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The $\mathbf{i}$-trails of Berenstein and Zelevinsky was introduced on page 5 (Definition 2.1) in this paper. It is defined as follows. Let $\gamma, \delta \in \mathfrak{h}^*$. Let ${\bf i}=(i_1, \ldots, i_l)$. Then $\pi=(\gamma = \gamma_0, \gamma_1, \ldots, \gamma_l=\delta)$ is called an ${\bf i}$-trail if for $k=1,\ldots, l$, $\gamma_{k-1}-\gamma_{k} = c_k\alpha_{i_k}$ for some $c_k \in \mathbb{Z}_{\geq 0}$ and $e_{i_1}^{c_1} \cdots e_{i_l}^{c_l}$ is a non-zero linear map from $V(\delta)$ to $V(\gamma)$.

One main result in the paper is a polyhedral formula for tensor product multiplicities (Theorems 2.2 and 2.3 in the above paper).

My question is: let $\mathfrak{g}$ be the simple complex Lie algebra of type $G_2$. Let $\lambda=k_1 \omega_1 + l_1 \omega_2$ and $\mu=k_2 \omega_1 + l_2 \omega_2$, where $\omega_1, \omega_2$ are fundamental weights. Are there some polyhedral formula which describe the decomposition of $V(\lambda) \otimes V(\mu)$ into irreducible $\mathfrak{g}$-modules using only $k_1, k_2, l_1, l_2$? The formula will be of the form $$ V(\lambda) \otimes V(\mu) = \oplus_{(r_1, r_2) \in A} V(r_1 \omega_1 + r_2 \omega_2), \quad (1) $$ where $$A = \left\{(r_1, r_2) \in \mathbb{Z}_{\geq 0}^2 \middle\vert \begin{aligned} r_1 &=g_1(k_1,k_2,l_1,l_2),\\ r_2 &=g_1(k_1,k_2,l_1,l_2),\\ 0\ &\le f_1(k_1,k_2,l_1,l_2),\\ &\quad\ldots,\\ 0\ &\le f_m(k_1,k_2,l_1,l_2) \end{aligned} \right\},$$ and where $f_1,\ldots, f_m,g_1,g_2$ are some (linear) polynomials. Maybe use the definition of ${\bf i}$-trail, we can translate Theorem 2.2 in the paper to the form $(1)$. But I don't know how to verify that $e_{i_1}^{c_1} e_{i_2}^{c_2}$ is a non-zero linear map from $V(s_i \omega_i^{\vee})$ to $V(w_0 \omega_i^{\vee})$ (in the case of type $G_2$, $w_0 = s_1s_2s_1s_2s_1s_2$). Are there some references which have some formula similar to $(1)$? Thank you very much.

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There is a bit of a mistake in the question. If a formula like (1) held, then tensor product multiplicities for $G_2$ could only be at most 1.

On the other hand, Theorem 2.2 of the paper does give a polyhedral formula of the form: $$ V(\lambda)\otimes V(\mu) = \oplus_{(t_1, t_2, t_3, t_4, t_5, t_6) \in A} V(\lambda + \mu - \sum_{k=1}^6 t_k \beta_k)$$ where $ \beta_1, \dots, \beta_6 $ are the positive roots of $G_2$ (ordered using a reduced word for $ w_0$) and $ A $ is a polyhedron.

In Theorem 2.2, $ A$ is described using $\mathbb i$-trails. I agree that this description of $ A $ is a bit confusing. However, there is a simpler description given in Corollary 3.4 of the same paper. $$ A = \{ (t_1, \dots, t_6) : t_k \ge 0, t_1 \le \langle \lambda, \alpha_1^\vee \rangle, p_1 \le \langle \lambda, \alpha_2^\vee \rangle, t_6 \le \langle \mu, \alpha_2^\vee \rangle, p_6 \le \langle \mu, \alpha_1^\vee \rangle \} $$ where $p_1, p_6 $ are computed from $ t_1, \dots, t_6 $ using the tropicalization of the formula in Proposition 7.1.(4). Since $p_1, p_6 $ are tropical expressions in $ t_1, \dots, t_6 $, the above description of $ A $ consists only of inequalities. Unfortunately, the formulas for $ p_1, p_6 $ are pretty complicated, but this is an explicit description of the polyhedron.

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