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For integral polytopes, it is conjectured (T. Hibi), that if the $h^*$-vector is symmetric, then it is also unimodal (increasing, then non-decreasing).

A non-integral polytope do not, in general, have a polynomial Ehrhart function. However, if it does, we can compute the corresponding $h^*$-vector.

Now, are there explicit examples where the conjecture mentioned above is false, published somewhere? That is, that the conjecture does not extend to non-integral polytopes of this special kind above?

I stumbled upon such an example in my current research, and I am just wondering if it is worth mentioning. The polytope in question is the Gelfand-Tsetlin polytope, $GT(\lambda,w)$ for $\lambda=(2,2,2)$ and $w=(1,1,1,1,1,1)$.

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  • $\begingroup$ What is an $h^*$-vector? What is unimodal? What is an Ehrhart function? What is the Gelfand-Tsetlin polytope? What is ... ? $\endgroup$
    – Tom De Medts
    Commented Mar 26, 2014 at 16:01
  • $\begingroup$ The definition of GT-polytope not super-hard, you can find it in any paper if you google for this term, but it is lengthy, and is not essential for the question... $\endgroup$
    – Per Alexandersson
    Commented Mar 26, 2014 at 20:40

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The conjecture is false for integral polytopes in dimension $\ge 6$; see Mircea Mustata and Sam Payne, Ehrhart polynomials and stringy Betti numbers, Math. Ann. 333 (2005), no. 4, 787-795 (arXiv version), and Sam Payne, Ehrhart series and lattice triangulations, Discrete Comput. Geom. 40 (2007), 365-376 (arXiv version). For a rational version of Hibi's palindromic theorem, you might want to check out this paper by Matthew Fiset and Alexander Kasprzyk. The unimodality conjecture for integrally closed reflexive polytope is still open; see, e.g., this recent paper by Ben Braun and Robert Davis.

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