Let $(\lambda_1, \cdots , \lambda_d) \vdash k$ be a partition of $k$ of length $d$. Suppose we have $k$ non-zero vectors $\alpha^{(j)} = (\alpha^{(j)}_1, \cdots , \alpha^{(j)}_d) \in \mathbb{Z}^d$, such that $\sum_{i=1}^d \alpha_i^{(j)} = 0$ and $\alpha_1^{(j)} \geq \cdots \geq \alpha_d^{(j)}$, for $1 \leq j \leq k$. Does the origin belongs to the convex-hull of the following set? \begin{equation} \{ ( \underbrace{\alpha_1^{(j_1)}, \cdots, \alpha_1^{(j_1)}}_{\lambda_1}, \cdots , \underbrace{\alpha_d^{(j_d)}, \cdots , \alpha_d^{(j_d)}}_{\lambda_d} ) \in \mathbb{Z}^k: \, j_i \in \mathcal{J}_i \subset [k], \, |\mathcal{J}_i| = \lambda_i, \, 1 \leq i \leq d \}. \end{equation} In a simpler version that I asked here, the answer is no.
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3$\begingroup$ If I understand your question correctly, the answer should still be no for the same reason ($\alpha_d^{(j)}<0$ for all $j$), unless the zero vector is one of your alpha vectors. $\endgroup$– Yoav KallusCommented Aug 12, 2017 at 15:47
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$\begingroup$ You mean that the sets $\mathcal{J}_i$ are fixed? That they form a partition of $[k]$, and only $j_i\in \mathcal{J}_i$ vary? $\endgroup$– Fedor PetrovCommented Aug 19, 2017 at 10:52
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$\begingroup$ @FedorPetrov Yes, that is what I meant. $\endgroup$– SMDCommented Aug 20, 2017 at 2:38
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