The shuffle conjecture for the character of diagonal coinvariants

Fix nn and let δn=(n1,n2,,1,0)\delta_n=(n-1,n-2,\ldots,1,0). For a partition λδn\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)=λδn  TSSYT(λ+(1n)/λ)tδn/λqdinv(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=λδn  TSSYT(λ+(1n)/λ,μ)tδn/λqdinv(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.

Sources & referencesView supporting material

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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