathrm{diag}(1,1,1,-3)$ in the $SU(4)$ fundamental representation, and $T_{3R}=\mathrm{diag}(1/2,-1/2)$ in the second $SU(2)$. These commute with $S{U}_w(2)\times S{U}_c(3)$. The integer normalization in the paper isExplicitly, this circle group embeds in the covering group byBoth nontrivial blocks have determinant one, and the map is injective since the first three entries of the $SU(4)$ block recover $z$.

On $(2,1,4)$, $T_{3R}=0$, so the branching of a representation gives $(2,3{)}_1\oplus (2,1{)}_{-3}$. On $(1,2,\overline{4})$, the U(1) charges under $X$ are $-1$ on $\overline{3}$ and $3$ on $1$, while $6T_{3R}$ takes values $\pm 3$. HenceCombining the two summands givesThis is the Pati-Salam charge embedding, with $X=3(B-L)$. In terms of the common Standard Model convention $Q_{\mathrm{electric}}=T_{3L}+Y_{\mathrm{SM}}$, the paper's integer hypercharge is $Y=6Y_{\mathrm{SM}}$.

The extra neutral singlet can be interpreted as the left-handed charge conjugate ${\nu }_R^c$ of a right-handed neutrino. It is a sterile neutrino under the Standard Model gauge group, and completes the sixteen Weyl fermions of one family in this half-spin representation. Its neutrality permits a Majorana mass term without violating the remaining Standard Model gauge group.