This question is motivated by

  1. Why do combinatorial abstractions of geometric objects behave so well?
  2. The algebraic geometry of Kazhdan-Lusztig-Stanley polynomials

Kazhdan-Lusztig-Stanley polynomials (KLS) are vast generalizations of the classical Kazhdan-Lusztig polynomials, whose special values have deep meaning in representation theory [1]. They also include the matroid analogue studied in the past ten years. (Aside: they also include general zeta functions.)

KLS bridge combinatorics and algebraic geometry. While the nonnegative coefficients of the KLS can be interpreted as the dimension of suitable cohomologies of certain perverse sheaves [2], it seems to be less transparent in pure combinatorial settings.

I hope to understand KLS more from its combinatorial perspective, without any interference by the geometrical side. However, the definition of KLS [2] is done algebraically on the deformed dual of the underlying poset, making its meaning less transparent.


  • Why did combinatorialists consider KLS in their point of view?
  • Any baby examples of posets whose KLS shows rich combinatorial information right away?

Remark: I have no background in combinatorics. Being aware of that KLS also generalizes (in some sense) many combinatorial invariants (h-vector, g-polynomials).. I'd hope the answer can be pedagogical, and show the easiest nontrivial example.


  • 3
    $\begingroup$ Well, the obvious reference is Stanley's original paper, Subdivisions and local $h$-vectors. Is this paper unsatisfactory for your purposes? $\endgroup$ – Timothy Chow Oct 1 at 21:18
  • 1
    $\begingroup$ The Proudfoot paper is very readable and contains many nice examples. It maybe doesn't offer much insight into why you would define these general constructions; but in some sense the examples motivate the machinery. $\endgroup$ – Sam Hopkins Oct 2 at 4:00
  • $\begingroup$ @TimothyChow Stanley's original paper is nice. But I still feel there's a deep rabbit hole. I have tried to summarize what I should ask, and maybe the question should have been: what's the combinatorial meaning of the h-vector? Sure, h-vector is just a shift of the f-vector, whose meaning is clear. But that dims its meaning, and I couldn't find explanation except from the point of view of intersection cohomology of some toric variety. $\endgroup$ – Student Oct 2 at 7:24
  • $\begingroup$ @Student : I don't think there's any intuitively satisfying purely combinatorial way of understanding the h-vector. Historically, the h-vector was discovered when people were trying to characterize which vectors can arise as f-vectors. There are some linear constraints. These linear constraints can be expressed in a particularly simple way if one changes variables from f to h (the Dehn-Sommerville equations, or in geometric terms, Poincare duality). If your intuition tells you that the h-vector must have some direct combinatorial meaning, then I think you can regard that as an open problem. $\endgroup$ – Timothy Chow Oct 2 at 12:57
  • 1
    $\begingroup$ Note that one of the linear conditions can be regarded as saying something about the Euler characteristic. So you might start by asking, what is the combinatorial meaning of the Euler characteristic? The conventional wisdom is that the Euler characteristic is best understood topologically and not combinatorially. But if you can find a convincing combinatorial interpretation of the Euler characteristic, then perhaps you can generalize that to a combinatorial interpretation of the h-vector. $\endgroup$ – Timothy Chow Oct 2 at 13:01

This is maybe more addressed at the comments and the discussion with Timothy Chow, but I just wanted to point out that, at least in a certain context, there is a very very "concrete" description of the $h$-vector. Say $\mathcal{P}$ is a simple (convex, full-dimensional) polytope in $\mathbb{R}^n$. Then let $\phi$ be a generic enough linear functional on $\mathbb{R}^n$. Use $\phi$ to orient the $1$-skeleton of $\mathcal{P}$: orient an edge $uv$ from $u$ to $v$ if $\phi(u) < \phi(v)$ (since $\phi$ is generic there will not be ties). Then if $h=(h_0,h_1,\ldots,h_n)$ is the $h$-vector of $\mathcal{P}$ (defined in the usual way as a transform of the $f$-vector), we have that $$ h_i = \# (\textrm{vertices $v$ with indegree $=i$})$$ according to our orientation of the $1$-skeleton. So for instance this explains that the $h_i$ are positive, that $h_0+h_1+\cdots+h_n$ is the number of vertices; also we will have a $h_i=h_{n-i}$ symmetry which swaps indegree according to $\phi$ for outdegree according to $-\phi$, etc.

Incidentally, I don't know who to attribute this simple but nice perspective on the $h$-vector to; to me it is folklore.

EDIT: As Richard notes in the comments this perspective is the same as the idea of a line shelling for a simplicial polytope, which I guess was assumed by Schläfli in his proof of the Euler-Poincaré formula and formally established by Bruggesser and Mani.

| cite | improve this answer | |
  • 2
    $\begingroup$ This description of the $h$-vector is just the dual to the description of the $h$-vector of a simplicial polytope via a line shelling. $\endgroup$ – Richard Stanley Oct 2 at 14:22
  • $\begingroup$ Thanks for the explanation in certain context. I provide another one here: A combinatorial interpretation of the h- and gamma-vectors of the cyclohedron-[Pol Gomez Riquelme]. Not sure if it can be interpreted as the same thing. $\endgroup$ – Student Oct 2 at 14:42
  • $\begingroup$ @Student: that's an interpretation for a very special class of simple polytopes. But let me mention that the orientation/indegree picture does work very nicely in some special cases. E.g.: the permutohedron- its oriented $1$-skeleton is the Hasse diagram of the weak order on the symmetric group and we see that the $h$-vector gives the Eulerian numbers (permutations according to descents). Or the associahedron- its $1$-skeleton becomes the Tamari lattice and $h$-vector is Narayana numbers (Dyck paths by number of valleys). If I need an "intuitive" picture of the $h$-vector, I use orientations. $\endgroup$ – Sam Hopkins Oct 2 at 14:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.