The shuffle conjecture for the character of diagonal coinvariants

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Fix nn and let δn=(n−1,n−2,…,1,0)\delta_n=(n-1,n-2,\ldots,1,0). For a partition λ⊆δn\lambda\subseteq\delta_n, let TT range over semistandard tableaux of skew shape (λ+(1n))/λ(\lambda+(1^n))/\lambda, and let dinv⁡(T)\operatorname{dinv}(T) be the number of d-inversions defined from the diagonals of the tableau. Define

Dn(z;q,t)=∑λ⊆δn  ∑T∈SSYT⁡(λ+(1n)/λ)t∣δn/λ∣qdinv⁡(T)zT.D_n(z;q,t)=\sum_{\lambda\subseteq\delta_n}\;\sum_{T\in\operatorname{SSYT}(\lambda+(1^n)/\lambda)}t^{|\delta_n/\lambda|}q^{\operatorname{dinv}(T)}z^T.

Here ∇\nabla is the nabla operator on symmetric functions, ene_n is the elementary symmetric function, hμh_\mu is the complete homogeneous symmetric function, and ⟨  ⟩\langle\,\ \rangle is the Hall inner product. The shuffle conjecture. One has

∇en(z)=Dn(z;q,t).\nabla e_n(z)=D_n(z;q,t).

Equivalently, for every partition μ\mu,

⟨∇en,hμ⟩=∑λ⊆δn  ∑T∈SSYT⁡(λ+(1n)/λ,μ)t∣δn/λ∣qdinv⁡(T).\langle\nabla e_n,h_\mu\rangle=\sum_{\lambda\subseteq\delta_n}\;\sum_{T\in\operatorname{SSYT}(\lambda+(1^n)/\lambda,\mu)}t^{|\delta_n/\lambda|}q^{\operatorname{dinv}(T)}.

This gives a tableau formula for the character of the diagonal coinvariant ring. The source presents it as the main conjecture; no resolution is stated in the supplied text.

References

Primary source

J. Haglund, M. Haiman, N. Loehr, J. B. Remmel and A. Ulyanov, “A Combinatorial Formula for the Character of the Diagonal Coinvariants”, arXiv:math/0310424 (2004).

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