= Solution
A <G2 (mathematics)> <multiplet> containing the known <meson octet> must restrict to a <representation> containing $\mathbf8$. Among the two fundamental <representations> just computed, $\mathbf7$ fails this test, while the <Adjoint representation> $\mathbf{14}$ contains the required octet. Thus the minimal such choice is
$$
\boxed{\text{meson octet}\subset\mathbf{14}=(1,0).}
$$
Larger <representations> containing an octet would be other model choices; the octet alone does not force a unique arbitrarily large <representation>.
A dual <vector> must be pulled back by $P^T$, since $q_A^TPw=(P^Tq_A)^Tw$. Applying this to the <electric charge> and <flavour hypercharge> <vectors> gives
$$
\boxed{Q_{G_2}=P^T\frac13\begin{pmatrix}2\\1\end{pmatrix}
=\begin{pmatrix}1\\2/3\end{pmatrix},\qquad
Y_{G_2}=P^T\frac13\begin{pmatrix}1\\2\end{pmatrix}
=\begin{pmatrix}1\\1/3\end{pmatrix}.}
$$
Thus a <G2 root system> <weight> $(a,b)$ has $(q,y)=(a+2b/3,a+b/3)$. This pullback ensures that every known octet <meson> retains its values from <3/b/solution>. The extra $\mathbf3\oplus\overline{\mathbf3}$ gives six additional <mesons>, one at each of
$$
\boxed{(q,y)=\left(\tfrac23,\tfrac13\right),\
\left(-\tfrac13,\tfrac13\right),\
\left(-\tfrac13,-\tfrac23\right),\
\left(-\tfrac23,-\tfrac13\right),\
\left(\tfrac13,-\tfrac13\right),\
\left(\tfrac13,\tfrac23\right).}
$$
For instance the first three come from the <G2 root system> <weights> $(0,1)$, $(1,-2)$ and $(-1,1)$; the other three come from their negatives. The <weight diagram> displays the preserved octet and the new states.
\Image[/past-exam-of-the-groups-and-representations-course-of-the-university-of-oxford/2021-g2-meson-charges.png]
{title=The meson octet and its minimal G2 extension in electric charge and flavour hypercharge coordinates}
The fractional <electric charges> make these additional states qualitatively different from ordinary <quark>-<antiquark> <mesons>, whose <electric charges> are integers. This is a consequence of the proposed <symmetry> extension, not evidence for such particles. If “known mesons” also includes a flavour singlet from $\mathbf3\otimes\overline{\mathbf3}$, that singlet can be accommodated by a separate <G2 (mathematics)> <trivial representation>: the minimal octet-plus-singlet construction is then $\mathbf{14}\oplus\mathbf1$, still with six new states. A demand for one irreducible <G2 (mathematics)> <multiplet> containing the entire nonet is a different extra assumption; $\mathbf{14}$ alone does not contain that singlet.
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