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The Adjoint representation of G2 has one weight at each of its twelve roots and two at zero. Its six long roots form the A2 root system; together with the two zero weights they give the Adjoint representation of SU(3). The remaining six short roots form the weight systems of its defining representation and dual representation. Hence restriction to the long-root SU(3) gives $14=8\oplus 3\oplus \overline{3}$. This is a branching of a representation; the complement is invariant under the subgroup, not under the full G2 action.

The $G_2$ root system has six short and six long roots. For a short simple root ${\alpha }_1$ and a long simple root ${\alpha }_2$, its positive roots are

These are the root data associated with G2.

A G2 multiplet containing the known meson octet must restrict to a representation containing $8$. Among the two fundamental representations just computed, $7$ fails this test, while the Adjoint representation $14$ contains the required octet. Thus the minimal such choice isLarger 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 $q_A^{T}Pw=(P^T{q}_A{)}^{T}w$. Applying this to the electric charge and flavour hypercharge vectors givesThus 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 solution. The extra $3\oplus \overline{3}$ gives six additional mesons, one at each ofFor 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.

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 $3\otimes \overline{3}$, that singlet can be accommodated by a separate G2 trivial representation: the minimal octet-plus-singlet construction is then $14\oplus 1$, still with six new states. A demand for one irreducible G2 multiplet containing the entire nonet is a different extra assumption; $14$ alone does not contain that singlet.

Let ${\rho }_3(P)=P\otimes P\otimes P$ act on the alternating coefficient tensor $\phi$, and setThe coefficient convention here is the one in the source's hint; using the usual covariant pullback gives the same stabilizer subgroup, since an invertible transformation preserves a differential form if and only if its inverse does. The identity matrix fixes $\phi$. If $P,Q$ fix it, then ${\rho }_3(PQ)\phi ={\rho }_3(P){\rho }_3(Q)\phi =\phi$. Multiplying ${\rho }_3(P)\phi =\phi$ by ${\rho }_3(P^{-1})$ shows that $P^{-1}$ fixes it as well. Thus $H$ is a subgroup of $SO(7)$. Its defining equations are polynomial, so it is closed and hence a Matrix Lie group.

For $P(t)=I+tT+O(t^2)\in H$, differentiating $P^{T}P=I$ gives $T^T=-T$. Differentiating the three factors in ${\rho }_3(P(t))\phi =\phi$ givesThe skew-symmetric matrix condition records the ambient special orthogonal group; it must be included when differentiating that group's definition.

These linear conditions are sufficient as well as necessary. On the tensor product, the infinitesimal action isIts three terms commute, so ${\rho }_3(e^{tT})=e^{tD(T)}$. If $D(T)\phi =0$, every positive power of $D(T)$ annihilates $\phi$, giving $e^{tD(T)}\phi =\phi$. Also $T\in \mathfrak{so}(7)$ implies $e^{tT}\in SO(7)$. Therefore $e^{tT}\in H$, proving the asserted Lie algebra description. For the given positive three-form this is the compact G2 Lie algebra ${\mathfrak{g}}_2$.

On ${\mathbb{C}}^3\oplus \mathbb{R}$, take $\omega =(i/2){\sum }_{j}d{z}_j\wedge d{\bar{z}}_j$ and $\Omega =d{z}_1\wedge d{z}_2\wedge d{z}_3$. The special unitary group $SU(3)$ preserves the Kähler form $\omega$ by unitarity and preserves $\Omega$ by its unit determinant. Fixing $y$ therefore embeds it in the compact G2 stabilizer of $\varphi$. The complexified seven-dimensional representation restricts as $3\oplus \bar{3}\oplus 1$.