e-log-concavity conjecture for chromatic quasisymmetric functions

At least 1 year old · documented by

Let G=([n],E)G=([n],E) be a natural unit interval graph, and write

XG(x;q)=∑λcλ(q)eλ.X_G(\bm x;q)=\sum_\lambda c_\lambda(q)e_\lambda.

A polynomial cλ(q)c_\lambda(q) with coefficients aℓa_\ell is ee-log-concave when its coefficients satisfy aℓ2≥aℓ+1aℓ−1a_\ell^2\geq a_{\ell+1}a_{\ell-1} for every relevant index ℓ\ell.

ee-log-concavity conjecture. Let GG be a natural unit interval graph. Then XG(x;q)X_G(\bm x;q) is ee-log-concave.

This is proposed as a strengthening of ee-unimodality. The source reports computer verification for every natural unit interval graph with at most 1010 vertices, but no general proof.

Equivalent formulations 1Other wordings

Other statements of this same problem, merged from separate entries. Each is equivalent to the statement above — proving any one settles them all.

  1. e-log-concavity conjecture for chromatic quasisymmetric functions

    Let GG be a natural unit interval graph, and write

    XG(x;q)=∑μcμ(q)eμ.X_G(\bm x;q)=\sum_\mu c_\mu(q)e_\mu.

    A polynomial a(q)=∑k=ijakqk∈N[q]a(q)=\sum_{k=i}^j a_kq^k\in\mathbb N[q] is log-concave when ak2≥ak−1ak+1a_k^2\geq a_{k-1}a_{k+1} for every kk. The e-log-concavity conjecture. The function XG(x;q)X_G(\bm x;q) is ee-log-concave: for every partition μ\mu, the coefficient cμ(q)c_\mu(q) of eμe_\mu is a log-concave polynomial. The author reports computer verification for every natural unit interval graph with at most 1010 vertices. Whether this holds in general remains open.

    source: Foster Tom, “A signed e-expansion of the chromatic quasisymmetric function”, arXiv:2311.08020 (2024).

References

Primary source

Bruce E. Sagan and Foster Tom, “Chromatic symmetric functions and change of basis”, arXiv:2407.06155 (2024).

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.