IDP polytopes, hypersimplices, projective spaces

I wanted to update on 3 small new results.

First, Luis Ferroni used LLM (and a lot of persistence and cleverness) to prove something I long suspected: that IDP lattice polytopes do NOT have unimodal $h^\ast$ vectors. Unimodality is nice because it makes the sequence especially simple: it rises, then falls again… a dromedary, instead of a camel. And that IDP polytopes (that is, polytopes which are generated by atoms on the same level, that is, level one) satisfy it was a (I thought unlikely, but long standing) conjecture of Stanley….

Luis found, quite amazingly, that these are rather nice polytopes: smooth Cayley polytopes of rectangular prisms. This complements our work on IDP polytopes, proving among other things monotonicity of the $h^\ast$ vector in the second half. Congratulations Luis!

The paper is already transparent and good, so you should read it.

Second small result that this makes more relevant: Mingzhi Zhang and I found a way to prove that hypersimplices, a very nice and simple class of polytopes, are “eventually” unimodal at least; if you fix rank, then increasing the dimension leads you to unimodality. The proof is… really just a little analysis. It will probably appear soon on HAL.

Finally, you might remember some time ago, Sergey Avvakumov, Roman Karasev and I constructed subexponential size triangulations of $\mathbb{RP}^n$ ? Well, when we submitted it to Crelle, we got the following report

Well, suck it referee. Turns out, our construction was essentially tight! The paper is here, but since it is a little heavily AI generated (down to the title…) let me try to give a human perspective. It is essentially a method of Gromov, which I encourage you to learn (just as I encourage you to look at Martin’s paper on Wachspress geometry. Goose stuff). More on that later.

The question is this: How many vertices does a triangulation of $\mathbb{RP}^n$ need to have? Equivalently: Assume you have triangulation of $S^n$ with free involution (henceforth just “free”). How many vertices do you need? (you loose a factor of two…)

The key theorem is that a finite strongly regular complex $X$ with $v$ vertices and $f$ facets satisfies

$\displaystyle \mathrm{ind} X \le C\log v\log f.$

Here $\mathrm{ind} X$ is the least $k$ for which $X$ admits an equivariant map to the antipodal sphere $S^k$ ; in particular, $\mathrm{ind} S^d=d$ . A facet is an inclusion-maximal simplex. $C$ is an absolute constant, and logarithms are natural, because we don’t do artificial here (lets see where it goes). The reduction of the question to this is inequality (which the authors call “topological Figiel-Lindenstrauss-Milman) is elementary (covered below). The proof of the inequality (also described) owes to a marvellous idea of Gromov to uses traces to bound indices of matrices, and apply it to Morse functions. The authors fully acknowledge that this was LLM generated, but I want to make their proof a little more human-understood and transparent here, so I also cover it. (I also mention that the method, in physics, is also known as Birman-schwinger.)

1. Enlarge the lifted triangulation

Lift an $N$ -vertex triangulation of $\mathbb{R}P^d$ to an antipodal triangulation $K$ of $S^d$ , with $2N$ vertices. For each vertex $w$ , let $N[w]$ consist of $w$ and its neighbors. Form $L$ by filling each $N[w]$ with a full simplex.

Every simplex of $K$ lies in $L$ . No $N[w]$ contains an antipodal vertex pair: two edges from $w$ to the two lifts of the same base vertex would violate uniqueness of edge lifting. Thus $L$ still has a free involution, with $2N$ vertices and at most $2N$ facets. The theorem gives

$\displaystyle \begin{gathered} d=\mathrm{ind} K\le\mathrm{ind} L\le C\bigl(\log(2N)\bigr)^2, \\[4pt] N\ge\tfrac12\exp\bigl((d/C)^{1/2}\bigr). \end{gathered}$

$L$ need not have the homotopy type of a sphere. The equivariant inclusion $K\subseteq L$ is enough to preserve the required lower bound on its index.

2. Encode vertices and facets by a matrix

For a free simplicial complex with $n$ vertex orbits and $m$ facet orbits, place the vertices at $\pm e_j$ in $\mathbb{R}^n$ . Choose one facet from each orbit. Set $A_{ij}=1$ if that facet contains $e_j$ , $-1$ if it contains $-e_j$ , and $0$ otherwise.

On each facet, one coordinate of $Ax$ is $1$ or $-1$ . Since $|A_{ij}|\le1$ , the realized complex lies in

$\displaystyle R(A):=\bigl\{x\in\mathbb{R}^n:\|x\|_1=\|Ax\|_\infty=1\bigr\}.$

The norms are $\|x\|_1=\sum_j|x_j|$ and $\|Ax\|_\infty=\max_i|(Ax)_i|$ . It suffices to show $\mathrm{ind} R(A)\le C\log(2m)\log(2n)$ . This matrix bound holds whenever $|A_{ij}|\le1$ .

3. Smoothen the norms

Assume $m,n\ge2$ ; the other cases are immediate. Choose an even integer $p\ge4$ , an exponent $1<q<2$ , and $\varepsilon=1/(4n)$ . Define

$\displaystyle \begin{gathered} F(x)=\sum_{i=1}^{m}|(Ax)_i|^p, \\[4pt] G(x)=\sum_{j=1}^{n}(x_j^2+\varepsilon^2)^{q/2},\qquad\Sigma=\{G=1\}. \end{gathered}$

$F^{1/p}$ approximates $\|Ax\|_\infty$ ; $\Sigma$ is a smooth approximation to the $\ell_1$ -sphere when $q$ is close to $1$ . Radial rescaling sends $R(A)$ into $\{x\in\Sigma:F(x)^{1/p}\ge3/4\}$ . We will bound the Hessian at all critical points with $F^{1/p}\ge1/2$ ; the gap between the levels permits a later perturbation.

4. Lagrange gives a negative definite term

Fix a critical point $z$ of $F$ on $\Sigma$ . Write $E=F(z)$ , $y_i=(Az)_i$ , and $r_j=(z_j^2+\varepsilon^2)^{1/2}$ . Then $\sum_i|y_i|^p=E$ and $\sum_j r_j^q=1$ . For a tangent vector $h$ , let $H(h,h)$ be the second derivative of $F$ along $\Sigma$ .

At acritical point, $\nabla F(z)=\lambda\nabla G(z)$ , and

$\displaystyle H(h,h)=\nabla^2F(z)[h,h]-\lambda\nabla^2G(z)[h,h].$

Hence, $z$ gives $pE=\lambda\langle z,\nabla G(z)\rangle$ , with $0<\langle z,\nabla G(z)\rangle\le q$ , so $\lambda\ge pE/q$ . Differentiate! Then $\nabla^2G(z)[h,h]\ge q(q-1)\sum_j r_j^{q-2}h_j^2$ . Hence,

$\displaystyle \begin{gathered} \frac{H(h,h)}{p}\le(p-1)\sum_i|y_i|^{p-2}(Ah)_i^2 \\[2pt] {}-(q-1)E\sum_j r_j^{q-2}h_j^2. \end{gathered}$

The first term is positive semidefinite. The second is negative definite. Now, how many non-negative eigenvalues (with multiplicity are there?)

5. A trace estimate

Lemma. If a symmetric quadratic form $B$ satisfies $B\le P-\beta I$ , where $P$ is positive semidefinite and $\beta>0$ , then the number $k$ of nonnegative eigenvalues of $B$ is at most $\mathrm{tr}(P)/\beta$ . Indeed, choose orthonormal vectors $u_1,\ldots,u_k$ in its nonnegative eigenspace. Each satisfies $u_i^T P u_i\ge\beta$ . Completing them to an orthonormal basis gives $\mathrm{tr}(P)\ge k\beta$ . This also works when $B$ is defined only on a subspace.

Rescale our tangent coordinates by $u_j=r_j^{(q-2)/2}h_j$ . The negative term becomes $-(q-1)E\|u\|^2$ . The positive matrix $P$ has trace $(p-1)\sum_{i,j}|y_i|^{p-2}A_{ij}^2 r_j^{2-q}$ . Applying the lemma and using $|A_{ij}|\le1$ gives

$\displaystyle \begin{gathered} k\le\frac{p-1}{(q-1)E}\left(\sum_i|y_i|^{p-2}\right)\left(\sum_j r_j^{2-q}\right) \\[4pt] \le\frac{p-1}{q-1}\,m^{2/p}n^{2-2/q}E^{-2/p}. \end{gathered}$

The last inequality is the one and only Hölder, using $\sum|y_i|^p=E$ and $\sum r_j^q=1$ . Here $k$ counts nonnegative Hessian eigenvalues, including zero; it is not yet the equivariant index.

6. Morse theory

Take $p=2\lceil\log(2m)\rceil$ and $q=1+1/(2\log(2n))$ . Then $m^{2/p}$ and $n^{2-2/q}$ are bounded by absolute constants. At the relevant critical points $E^{1/p}\ge1/2$ , so $E^{-2/p}\le4$ . Thus

$\displaystyle k\le C\log(2m)\log(2n).$

For a superlevel set, Morse cell dimensions count positive Hessian directions. A small perturbation on the quotient $\Sigma/\{\pm1\}$ therefore gives a slightly larger superlevel with a CW model of this bounded dimension. Its double cover gives an equivariant CW model. A free $D$ -dimensional CW complex maps equivariantly to $S^D$ , proving the desired index bound.

On to more important business now.

Unsolicited Nonsense