Symplectic Balls in Symmetric Polar Products

1. From finite strips to phase space

Write $K=\{x:|b_j\cdot x|\le1,\ 1\le j\le m\}$, where the $b_j$ span $\mathbb R^n$. Using the lens $D$ and its inverse coordinate $g:D\to\mathbb D$ below, put $\ell_j(z)=b_j\cdot z$, $\Omega=\bigcap_j\ell_j^{-1}(D)$, $f_j=g(\ell_j)^k$, and $\tau=\sum_j|f_j|^2$. The domain $\Omega$ is bounded and the common zero is 0. On $\{\tau\le s\}$ each $\ell_j(z)$ lies in $F(\{|w|\le s^{1/(2k)}\})\Subset D$. Injectivity of the linear map $(\ell_j)_j$ makes these closed sublevels compact, verifying the hypotheses of Theorem 2.1.

Writing $z=u+ix$, the convention is $dd^c|z|^2=\sum du_j\wedge dx_j$. Thus target position is $q=x$ and the original momentum is $-u$. Correct it by

The identity $\partial_vJ_k=|(g^k)'|^2$ gives $\Phi_k^*\omega_0=dd^c(|z|^2+\tau)$. This local identity alone does not prove embedding. For two points with the same $x$, monotonicity on the horizontal slices gives

which proves global injectivity. The comparison uses the entire horizontal interval and remains valid even if a fiber of the sublevel domain is disconnected.

Proposition 4.1, Steps 1–2 · pp. 12–13

2. Polar control and cancelling dilations

For the support function $h_K$, one has $h_K(\pm b_j)\le1$. Proposition 3.2 therefore yields

Fix $K,m,C_K$ before increasing $k$. For any $\eta>0$, the image lies in $\operatorname{int}K\times S\operatorname{int}K^\circ$ with $S=(1+\eta)\pi k/4$ once $k$ is large enough. If $(1+\eta)c<4$, then $Sc<\pi k$, so Theorem 2.1 supplies $B^{2n}(Sc)$. Precomposing by $D_{\sqrt S}(z)=\sqrt S z$ and postcomposing by $L_S(q,p)=(q,p/S)$ cancels the factors $S$ and $S^{-1}$ in the symplectic forms. This gives $B^{2n}(c)\hookrightarrow U_K$.

Proposition 4.1, Steps 3–4; Equations (4.4)–(4.6) · p. 14

3. Finite approximation for an arbitrary body

For each $c<4$, choose $\lambda=1+\alpha>1$ with $\lambda c<4$. Lemma 5.1 takes the polar of the symmetric convex hull of a finite net in $K^\circ$ to obtain a finite strip body $K'$ satisfying

View $B^{2n}(\lambda c)\hookrightarrow\operatorname{int}K'\times\operatorname{int}(K')^\circ$ as an embedding into $\lambda\operatorname{int}K\times\operatorname{int}K^\circ$. A source dilation by $\sqrt\lambda$ and a position contraction by $1/\lambda$ again cancel their conformal factors. The choices are made in the order $c$, $\alpha$, $K'$ and its finite constraint list, $\eta$, then $k$. Each prescribed capacity uses one finite construction; no limiting embedding or smoothing of the convex boundary is required.

Lemma 5.1; proof of Theorem 1.1; Remark 5.2 · pp. 15–16

4. The supporting-cylinder upper bound

Choose $q_0\in K$ and $p_0\in K^\circ$ with $\langle q_0,p_0\rangle=1$. Complete $q_0$ by a basis of $\ker p_0$ to obtain a matrix $A$. The symplectic change $Q=A^{-1}q$, $P=A^Tp$ gives $|Q_1|,|P_1|<1$. Thus $U_K$ embeds into the cylinder over a square of area 4 in the first conjugate coordinate pair. An area-preserving map sends that square to the disk of area 4, and Gromov's nonsqueezing bounds the capacity of any embedded ball by 4. The lower bound is not inferred from the value of the Hofer–Zehnder capacity.

Proof of Theorem 1.1, upper bound · p. 16 [AAKO14, Remark 4.2, pp. 14–15] [Gro85, §0.3.A, Corollary, print p. 310 / PDF p. 4]

1. Logarithmic slope and the regularized maximum

Write the capacity as $c=\pi a$. Choose $0<a<k$, $a/k<b<1$, and $a<\mu<bk$. An increasing convex function $\chi$ makes $H=\chi(\log\tau)$ agree with $\tau$ near the outer part of the domain and satisfy $H\le bk\log|z|^2+C$ near 0. For $P_\delta=\mu\log(|z|^2+\delta)-M$, first arrange $P_\delta<H-2$ on an outer sphere, then $P_\delta>H+2$ on an inner ball whose radius is independent of $\delta$. The strict inequality $\mu<bk$ provides the latter comparison. Only afterwards is $\delta>0$ chosen small.

A regularized maximum produces a smooth plurisubharmonic $\widetilde H$, equal to $P_\delta$ inside and to $H$ outside. The support of $\widetilde H-\tau$ lies in some $\{\tau\le\sigma\}$ with $\sigma<1$, compact by hypothesis.

Theorem 2.1, Steps 1–2 · pp. 4–6

2. Moser transport and the radial capacity

The forms $\omega_s=dd^c(|z|^2+(1-s)\tau+s\widetilde H)$ remain nondegenerate. The solution of $\iota_{Y_s}\omega_s=-d^c(\widetilde H-\tau)$ has fixed compact support, so its flow exists for $0\le s\le1$ even though $V$ is noncompact, and satisfies $\phi_1^*\omega_1=\omega$. The manuscript proves the needed local version of Moser's deformation method; it does not apply the original compact-manifold theorem to a noncompact domain without a support argument.

Inside, $\Psi(t)=t+\mu\log(t+\delta)$ and $R(z)=\sqrt{\Psi'(|z|^2)}z$ carry the radial form to the standard one. The image of the radius-$r_1$ ball has capacity $\pi(r_1^2+\mu r_1^2/(r_1^2+\delta))>\pi a$. Pull the smaller standard ball back by $R^{-1}$, inclusion, and $\phi_1^{-1}$. Neither a regular-value condition on the sublevel boundary nor connectedness of $V$ is needed.

Theorem 2.1, Steps 3–4 · pp. 6–7 [Mos65, §4, Theorem 2 and deformation equation, print pp. 292–293 / PDF pp. 7–8]

1. Radial integration and the leading coefficient

Write $F(ir)=iT(r)$, with $T'(r)=8\operatorname{arctanh}r/(\pi^2r)\to\infty$. On a positive horizontal slice, write $w=re^{i\theta(r,t)}$. Its radius increases from $r_0=T^{-1}(|t|)$ to $R=|g(v+it)|$, and

The constant part $\pi/2$ of $1/A$ contributes $\frac\pi4(R^{2k}-r_0^{2k})$. A boundary limit at fixed $t$ alone would not control the remainder when $t=t_k\to\pm1$.

Lemma 3.1; Equation (3.11) · pp. 7–10

2. Moving endpoints and the order of limits

Split at $r_*\in(1/2,1)$. The inner contribution is bounded by $C_*kr_*^{2k-2}$. Outside, put $r_a=\max(r_0,r_*)$ and $L_* =\inf_{r_*\le r<1}T'(r)$. Then

For the remaining integral $I=k\int_{r_a}^1r^{2k-1}\sqrt{(1-r)/(r-r_a)}\,dr$, the substitution $y=k(1-r)$ and $H=k(1-r_a)$ yields $I\le1+\pi/e$ uniformly for every $H>0$. The excess is therefore at most

First choose $r_*\uparrow1$ to make the second term small; then hold that cutoff fixed and let $k\to\infty$. This permits the height and both endpoints to vary with $k$. Smoothness follows by differentiating over compact horizontal segments, and oddness and monotonicity follow from symmetry and nonnegativity of the density. These conclusions also enter the global embedding in §4.

Equations (3.15)–(3.21) · pp. 10–12

1. Interiors and the capacity limit

For a convex body $C$ containing 0 in its interior, $tC\subset\operatorname{int}C$ for $0<t<1$. Letting $t\uparrow1$ shows that its interior has the same volume. Thus $\operatorname{vol}_{2n}(U_K)=\operatorname{vol}_n(K)\operatorname{vol}_n(K^\circ)$. A symplectic embedding preserves $\omega_0^n/n!$, so $c^n/n!\le\operatorname{vol}_{2n}(U_K)$ for every $c<4$. Letting the number $c$ increase to 4 is sufficient; no capacity-4 embedding or convergence of embeddings is required.

2. Dimension one and sharpness

For $n=1$, $K=[-a,a]$ and $K^\circ=[-a^{-1},a^{-1}]$ have length product 4. In every dimension, the cube has volume $2^n$ and its crosspolytope polar has volume $2^n/n!$, by decomposing it into one simplex per orthant. Their product is $4^n/n!$. This example does not classify all other equality cases.

Corollary 5.3; Remark 5.4 · pp. 16–17

1. Strict slack and shrinking neighborhoods

Finiteness and strictness allow $c<4$ with $\sum_iR_i^n<c^n$ and $R_i+R_j<c$ for every pair. The external Theorem 1.1 packs the closed balls into $B^{2n}(c)$ for $n\ge3$, by embeddings defined on open neighborhoods. Since their compact images are disjoint and lie in the open ball, shrink these finitely many neighborhoods to preserve both disjointness and containment. Compose with $B^{2n}(c)\hookrightarrow U_K$ from Theorem 1.1. The external theorem's necessity statement concerns a ball target; this composition does not establish necessity for $U_K$.

Corollary 5.5 · p. 17 [OpenAI26p, Theorem 1.1, p. 2]