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” like a bird). How many vertices do you need? (yeah yeah you loose a factor of two… but go where the referee went)

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 (I am aware I might have one glass of wine too many as I write this… 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. Smooth 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. Curvature 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. The elementary 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. Moooorse 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