Log-concavity of matroids: characterization of equality?

Question

Let $M$ be a (loopless) matroid of rank $r$.

The characteristic polynomial $\chi_M(x)$ is defined by $\chi_M(x)=\sum_{F \in \mathcal{L}(M)}\mu(\hat{0},F) \cdot x^{\mathrm{rk}(F)}$, where $ \mathcal{L}(M)$ is the lattice of flats of $M$ and $\mu$ its Möbius function. It is known that the signs of the characteristic polynomial alternate, and the so-called Whitney numbers of the 1st kind $\omega_i$ are defined by $\chi_M(x) = \sum_{i=0}^{r} (-1)^i \omega_i x^i$.

In Hodge theory for combinatorial geometries, affirming a long-standing conjecture of Rota and others, Adiprasito-Huh-Katz (AHK) showed that these $\omega_i$ form a log-concave sequence, i.e., $\omega_i^2 \geq \omega_{i-1}\omega_{i+1}$.

Actually, they proved something stronger. It is known that $(1-x)$ is always a factor of $\chi_M(x)$, and the reduced Whitney numbers $\overline{\omega}_i$ are thus defined by $\chi_M(x)/(1-x)=\sum_{i=0}^{r-1} (-1)^i \overline{\omega}_i x^i$. AHK showed that in fact the reduced Whitney numbers form a log-concave sequence: $\overline{\omega}_i^2 \geq \overline{\omega}_{i-1}\overline{\omega}_{i+1}$ (which implies log-concavity of the $\omega_i$ by a straightforward argument).

In fact, an easy reduction using the truncation of the matroid $M$ shows that it is enough to prove log-concavity of the $\overline{\omega}_i$ in the "last-spot": i.e., that $\overline{\omega}_{r-2}^2 \geq \overline{\omega}_{r-3}\cdot\overline{\omega}_{r-1}$.

To prove this, AHK use the Chow ring $A(M) = A^0(M)\oplus A^1(M)\oplus \cdots\oplus A^{r-1}(M)$ of the matroid $M$: they show that $\overline{\omega}_k=\langle \alpha^{r-1-k}\beta^{k}\rangle$ for certain linear elements $\alpha,\beta\in A^1(M)$ of the Chow ring, where $\langle \cdot \rangle\colon A^{r-1}(M)\to \mathbb{R}$ is the canonical degree map isomorphism; and they deduce $\overline{\omega}_{r-2}^2 \geq \overline{\omega}_{r-3}\cdot\overline{\omega}_{r-1}$ from the so-called Kähler package for $A(M)$, in particular, the Hodge-Riemann relations, which imply that the relevant $2\times 2$ determinant is nonpositive.

Question: From the work of AHK (or elsewhere) is it possible to deduce when (i.e. for which matroids) we have an equality $\overline{\omega}_{r-2}^2 = \overline{\omega}_{r-3}\cdot\overline{\omega}_{r-1}$? (In other words, when the determinant of the $2\times 2$ HR relations matrix is $0$ rather than negative?)

Answer 1

I think the following shows it's never possible for there to be equality.

Indeed, Ardila-Denham-Huh https://arxiv.org/abs/2004.13116 recently showed for any matroid $M$ that $T_M(x,0)$ has log-concave coefficient sequence, and hence obviously $\frac{1}{x}T_M(x,0)$ has log-concave coefficient sequence (Note that $T_M$ always has $0$ constant term except when $M=\emptyset$).

Now, Lenz in Lemma 4.2 of https://arxiv.org/abs/1106.2944 showed that if $f(x)$ has non-negative log-concave coefficient sequence then $f(x+1)$ has strictly log-concave coefficient sequence.

Taking $f(x)=\frac{1}{x}T_M(x,0)$, we have $f(x+1)=\frac{1}{1+x}T_M(1+x,0)$ has strictly log-concave coefficient sequence. But this is exactly the reduced characteristic polynomial with all coefficients made to have positive signs.

Answer 2

I have nothing to add to Hunter's answer to your main question, but I thought it might be helpful to comment more generally on where the difficulty lies in extracting such information from a Kähler package.

Let us adopt the general setting in this review of Huh. Let $A(X)=\bigoplus_{q=0}^d A^q(X)$ be a graded algebra with Kähler cone $\mathrm{K}(X)\subset A^1(X)$, satisfying Poincare duality, hard Lefschetz (HL), and Hodge-Riemann (HR). A simple consequence of (HR) is that for any $\mathrm{L}_0,\ldots,\mathrm{L}_{d-2}\in\mathrm{K}(X)$ and $\eta\in A^1(X)$, we have $$(\eta \mathrm{L}_0\cdots \mathrm{L}_{d-2})^2\ge (\eta^2\mathrm{L}_1\cdots \mathrm{L}_{d-2})\, (\mathrm{L}_0^2\mathrm{L}_1\cdots \mathrm{L}_{d-2}).$$This is what gives rise to log-concavity in various applications. Moreover, (HL) implies that in this inequality, equality holds if and only if $\eta$ is proportional to $\mathrm{L}_0$. There are various ways to see this. For example, if one has equality in the above inequality, then the quadratic polynomial $$q(t) = ((\eta+t\eta') \mathrm{L}_0\cdots \mathrm{L}_{d-2})^2-((\eta+t\eta')^2\mathrm{L}_1\cdots \mathrm{L}_{d-2})\, (\mathrm{L}_0^2\mathrm{L}_1\cdots \mathrm{L}_{d-2})$$satisfies $q(0)=0$ and $q(t)\ge 0$ for all $t$ and $\eta'\in A^1(X)$. Computing $q'(0)=0$ shows that a linear combination of $\eta$ and $\mathrm{L}_0$ lies in the kernel of the map $\mathrm{L}_1\cdots \mathrm{L}_{d-2}:A^1(X)\to A^{d-1}(X)$. But (HL) states this map is a bijection, so its kernel is trivial, and thus $\eta$ and $\mathrm{L}_0$ are proportional.

The difficulty is that in many applications, the $\mathrm{L}_i$ of interest do not lie in $\mathrm{K}(X)$, but can only be approximated by elements of $\mathrm{K}(X)$. This is the case in AHK (see, e.g., the last section of this Notices paper), in the Khovanskii-Teissier inequality when one deals with nef classes rather than ample classes, etc. In such cases, the inequality still follows by approximation, but all information on the equality cases is lost. While no nontrivial equality cases turn out to appear in the particular case of AHK (as per Hunter's answer), many new equality cases can appear for the limiting inequalities in other situations.

As far as I know, the only structures of this kind where the equality cases of the limiting inequalities are fully understood (in the sense that there is a complete geometric/combinatorial characterization) are mixed discriminants and mixed volumes of convex polytopes. The last section of the latter paper contains some speculative remarks on what such results might look like more generally.

Answer 3

Not an answer, but a long comment. It is possible that your "when we have an equality" can have different answers depending what do you mean by that.

1. If you mean "give a combinatorial description" of what kind of matroids one can have, there is a decent chance that there is no characterization even for the case of graphs. What you are asking is a weaker version of "give a combinatorial interpretation of the difference" that I discuss in my inequalities note, which I conjecture is unlikely.

2. If you mean "give some equivalent condition" then this might be possible. For example, there is a recent paper by Shenfeld and van Handel (see Theorem 15.3) which proves that Stanley's log-concavity happens if and only if all three terms are equal. Maybe that's what you really want in this case.