Symplectic Balls in Symmetric Polar Products

Proof. Step 1: a logarithmic potential with the required slope. For a holomorphic tuple, differentiation along a complex line gives

The second expression is nonnegative by Cauchy–Schwarz. Hence \tau and, away from its zero, \log\tau are plurisubharmonic.

Fix 0<a<k; we will embed the ball of capacity \pi a. Choose

Choose numbers \log b<t_-<t_+<0 and a smooth nondecreasing cutoff \beta:\R\to[0,1] which is 0 for t\le t_- and 1 for t\ge t_+. Define \chi on (-\infty,0) by

choosing its additive constant so that \chi(t)=e^t for t\ge t_+. This is possible since \chi'=e^t there. Moreover, \chi'>0 and

because the transition occurs where e^t>b. For t\le t_-, the function \chi is affine with slope b.

It follows that

is smooth and plurisubharmonic on V\setminus\{0\} and satisfies H=\tau wherever \tau\ge s_0:=e^{t_+}. The vanishing assumption gives \tau(z)\le C_0|z|^{2k} near zero. Since \chi is eventually affine, there is a constant C_1 such that

Step 2: smooth the singularity without changing the form near the boundary. The strict inequality \mu<bk lets a smooth radial logarithm dominate H close to zero while remaining below it on an outer sphere. Write B_r=\{z\in\C^n:|z|<r\}, and choose r_2>0 with \overline{B}_{r_2}\subset V. For 0<\delta\le1, consider

For each fixed r>0, the radial calculation in Step 4 identifies (B_r,\ddc(|z|^2+P_\delta)) with a standard ball whose squared radius is r^2+\mu r^2/(r^2+\delta). Since \mu>a, this exceeds a for all sufficiently small \delta. We will therefore arrange that the modified potential agrees with P_\delta on a ball whose radius is fixed before \delta is chosen.

The functions P_\delta are plurisubharmonic by Equation (2.2), applied to the nowhere-zero holomorphic tuple (\sqrt\delta,z_1,\ldots,z_n). Choose M so that

The minimum is finite because f has no zero on this sphere. Then P_\delta<H-2 on |z|=r_2 for every 0<\delta\le1. On the other hand, Equation (2.3) gives

As z\to0, this lower bound tends to +\infty, uniformly in \delta. We may therefore fix 0<r_1<r_2, independently of \delta, such that

Fix from now on one positive \delta satisfying

The upper bound is positive because \mu>a.

We next replace the singularity of H by P_\delta using a regularized maximum, the standard gluing device in Richberg’s smoothing method; see [8, Section 2]. A related replacement of Kähler potentials appears in [23, Proposition 3.3] in a projective setting. We give the elementary two-function construction explicitly. Let \vartheta\in C_c^\infty((-1,1)) be nonnegative and even, with \int_{\R}\vartheta=1, and set

The function \rho is smooth and convex, with |\rho'|\le1, and equals |s| for |s|\ge1. Thus \operatorname{rmax} is smooth, convex, nondecreasing in both arguments, and equals \max(h,p) when |h-p|\ge1. Define \widetilde H to be \operatorname{rmax}(H,P_\delta) on B_{r_2}\setminus\{0\}, to be H outside B_{r_2}, and to take the value P_\delta(0) at zero. The outer comparison gives equality with H in a neighborhood of \partial B_{r_2}, and Equation (2.4) gives equality with P_\delta on B_{r_1}. Consequently \widetilde H is smooth and plurisubharmonic on all of V.

The difference \widetilde H-\tau has compact support in V. Indeed, set s_2=\max_{\overline B_{r_2}}\tau<1, choose \max(s_0,s_2)<\sigma<1, and put

This is a compact subset of V by hypothesis. Outside C a point lies outside \overline B_{r_2} and has H=\tau, so that \widetilde H-\tau vanishes there.

Step 3: transport the smoothed form. We now use Moser’s deformation argument [15, Section 4]. The support estimate above supplies a complete flow even though V need not be compact. Consider the symplectic forms

and write \widetilde\omega=\omega_1. Each form is symplectic because the two non-Euclidean potentials are plurisubharmonic. Define the smooth time-dependent vector field Y_s by

Both \gamma and every Y_s are supported in the fixed compact set C. Extending Y_s by zero outside V gives a smooth vector field on \C^n, with its coefficients and first derivatives bounded uniformly for s\in[0,1]. The ordinary differential equation for this field has a flow \varphi_s throughout this time interval. The ambient flow fixes \C^n\setminus V pointwise and hence restricts to diffeomorphisms of V. This argument applies even if V is disconnected and places no regularity assumption on its boundary.

Since d\omega_s=0, differentiation of pullbacks gives

Therefore

Step 4: read off the ball in the radial model. The deformation has reduced the problem to an explicit radial form. On B_{r_1} the new potential, up to the additive constant -M, is \Psi(|z|^2), where

Define the radial map

This is the radial case of the symplectic-coordinate construction in [13, proof of Theorem 1.1, Equation (23)]. In the pullback of \lambda_0 from Equation (2.1), the terms differentiating the radial factor cancel. Hence

The squared radial coordinate is strictly increasing, because

At zero the map is smooth and has derivative DR(0)=\sqrt{1+\mu/\delta}\,I, so its inverse is smooth there as well. Thus R is a diffeomorphism from B_{r_1} onto the standard ball B^{2n}(\pi Q_\delta), where

by Equation (2.5). If \iota:B_{r_1}\hookrightarrow V denotes inclusion, the map

is a smooth embedding, and its pullback of \omega is \omega_0 by Equation (2.7) and the radial pullback identity. Since a was arbitrary in (0,k), the theorem follows.□

Proof. The map and its angular derivative. The series in Equation (3.1) is absolutely summable on the closed disk. Hence F is bounded, odd, and commutes with conjugation. Termwise differentiation in the open disk gives

Here the logarithm is the branch on the right half-plane that is real on the positive real axis. Indeed,

This also shows that \arctan w=0 only at w=0. Thus F' has no zeros away from the origin, and the series gives F'(0)=8/\pi^2\ne0.

For w=re^{i\theta}, define

To obtain the last identity, the imaginary part of the fraction in Equation (3.4), divided by its real part, is 2r\cos\theta/(1-r^2), and its argument lies in (-\pi/2,\pi/2). On the right semicircle, 0<A(r,\theta)<2/\pi. The identities

therefore show that \Re F(w)>0 in the right half-disk: its radial derivative is positive, and its value tends to zero at the origin. They also show that \Im F(re^{i\theta}) increases strictly with \theta on each right semicircle.

Oddness and conjugation symmetry give F(ir)=iT(r) with real T. Differentiating Equation (3.4) on the imaginary axis gives Equation (3.3). In particular, T is strictly increasing, T(0)=0, and T'(r) diverges as r\uparrow1. Since F(r) is real, Equation (3.6) also gives

The integrand is bounded by 2/\pi and tends to 2/\pi at every \theta\in[0,\pi/2) as r\uparrow1. Dominated convergence therefore gives T(r)\to1.

Horizontal slices and global invertibility. Fix |t|<1 and let T(r_0)=|t|. For every r\in(r_0,1), the strictly increasing imaginary coordinate on the right semicircle ranges from -T(r) to T(r). There is consequently exactly one angle \theta(r,t) there with imaginary coordinate t. It depends smoothly on r, since its angular derivative is A>0. Writing wF'(w)=A+iB, implicit differentiation at fixed t gives

Thus

As r\downarrow r_0>0, the angle tends to the appropriate imaginary-axis endpoint of the semicircle, so v(r,t)\to0. If r_0=0, the same limit follows from F(0)=0. Boundedness of F now shows that v(r,t) maps (r_0,1) continuously and strictly increasingly onto an interval (0,V(t)), where 0<V(t)<\infty.

This parametrization proves injectivity on each positive horizontal slice. The reflection identity F(-\overline w)=-\overline{F(w)} gives the negative slices, and the strict monotonicity of T gives injectivity on the imaginary axis. The right half-disk, imaginary axis, and left half-disk have images with positive, zero, and negative real parts, respectively. Hence F is globally injective. Every point of its image has imaginary coordinate in (-1,1), and every such coordinate occurs on the imaginary axis. This proves Equation (3.2). Since F'\ne0, the local holomorphic inverses combine into a holomorphic inverse g:D\to\D.□

Proof. Step 1: express the slice integral in the disk radius. For any fixed v+it\in D, the horizontal segment from it to v+it is a compact subset of the open set D; the same is true for all sufficiently nearby endpoints and heights. The formula

then proves smoothness by differentiation under the integral. The integrand is smooth even at the zero of g, since |g|^{2k-2}=(|g|^2)^{k-1} and k is an integer. Equation (3.9) follows from the fundamental theorem of calculus. The symmetry g(-\overline\zeta)=-\overline{g(\zeta)} makes the density in Equation (3.8) even in its real coordinate. Thus J_k(-v,t)=-J_k(v,t), and it suffices to estimate v>0.

Fix such a point and put

Along its positive horizontal slice, the radius increases from r_0 to R. Since g'(F(w))=1/F'(w), Equation (3.7) gives

Indeed, |g'|^2\,dv=(r/A)\,dr. At the lower endpoint this is understood as an improper integral: perform the change of variables first on a segment beginning at a positive real coordinate, and then let that coordinate decrease to zero. The original integrand is smooth on the full compact segment, so its integral is finite; the nonnegative transformed integrals converge to the same value. This argument also covers t=0, when r_0=0.

The leading constant comes from splitting 1/A into \pi/2 and the nonnegative excess 1/A-\pi/2, since 0<A<2/\pi. The constant part in Equation (3.11) contributes \frac{\pi}{4}(R^{2k}-r_0^{2k}). It therefore suffices to bound the remaining weighted integral uniformly over all heights and endpoints by a quantity tending to zero as k\to\infty.

Step 2: estimate the loss near the two tips. We next control the possible small denominator near the ends of the right semicircle. Suppose 1/2\le r<1 and set

For every X>0,

For X\le1, this follows from \arctan X\ge X/2; for X\ge1, use \pi/2-\arctan X=\arctan(1/X)\le1/X and \arctan X\ge\pi/4. Combining Equation (3.12) with \sin s\ge2s/\pi and (1+r)/(2r)\le3/2, we obtain

On the other hand, the vertical gap to the semicircle endpoint is

Here we used \arctan y\le y and \sin\xi\le\xi. The last inequality supplies the lower bound s\ge(\pi/\sqrt2)\sqrt{\Delta(r,\theta)(1-r)}. Consequently, with the absolute constant C_0=3\sqrt2\pi^2/8, Equation (3.13) implies

Step 3: separate a compact part from the tip region. Fix a radius r_*\in(1/2,1). The part of the original horizontal integral for which |g|\le r_* is uniformly bounded, after division by k, by

To see this, the radius is strictly increasing along the positive slice, so the relevant portion is exactly its initial segment ending at radius \min(R,r_*), when r_0\le r_*, and is empty otherwise. Every point of this segment lies in F(\{|w|\le r_*\}), and its horizontal length is at most \operatorname{diam}(D). This includes the case R\le r_*, when it is the whole integral.

If an outer portion remains, put

Since T(r_a)\ge T(r_0)=|t|, for r>r_a we have

Equations (3.11) and (3.15) therefore bound the outer contribution by

The extension of the error integral from R to 1 uses its nonnegativity and removes any dependence on the distance R-r_a.

The last integral has an absolute bound, including when its lower endpoint depends on k. Write

The substitution y=k(1-r) gives

since \log r\le-(1-r) and 2k-1\ge k. On [0,H/2] the square-root factor is at most one. On [H/2,H], bound the exponential by e^{-H/2} and use

It follows that

The same integral without its exponential factor also gives I(k,r_a)\le\pi H/2. These estimates hold for every H>0, covering both H\downarrow0 and H\to\infty.

Combining Equations (3.16), (3.18), and (3.19), and using symmetry when v<0, we have, for every v+it\in D,

When there is no outer portion, the first term alone bounds the excess, so the same displayed inequality remains valid. It is also valid at v=0.

Step 4: make the error uniform. The cutoff radius will be chosen before the power k, so the estimates above allow the slice height and both radial endpoints to move with k. Define explicitly

Equation (3.20) makes this supremum finite for each k. For fixed r_*, its first error term tends to zero as k\to\infty. Moreover, Equation (3.3) gives

Given \varepsilon>0, first choose r_* so that the second term in Equation (3.20) is less than \varepsilon/2. For this fixed r_*, choose N such that C_*k r_*^{2k-2}<\varepsilon/2 for every k\ge N. The resulting bound \epsilon_k<\varepsilon holds simultaneously on all of D. In particular it allows the height t, the initial radius r_0, and the final radius R all to vary with k, with no separation from either endpoint. This proves \epsilon_k\to0 and Equation (3.10).□

Proof. Step 1: encode the strips by holomorphic functions. The body K and its finite list of defining vectors are fixed throughout the proof. Extend each real functional b_j\cdot x complex-linearly to \ell_j(z)=b_j\cdot z. Let F:\D\to D and g=F^{-1} be the maps of Lemma 3.1, and define

This is an open neighborhood of the origin. The spanning condition makes the linear map z\mapsto(\ell_1(z),\ldots,\ell_m(z)) injective, so there is a constant C_b such that

Since D is bounded, \Omega is bounded as well; neither set depends on the integer k chosen below. The same linear bound makes K compact, and a sufficiently small Euclidean ball lies in all its defining strips. In particular, 0\in\interior(K).

For each integer k\ge2, put

The tuple f=(f_1,\ldots,f_m) is holomorphic on \Omega. Since g has its only zero at the origin, the common zero set of f is

Holomorphicity at the origin gives |f(z)|=O(|z|^k). To verify the compactness hypothesis of Theorem 2.1, fix 0<s<1. On \{\tau\le s\}, for every j we have

a compact subset of D. The linear bound above makes this sublevel bounded. Moreover, the limit of any convergent sequence in it still has all its images under the functionals \ell_j in that compact subset of D, hence belongs to \Omega and satisfies \tau\le s by continuity. Thus the sublevel is compact in \Omega. Theorem 2.1 now supplies symplectic embeddings into (V,\omega) of every standard ball of capacity less than \pi k.

Step 2: realize the form by a globally injective map. Write z=u+ix and use the signed integrals J_k of Proposition 3.2 to define

The signed integrals J_k are smooth on D by Proposition 3.2, so \Phi_k is smooth.

Set v_j=b_j\cdot u and t_j=b_j\cdot x. Pulling back the target form gives

Indeed, the terms containing \partial_tJ_k vanish because dt_j\wedge dt_j=0, while

This is precisely the coefficient of \ddc|g(v+it)^k|^2 in front of dv\wedge dt, with the convention for \dc fixed above.

The map \Phi_k is globally injective. To see this, equality of its q coordinates first fixes x. For two possible real coordinates u_1,u_2 at that x, write

and denote the corresponding momenta by p_1,p_2. Both v_{j1} and v_{j2} belong to the same horizontal interval of D at height t_j. Since \partial_vJ_k\ge0 on this entire interval,

Thus equal momenta force u_1=u_2. This comparison applies to any two points of the fiber, even if that fiber of V is disconnected. Equation (4.3) and nondegeneracy of \omega show that \Phi_k is a local diffeomorphism. It is therefore open, and its injectivity gives a smooth inverse onto its image. Consequently \Phi_k is a symplectic embedding.

Step 3: control the momentum by the polar body. We next bound its image. All imaginary parts of points of D lie in (-1,1), so |b_j\cdot x|<1 for every j. The finite system of strict inequalities implies q=x\in\interior(K). Introduce the support function

It satisfies h_K(\pm b_j)\le1, and \interior(K^\circ)=\{p:h_K(p)<1\}. Because \Omega is bounded, there is a finite constant C_K, independent of k, such that h_K(-u)\le C_K on \Omega. Subadditivity and positive homogeneity of h_K, followed by Proposition 3.2, give

Here \epsilon_k\ge0 tends to zero, uniformly in the planar argument. The number m and the constant C_K are fixed before k varies. It follows that, for any \eta>0 and all sufficiently large k,

Since \tau<1 on V, Equation (4.4) then gives the strict inclusion

Step 4: rescale the ball and the momentum together. Finally, fix 0<c<4. Choose \eta>0 with (1+\eta)c<4, and then fix an integer k large enough for Equation (4.5). The resulting constant S satisfies Sc<\pi k, so there is a symplectic embedding

Let D_{\sqrt S}(z)=\sqrt S\,z and L_S(q,p)=(q,p/S). The composition

is a smooth embedding. Since D_{\sqrt S} maps B^{2n}(c) onto B^{2n}(Sc) and

its pullback of \omega_0 is exactly \omega_0. This proves the proposition.□

Proof. First, 0\in\interior K. Indeed, if the Euclidean ball B(y,r) is contained in K, then so is B(-y,r), and their midpoints contain B(0,r). Since K is also bounded, there are r,R>0 with B(0,r)\subset K\subset B(0,R). The definition of the polar gives

Thus K^\circ is a compact, origin-symmetric convex body with nonempty interior.

We will use the bipolar identity K^{\circ\circ}=K. To recall its justification here, if x\notin K, separation gives a nonzero vector v such that

The normalized vector v/h_K(v) belongs to K^\circ and pairs with x to a value greater than 1, so x\notin K^{\circ\circ}. The reverse inclusion follows directly from the definition. The same argument applies to every convex body containing the origin in its interior.

Put \lambda=1+\alpha and

Choose d>0 with d\le\alpha\rho/(1+\alpha), and choose a finite d-net \{b_1,\ldots,b_m\} in the compact set K^\circ, with all net points in K^\circ. Let

Symmetry of K^\circ gives P\subset K^\circ. For every unit vector v, a maximizer of \ip{v}{p} over p\in K^\circ lies within distance d of a net point. Consequently

The characterization of inclusion by support functions, which follows from convex separation, now gives

In particular, P has interior, so its generating vectors span \R^n. Taking polars and using the bipolar identity yields

Finally, the definition of P shows that P^\circ is exactly the finite strip body K' in the statement.□

Proof of Theorem 1.1. Fix 0<c<4. Choose \alpha>0 such that \lambda c<4, where \lambda=1+\alpha, and choose one body K' from Lemma 5.1. Proposition 4.1 gives a smooth symplectic embedding

If one body is contained in another, every interior ball in the first is contained in the second. Thus Equation (5.1) implies

even when their boundaries meet. Regard e as an embedding into \lambda\interior K\times\interior K^\circ. Define

The composition

is a smooth embedding. Since

its pullback of \omega_0 equals \omega_0. This proves c_G(U_K)\ge4.

For the upper bound, we use the supporting-slab construction of Artstein-Avidan, Karasev, and Ostrover [1, Remark 4.2]. Choose q_0\in K and p_0\in K^\circ with \ip{q_0}{p_0}=1. Such a pair is obtained by maximizing any nonzero linear functional on K and normalizing its maximum to 1. Complete q_0 to a basis of \R^n by vectors in \ker p_0, and let A be the matrix of this basis. The linear change

is symplectic and has first coordinates Q_1=\ip{p_0}{q} and P_1=\ip{q_0}{p}. Both functionals are nonzero, so symmetry, polarity, and the interior condition give |Q_1|<1 and |P_1|<1 on U_K. Thus this change embeds U_K into (-1,1)^2\times\R^{2n-2}, with the square in its first conjugate coordinate pair.

For completeness, this open square is area-preservingly diffeomorphic to the disk of area 4. Put R=2/\sqrt\pi, h(X)=\sqrt{R^2-X^2} for -R<X<R, and

Since \int_{-R}^{R}h(s)\,ds=2 and H'=h>0, H is a smooth diffeomorphism from (-R,R) onto (-1,1). The map

has image B^2(4) and Jacobian X'(u)h(X)=1. Applying it to the first conjugate pair gives a symplectic embedding U_K\hookrightarrow B^2(4)\times\R^{2n-2}. Gromov’s nonsqueezing theorem [5] therefore bounds the capacity of every ball in U_K by 4. Combined with the lower bound, this proves c_G(U_K)=4.□

Proof. For n=1, write K=[-a,a] with a>0. Then K^\circ=[-a^{-1},a^{-1}], and the product of their lengths is 4. Assume henceforth that n\ge2.

We first justify that taking interiors does not change the volumes in the statement. If a convex body C\subset\R^n contains B(0,r), then convexity gives, for 0<t<1,

Hence tC\subset\interior C, and

Letting t\uparrow1 proves \vol_n(\interior C)=\vol_n(C). Apply this to K and K^\circ, both of which contain a neighborhood of the origin. The product formula for Lebesgue measure then gives

For every 0<c<4, Theorem 1.1 supplies a smooth symplectic embedding f:B^{2n}(c)\to U_K. Taking the nth exterior power of f^*\omega_0=\omega_0 shows that \det Df=1, so the change of variables formula gives

Here the first equality follows from the Euclidean ball volume \pi^nR^{2n}/n! at radius R=\sqrt{c/\pi}. Combining this with Equation (5.3) and letting c\uparrow4 proves the claimed inequality.□

Proof. Strictness and finiteness permit a choice of 0<c<4 such that \sum_i R_i^n<c^n and R_i+R_j<c for every i<j. The ball-packing theorem [17, Theorem 1.1] embeds these closed balls into B^{2n}(c), with embeddings defined on neighborhoods. Their compact images are pairwise disjoint and lie in the open target, so the neighborhoods can be shrunk to have pairwise disjoint images contained in B^{2n}(c). Composing with the embedding in Theorem 1.1 proves the assertion.□