Symmetric Volume Descent from U4 and a Four-Dimensional Lower Bound

The result and the geometric idea

The body $U_4$ has the symmetries of a regular four-simplex [1]; write $U_4^{(1)}$ for its width-one normalization. The companion preprint [5], Theorem 1.1 establishes that it has smaller volume than the other members of the finite collection studied there. Here we show that it can be deformed, while retaining those symmetries and constant width, to have smaller volume.

Width is a linear constraint. The support function $h(u)$ gives the signed distance from the origin to the supporting hyperplane with outward unit normal $u$. Width one means $h(u) + h(-u) = 1$. Thus moving one supporting plane by $\varphi(u)$ and its opposite by $-\varphi(u)$ preserves width exactly. In support-function language, the perturbation must be odd.

Research development and supervision. The initial idea of a volume-decreasing deformation of $U_4$ was developed by the author and communicated to Professor Andriy Prymak on August 7, 2026. Professor Prymak subsequently examined the proposed approach and contributed further mathematical details and expanded, AI-assisted expositions of the argument. The present manuscript was developed through this exchange and the author’s subsequent investigations and verification.

Affiliations. China University of Geosciences (Wuhan), China; University of Manitoba, Canada (Mitacs Globalink Research Intern, 2026).

Acknowledgements. Most of this work was completed during the author’s 2026 Mitacs Globalink Research Internship at the University of Manitoba. The author gratefully acknowledges Mitacs for its support and thanks Professor Andriy Prymak for his supervision, mathematical contributions, and generous guidance in developing more effective research and learning methods.

AI assistance and manuscript preparation. The manuscript combines human-written material with AI-assisted drafting, organization, revision, and editorial refinement. OpenAI’s GPT-5.6 Sol, GPT-5.6 Sol, and Astra assisted with selected proof steps, auxiliary checks, code development, and manuscript preparation. Astra was used through Codex; Claude provided limited editorial assistance with the \LaTeX{} manuscript. The author takes responsibility for the mathematical content and conclusions.

Verification: all parts checked and understood. The author has checked and understood all parts of this manuscript, including the mathematical arguments and supporting calculations. This statement describes the author’s own review and does not assert formal verification or independent peer review.

Convexity is the constraint that matters. An odd function need not describe a convex body. The matrix $A_h=\nabla^2h+hI$ describes its curvature radii and must satisfy $0\preceq A_h\preceq I$. Here the spectral endpoints are the eigenvalues $0$ and $1$. At $U_4$ some radii are already at an endpoint. The perturbation must move these radii in the permissible direction. This is why an arbitrary asymmetric bump, or a bump on just one branch, does not settle the problem.

Why two coupled pieces succeed. We place a favorable cap on a region with two positive and three negative coordinates. Matching the opposite support branch forces a compensating piece on the region with three positive and two negative coordinates. There is more room for the favorable piece. For localization parameter $\delta$, its contribution has size at least $c\delta^{9/2}$, whereas the loss is at most $C\delta^{13/2}$. The loss-to-gain ratio is therefore $O(\delta^2)$ and tends to zero. The proof therefore needs no optimized numerical choice of parameters.

From a direction to a body. We finally add a motion of order $\delta^5$ towards the ball. It supplies strict curvature margins while costing less than the favorable $\delta^{9/2}$ term. A compactness lemma then produces an actual support function for small positive time. This step certifies global convexity, not just the constant-width identity or convexity at sampled directions.

The lower-bound part is independent. § 5 starts again with an arbitrary width-one body, not a perturbation of $U_4$. It gives

and an independent spectral refinement above 0.17818404. Corollary 5.4 then applies Jiang’s three-dimensional bound [6] to all hyperplane projections and yields

The latter improvement uses an external theorem; the four-dimensional spectral proof is independent of it. Neither argument identifies a minimizer. Detailed optimization of tiny volume reductions is omitted.

Support functions and convexity

Support functions, constant width, and curvature radii

A convex body is a compact convex subset of $\mathbb{R}^n$ with nonempty interior. Its support function is

and its positively one-homogeneous extension is

The body can be recovered from its support function through

If $a\in\mathbb{R}^n$, then $h_{K+a}(u)=h_K(u)+\langle a,u\rangle$; thus first spherical harmonics represent translations. The width of $K$ in direction $u$ is

Accordingly, $K$ has constant width $d>0$ if and only if

or, equivalently,

The antipodal identity is the basic constraint used throughout the paper. It gives both the regularity of constant-width support functions and the relation between their curvature-radius operators at antipodal points.

Proof. The homogeneous support function $H=H_K$ is convex. By (3), $H(-x)=d\lVert x\rVert-H(x)$, and $H(-\cdot)$ is also convex. Hence, in the sense of distributions,

On an annulus $a\leq\lVert x\rVert\leq b$ with $a>0$,

so

Thus $H$ is locally semiconvex and semiconcave on $\mathbb{R}^n\setminus\{0\}$, whence $H\in C^{1,1}_{\mathrm{loc}}$ there. Restricting to $S^{n-1}$ gives $h\in C^{1,1}(S^{n-1})$.

At a point of twice differentiability, the standard identity for a one-homogeneous extension gives

Restricting the distributional Hessian inequalities to $u^\perp$ yields $0\preceq A_h(u)\preceq dI$. Differentiating (3) twice at $u$ and $-u$ gives $A_h(-u)=dI-A_h(u)$ almost everywhere. □

For a smooth strictly convex body, the eigenvalues of $A_h$ are the principal radii of curvature. Lemma 2.1 shows that the same terminology is meaningful almost everywhere for every body of constant width: the principal radii lie in $[0,d]$, and antipodal radii add to $d$.

We also need the support-function volume formula in precisely the nonsmooth regularity class that occurs for $U_4$. Let $S_{n-1}(K,\cdot)$ denote the surface-area measure of $K$ on $S^{n-1}$, and let $\sigma$ denote ordinary, unnormalized spherical surface measure. The Minkowski integral formula [2 Section 5.1] is

For $C^{1,1}$ support functions the measure has an almost-everywhere density.

Proof. Because $h=h_K$ is $C^{1,1}$, the reverse Gauss parametrization

is Lipschitz. Rademacher’s theorem gives differentiability almost everywhere, and Aleksandrov’s theorem gives twice differentiability of $h$ almost everywhere. If $v\in T_uS^{n-1}$ is a tangent vector, differentiation of (9), followed by cancellation of the normal terms, gives

Since $A_h(u) \succeq0$ almost everywhere, the tangential Jacobian is

Almost every boundary point of a convex body has a unique outer normal. Since differentiability of $h$ makes each support face a singleton, $X_h$ has multiplicity one at these regular boundary points. The exceptional boundary set has zero $(n-1)$-dimensional measure. The area formula [3] therefore accounts for collapsed normal regions as zero Jacobian and gives, for every Borel set $\omega\subset S^{n-1}$,

Substitution into (6) yields (8). This argument is the $C^{1,1}$ realization of Schneider’s general measure-valued formula and does not require a globally $C^2$ boundary.

The support geometry of $U_4$

We recall the support formula and curvature spectrum from [5 Section 3 and Proposition 3.1], where the latter is proved. Only the notation and statements needed for the descent argument are retained here.

The support function on sign cells

3.1. The support function on sign cells. Identify $H_4=1_5^{\perp}\subset\mathbb{R}^5$ with $\mathbb{R}^4$ and let $S(H_4)\cong S^3$. For a nonempty proper subset $P$ of $\{1,\ldots,5\}$, the corresponding open sign cell is

For $u\in S(H_4)$ put

The projection construction of Arman–Bondarenko–Prymak–Radchenko [1] gives the following width-two support function:

Write

The branch-switch set $r=s$ and the coordinate hyperplanes have spherical measure zero.

If $u\in\mathcal{D}$ had four positive coordinates, the unique negative coordinate would have absolute value $\mu$, hence $s=\mu>r$ in the interior of the cell. Therefore

and consequently $\det Q_H=0$ almost everywhere on $\mathcal{D}$.

For the rest of the descent argument we use width one, so $h=H/2$ and $A_h=Q_H/2$ on the first branch. In particular $\det A_h=0$ there. The zero eigenvalues describe the directions along which the initial body has no curvature margin; the construction below is adapted to these directions.

A volume-decreasing deformation of $U_4$

The initial idea of a volume-decreasing deformation of $U_4$ was proposed by the author and communicated to Prymak on August 7, 2026. The expanded exposition here draws on the subsequent research notes supplied by Prymak [4], specifically the coupled cap formulas and branch matching, the endpoint curvature checks, the comparison of the positive and negative first-variation contributions, and strictification towards the ball. The expanded notes also prove a dimension-independent endpoint tangent-cone criterion and an explicit symmetric descent theorem for all $n \ge4$; their Theorem 18 includes the four-dimensional nonminimality conclusion. Here we give a detailed four-dimensional exposition of the construction and feasibility proof, without invoking the higher-dimensional theorem as a black box.

This section proves Theorem 1.1. We first derive the endpoint curvature constraints and pair antipodal points in the volume variation. We then construct the two matching pieces, check their curvature signs, and compare their contributions. The last step adds a smaller motion towards the ball and applies compactness to obtain a globally feasible path. The geometric reasons for these steps were described in § 1.

Local minimizers, admissible paths, and endpoint constraints

Let $\mathcal{W}_{4,1}$ be the class of four-dimensional convex bodies of constant width one. Since translations preserve width and volume, we work modulo translations.

Hausdorff convergence of convex bodies is equivalent to uniform convergence of their support functions. Let $h$ be the width-one support function of $U_4$. For the affine paths used here, a candidate support function has the form

with $h,\varphi\in C^{1,1}$ and continuous second derivatives on each closed smooth piece. The width condition forces $\varphi$ to be odd.

Write

Here $B_\varphi$ is the linearized curvature-radius operator: $A_{h_t}=A_h+tB_\varphi$. If (4.1) is a path of width-one support functions, then $0 \preceq A_{h_t} \preceq I$. Let $P_0$ and $P_1$ be the orthogonal projections onto $\ker A_h$ and $\ker(I-A_h)$.

Proof. If $v \in\ker A_h(u)$, then

so dividing by $t>0$ and letting $t \downarrow0$ gives $\langle B_\varphi v,v\rangle\ge0$. Similarly, for $v \in\ker(I-A_h(u))$,

which yields the upper inequality. $\square$

These pointwise inequalities are only necessary; gluing across sign and branch interfaces is part of the global construction below.

First variation of volume

The first mixed-volume functional, evaluated on a $C^{1,1}$ velocity, is

Here $S^3$ is the unit sphere in $H_4=1_5$. For a feasible affine path this equals the derivative of volume at zero, by Minkowski’s first-variation formula. Before feasibility is proved we use it only as the displayed linear functional.

Let

be the first support branch. The spectrum calculation in Proposition 3.1 shows that $\det A_h=0$ almost everywhere on $D$. Since width one gives $A_h(-u)=I-A_h(u)$ and an admissible velocity is odd,

Thus

A feasible direction with a positive integral on the right therefore gives negative first volume variation.

Leaf-adapted parametrization and a localized perturbation

On the first branch in $H_4=1_5$, the homogeneous width-one support function is

Fix a smooth sign cell with positive coordinate set $P$ and put $r=\|x_+\|$. Write

Then

On the open cone over the interior of a first-branch sign cell, let $L_p$ be a fiber of the map $x\mapsto p=x_+/\|x_+\|$. Its tangent space is

where the Hessian is restricted to $H_4$ and $w_P$ is the component of $w$ on the positive coordinate indices. Thus these fibers are the integral leaves of the Hessian kernel. At fixed $r$ and $p$, variations of $q$ satisfying $\sum q_j=m$ lie in a slice of one such leaf; the unperturbed homogeneous support function has vanishing second derivative in these directions. Here null refers to the support Hessian and, on the sphere, to the curvature-radius operator.

Here $m$ is normalized: on the unit sphere it equals $\mu(u)/r(u)$, where $\mu$ is the unnormalized coordinate sum from Section 3 (denoted by $m(u)$ in [5 Section 3]). The symbol $\widehat h$ denotes the homogeneous width-one support, whereas $H$ in (10) has width two. The branch-switch set is $\|q\|=1$. The first branch has one, two, or three positive coordinates. The interface used below has two positive and three negative coordinates, or the antipodal sign pattern.

The reason for choosing this interface is visible in the negative-coordinate sections. Put $k=|P|$ and $\ell=5-k$, and suppress the zero coordinates of $q$. At fixed $r,p$ (hence fixed $m$), write

Thus the section is given by $\|z\|^2\leq1-m^2/\ell$, with the coordinate conditions $q_j>0$ still imposed. For $k=2$, its limiting radius as $m\uparrow\sqrt{2}$ is $1/\sqrt{3}$, and the section retains nonempty two-dimensional relative interior around $(\sqrt{2}/3)_{13}$. For $k = 3$, the radius tends to zero and the section contracts to the single point $(1,1)/\sqrt{2}$. This contrast motivates placing the positive perturbation on the two-positive cells and its compensating continuation on the three-positive cells near $m = \sqrt{2}$; the weighted volume comparison is proved below.

Fix

and define

The cutoff $\chi_\delta$ restricts the perturbation to $m > \sqrt{2}-\delta$, near $m = \sqrt{2}$ on the active branches. We use cap as shorthand for this localized perturbation. The function $F$ is convex, permutation invariant, and

The numerical-looking constant $4/5$ is chosen for a simple exact geometric reason.

Proof. Let $(a,b)$ be a nonnegative unit two-vector with $a+b=m$. Solving $a^2+b^2=1$ and $a+b=m$ gives

For $m>1$ this is strictly decreasing, and

On the support of $\chi_\delta$ we have $m \ge m_0 > 7/5$, hence each coordinate is strictly below $4/5$ when the configuration approaches the coordinate boundary that would change the sign pattern. The same upper bound applies to a nonnegative two-vector of norm at most one: at fixed sum $m$, its largest coordinate is maximized when the norm is one. This covers the negative leaf when one of its three coordinates vanishes. On a positive three-vector boundary the norm is already one. The remaining one-positive boundary has $m = 1 < m_0$ and is excluded by $\chi_\delta$ itself. The strict gap $m_0 > 7/5$ makes these statements hold on neighborhoods of the interfaces. The remaining $3 \leftrightarrow4$ interface cannot meet the active first branch: there $q \to(m,0)$ up to permutation, contradicting $\lVert q\rVert\le1$ and $m \ge m_0 > 1$. Therefore every positive-part factor $(y_i-4/5)_+$ is identically zero on a neighborhood of that interface; the same holds for its first derivative. $\square$

On a two-positive first-branch cell define

For a positive unit three-vector $p$ define

and on a three-positive cell set

On one-positive cells set $G_\delta=0$, and define the second support branch by oddness, $G_\delta(-x)=-G_\delta(x)$.

Finally put

Matching across the mixed branch

The correction $\Gamma_\delta(\lVert q\rVert^2-1)$ in (24) is designed exactly so that the two branch formulas match to first order.

Verification of the endpoint curvature inequalities

We now verify Lemma 4.2 for the constructed velocity. The calculation is most transparent in homogeneous leaf coordinates.

dependence of (22) is through the convex function $F(q)$, so the corresponding homogeneous Hessian is positive semidefinite. On a three-positive cell the $q$-dependent correcting term is

whose $q$ Hessian is $2r\Gamma_\delta(p)I \succeq0$ by (4.12). This proves the lower endpoint inequality.

For the upper endpoint we use the explicit principal radii. On a width-one two-positive cell the unique nonzero radius, denoted by $\lambda_2$, is nonsaturated in the interior and equals

It reaches $1$ only at the mixed branch endpoint $m=\sqrt{2}$, $\lVert q\rVert=1$. Here the positive unit vector is $p=(1,1)/\sqrt{2}$, and the saturated vector $w$ is supported on its two coordinates with zero sum. Along $x+tw$ the negative coordinates and their sum are fixed, while $r'(0)=0$ and $r''(0)=\lVert w\rVert^2/r$. Thus $G_{2,\delta}=\Psi(r)$, and $\Psi'=-2\Gamma_\delta$ from the matching calculation yields

On a three-positive cell the two positive radii are

The second is at most $4/5$, so only $\lambda_{3,1}$ can saturate at $1$, and it does so exactly on $\lVert q\rVert=1$. Its saturated vector $w$ is supported on the three positive coordinates and satisfies $\sum w_i=0$ and $p\cdot w=0$. To display the cancellation, put $L(p)=\chi_\delta(m)F(p)$, where $m=\sum p_i$, so that $p\cdot\nabla L-L=2\Gamma_\delta(p)$. Along $x+tw$ at the interface,

Consequently the two terms in (24) have second derivatives

Here $D^2L(p)[w,w]=\chi_\delta(m)D^2F(p)[w,w]$ because $\sum w_i=0$. Adding gives

The endpoint inequalities hold for the continuous one-sided Hessian extensions on the closures of the smooth pieces. On first-branch mixed closures, the positive radii stay bounded away from zero, so no new lower kernel appears; the lower quadratic forms retain their signs. The only newly saturated upper directions are those checked in (29) and (31). Near coordinate-sign interfaces the cap vanishes identically, so rank changes there impose no further condition. At the cutoff $m=m_0$ and hinge thresholds $y_i=c$, the newly activated terms, including their corrections, have vanishing derivatives through order two. On one-positive cells the cap is zero. These observations also cover intersections of the interfaces. The antipodal cases follow from oddness and $A_h(-u)=I-A_h(u)$. □

The sign of the first variation

Define

where the two terms come from the two-positive and three-positive cells. By construction $I_{2,\delta}\geq0$ and $I_{3,\delta}\leq0$.

Proof. On a two-positive unit vector write

$p(\theta) = (\cos\theta,\sin\theta)$, $m(\theta) = \cos\theta+ \sin\theta= \sqrt{2}\cos(\theta-\pi/4)$.

Use $q_1,q_2$ as independent leaf coordinates and set $q_3=m-q_1-q_2$. The parametrization

$u=(p,-q)/\sqrt{1+\lVert q\rVert^2}$

has spherical measure

For clarity, in a cell with $\ell$ negative coordinates write $q=(m/\ell)\mathbf{1}_{\ell}+z$, $\sum z_i=0$. On the slice $\lVert p\rVert=1$ the Gram determinant relative to $d\sigma(p)\,dz$ is $(5-m^2)/\ell$. Radial projection to the sphere multiplies its area element by $(1-m^2/5)^{-1/2}(1+\lVert q\rVert^2)^{-2}$. Thus the resulting density is $\sqrt{5/\ell}(1+\lVert q\rVert^2)^{-2}d\sigma(p)\,dz$; the displayed formula uses $dz=\sqrt{3}\,dq_1dq_2$. Taylor expansion at $\pi/4$ gives

Thus the angular set on which $m\ge\sqrt{2}-\delta/2$ has length comparable to $\sqrt{\delta}$. There

Such a negative-leaf patch exists explicitly: at $m=\sqrt{2}$ take

All coordinates are positive, $\lVert q\rVert<1$, and $F(q)>0$. A small patch around it varies continuously for nearby $m$. Choose such a compact interior patch of the negative leaf for which $F(q)>0$ and which stays away from all coordinate and branch boundaries. On this fixed patch the spherical Jacobian and $\det(I-A_h)$ have positive lower bounds independent of $\delta$. Integrating over the angular cap and this fixed leaf patch therefore yields

For the three-positive contribution, the negative leaf has two coordinates. Write

Let $\alpha$ be the angular coordinate on the circle $\{p\in S^2:\sum p_i=m\}$. Then $d\sigma_{S^2}(p)=dm\,d\alpha/\sqrt{3}$ and $dz=\sqrt{2}\,dd$, so the same radial calculation gives

This density is bounded uniformly for $m_0\le m\le\sqrt{2}$; the range of $\alpha$ has length at most $2\pi$. A direct calculation gives

Let

On the active cap, $0\le x\le\delta$. From (20) and (23),

Writing $\eta=\eta(m)$, we have $\eta^2=y(\sqrt{2}+m)/4=O(y)$. By (30), $\lambda_{3,2}\le4/5$ and

Thus $\det(I-A_h)$ is bounded above by a constant times $\eta^2-d^2$. The bounded density just computed and the identities

give, with constants independent of small $\delta$,

Rescaling $x=\delta s$ gives

because both beta-type integrals in $s\in[0,1]$ are finite. Since $13/2>9/2$, the positive two-cell contribution dominates for sufficiently small $\delta$.

By (15),

for all sufficiently small $\delta$.

Strict curvature margins and construction of an admissible support path

Endpoint feasibility is non-strict, so it is not by itself a uniform finite-$t$ convexity certificate. We create a small spectral margin by moving slightly toward the width-one ball. Let

This function is odd, and

Define the modified perturbation direction

The tilde denotes the modified perturbation direction, now with a strict spectral margin. On the two saturated subspaces,

Since $|\delta V_h(b)|\le2\pi^2\|b\|_\infty$, the added term changes the first variation by $O(\delta^5)$ with a constant independent of $\delta$, whereas the negative contribution in (35) has magnitude at least a constant multiple of $\delta^{9/2}$. Since $5>9/2$,

for all sufficiently small $\delta$.

We use one elementary compactness lemma to pass from the infinitesimal spectral inequalities to a genuine path.

Proof. Suppose the lower inequality failed for a sequence $t_j\downarrow0$. Then there are points $x_j$ and unit vectors $v_j$ with

By compactness, after passing to a subsequence, $x_j\to x$ and $v_j\to v$. Boundedness of $C$ and $A\succeq0$ give

Taking limits yields $v\in\ker A(x)$ and $\langle C(x)v,v\rangle\le0$, contradicting the strict positive bound. Apply the same argument to $I-A$ and $-C$ for the upper inequality. ∎

For fixed $\delta$, partition the sphere by coordinate signs, support branches, and cutoff and hinge thresholds, and take the finitely many piece closures. On each closure, $A_h$ and $B_{\varphi_\delta}$ have continuous one-sided extensions: on the first branch $r=\lVert u_+\rVert\ge1/\sqrt{2}$, and the positive-part formulas are smooth on their respective pieces; the opposite branch follows by antipodality. Lemma 4.6 covers these extensions, including newly saturated boundary directions. Thus $C=B_{\varphi_\delta}+\delta^5(\frac{1}{2}I-A_h)$ satisfies the strict endpoint conditions with $\eta=\delta^5/2$ on every closed piece. Using finitely many compact patches with local orthonormal tangent frames, apply Lemma 4.8 and take the minimum of the resulting positive times to obtain a common $t_\delta>0$.

Proof. Proposition 4.5 shows that $\varphi_\delta$ is odd, $S_5$-invariant and $C^{1,1}$. The ball direction $b=\frac{1}{2}-h$ has the same properties, so (37) implies that $\widetilde{\varphi}_\delta$ does as well. Hence

The compactness argument above gives

on every smooth piece for $0\le t<t_\delta$. Since $h_{\delta,t}\in C^{1,1}$, nonnegativity of the homogeneous curvature-radius operator is the support-function convexity criterion. More explicitly, the homogeneous extension has positive semidefinite Hessian almost everywhere away from the origin. Its $C_{\mathrm{loc}}^{1,1}$ regularity rules out negative singular Hessian terms at the interfaces, so its restriction to every segment avoiding the origin is convex. On a line through the origin, for a unit vector $u$, the extension equals $\tau h_{\delta,t}(u)$ for $\tau\ge0$ and $-\tau h_{\delta,t}(-u)$ for $\tau\le0$. The right derivative minus the left derivative is $h_{\delta,t}(u)+h_{\delta,t}(-u)=1$. Thus the restriction to these lines is also convex. Hence $h_{\delta,t}$ is the support function of a convex body. The antipodal identity gives width one, and the construction is $S_5$-invariant. Finally,

and the coefficient is strictly negative by (39). Therefore the volume is smaller for all sufficiently small positive $t$. □

Equivalently, the deformed body can be written directly as

Thus the result is not merely an abstract existence theorem: the formulas (22), (24), and (37) define an explicit nearby family. The choices “$\delta$ sufficiently small” and “$t$ sufficiently small” have been justified analytically, without assigning numerical endpoints to the allowable ranges. Thus the theorem proves strict descent and global feasibility; it supplies no six-decimal volume for a specified optimized member of this family.

Remark 4.10 (One-sided and two-sided questions). The path only uses positive $t$. A two-sided Minkowski decomposition asks for feasibility in both directions, which is a stronger requirement. Restrictions on two-sided perturbations therefore do not rule out the one-sided descent just constructed.

A lower bound for all four-dimensional constant-width bodies

In this independent section, $p$ denotes the support function of an arbitrary body $K$ of constant width one, and $S_3(K)$ denotes its surface area. We first obtain a surface-area lower bound from the curvature constraint, then convert it to volume using an inscribed ball.

A spectral surface-area estimate

5.1. A spectral surface-area estimate. First suppose $p$ is smooth and write $p=1/2+f$. Constant width makes $f$ odd. A translation removes its degree-one spherical harmonic without changing $\nabla^2f+fI$. Set

The eigenvalues of $A$ lie in $[-1/2,1/2]$. Here $A$ is the centered curvature matrix: the support curvature matrix from the preceding sections is $A_p=\frac{1}{2}I+A$. Throughout this section, $d\sigma$ is unnormalized spherical surface measure, so $\sigma(S^3)=2\pi^2$. Our norm conventions are

where $\lVert A(u)\rVert_F$ is the pointwise Frobenius (Hilbert–Schmidt) norm. Thus $\lVert A\rVert_2^2$ is a spherical $L^2$ norm of a matrix field.

Proof. Oddness gives $A(-u)=-A(u)$ after identifying antipodal tangent spaces. In the expansion of $\det(\frac{1}{2}I+A)$, the linear and cubic terms therefore integrate to zero. If $\sigma_2(A)$ is the second elementary symmetric function of its eigenvalues, then

The identity $2\sigma_2(A)=(\operatorname{tr}A)^2-\lVert A\rVert_F^2$, integration by parts, and the integrated Bochner formula on $S^3$ give

This proves the area identity since $\lvert S^3\rvert=2\pi^2$. On a degree-$\ell$ harmonic, $-\Delta f_\ell=\lambda_\ell f_\ell$. Expanding $\lVert\nabla^2f+fI\rVert_2^2$ gives $(\lambda_\ell^2-4\lambda_\ell+3)\lVert f_\ell\rVert_2^2$. Orthogonality of distinct degrees finishes the proof. $\square$

Since the first remaining eigenvalue is $\lambda_3=15$,

Consequently

From surface area to volume

Jung’s theorem places a diameter-one body in a ball of radius at most $R_J=\sqrt{2/5}$. Choose its centre as the origin. Then $p(u)\le R_J$, and width one implies $p(u)=1-p(-u)\ge1-R_J=:r_J$. Thus the concentric ball of radius $r_J$ lies inside $K$. Minkowski’s formula yields

Combining this with (42) gives

The translation used here may differ from the one used to remove the degree-one harmonic; surface area and volume are translation invariant.

The smoothness restriction can be removed by averaging the support function against a smooth approximate identity on the rotation group, then mixing with a vanishing amount of the radius-$1/2$ ball if strict curvature is needed. These operations preserve width one and converge in Hausdorff distance. Continuity of volume and surface area proves the same bounds for every constant-width body.

An explicit improvement of the preceding bound

There are two sources of nonsharpness in the preceding estimate: the curvature eigenvalues need not all have maximal absolute value, and harmonics of degree at least five satisfy a stronger spectral inequality. An elementary great-circle argument shows that these two losses cannot both be too small.

The improvement over (5.2) exceeds $4.69\cdot10^{-5}$; a large gap to the upper constructions remains.

Proof. First use a smooth support $p=1/2+f$ and remove its degree-one part. Write $A=\nabla^2f+fI$, so its three eigenvalues are in $[-1/2,1/2]$. Let $f_3$ be the degree-three part and set

Lemma 5.1 gives, degree by degree,

On the other hand, $Z=\sum_{\ell\ge5,\ \ell\ \mathrm{odd}}(\lambda_\ell-3)^2\lVert f_\ell\rVert_2^2$. For $\lambda_\ell\ge35$,

the ratio is increasing and equality holds at degree five. Consequently

Set $g=\operatorname{tr}A_3=-12f_3$. Then

There are 16 degree-three harmonics on $S^3$. For an orthonormal basis $Y_j$, the tensor $T(u)=\sum_j\nabla Y_j(u)\otimes\nabla Y_j(u)$ is unchanged by rotations fixing $u$, so it is a scalar multiple of the identity on $T_uS^3$. Its trace is constant and its integral is $16 \cdot15$. Since $|S^3| = 2\pi^2$ and the tangent space has dimension three,

Write $g = \sum_j c_jY_j$. For every unit tangent vector $\mathbf{v}$, Cauchy–Schwarz gives

Taking the supremum over $\mathbf{v}$ proves the global Lipschitz bound

Let $a=9/50$, $r=1/40$, and put

On every great circle $g$ is odd under a half turn, so it has a zero and an antipodal zero. Around these two zeros, the four half-arcs of length $r=1/40$ lie in $\Omega_a$, because $(36/5)r=a$. These arcs do not overlap, since $r<\pi/2$. Their total circle length is $4r$. Averaging over rotations of one fixed great circle and applying Fubini gives the spherical measure estimate

This argument does not require a choice of a common zero for different circles.

For any three eigenvalues in $[-1/2,1/2]$,

To verify it, maximize $\sum\lambda_i^2-(\sum\lambda_i)^2=-2\sum_{i<j}\lambda_i\lambda_j$ on the cube; it is affine in each coordinate separately and its vertex maximum is $1/2$. Choose $b=27/100$. By Chebyshev’s inequality, $|\operatorname{tr}A_R|\le b$ outside a set of measure at most $Z/b^2$. On the remaining portion of $\Omega_a$,

Here $d$ is a positive scalar gap, unrelated to the width parameter in § 2. The deficit is nonnegative everywhere; hence

These rational choices obey $(5/224)/C=729/21280<1/28$. Combining this inequality with (46) and (48) gives

Combining Jung’s theorem with Minkowski’s volume formula, as above, gives $4\operatorname{Vol}_4(K)\ge r_JS_3(K)$ and proves the volume assertion. We refer to this combination as the Jung–Minkowski estimate. Rotational smoothing preserves width one and converges in the convex-body topology, so continuity of surface area and volume extends the inequalities to arbitrary constant-width bodies. The rational comparisons and the displayed decimal lower endpoint can be checked by the accompanying 192-bit Arb script certify_lower_gap.py; this numerical check is separate from the analytic proof just given. □

Remark 5.3 (The lower-bound constant). The exact right-hand side of (45) equals

This improves (43) without using a three-dimensional volume lower bound. It is a lower-bound constant, not an approximation to the unknown minimum. The argument uses the simple rational thresholds $a=9/50$ and $b=27/100$ without optimizing them. Its longer spectral proof records an independent refinement; numerically, the projection bound in Corollary 5.4 is stronger.

A stronger bound from three-dimensional projections

Jiang’s Theorem 1.1 in [6] states that every three-dimensional convex body of constant width one has volume at least $1651/4000$. This external result gives a short improvement of the four-dimensional bound.

The constant satisfies

This corollary uses only the stated constant $1651/4000$ from Jiang’s preprint. It does not require extending that paper’s contact-geometry argument to four dimensions.

Discussion

Theorem 4.9 proves that $U_{4}^{(1)}$ is not a local volume minimizer, even with all simplex symmetries retained. The proof gives a formula for the deformation and justifies a positive parameter range without quantifying its endpoints. It does not determine an optimal member of this family or an optimal symmetric body.

The spectral bound in Theorem 5.2 and the literature-dependent projection bound in Corollary 5.4 remain well below $\operatorname{Vol}_{4}(U_{4}^{(1)}) \approx0.221836$, whose six-decimal rounding is certified in [5 equation (1.4)]. This is the volume of the original body, not a numerical evaluation of the deformed bodies. A useful quantitative descent would need a larger certified gain that justifies its proof’s complexity. Such an optimization is not pursued here. Neither the descent nor the lower bounds decide whether a global minimizer retains simplex symmetry.

Code for checking only the displayed lower-bound constants is available in the computational supplement (manuscript2/lower_bound_checks/).