Polar bodies and the volume product

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

For an origin-symmetric convex body \(K\subset\mathbf R^n\), the volume product \(\operatorname{vol}(K)\operatorname{vol}(K^\circ)\) does not change under invertible linear maps. Mahler conjectured that it is at least \(4^n/n!\), the value for the cube. Iriyeh and Shibata proved this for \(n=3\) [IS]. In every dimension the inequality was known up to a factor \(c^n\) (Bourgain and Milman; see [Ku]), and Kuperberg proved it up to the factor \((\pi/4)^n\gamma_n\), where \(\gamma_1=4/\pi\) and \(\gamma_n\to\sqrt2\) [Ku]. Artstein-Avidan, Karasev and Ostrover computed a symplectic capacity of \(K\times K^\circ\) to be \(4\) and observed that Viterbo's volume–capacity conjecture would therefore give the inequality, because symplectic maps preserve volume [AKO]. Viterbo's conjecture fails for some convex domains [HKO]; a ball of capacity close to \(4\) inside \(K\times K^\circ\) still suffices. OpenAI proved that every ball of capacity less than \(4\) embeds symplectically into \(\operatorname{int}K\times\operatorname{int}K^\circ\) [OpenAI-SB]. This course proves that theorem (Theorem B) and the symmetric Mahler inequality (Theorem A).

The theorems

A convex body is a compact convex subset of \(\mathbf R^n\) with nonempty interior. It is origin-symmetric if \(K=-K\). Its polar is \[ K^\circ=\{p\in\mathbf R^n:\langle q,p\rangle\le1\ \forall q\in K\}. \]

Theorem A (symmetric Mahler inequality). For every \(n\ge1\) and every origin-symmetric convex body \(K\subset\mathbf R^n\), \[ \operatorname{vol}_n(K)\,\operatorname{vol}_n(K^\circ)\ge\frac{4^n}{n!}. \] Equality holds for the cube \([-1,1]^n\) (Example 1.7).

Write the points of \(\mathbf R^{2n}=\mathbf R^n_q\times\mathbf R^n_p\) as \((q,p)\), and let \(J\) be the matrix of the standard symplectic form \(\omega_0=\sum_jdq_j\wedge dp_j\), that is, \(\omega_0(X,Y)=X^{\mathsf T}JY\) with \[ J=\begin{pmatrix}0&I_n\\-I_n&0\end{pmatrix}. \] A smooth map \(e\) from an open set \(U\subseteq\mathbf R^{2n}\) to \(\mathbf R^{2n}\) is symplectic if \(De(z)^{\mathsf T}J\,De(z)=J\) for every \(z\in U\). For \(c>0\), the ball of capacity \(c\) is \[ B^{2n}(c)=\{(q,p):\ \pi(|q|^2+|p|^2)<c\}, \] the open Euclidean ball of radius \(\sqrt{c/\pi}\); its volume is \(c^n/n!\).

Theorem B (OpenAI). Let \(n\ge2\), let \(K\subset\mathbf R^n\) be an origin-symmetric convex body and \(U_K=\operatorname{int}K\times\operatorname{int}K^\circ\). For every \(0<c<4\) there is an injective symplectic map \(B^{2n}(c)\to U_K\), which is a diffeomorphism onto an open subset of \(U_K\).

Theorem B gives Theorem A for \(n\ge2\) because symplectic maps preserve volume (Lemma 2.4); the deduction is in the last lesson. The capacity \(4\) cannot be exceeded: a linear symplectic change of coordinates maps \(U_K\) into \((-1,1)^2\times\mathbf R^{2n-2}\) [AKO, Remark 4.2], and Gromov's nonsqueezing theorem then bounds the capacity of every embedded ball by \(4\). This upper bound is not used in the course and nonsqueezing is not proved here.

1. Polar bodies and support functions

The support function of a compact set \(C\subset\mathbf R^n\) is \(h_C(p)=\max_{y\in C}\langle p,y\rangle\). Thus \(K^\circ=\{p:h_K(p)\le1\}\). We write \(B(x,r)\) for the open Euclidean ball.

Lemma 1.1. Let \(K\) be an origin-symmetric convex body. There are \(0<r\le R\) with \(B(0,r)\subseteq K\subseteq\overline{B(0,R)}\). Then \(B(0,1/R)\subseteq K^\circ\subseteq\overline{B(0,1/r)}\), \(K^\circ\) is an origin-symmetric convex body, and \[ \operatorname{int}K^\circ=\{p:\ h_K(p)<1\}. \]

Proof. Let \(B(y,r)\subseteq K\). By symmetry \(B(-y,r)\subseteq K\), and by convexity the midpoints \(\frac12(a+b)\) with \(a\in B(y,r)\), \(b\in B(-y,r)\) fill \(B(0,r)\), which therefore lies in \(K\). Boundedness gives \(R\). If \(|p|<1/R\), then \(\langle q,p\rangle\le|q||p|<1\) for \(q\in K\). If \(p\ne0\) lies in \(K^\circ\), the point \(q=tp/|p|\) with \(t<r\) lies in \(K\), so \(t|p|\le1\) for all \(t<r\), and \(|p|\le1/r\). The set \(K^\circ\) is an intersection of closed half-spaces, hence closed and convex, and it is symmetric because \(K\) is.

The support function satisfies \(|h_K(p)-h_K(p')|\le R|p-p'|\), so \(\{h_K<1\}\) is open; it lies in \(K^\circ\) and hence in the interior. If \(h_K(p)=1\), the points \((1+\varepsilon)p\) have \(h_K=1+\varepsilon>1\) and lie outside \(K^\circ\), so \(p\) is not an interior point. \(\square\)

Lemma 1.2 (separation). Let \(C\subset\mathbf R^n\) be a nonempty compact convex set and \(x\notin C\). There is a vector \(v\) with \(\langle v,x\rangle>h_C(v)\).

Proof. Let \(c_0\) be a point of \(C\) nearest to \(x\), and \(v=x-c_0\ne0\). For \(y\in C\) and \(0<s\le1\), the point \(c_0+s(y-c_0)\) lies in \(C\), so \(|v-s(y-c_0)|^2\ge|v|^2\); dividing by \(s\) and letting \(s\to0\) gives \(\langle v,y-c_0\rangle\le0\). Hence \(\langle v,y\rangle\le\langle v,c_0\rangle=\langle v,x\rangle-|v|^2\) for all \(y\in C\). \(\square\)

Lemma 1.3. (a) For nonempty compact convex sets \(C\) and \(C'\), \(C\subseteq C'\) if and only if \(h_C\le h_{C'}\).

Let \(C\) and \(C'\) be convex bodies containing \(0\) in their interiors.

(b) \(C^{\circ\circ}=C\).

(c) If \(C\subseteq C'\), then \(C'^\circ\subseteq C^\circ\).

Proof. (a) Inclusion gives the inequality of maxima. Conversely, if \(x\in C\setminus C'\), Lemma 1.2 gives \(v\) with \(\langle v,x\rangle>h_{C'}(v)\), while \(\langle v,x\rangle\le h_C(v)\).

(b) If \(x\in C\), then \(\langle x,p\rangle\le1\) for every \(p\in C^\circ\), so \(x\in C^{\circ\circ}\). If \(x\notin C\), Lemma 1.2 gives \(v\) with \(\langle v,x\rangle>h_C(v)\); here \(h_C(v)>0\) because \(C\) contains a ball about \(0\). The vector \(v/h_C(v)\) lies in \(C^\circ\) and pairs with \(x\) to a value greater than \(1\), so \(x\notin C^{\circ\circ}\).

(c) follows from the definition. \(\square\)

Lemma 1.4 (strip bodies). Let \(b_1,\dots,b_m\in\mathbf R^n\) span \(\mathbf R^n\), let \(P\) be the convex hull of \(\pm b_1,\dots,\pm b_m\), and put \[ K'=\bigcap_{j=1}^m\{x\in\mathbf R^n:|\langle b_j,x\rangle|\le1\}. \] Then \(K'\) is an origin-symmetric convex body, \(K'=P^\circ\) and \((K')^\circ=P\).

Proof. A linear functional attains its maximum over \(P\) at one of the points \(\pm b_j\), so \(P^\circ=K'\). Since the \(b_j\) span \(\mathbf R^n\), the linear map \(x\mapsto(\langle b_j,x\rangle)_j\) is injective, so there is \(C_b\) with \(|x|\le C_b\max_j|\langle b_j,x\rangle|\); thus \(K'\) is bounded, and it contains the ball \(B(0,1/\max_j|b_j|)\). It is closed, convex and symmetric. The set \(P\) contains a basis of vectors together with their negatives, hence a neighbourhood of \(0\), so Lemma 1.3(b) gives \((K')^\circ=P^{\circ\circ}=P\). \(\square\)

We call \(K'\) the strip body of \(b_1,\dots,b_m\).

Lemma 1.5 (finite polar approximation). Let \(K\) be an origin-symmetric convex body and \(\alpha>0\). There are vectors \(b_1,\dots,b_m\) spanning \(\mathbf R^n\) whose strip body \(K'\) satisfies \[ K\subseteq K'\subseteq(1+\alpha)K,\qquad(K')^\circ\subseteq K^\circ. \]

Proof. Put \(\lambda=1+\alpha\). By Lemma 1.1, \(K^\circ\) is an origin-symmetric convex body, so \(\rho=\min_{|v|=1}h_{K^\circ}(v)>0\). Let \(d=\alpha\rho/\lambda\), and choose finitely many points \(b_1,\dots,b_m\in K^\circ\) such that every point of the compact set \(K^\circ\) lies within distance \(d\) of one of them. Let \(P\) be the convex hull of the \(\pm b_j\); then \(P\subseteq K^\circ\). For a unit vector \(v\), a point of \(K^\circ\) where \(\langle v,\cdot\rangle\) is largest lies within \(d\) of some \(b_j\), so \[ \begin{aligned} h_P(v)&\ge h_{K^\circ}(v)-d\\ &\ge h_{K^\circ}(v)-\tfrac\alpha\lambda h_{K^\circ}(v)\\ &=\lambda^{-1}h_{K^\circ}(v). \end{aligned} \] Both sides are positively homogeneous, so \(h_P\ge h_{\lambda^{-1}K^\circ}\) everywhere, and Lemma 1.3(a) gives \(\lambda^{-1}K^\circ\subseteq P\subseteq K^\circ\). In particular \(P\) has interior points, so the \(b_j\) span \(\mathbf R^n\). Taking polars (Lemma 1.3(b), (c)), with \((\lambda^{-1}K^\circ)^\circ=\lambda K\), gives \(K\subseteq P^\circ\subseteq\lambda K\), and \(P^\circ=K'\), \((K')^\circ=P\subseteq K^\circ\) by Lemma 1.4. \(\square\)

Lemma 1.6. If \(C\) is a convex body with \(B(0,r)\subseteq C\), then \(tC\subseteq\operatorname{int}C\) for \(0<t<1\), and \(\operatorname{vol}(\operatorname{int}C)=\operatorname{vol}(C)\).

Proof. For \(x\in C\) and \(|y|<(1-t)r\), the point \(tx+y=tx+(1-t)\frac{y}{1-t}\) lies in \(C\) by convexity; so \(tx\) is an interior point. Hence \(t^n\operatorname{vol}(C)\le\operatorname{vol}(\operatorname{int}C)\le\operatorname{vol}(C)\) for all \(t<1\). \(\square\)

Example 1.7. The cube \([-1,1]^n\) has volume \(2^n\). Its polar is the cross-polytope \(\{p:\sum_j|p_j|\le1\}\), whose intersection with each of the \(2^n\) closed coordinate orthants is a simplex of volume \(1/n!\); these simplices overlap only in sets of measure zero. So the volume product of the cube is \(2^n\cdot2^n/n!=4^n/n!\).

2. Plan of the course

The proof of Theorem B combines two scales. For a strip body \(K\) and a large integer \(k\), the functions \(f_j(z)=g(\langle b_j,z\rangle)^k\) on a domain in \(\mathbf C^n\), with a fixed conformal map \(g\), vanish to order \(k\) at the origin. Lesson 4 shows that the domain \(\{\sum_j|f_j|^2<1\}\), with the symplectic form of the potential \(|z|^2+\sum_j|f_j|^2\), contains balls of every capacity below \(\pi k\). Lessons 5 and 6 study the conformal map and prove a uniform estimate for integrals along horizontal slices. Lesson 7 maps the domain symplectically into \(\operatorname{int}K\times S\operatorname{int}K^\circ\) with \(S\approx\pi k/4\), rescales, and passes to general \(K\) by Lemma 1.5. Lessons 2 and 3 supply the tools: forms, symplectic maps and Moser's method; and plurisubharmonic potentials.

3. Exercises

1.1 (easy). Show that \(([-1,1]^n)^\circ\) is the cross-polytope, and that the polar of the ellipsoid \(\{x:\sum_jx_j^2/a_j^2\le1\}\) is \(\{p:\sum_ja_j^2p_j^2\le1\}\).

1.2 (easy). Show that \((AK)^\circ=A^{-\mathsf T}K^\circ\) for an invertible matrix \(A\), and deduce that the volume product is invariant under invertible linear maps.

1.3 (medium). Compute the volume product of the regular hexagon with vertices at distance \(1\) from the centre, and compare it with \(4^2/2!=8\).

1.4 (easy). Show that the volume product of the Euclidean unit ball in \(\mathbf R^n\) is \(\kappa_n^2\), with \(\kappa_n=\operatorname{vol}(B(0,1))\), and check \(\kappa_2^2=\pi^2>8\).

4. Solutions

1.1. \(h_{[-1,1]^n}(p)=\sum_j|p_j|\), so the polar is \(\{\sum_j|p_j|\le1\}\). For the ellipsoid \(E\), the substitution \(x_j=a_jy_j\) gives \(h_E(p)=\max_{|y|\le1}\sum_ja_jp_jy_j\), and this maximum is \((\sum_ja_j^2p_j^2)^{1/2}\).

1.2. \(\langle Aq,p\rangle=\langle q,A^{\mathsf T}p\rangle\), so \(p\in(AK)^\circ\) if and only if \(A^{\mathsf T}p\in K^\circ\). The volumes change by the factors \(|\det A|\) and \(|\det A|^{-1}\).

1.3. The hexagon \(H\) is the union of six equilateral triangles of side \(1\), of area \(\frac{3\sqrt3}2\). Its polar is the regular hexagon whose sides lie on the lines \(\langle v,p\rangle=1\) for the six vertices \(v\); its inradius is \(1\), its circumradius \(2/\sqrt3\), and its area \(\frac{3\sqrt3}2\cdot\frac43=2\sqrt3\). The product is \(9>8\).

1.4. The unit ball is its own polar. \(\kappa_2=\pi\) and \(\pi^2\approx9.87\).

References