Power-sum equality for orientation posets

At least 6 years old · documented by

Let a\mathbf{a} be an area sequence of length nn, let O(a)O(\mathbf{a}) be the set of orientations associated with a\mathbf{a}, let P(θ)P(\theta) be the poset obtained from the ascending edges of θ\theta, and let B(θ)B(\theta) be the disjoint union of chains whose lengths are given by π‾(θ)\overline{\pi}(\theta). For a partition λ⊢n\lambda\vdash n, let Oλ∗(P)\mathcal{O}^{\ast}_{\lambda}(P) denote the set of order-preserving surjections of type λ\lambda whose fibers have unique minimal elements. Equivalent power-sum conjecture. For every area sequence a\mathbf{a} of length nn and every partition λ⊢n\lambda\vdash n,

∑θ∈O(a)qasc⁡(θ)∣Oλ∗(P(θ))∣=∑θ∈O(a)qasc⁡(θ)∣Oλ∗(B(θ))∣.\sum_{\theta\in O(\mathbf{a})}q^{\operatorname{asc}(\theta)}\left|\mathcal{O}^{\ast}_{\lambda}(P(\theta))\right|=\sum_{\theta\in O(\mathbf{a})}q^{\operatorname{asc}(\theta)}\left|\mathcal{O}^{\ast}_{\lambda}(B(\theta))\right|.

This is stated as an equivalent reformulation of the main LLT identity after comparing power-sum coefficients; the source gives no resolution status.

References

Primary source

Per Alexandersson, “LLT polynomials, elementary symmetric functions and melting lollipops”, arXiv:1903.03998 (2019).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.