Archive for July 6th, 2009

A conjecture on the Uniqueness of MV-Polytopes

July 6, 2009

Fix a lattice L with \langle , \rangle and a partial order \le. Define
Q^+=\{x\in L|x\ge 0\} = \text{``positive root cone''}
L^+=\{x\in L|\langle x,y\rangle \ge 0 \forall y\in Q^+\} = \text{``dominant weights''}
(Alternatively we could fix Q^+ and define x\ge y \iff x-y\in Q^+. Similarly we might fix L^+ without reference to any inner product.)

We say that a set of “characters” \{\chi_\lambda:L\rightarrow\mathbb{N}\}_{\lambda\in L^+} and a set of subsets \{M^\lambda_\mu \subset L\}_{\lambda,\mu\in L} satisfy the “tensor property” if:

  1. \chi_\lambda(\mu)=\#\{m\in M^\lambda_\mu | m\subset \text{supp} \chi_\lambda\}
  2. \chi_{\lambda_1} * \chi_{\lambda_2}=\sum c^{\lambda_1,\lambda_2}_\mu \chi_\mu where
    c^{\lambda_1,\lambda_2}_\mu= \#\{ m \in  M^{\lambda_1+\lambda_2}_\mu  | m\subset ((\text{supp}\chi_{\lambda_1})+\lambda_2)\cap(\mu+\lambda_2-\text{supp}\chi_{\lambda_2}))\}

(Here * indicates convolution and \mathbb{N} includes zero.)

Suppose we also require the following about \{M^\lambda_\mu\}

  1. M^\lambda_\mu\neq\o \iff \lambda\ge\mu
  2. m\in M^\lambda_\mu\Rightarrow \mu\le x\le\lambda \forall x\in m \text{ and } \mu,\lambda\in m
  3. m\in M^\lambda_\mu\Rightarrow m+a\in M^{\lambda+a}_{\mu+a} \forall a\in L \text{ and } (-m)\in M^{-\mu}_{-\lambda}
  4. (*)\sigma a\in M^{\sigma\lambda}_{\sigma\mu} for any isotopy, \sigma of Q^+.

(I’m not sure if the last condition is worded correctly, but I’d like the collection to be invariant under permutations of, for example, minuscule weights.)

And, fixing a group W acting on L for which L^+ is a fundamental domain, suppose we require that:

  1. \chi_\lambda(\mu)=\chi_\lambda(w\mu) \forall w\in W

Conjecture For L,Q^+,L^+,W there is only one \{M^\lambda_\mu\}_{\lambda,\mu\in L} (and corresponding \{ \chi_\lambda\}_{\lambda\in L^+}) satisfying the tensor property and all the above requirements.

For L,Q^+,L^+,W coming from \mathfrak{sl_3} or \mathfrak{sp_4} it seems to be true.