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.


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.