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\title{Fixed points of the polar projection body operator\\ and its square}
\author{Sihan Chen}
\date{November 12, 2026}
\hypersetup{pdftitle={Fixed points of the polar projection body operator and its square},
            pdfauthor={Sihan Chen}}

\begin{document}
\maketitle

\begin{abstract}
Let $TK=(\Pi K)^\circ$ be the polar projection body of a convex body
$K$ in $\R^n$. In dimension $n\ge3$, we prove that $TK$ is a positive
dilate of $K$ exactly when $K$ is an origin-centred ellipsoid,
settling Petty's polar projection conjecture. The same classification
holds under the weaker assumption that $T^2K$ is a positive dilate of
$K$. The proof uses a distributional Laplacian identity to compare two
integrals of central section volumes. Equality supplies contractive
hyperplane projections with every kernel direction, and James's theorem
gives ellipsoidal rigidity. The argument applies to arbitrary convex
bodies and yields the translative constant volume conjecture of
Horv\'ath and L\'angi.
\end{abstract}

\section{The polar projection problem}

A convex body is a compact convex set with nonempty interior. Write
$|\cdot|_d$ for $d$-dimensional volume, $|K|=|K|_n$, and
$\kappa_d=|B_2^d|_d$. Let $h_K$ be the support function of $K$ and
$S_K$ its surface area measure, obtained by pushing forward boundary
area under the almost everywhere defined outer unit normal $\nu_K$.
We use the normalization of Cauchy's projection formula
\begin{equation}\label{eq:cauchy}
 h_{\Pi K}(x)
 =\frac12\int_{S^{n-1}}|\ip{x}{u}|\,dS_K(u)
 = |x|\,|P_{x^\perp}K|_{n-1}\qquad(x\ne0),
\end{equation}
where $P_H$ denotes orthogonal projection onto the hyperplane $H$.
See \cite[Lemma 2.1.2(ii)]{BFR}. For a body $C$ containing the
origin in its interior, put
\[
 C^\circ=\{y:\ip{x}{y}\le1\text{ for all }x\in C\},
 \qquad p_C(x)=\inf\{s>0:x\in sC\}=h_{C^\circ}(x).
\]
Define $TK=(\Pi K)^\circ$. This is an origin-symmetric convex body,
and $T(sK)=s^{1-n}TK$ for $s>0$.

Petty's polar projection problem \cite{Petty} asks whether
$TK=aK$ for some $a>0$ characterizes origin-centred ellipsoids in
dimension $n\ge3$ \cite[Conjecture 2]{HL22}. We prove the classification
also for a return after two applications, with an arbitrary positive
dilation.

\begin{theorem}[Petty's conjecture and the second iterate]
\label{thm:classification}
Let $n\ge3$ and let $K\subset\R^n$ be a convex body. The following
are equivalent.
\begin{enumerate}
 \item $K$ is an ellipsoid centred at the origin.
 \item $TK=aK$ for some $a>0$.
 \item $T^2K=bK$ for some $b>0$.
\end{enumerate}
When these conditions hold,
\[
 a=\frac{\kappa_n}{\kappa_{n-1}|K|},
 \qquad b=a^{2-n}.
\]
\end{theorem}

The fixed point and the return condition each imply $K=-K$ because
$T$ takes values in origin-symmetric bodies. The proof first compares
the central sections of an origin-symmetric $K$ and $L=TK$. Under
$T^2K=bK$, the same comparison for $L$ reverses the inequality
after both sides are divided by $b^{n-1}$.

\section{A flux identity for the polar projection body}
\label{sec:flux}

For a body $C$ containing the origin in its interior and a unit vector
$u$, put $A_C(u)=|C\cap u^\perp|_{n-1}$. This is continuous in $u$ by
the radial formula for section volume.

\begin{lemma}\label{lem:flux}
Let $n\ge2$, let $K$ be a convex body, and set $L=TK$. Then
\begin{equation}\label{eq:flux}
  \int_{S^{n-1}}\frac{1}{h_L(u)}\,dS_L(u)
  =\int_{S^{n-1}}A_L(u)\,dS_K(u).
\end{equation}
\end{lemma}

\begin{proof}
By polarity and \eqref{eq:cauchy}, $p_L=h_{\Pi K}$. In the sense of
distributions, $\Delta|\ip{x}{u}|=2\mathcal H^{n-1}\!\restriction u^\perp$.
This follows by writing $x=z+tu$ with $z\in u^\perp$ and using
$(|t|)''=2\delta_0$. Integration against compactly supported smooth
test functions and Fubini's theorem therefore give the positive measure
\begin{equation}\label{eq:laplacian}
 M=\Delta p_L
   =\int_{S^{n-1}}\mathcal H^{n-1}\!\restriction u^\perp\,dS_K(u).
\end{equation}
The measures in the integral have uniformly bounded mass on each
compact set. In particular $M(L)=\int A_L\,dS_K$. Also, $M(\partial L)=0$.
Indeed, each $u^\perp$ meets $\operatorname{int} L$, so
$\partial L\cap u^\perp$ is the relative boundary of $L\cap u^\perp$
and has zero $(n-1)$-dimensional volume. To see this, if a point of
$\partial L\cap u^\perp$ were interior to the section, a supporting
normal there would have to be parallel to $u$. Its support value would
then be zero, contrary to $0\in\operatorname{int} L$.

The boundary of $L$ is Lipschitz and has a unique outer unit normal
at almost every point \cite[Lemma 1.5.6]{BFR}. At such a point $y$,
write $u=\nu_L(y)$. Since $p_L=h_{L^\circ}$, its subgradients at $y$
are exactly the $g\in L^\circ$ with $\ip{g}{y}=1$. They are the
supporting normals scaled to take value one at $y$. Thus the
subdifferential is a singleton and
\begin{equation}\label{eq:boundary-gradient}
 \nabla p_L(y)=\frac{u}{h_L(u)}.
\end{equation}

To integrate \eqref{eq:laplacian}, let $\varphi_\varepsilon$ be a
nonnegative smooth mollifier of integral one supported in
$\varepsilon B_2^n$, and put $p_\varepsilon=p_L*\varphi_\varepsilon$.
The integral of $M*\varphi_\varepsilon$ over $L$ tends to $M(L)$:
indeed,
\[
 \int_L\Delta p_\varepsilon(x)\,dx
 = \int_{\R^n}\left(\int_L \varphi_\varepsilon(x-z)\,dx\right)dM(z),
\]
and the inner integral is bounded by one and tends to $\mathbf1_L(z)$
off $\partial L$. Its support lies in the fixed compact set $L+B_2^n$
when $\varepsilon\le1$. This proves convergence since $M(\partial L)=0$.

Choose $r>0$ with $rB_2^n\subset L$. All subgradients of $p_L$
have norm at most $1/r$. At a point $y$ of differentiability, any
limit of subgradients at points approaching $y$ belongs to
$\partial p_L(y)$ by the subgradient inequality. Since this set is
a singleton, nearby gradients converge to $\nabla p_L(y)$ wherever
they exist. It follows that
$\nabla p_\varepsilon(y)\to\nabla p_L(y)$ at the boundary points
used in \eqref{eq:boundary-gradient}, and
$|\nabla p_\varepsilon|\le 1/r$.
The divergence theorem for the smooth field $\nabla p_\varepsilon$
on $\operatorname{int} L$ \cite[Theorem 2.1.4]{BFR} now gives
\[
 M(L)=\lim_{\varepsilon\downarrow0}
 \int_{\partial L}\ip{\nabla p_\varepsilon(y)}{\nu_L(y)}
                   \,d\mathcal H^{n-1}(y)
 =\int_{S^{n-1}}\frac{1}{h_L(u)}\,dS_L(u).
\]
The last limit uses dominated convergence and the definition of $S_L$.
Comparing with the value of $M(L)$ from \eqref{eq:laplacian}
proves \eqref{eq:flux}.
\end{proof}

\section{The section comparison and its equality case}
\label{sec:comparison}

\begin{proposition}\label{prop:comparison}
Let $n\ge2$, let $K=-K$ be a convex body, and put $L=TK$. Then
\begin{equation}\label{eq:comparison}
 \int_{S^{n-1}}A_K(u)\,dS_L(u)
 \ \le\ \int_{S^{n-1}}A_L(u)\,dS_K(u).
\end{equation}
If $n\ge3$ and equality holds, then $K$ is an origin-centred ellipsoid.
\end{proposition}

\begin{proof}
For a unit vector $u$, choose a support point $y\in\partial L$ with
$\ip{y}{u}=h_L(u)>0$. The map
\[
 Q_{y,u}z=z-\frac{\ip{z}{u}}{h_L(u)}y
\]
projects onto $u^\perp$ along $\R y$. Its restriction from $y^\perp$
to $u^\perp$ is inverse to the restriction of orthogonal projection
onto $y^\perp$ from $u^\perp$. The latter map has absolute Jacobian
$\ip{y/|y|}{u}=h_L(u)/|y|$, the cosine of the angle between the normals.
This follows by using the common $(n-2)$-dimensional subspace of the
hyperplanes, with the parallel case giving Jacobian one. As $Q_{y,u}$
kills $y$, it acts on $K$ by first orthogonally projecting onto
$y^\perp$ and then applying this inverse. Hence
\begin{equation}\label{eq:oblique-volume}
 |Q_{y,u}K|_{n-1}
 =\frac{|y|\,|P_{y^\perp}K|_{n-1}}{h_L(u)}
 =\frac{h_{\Pi K}(y)}{h_L(u)}
 =\frac{1}{h_L(u)}.
\end{equation}
The last equality uses $y\in\partial L$ and $p_L=h_{\Pi K}$.

Since $Q_{y,u}$ fixes $u^\perp$, the image $Q_{y,u}K$ contains
$K\cap u^\perp$. This gives $A_K(u)\le 1/h_L(u)$. Integrating this
inequality with respect to $S_L$ and using Lemma~\ref{lem:flux}
proves \eqref{eq:comparison}.

Suppose $n\ge3$ and equality holds. Then $A_K(u)=1/h_L(u)$ for
$S_L$-almost every $u$. For every support point $y$ of $L$ in such a direction, the two
convex bodies $Q_{y,u}K$ and $K\cap u^\perp$ have equal volume in
$u^\perp$. The inclusion gives $Q_{y,u}K=K\cap u^\perp$, so
$Q_{y,u}K\subset K$. By the definition of $S_L$, this conclusion
holds with $u=\nu_L(y)$ on a set $G\subset\partial L$ of full
boundary area. Every nonempty relatively open patch of $\partial L$
has positive boundary area, as seen by orthogonally projecting a
local Lipschitz graph onto its base. Thus $G$ is dense in $\partial L$
even when $S_L$ is supported on a proper subset of the sphere.

Fix $v\ne0$ and set $y=v/p_L(v)\in\partial L$. Take $y_j\in G$
with $y_j\to y$, and write $u_j=\nu_L(y_j)$ and
$g_j=u_j/h_L(u_j)\in L^\circ$. By compactness of $L^\circ$ a
subsequence of $g_j$ converges to $g$, with $\ip{g}{y}=1$.
The maps $z\mapsto z-\ip{g_j}{z}y_j$ converge to
$Qz=z-\ip{g}{z}y$. Closedness of $K$ gives $QK\subset K$.
The identity $\ip{g}{y}=1$ makes $Q$ a projection with range
$\ker g$ and kernel $\R v$. Thus for every $z\in\ker g$ and
$t\in\R$,
\[
             p_K(z)=p_K(Q(z+tv))\le p_K(z+tv).
\]
In Birkhoff's terminology, the hyperplane $\ker g$ is orthogonal to
$v$: each element of the hyperplane is in the first argument of the
orthogonality relation. James's characterization of real inner
product spaces \cite[Theorem 5, p.~565]{James} requires just such a
hyperplane for each $v\ne0$ in dimension at least three. Therefore
$p_K$ is an inner product norm and its unit ball $K$ is an
origin-centred ellipsoid.
\end{proof}

The pointwise comparison $A_K(u)\le 1/h_{TK}(u)$ is the classical
inclusion $IK\subseteq \Pi K$, where the intersection body $IK$ has
radial function $A_K$ \cite{MakaiMartini}. Formula~\eqref{eq:oblique-volume}
identifies equality in that comparison with a contractive projection.

\section{Fixed points and two applications of the operator}

\begin{proof}[Proof of Theorem~\ref{thm:classification}]
Assume $T^2K=bK$ and write $L=TK$. Both $K$ and $L$ are
origin symmetric. Proposition~\ref{prop:comparison} gives
\eqref{eq:comparison}. Applied to $L$ and $TL=bK$, it gives
\begin{align*}
 b^{n-1}\int_{S^{n-1}} A_L\,dS_K
   &=\int_{S^{n-1}} A_L\,dS_{bK}\\
   &\le\int_{S^{n-1}} A_{bK}\,dS_L
     = b^{n-1}\int_{S^{n-1}} A_K\,dS_L .
\end{align*}
Here $S_{bK}=b^{n-1}S_K$ and $A_{bK}=b^{n-1}A_K$.
Dividing by $b^{n-1}$ gives the reverse of \eqref{eq:comparison}.
Its equality case shows that $K$ is an ellipsoid. Also, if $TK=aK$,
the homogeneity of $T$ gives $T^2K=a^{2-n}K$, so the fixed point
condition is covered by the same argument.

For the converse and the constants, recall the segment formula
\begin{equation}\label{eq:segment}
 |C+[0,v]|=|C|+h_{\Pi C}(v)
\end{equation}
for every convex body $C$ and $v\in\R^n$ \cite[Lemma 1]{HL22}.
For $v\ne0$, slicing along the lines parallel to $v$ adds $|v|$
to the length of each nonempty fibre, which gives \eqref{eq:segment}
by \eqref{eq:cauchy}. If $K=UB_2^n$ for $U\in GL(n)$, a change of
variables by $U$ in \eqref{eq:segment} yields
\[
 h_{\Pi K}(v)=|\det U|\,h_{\Pi B_2^n}(U^{-1}v)
            =\kappa_{n-1}|\det U|\,|U^{-1}v|.
\]
Thus $TK=(\kappa_{n-1}|\det U|)^{-1}K$. Since
$|K|=\kappa_n|\det U|$, this gives $a$, and homogeneity gives $b$.
\end{proof}

\begin{remark}[Dimension two]
In the plane, for $K=-K$ and a rotation $J$ through $\pi/2$,
brightness is perpendicular width, so $\Pi K=2JK$.
Then $TK=\tfrac12 JK^\circ$ and $T^2K=K$ for every such $K$.
Nonellipsoidal one-application fixed points up to dilation also
exist in the plane. For instance a regular hexagon of circumradius
one satisfies $K^\circ=(2/\sqrt3)JK$ and $TK=K/\sqrt3$.
See \cite[Theorem 2]{HL22} for the connection with Radon curves.
\end{remark}

\section{The translative constant volume conjecture}

\begin{corollary}[Horv\'ath--L\'angi]\label{cor:translative}
Let $n\ge3$ and let $K\subset\R^n$ be a centrally symmetric convex
body. Suppose the volume of the convex hull of any two touching
translates of $K$ is independent of the pair of translates. Then
$K$ is an ellipsoid. Conversely, ellipsoids have this property.
\end{corollary}

\begin{proof}
Translate $K$ so that $K=-K$. Two translates touch when they
intersect and their interiors are disjoint. By translation invariance
it suffices to use $K$ and $K+v$, which touch exactly for
$v\in\partial(K-K)=\partial(2K)$. Moreover, convexity and
\eqref{eq:segment} give
\[
 \big|\operatorname{conv}(K\cup(K+v))\big|
    =|K+[0,v]|=|K|+h_{\Pi K}(v).
\]
Thus $h_{\Pi K}$ is constant, say $d>0$, on $\partial(2K)$.
On that boundary $p_K(v)=2$, so positive homogeneity gives
$h_{\Pi K}=(d/2)p_K$ everywhere. Since $p_{TK}=h_{\Pi K}$, this
is $TK=(2/d)K$. Theorem~\ref{thm:classification} gives the result.
For an ellipsoid, the same identities and $TK=aK$ imply constancy.
\end{proof}

Horv\'ath and L\'angi posed the conjecture in
\cite[Section 4, Conjecture 2]{HL14} and established its equivalence
with Petty's polar projection problem in \cite[Theorem 3]{HL22}.

\appendix
\section{Earlier approaches}

For $TK=aK$, the second moment of the boundary functionals under
cone measure gives $\beta E\subset K\subset\sqrt{\beta}E$ for an
ellipsoid $E$, with $\beta=2/(na|K|)$. This yields a Banach--Mazur
bound of order $n^{1/4}$. The associated embedding into $L^1$ has
relative projection constant $n\beta$. The absolute bound in
\cite{DL} allows $\beta$ above the ellipsoidal value, and general
subspaces of $L^1$ can exceed the Euclidean projection constant
\cite{CL}. The sign kernel calculation gives a squared residual
$1-n\beta^2$, positive even for ellipsoids.

For two rotational blocks, estimates at the axes and an averaging
identity from affine shears forced an ellipsoidal profile.
Fractional moments of the normals reduced two arbitrary block norms
to the rotational case. For homothetic ellipsoidal parallel sections,
a contraction of the section centres forced their collinearity.

\section*{Use of AI}

OpenAI's ChatGPT was used to explore proof strategies, develop and
check mathematical arguments, and draft and revise the exposition.

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\end{thebibliography}
\end{document}