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\title{Improved and sharp bounds for illumination}
\author{Working research draft}
\date{1 November 2026}
\begin{document}
\maketitle
\begin{abstract}
We prove improved ordinary illumination bounds and exact ordinary
and fractional values for several classes of convex bodies.
Every full-dimensional complex zonotope in $\C^2$ has illumination
number five, except complex affine images of the bidisc, whose
number is seven. For complex $1$-symmetric non-polydiscs in $\C^3$,
an explicit construction gives twelve illuminating directions.
For real $1$-symmetric non-cubes in $\R^n$, connected domination in
the cube graph gives the upper bound
$2^{n-1}+\gamma_c(Q_{n-1})+n-1
=2^{n-1}(1+(1+o(1))/n)$; an explicit family has illumination
number $2^{n-1}$.
We also give a product boundary measure proving fractional lower
bounds $2^n$ for bodies between specified inner and outer bodies,
and attain these bounds along Minkowski interpolations with a
positive inner summand. Finally, the least size $U_n$ of a single
family illuminating every normalized complex $1$-unconditional
body satisfies $U_n\sim3^{n+1}/\sqrt{\pi n}$.
\end{abstract}

\noindent\textit{Draft status.} The proofs included here have been
checked internally; external review and literature priority remain
outstanding. The finite covering assertions have exact, reproducible
certificates. Bounds for an individual body and for a single family
serving all normalized bodies are kept separate. The citation audit
identifies established inputs and earlier special cases; it does not
certify novelty of the remaining results.

\section{Introduction}\label{sec:introduction}

Illumination asks for directions that move every boundary point
of a convex body into its interior. The ordinary illumination
number $\ill(K)$ counts directions. Its fractional counterpart
$\ill^*(K)$ minimizes the mass of a measure giving at least unit
mass to the directions illuminating each boundary point.
We study exact values, improvements under coordinate symmetry,
and the additional cost of choosing directions independently of
the body.

Nasz\'odi introduced fractional illumination with measures of finite
support~\cite[Definition~1]{Naszodi}. We use the Borel-measure
formulation of Rotem, Schejter, and Slomka~\cite[Section~1.2]{RSS}.
Their polydisc values
$\ill(\mathbb D^n)=2^{n+1}-1$ and $\ill^*(\mathbb D^n)=2^n$
are established inputs~\cite[Theorems~1.3--1.4]{RSS}.
The ordinary formula was asserted earlier by Boltyanski and
Martini~\cite{BM}; we use the corrected construction in
\cite[Proposition~2.7 and Remark~2.8]{RSS}.

Theorem~\ref{thm:zonotopes-two} determines the ordinary number
for every full-dimensional complex zonotope in $\C^2$:
\[
 \ill(Z)=7\quad\text{for a complex affine bidisc},\qquad
 \ill(Z)=5\quad\text{otherwise}.
\]
The proof reduces to three generating discs. Five explicit
directions illuminate that model; the only nonsmooth boundary
points lie on three circles, on which illumination is verified
by a finite table of open intervals. In particular, six never
occurs in this class. The previously published
\cite[Theorem~1.8]{RSS} gives at most six directions in this
dimension; the five-direction construction below sharpens that bound.

Theorem~\ref{thm:symmetric-three} gives twelve directions for
every complex $1$-symmetric non-polydisc in $\C^3$.
The construction uses twelve rows of thirteenth roots of unity,
with one distinguished coordinate in each row. A finite covering
check gives a common strictly inward margin, and a supporting-normal
argument determines the coordinate magnitudes. The maximum is six
in dimension two and lies in $[9,12]$ in dimension three
(Corollary~\ref{cor:complex-symmetric-extrema}); sharpness of
twelve is not established. The earlier bound fourteen and the
geometric reduction are in the coauthored manuscript
\cite[Theorem~1.5 and Proposition~2.8]{RSSSym}.

For real $1$-symmetric bodies, write $M_n$ for the supremum of
$\ill(K)$ over bodies in $\R^n$ that are not affine cubes.
Tikhomirov proved $M_n<2^n$ in sufficiently large
dimensions~\cite[Theorem~1]{Tikhomirov}. Sun and Vritsiou obtained
$M_n\le2^n-2$ for every $n\ge3$~\cite[Theorem~23]{SVSym}.
The coauthored manuscript~\cite[Theorem~1.4]{RSSSym} gives
$M_n\le3\cdot2^{n-2}+1$ for $n\ge4$.
Proposition~\ref{prop:real-connected-domination} proves
\[
 M_n\le2^{n-1}+\gamma_c(Q_{n-1})+n-1,
\]
where $\gamma_c$ is the connected domination number of the cube
graph. A connected dominating set in $Q_{n-1}$ gives a vertex
cover in $Q_n$ with an adjacent successor assignment, and this
produces illuminating directions. The asymptotic estimate of
Griggs~\cite{Griggs} yields
$M_n\le2^{n-1}(1+(1+o(1))/n)$.
Proposition~\ref{prop:real-symmetric-examples} gives the explicit
non-cubes
\[
 P_n=\left\{x\in[-1,1]^n:\sum_{j=1}^n|x_j|\le n-1\right\},
 \qquad \ill(P_n)=2^{n-1}.
\]

The fractional statements concern convex bodies between
$H=\mathbb D^p\times B^r$ and
$E=\mathbb D^p\times\prod_{j=1}^rL(a_j,b_j)$,
where $p+2r=n$, $B$ is the complex two-dimensional $\ell_1$ ball,
and the sets $L(a_j,b_j)$ are defined in
\eqref{eq:planar-equality-envelope}. If $a_jb_j\ge1/2$, then
every $H\subseteq C\subseteq E$ satisfies $\ill^*(C)\ge2^n$,
without a symmetry assumption on $C$.
Proposition~\ref{prop:fractional-equality-minkowski} attains this
lower bound for $(1-t)H+tC$ whenever $0\le t<1$. The upper
bound for an arbitrary endpoint $C$ is not asserted.
The inner bodies are the Hanner equality cases in the companion
paper~\cite{Families}; the planar envelopes also occur in the
planar classification in~\cite[Section~7]{Unconditional}.
Remark~\ref{rem:fractional-conjecture} explains why these examples
contradict the equality clause, but not the upper bound, of
\cite[Conjecture~1.2]{RSS}.

Appendix~\ref{app:simultaneous} treats the distinct quantity
$U_n$: the least number of directions in one family that
illuminates every normalized complex $1$-unconditional body.
A lower bound obtained from different bodies matches the
support construction asymptotically. This proves
$U_n\sim3^{n+1}/\sqrt{\pi n}$, without giving the same lower
bound for any one body's illumination number. The common families
in \cite[Theorem~D]{SVSym} and \cite[Theorem~1.6]{RSSSym}
serve the real and complex $1$-symmetric classes, respectively;
they do not give this result for all normalized complex
$1$-unconditional bodies.

\section{Notation and preliminary facts}\label{sec:preliminaries}

Write $[n]=\{1,\ldots,n\}$, $\mathbb D=\{z\in\C:|z|\le1\}$,
and $\R_+=[0,\infty)$. A convex body is a compact convex set with
nonempty interior in the underlying real space. It is complex
$1$-unconditional if it is invariant under independent rotations of
its coordinates. For such a body, its Minkowski norm is denoted by $\norm{\cdot}$.
For $z\in\C^n$, write $|z|=(|z_1|,\ldots,|z_n|)$ and
$\supp(z)=\{j\in[n]:z_j\ne0\}$. We also write
\[
 M_x=\{j\in[n]:|x_j|=\|x\|_\infty\},\qquad
 Z_x=\{j\in[n]:x_j=0\}.
\]

A nonzero vector $d$ illuminates $x\in\partial K$ if
$x+td\in\operatorname{int}K$ for some $t>0$. The least cardinality of
a family illuminating every boundary point is $\ill(K)$.
For real bodies the same definition uses $\R^n$.
A finite nonnegative Borel measure $\mu$ on $\C^n\setminus\{0\}$
fractionally illuminates $K$ if
\[
 \mu\{d:d\text{ illuminates }x\}\ge1\qquad(x\in\partial K).
\]
The infimum of the total masses of such measures is $\ill^*(K)$.

We call $K$ normalized when $e_j\in\partial K$ for every $j\in[n]$.
Positive diagonal rescaling gives this normalization, and illumination
numbers are invariant under translations and invertible real linear maps. Coordinate
contractions preserve membership, as proved below; consequently a
normalized body satisfies $K\subseteq\mathbb D^n$ and contains each
unit coordinate disc.

\begin{lemma}\label{lem:section-contraction}
For a complex $1$-unconditional convex body $K$, the following statements hold.
\begin{enumerate}
\item If $|w_j|\le|z_j|$ for every $j$, then $\norm{w}\le\norm{z}$.
\item For every real linear subspace $E\subseteq\C^n$,
\[
 \operatorname{relint}(K\cap E)=E\cap\operatorname{int}K.
\]
\end{enumerate}
\end{lemma}
\begin{proof}
Invariance under all coordinate rotations implies central symmetry.
Because $K$ has nonempty interior, its center $0$ is interior: if
an open Euclidean ball of radius $r$ around $a$ lies in $K$, the same
is true around $-a$, and convexity puts the ball of radius $r$ around
$0$ in $K$. Its Minkowski functional is therefore a norm, with
$\operatorname{int}K=\{z:\norm{z}<1\}$.

Every number $\lambda\in\C$ with $|\lambda|\le1$ is a convex
combination of two unit complex numbers. Explicitly, if
$\lambda=se^{\ii\theta}$, $0\le s\le1$, then
\[
 \lambda=\frac{1+s}{2}e^{\ii\theta}
          +\frac{1-s}{2}(-e^{\ii\theta}).
\]
Thus contracting a single coordinate by $\lambda$ preserves membership
in $K$, by convexity and rotation invariance. Applying these contractions
one coordinate at a time proves coordinatewise monotonicity of membership.
Apply the same fact to every dilation of $K$ to obtain the norm inequality
in part~1. Coordinates with $z_j=0$ also have $w_j=0$, so cause no exception.

The inclusion from right to left in part~2 follows by intersecting an
ambient open ball with $E$. For the reverse inclusion, let
$y\in\operatorname{relint}(K\cap E)$. If $y=0$ there is nothing to prove.
Otherwise relative interior gives an $s>0$ with $(1+s)y\in K\cap E$.
Consequently $\norm{y}\le(1+s)^{-1}<1$, as required.

\end{proof}

We use the real pairing
\[
 [u,z]=\operatorname{Re}\sum_{j\in[n]}\overline{u_j}z_j,
 \qquad h_K(u)=\max_{z\in K}[u,z].
\]
A direction $d$ illuminates $x\in\partial K$ if and only if
\begin{equation}\label{eq:normal-criterion}
 [u,d]<0\quad\text{for every nonzero }u\text{ with }[u,x]=h_K(u).
\end{equation}
Here is a proof that also covers nonsmooth boundary points. If $x+td\in\operatorname{int}K$
for some $t>0$, then $[u,x+td]<h_K(u)=[u,x]$ for every nonzero supporting
normal at $x$, proving necessity. For sufficiency, suppose no positive step enters
$\operatorname{int}K$. For each integer $j\ge1$, separation at
$x+j^{-1}d\notin\operatorname{int}K$ supplies a Euclidean unit vector $u^j$
with $[u^j,x+j^{-1}d]\ge h_K(u^j)$. Since
$h_K(u^j)\ge[u^j,x]$, we have $[u^j,d]\ge0$.
Pass to a convergent subsequence of these unit vectors, with limit $u$.
The support function is continuous, so $[u,x]\ge h_K(u)$; the reverse
inequality holds because $x\in K$. Thus $u$ is a nonzero supporting normal
at $x$ with $[u,d]\ge0$, contradicting \eqref{eq:normal-criterion}.

For a normalized unconditional body, put $S=\supp(x)$ and $a_j=|u_j|$ for a supporting normal $u$ at $x$. Independent coordinate rotations give $u_j=|u_j|x_j/|x_j|$ whenever $j\in S$:
the maximum of $[u,\cdot]$ on the coordinate-rotation orbit of $x$ is
$\sum_j|u_j||x_j|$, and $x$ already attains that maximum. Thus, writing $h=h_K(u)>0$,
\begin{equation}\label{eq:normalized-normal-estimates}
 h=\sum_{j\in S}a_j|x_j|,
 \qquad a_j\le h\quad(j\in[n]),
 \qquad h\le\sum_{j\in S}a_j.
\end{equation}
The middle inequality follows from the coordinate discs in $K$; the last uses $K\subseteq\mathbb D^n$.

The illumination relation is open in the boundary point
and the direction: for fixed $t>0$, the condition
$x+td\in\operatorname{int}K$ is open, and illumination is its
union over $t>0$. Consequently all integrals of this relation
below are measurable, and Tonelli's theorem applies.
The normal criterion has the same proof over $\R^n$, using
the usual real inner product.

\section{Complex zonotopes in two dimensions}\label{sec:zonotopes-two}

The result in this section concerns complex zonotopes and does not
assume invariance under independent coordinate rotations.
Up to translation, a complex zonotope in $\C^2$ has the form
\[
 Z=\sum_{j\in[N]}\mathbb D v^j,
 \qquad v^j\in\C^2.
\]
It is full-dimensional precisely when its generators span $\C^2$
over $\C$.

\begin{theorem}\label{thm:zonotopes-two}
Let $Z\subset\C^2$ be a full-dimensional complex zonotope. Then
\[
 \ill(Z)=
 \begin{cases}
 7,&\text{if $Z$ is a translate of a complex-linear image of $\mathbb D^2$,}\\
 5,&\text{otherwise.}
 \end{cases}
\]
In particular, no full-dimensional complex zonotope in $\C^2$ has
illumination number $6$.
\end{theorem}

\begin{proof}
Translation does not affect illumination, so assume that $Z$ is
centered at the origin. The bidisc case follows from the exact
polydisc theorem \cite[Theorem~1.4]{RSS} and linear invariance.
We prove the other case by constructing five directions for
\begin{equation}\label{eq:zonotope-model}
 K_0=\mathbb D e_1+\mathbb D e_2+\mathbb D(e_1+e_2).
\end{equation}
The reduction to this model is the two-dimensional instance of
\cite[Lemma~3.4]{RSS}, using the numbering of the published version.
We include the argument to specify how the directions are transferred.

First, every set illuminating a full-dimensional convex body $K$
also illuminates $K+L$ for any nonempty compact convex set $L$.
Indeed, write $x=y+z\in\partial(K+L)$ with $y\in K$ and $z\in L$.
Then $y\in\partial K$, since
$\operatorname{int}K+L\subseteq\operatorname{int}(K+L)$.
If $y+td\in\operatorname{int}K$, then
$x+td=(y+td)+z\in\operatorname{int}(K+L)$.

Next, positive rescaling of individual generators preserves the same
illuminating sets. For $r_j>0$ and $r=\min_{j\in[N]}r_j$, write
\[
 \sum_{j\in[N]}r_j\mathbb D v^j
 =r\sum_{j\in[N]}\mathbb D v^j
   +\sum_{j\in[N]}(r_j-r)\mathbb D v^j.
\]
The preceding observation and dilation invariance prove one
implication; applying it to the rescaled generators with factors
$r_j^{-1}$ proves the reverse implication. Multiplication of a
generator by a unit complex number leaves its disc unchanged, so
the assertion also holds for nonzero complex rescalings.
These are Fact~3.2 and Lemma~3.3 of the published version of \cite{RSS}.

Discard zero generators and merge generators on the same complex
line. If exactly two lines remain, full dimensionality makes $Z$
a complex-linear image of $\mathbb D^2$. Otherwise choose generators
$v^1,v^2,v^3$ on three distinct complex lines and write
\[
 v^3=\alpha v^1+\beta v^2,\qquad \alpha\beta\ne0.
\]
The complex-linear map $T$ defined by
$Te_1=\alpha v^1$ and $Te_2=\beta v^2$ is invertible, and
\[
 TK_0=\mathbb D(\alpha v^1)+\mathbb D(\beta v^2)+\mathbb D v^3.
\]
Thus, if a set $V$ illuminates $K_0$, the set $TV$ illuminates
$TK_0$, then $\mathbb D v^1+\mathbb D v^2+\mathbb D v^3$
by generator rescaling, and finally $Z$ by Minkowski addition.

For $x\in\partial K_0$, define its normal cone by
\[
 N_{K_0}(x)=\{u\in\C^2:[u,y-x]\le0\text{ for every }y\in K_0\}.
\]
We call $x$ smooth if this cone is a single ray. The support function is
\begin{equation}\label{eq:zonotope-support}
 h_{K_0}(u)=|u_1|+|u_2|+|u_1+u_2|.
\end{equation}
Indeed, writing $x=(z_1+z_3,z_2+z_3)$ with $z_j\in\mathbb D$,
the three independent maxima in $[u,x]$ are the three terms on
the right-hand side.

Fix such a representation of a boundary point $x$.
Each term is at most its corresponding support value, so equality
in \eqref{eq:zonotope-support} holds exactly when equality holds
in all three terms. Consequently $u\in N_{K_0}(x)$ if and only if
\begin{equation}\label{eq:zonotope-normal-equations}
 u_1=t_1z_1,\qquad
 u_2=t_2z_2,\qquad
 u_1+u_2=t_3z_3,\qquad t_j\ge0,
\end{equation}
where $t_j=0$ is required if $|z_j|<1$.

If $|z_1|<1$, then $u_1=0$ and $u_2$ lies on the fixed ray
$\R_+z_2$; the case $|z_2|<1$ is analogous. If $|z_3|<1$,
then $u_2=-u_1$ and $u_1$ lies on $\R_+z_1$. In each case the
normal cone is contained in one ray. Since a boundary point has
a nonzero normal, it is smooth.

Suppose now that $|z_j|=1$ for every $j\in[3]$. The parameters
in \eqref{eq:zonotope-normal-equations} satisfy
\[
 t_1z_1+t_2z_2-t_3z_3=0.
\]
If the three numbers $z_j$ do not lie on one real line through
the origin, the real-linear map on the left has rank two.
Its kernel is one-dimensional, and its intersection with
$\R_+^3$ is at most a ray. This again makes $x$ smooth.

In the remaining case, write
$z_3=q$, $z_1=s_1q$, and $z_2=s_2q$, where $|q|=1$
and $s_1,s_2\in\{-1,1\}$. The choice $s_1=s_2=-1$ gives
$-t_1-t_2=t_3$, so all three parameters vanish; this is
impossible at a boundary point. The other choices give
\[
 A(q)=q(2,2),\qquad B(q)=q(2,0),\qquad C(q)=q(0,2).
\]
Writing $\operatorname{pos}\{u,v\}=\{au+bv:a,b\ge0\}$,
equation \eqref{eq:zonotope-normal-equations} yields
\begin{equation}\label{eq:zonotope-cones}
\begin{aligned}
 N_{K_0}(A(q))&=q\,\operatorname{pos}\{e_1,e_2\},\\
 N_{K_0}(B(q))&=q\,\operatorname{pos}\{e_1,e_1-e_2\},\\
 N_{K_0}(C(q))&=q\,\operatorname{pos}\{e_2,e_2-e_1\}.
\end{aligned}
\end{equation}
Each cone has dimension two. These three circles are therefore
exactly the nonsmooth boundary points of $K_0$.

Choose the five directions
\begin{equation}\label{eq:zonotope-five-directions}
\begin{aligned}
 d^1&=(1-\ii,-\ii),&
 d^2&=(1,1+\ii),&
 d^3&=(-1-\ii,-1),\\
 d^4&=(\ii,-1+\ii),&
 d^5&=(-1+\ii,1-\ii).
\end{aligned}
\end{equation}
They satisfy $\sum_{j\in[5]}d^j=0$, and the determinant of
their first four real coordinate rows is
\[
 \det\begin{pmatrix}
 1&-1&0&-1\\
 1&0&1&1\\
 -1&-1&-1&0\\
 0&1&-1&1
 \end{pmatrix}=5.
\]
For every nonzero $u\in\C^2$, some $j\in[5]$ therefore satisfies
$[u,d^j]<0$. Otherwise all five inner products would be nonnegative
with sum zero, and hence all would vanish, contradicting real
spanning. By \eqref{eq:normal-criterion}, these five directions
illuminate every smooth boundary point.

For $d=(a,b)$, equations \eqref{eq:normal-criterion} and
\eqref{eq:zonotope-cones} give the following necessary and
sufficient conditions:
\begin{equation}\label{eq:zonotope-inequalities}
\begin{array}{c|cc}
 &\multicolumn{2}{c}{\text{conditions for illumination by }d}\\
 A(q)&\operatorname{Re}(\overline q a)<0&
       \operatorname{Re}(\overline q b)<0\\
 B(q)&\operatorname{Re}(\overline q a)<0&
       \operatorname{Re}(\overline q(a-b))<0\\
 C(q)&\operatorname{Re}(\overline q b)<0&
       \operatorname{Re}(\overline q(b-a))<0
\end{array}
\end{equation}
Put $q=\exp(\ii\pi s/4)$, where $s\in\R/8\mathbb Z$.
For each direction in \eqref{eq:zonotope-five-directions}, the
sets of parameters satisfying \eqref{eq:zonotope-inequalities}
are the following open intervals, interpreted modulo $8$:
\begin{equation}\label{eq:zonotope-arcs}
\begin{array}{c|ccc}
 &A&B&C\\ \hline
 d^1&(1,4)&(2,5)&(0,2)\\
 d^2&(3,6)&(2,4)&(4,7)\\
 d^3&(-1,2)&(0,3)&(6,8)\\
 d^4&(5,8)&(4,6)&(6,9)\\
 d^5&\varnothing&(5,9)&(1,5)
\end{array}
\end{equation}
Every entry follows from
\[
 \operatorname{Re}(e^{-\ii t}z)<0
 \quad\Longleftrightarrow\quad
 t\in(\arg z+\pi/2,\arg z+3\pi/2)\pmod{2\pi}
 \qquad(z\ne0).
\]
For example, the two inequalities for $A(q)$ and $d^1$
give $s\in(1,5)$ and $s\in(0,4)$, whose intersection is $(1,4)$.
For the fifth direction these two half-circles are opposite,
so their intersection is empty.

In column $A$, the intervals in rows $3,1,2,4$ have union
$(-1,8)$ on the real line. In column $B$, rows $3,1,4,5$
have union $(0,9)$, and in column $C$, rows $1,5,2,4$
have union $(0,9)$. Each union has length $9>8$ and therefore
maps onto $\R/8\mathbb Z$. Successive intervals overlap
strictly, so their endpoints cause no exception. Hence the
five directions illuminate every point of all three circles.
Together with the smooth case, this proves $\ill(K_0)\le5$.
The reduction above gives $\ill(Z)\le5$.

Finally, any illuminating set of a full-dimensional convex body
in $\R^4$ has at least five members. For every nonzero $u$,
choose a point maximizing $[u,x]$; a direction illuminating
that point must have negative inner product with $u$.
By separation, the origin must therefore lie in the
full-dimensional interior of the convex hull of the directions.
The convex hull of at most four points has affine dimension
at most three, which is impossible. Thus $\ill(Z)\ge5$,
completing the proof.
\end{proof}

The value $5$ in Theorem~\ref{thm:zonotopes-two} concerns complex
zonotopes. It does not replace the sharp bound $6$ for general
complex $1$-unconditional bodies in $\C^2$ from
the proof of Corollary~\ref{cor:complex-symmetric-extrema}.

\section{$1$-symmetric bodies in dimension three}\label{sec:symmetric-three}

A complex $1$-unconditional body is \emph{$1$-symmetric} if it is
also invariant under every coordinate permutation. The bound
$\ill(K)\le14$ for non-polydiscs in this class follows from
\cite[Theorem~1.5]{RSSSym}. The following construction improves it.
It uses the reduction underlying \cite[Proposition~2.8]{RSSSym};
we include a complete geometric proof using supporting normals.

\begin{theorem}\label{thm:symmetric-three}
Every complex $1$-symmetric convex body $K\subset\C^3$ which is
not a complex affine image of $\mathbb D^3$ satisfies
\[
 \ill(K)\le12.
\]
\end{theorem}
\begin{proof}
Put $\zeta=\exp(2\pi\ii/13)$ and define
\begin{equation}\label{eq:symmetric-three-rows}
 v^r=(\zeta^r,\zeta^{3r},\zeta^{4r})\quad(r\in[12]),\qquad
 (f(1),\ldots,f(12))=(2,1,1,3,1,2,2,1,3,1,3,2).
\end{equation}
We first verify that, with $\beta=\sin(\pi/26)>0$, for every
nonempty proper $S\subset[3]$ and every $(z_j)_{j\in S}$ with
$|z_j|=1$, there is $r\in[12]$ such that
\begin{equation}\label{eq:symmetric-three-cover}
 f(r)\in S,\qquad
 \operatorname{Re}(\overline{z_j}v_j^r)\le-\beta
 \quad(j\in S).
\end{equation}

All residues in the following verification are taken modulo $13$.
Set $(a_1,a_2,a_3)=(1,3,4)$ and
$B_s=\{s,s+1,\ldots,s+5\}$. A nonempty residue set meets every
$B_s$ if and only if its largest cyclic gap is at most six: a
gap of length $g$ leaves exactly $g-1$ consecutive missing residues.
For singleton supports, the sets
$\{a_jr:f(r)=j\}$ and their largest gaps are
\[
\begin{array}{c|c|c}
 j&\{a_jr:f(r)=j\}&\text{largest gap}\\\hline
 1&\{2,3,5,8,10\}&5\\
 2&\{3,5,8,10\}&6\\
 3&\{3,5,10\}&6
\end{array}
\]
For $i<j$, define
\[
 R_{ij}(s)=\{a_jr:a_ir\in B_s,\ f(r)\in\{i,j\},\ r\in[12]\}.
\]
Their largest cyclic gaps, calculated directly from
\eqref{eq:symmetric-three-rows}, are
\[
\begin{array}{c|rrrrrrrrrrrrr}
 (i,j)\backslash s&0&1&2&3&4&5&6&7&8&9&10&11&12\\\hline
 (1,2)&6&6&6&4&4&4&6&6&6&6&6&6&6\\
 (1,3)&4&4&4&5&6&5&4&4&4&5&6&6&5\\
 (2,3)&6&6&6&6&6&4&6&6&6&6&6&6&6
\end{array}
\]
Every entry is at most six. Thus for any prescribed blocks
$B_{s_j}$, $j\in S$, a row satisfies
$f(r)\in S$ and $a_jr\in B_{s_j}$ for every $j\in S$.
This finite assertion comprises $3\cdot13+3\cdot13^2=546$ cases;
the accompanying verifier checks all of them with integer arithmetic.

For each unit $z$, the closed arc
$\{w:|w|=1,\ \operatorname{Re}(\overline z w)\le-\beta\}$
has angular length $12\pi/13$, exactly six spacings between
successive thirteenth roots of unity. It therefore contains six
consecutive roots. Applying the finite assertion to one such
block in each coordinate proves \eqref{eq:symmetric-three-cover},
including all endpoint cases.

Scale $K$ so that $e_j\in\partial K$ for $j\in[3]$. This uses
one common factor, since $K$ is $1$-symmetric. Then
$K\subset\mathbb D^3$ and there is a unique $\rho>0$ with
$\rho(1,1,1)\in\partial K$. Coordinate contractions give
$\rho\mathbb D^3\subset K$, so $0<\rho<1$: equality would give
$K=\mathbb D^3$. Fix
\begin{equation}\label{eq:symmetric-three-amplitudes}
 \Delta=\frac{\beta}{6},\qquad
 \delta=\min\left\{\frac{\Delta}{2},
             \frac{\beta\Delta(1-\rho)}{2\rho}\right\}.
\end{equation}
For every $r\in[12]$ define a direction by
\begin{equation}\label{eq:symmetric-three-directions}
 d_j^r=
 \begin{cases}
 v_j^r,&j=f(r),\\
 \delta v_j^r,&j=3\ne f(r),\\
 \Delta v_j^r,&j\ne f(r),\ j\ne3.
 \end{cases}
\end{equation}
These twelve nonzero directions are fixed before choosing a
boundary point.

Let $x\in\partial K$. For any nonzero supporting normal $u$ at
$x$, write $b_j=|u_j|$ and $t_j=|x_j|$. Maximizing independently
over coordinate rotations shows that
\[
 u_j=b_j\frac{x_j}{|x_j|}\quad(x_j\ne0),\qquad
 h_K(u)=\operatorname{Re}\sum_{j=1}^3\overline{u_j}x_j
       =\sum_{j=1}^3b_jt_j>0.
\]
There is no restriction on the argument of $u_j$ when $x_j=0$.
Swapping any two coordinate moduli of $x$ and choosing their
arguments to maximize the same functional gives
\begin{equation}\label{eq:symmetric-three-order}
 (b_i-b_j)(t_i-t_j)\ge0.
\end{equation}
Indeed, the swapped point belongs to $K$, and subtracting its
functional value from $h_K(u)$ gives the displayed inequality.

Suppose first that $M_x\ne[3]$. Apply
\eqref{eq:symmetric-three-cover} with $S=M_x$ and
$z_j=x_j/|x_j|$, and put $k=f(r)\in M_x$ for the chosen row.
Write $\alpha_j=|d_j^r|$. Then $\alpha_k=1$ and
$0<\alpha_j\le\Delta$ for $j\ne k$. For $j\notin M_x$,
\eqref{eq:symmetric-three-order} gives $b_j\le b_k$, including
when $x_j=0$. Consequently
\begin{align*}
 \operatorname{Re}\sum_{j=1}^3\overline{u_j}d_j^r
 &\le-\beta b_k-\beta\sum_{j\in M_x\setminus\{k\}}\alpha_jb_j
       +\Delta\sum_{j\notin M_x}b_j\\
 &\le-(\beta-3\Delta)b_k
       -\beta\sum_{j\in M_x\setminus\{k\}}\alpha_jb_j<0.
\end{align*}
For the strict inequality, $\beta-3\Delta=\beta/2>0$.
If $b_k=0$, all coefficients outside $M_x$ vanish, and some
coefficient on $M_x\setminus\{k\}$ is positive because $u\ne0$.
The row was chosen using $x$ alone, so this holds for every
supporting normal at $x$.

If $M_x=[3]$, all coordinate moduli equal $\rho$. Apply
\eqref{eq:symmetric-three-cover} with $S=\{1,2\}$ and
$z_j=x_j/|x_j|$ for $j\in\{1,2\}$. Then
$k=f(r)\in\{1,2\}$, $\alpha_1,\alpha_2\ge\Delta$, and
$\alpha_3=\delta$. The third unit coordinate disc is contained
in $K$, hence
\[
 b_3\le h_K(u)=\rho(b_1+b_2+b_3),\qquad
 b_3\le\frac{\rho}{1-\rho}(b_1+b_2).
\]
In particular $b_1+b_2>0$. By
\eqref{eq:symmetric-three-amplitudes},
\[
 \operatorname{Re}\sum_{j=1}^3\overline{u_j}d_j^r
 \le-\beta\Delta(b_1+b_2)+\delta b_3
 \le-\frac{\beta\Delta}{2}(b_1+b_2)<0.
\]
Again the chosen row works for every supporting normal. The
normal criterion \eqref{eq:normal-criterion} now shows that the
directions \eqref{eq:symmetric-three-directions} illuminate $K$.
\end{proof}

\begin{corollary}\label{cor:complex-symmetric-extrema}
The largest illumination number among complex $1$-symmetric
bodies that are not complex affine polydiscs is six in $\C^2$,
and lies between nine and twelve in $\C^3$.
\end{corollary}
\begin{proof}
First let $K\subset\C^2$ be any normalized complex
$1$-unconditional body with $K\ne\mathbb D^2$.
Three equally spaced unit directions on each coordinate axis
illuminate $K$. Indeed, if $x$ has only one nonzero coordinate,
then it belongs to the unit circle on that axis. For every
nonzero supporting normal the coefficient on that axis is
positive, since its support value is positive. Thus a strictly
inward direction on that axis illuminates $x$.
If $x=(r_1\xi_1,r_2\xi_2)$ has $r_1,r_2>0$, at least one
modulus, say $r_j$, is less than one: otherwise coordinate
contractions imply $K=\mathbb D^2$. Let $k\ne j$.
Write each supporting normal as $u_l=c_l\xi_l$, $c_l\ge0$.
If $c_k=0$, then $c_j>0$ and
$h_K(u)=r_jc_j<c_j$, contradicting the unit coordinate disc
in $K$. Hence $c_k>0$ for every nonzero normal. Any direction
on axis $k$ with $\operatorname{Re}(\overline{\xi_k}d_k)<0$
illuminates $x$. Three equally spaced points meet every open
semicircle, proving the upper bound six.

For the lower examples, put
$B_1^n=\{z\in\C^n:\sum_j|z_j|\le1\}$.
At $\xi e_j$, $|\xi|=1$, illumination by $d$ is equivalent to
\[
 \operatorname{Re}(\overline\xi d_j)+\sum_{k\ne j}|d_k|<0.
\]
Necessity follows by differentiating the norm along the ray;
sufficiency follows from the strict negative derivative.
This requires $|d_j|>\sum_{k\ne j}|d_k|$, so a direction
serves at most one axis circle. Its illuminated subset of
that circle is contained in an open semicircle. At least
three directions are needed for each circle, giving
$\ill(B_1^n)\ge3n$. Conversely, take three equally spaced
unit directions on each axis. At any boundary point select
a nonzero coordinate and a strictly inward direction on
that axis; the sum of moduli then decreases for a sufficiently
small positive step. Hence $\ill(B_1^n)=3n$.
These bodies are $1$-symmetric. For $n=2,3$, their values
$3n$ differ from the polydisc value $2^{n+1}-1$~\cite{RSS},
so they are not complex affine polydiscs. The planar bound
and Theorem~\ref{thm:symmetric-three} finish the proof.
\end{proof}

The interval $[9,12]$ does not establish that twelve is sharp.

\section{Real $1$-symmetric bodies}\label{sec:real-symmetric-improvement}

A real $1$-symmetric convex body is invariant under all
coordinate sign changes and permutations. Let
\[
 M_n=\sup\{\ill(K):K\subset\R^n\text{ is $1$-symmetric and is not an affine cube}\}.
\]
Let $Q_d$ be the graph on $\{0,1\}^d$ whose edges join vertices
differing in exactly one coordinate. A set dominates a graph
if every vertex outside it has a neighbor in it. Write
$\gamma_c(Q_d)$ for the minimum size of a dominating set whose
induced graph is connected.

\begin{proposition}\label{prop:real-connected-domination}
For every $n\ge3$,
\begin{equation}\label{eq:real-connected-domination}
 M_n\le2^{n-1}+\gamma_c(Q_{n-1})+n-1.
\end{equation}
Consequently,
\begin{equation}\label{eq:real-asymptotic-error}
 M_n\le2^{n-1}\left(1+\frac{1+o(1)}n\right)
 \qquad(n\longrightarrow\infty).
\end{equation}
\end{proposition}

The direction construction builds on Sun--Vritsiou's
large-coordinate and small-coordinate argument
\cite[Lemma~12]{SVSym}. Lemma~\ref{lem:real-proper-support}
is the real reduction already stated in
\cite[Proposition~2.9]{RSSSym}; we give a supporting-normal proof.
The subsequent cube assignment supplies the connected-domination
bound. For normalized bodies satisfying
$\frac12(1,\ldots,1)\notin\operatorname{int}K$,
Tikhomirov's construction already has
$2^{n-1}+O(2^n/n)$ directions in large dimensions
\cite[Section~4, Proposition~10]{Tikhomirov}.
That subclass estimate does not cover every body defining $M_n$.

We first give the geometric and graph arguments needed to construct the directions.

\begin{lemma}\label{lem:real-proper-support}
Let $n\ge2$ and let $\mathcal B\subset\{-1,1\}^n\times[n]$
be finite. Suppose that for every nonempty proper $S\subset[n]$
and every $a\in\{-1,1\}^S$ there is $(b,k)\in\mathcal B$ with
\begin{equation}\label{eq:real-proper-support}
 k\in S,\qquad b_j=-a_j\quad(j\in S).
\end{equation}
If $K\subset\R^n$ is $1$-symmetric and is not an affine cube,
then $\ill(K)\le|\mathcal B|$.
\end{lemma}
\begin{proof}
Scale $K$ by one common factor so that $e_j\in\partial K$ for
every $j\in[n]$. Coordinate contractions follow from convexity
and sign invariance. Thus $K\subset[-1,1]^n$, and there is a
unique $\rho>0$ with $\rho(1,\ldots,1)\in\partial K$.
Moreover, $\rho[-1,1]^n\subset K$, whence $0<\rho<1$;
equality would give $K=[-1,1]^n$. Set
\begin{equation}\label{eq:real-symmetric-amplitudes}
 \Delta=\frac1{2n},\qquad
 \delta=\min\left\{\frac\Delta2,
                  \frac{\Delta(1-\rho)}{2\rho}\right\}>0.
\end{equation}
For every $(b,k)\in\mathcal B$ define $d^{b,k}\in\R^n$ by
\begin{equation}\label{eq:real-symmetric-directions}
 d^{b,k}_j=
 \begin{cases}
  b_j,&j=k,\\
  \delta b_j,&j=n\ne k,\\
  \Delta b_j,&j\ne k,\ j\ne n.
 \end{cases}
\end{equation}
These directions are fixed before choosing a boundary point.

Fix $x\in\partial K$. For a nonzero supporting normal $u$ at
$x$, put $c_j=|u_j|$ and $t_j=|x_j|$. Maximizing over the
coordinate sign changes gives
\[
 u_j=c_j\operatorname{sign}(x_j)\quad(x_j\ne0),\qquad
 h_K(u)=\sum_{j=1}^n c_jt_j>0.
\]
At a zero coordinate of $x$, the sign of $u_j$ is unrestricted.
Exchanging the moduli of coordinates $i,j$ of $x$ and then
choosing signs that maximize the same functional gives
\begin{equation}\label{eq:real-normal-order}
 (c_i-c_j)(t_i-t_j)\ge0.
\end{equation}
Indeed, the exchanged point belongs to $K$, and subtracting
its functional value from $h_K(u)$ gives this inequality.

Suppose $M_x\ne[n]$. Choose $(b,k)$ using
\eqref{eq:real-proper-support} with $S=M_x$ and
$a_j=\operatorname{sign}(x_j)$. Write
$\alpha_j=|d^{b,k}_j|$. Then $\alpha_k=1$,
$0<\alpha_j\le\Delta$ for $j\ne k$, and
\eqref{eq:real-normal-order} gives $c_j\le c_k$ for
$j\notin M_x$, including the zero coordinates of $x$.
Consequently
\begin{align*}
 \langle u,d^{b,k}\rangle
 &\le-c_k-\sum_{j\in M_x\setminus\{k\}}\alpha_jc_j
       +\Delta\sum_{j\notin M_x}c_j\\
 &\le-(1-n\Delta)c_k
       -\sum_{j\in M_x\setminus\{k\}}\alpha_jc_j<0.
\end{align*}
For strictness, $1-n\Delta=1/2$. If $c_k=0$, every
coefficient outside $M_x$ vanishes, and some coefficient on
$M_x\setminus\{k\}$ is positive because $u\ne0$.

If $M_x=[n]$, then every $t_j=\rho$. Choose $(b,k)$ using
\eqref{eq:real-proper-support} with $S=[n-1]$ and
$a_j=\operatorname{sign}(x_j)$. Thus $k\ne n$,
$\alpha_j\ge\Delta$ for $j<n$, and $\alpha_n=\delta$.
Since $[-e_n,e_n]\subset K$,
\[
 c_n\le h_K(u)=\rho\sum_{j=1}^n c_j,
 \qquad
 c_n\le\frac\rho{1-\rho}\sum_{j=1}^{n-1}c_j.
\]
The sum on the right is positive. Hence
\[
 \langle u,d^{b,k}\rangle
 \le-\Delta\sum_{j=1}^{n-1}c_j+\delta c_n
 \le-\frac\Delta2\sum_{j=1}^{n-1}c_j<0.
\]
In both cases the pair $(b,k)$ depends on $x$ alone, and the
inequality holds for every nonzero supporting normal at $x$.
The supporting-normal criterion for illumination therefore
shows that the directions \eqref{eq:real-symmetric-directions}
illuminate $K$.
\end{proof}

Write $v\oplus e_j$ for the vertex obtained by changing
coordinate $j$ of $v$. A vertex cover meets every edge;
equivalently, its complement is independent.

\begin{lemma}\label{lem:real-cube-assignment}
Let $A\subset\{0,1\}^n$ be a vertex cover, and let
$T:A\to A$ satisfy
\[
 T(v)=v\oplus e_{k(v)}\qquad(v\in A)
\]
for some $k(v)\in[n]$. Suppose every directed cycle of $T$
uses every coordinate direction. Then the family
\[
 \mathcal B(A,T)=
 \left\{\big((2v_j-1)_{j\in[n]},k(v)\big):v\in A\right\}
\]
satisfies \eqref{eq:real-proper-support} and has cardinality
$|A|$.
\end{lemma}
\begin{proof}
Fix $S$ and $a$ as in \eqref{eq:real-proper-support}. The
vertices satisfying $2v_j-1=-a_j$ for every $j\in S$ form a
proper face $F$ of positive dimension. It contains an edge,
so $A\cap F\ne\varnothing$. If $k(v)\notin S$ for every
$v\in A\cap F$, then $T(A\cap F)\subset A\cap F$.
Iterating $T$ from any vertex of this finite nonempty set
would give a directed cycle contained in $F$. Such a cycle
cannot use any coordinate in $S$, a contradiction. Thus some
$v\in A\cap F$ has $k(v)\in S$, which proves the assertion.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{prop:real-connected-domination}]
For a binary vector $y$, let $p(y)$ be the sum of its coordinates
modulo two, and let $O_n=\{v\in\{0,1\}^n:p(v)=1\}$.
Choose a minimum connected dominating set $D$ in $Q_{n-1}$.
Translate it by a binary vector so that $0\in D$, and put
\[
 E_D=\{(y,p(y)):y\in D\}\subset Q_n.
\]
Every odd vertex is $(z,1-p(z))$. If $z\in D$, its
last-coordinate neighbor belongs to $E_D$. Otherwise choose
a neighbor $y\in D$ of $z$. Then $p(y)=1-p(z)$, so
$(y,p(y))$ is adjacent to the given odd vertex.
Adjacent elements of $D$ lift to even vertices at distance
two, joined by a path with an odd intermediate vertex.
Since $D$ is connected and every odd vertex has a neighbor
in $E_D$, the induced graph on $A_0=O_n\cup E_D$ is connected.

In the ordinary coordinate order, define
\begin{equation}\label{eq:real-full-coordinate-cycle}
 v^i=\sum_{h=1}^{i}e_h\quad(0\le i\le n),\qquad
 v^{n+i}=\sum_{h=i+1}^{n}e_h\quad(1\le i\le n-1).
\end{equation}
The vertices $v^0,\ldots,v^{2n-1}$ are distinct. In this order,
closed by $v^{2n}=v^0$, they form a cycle with successive
coordinate directions $1,\ldots,n,1,\ldots,n$.
Let $V$ be its vertex set and put $A=A_0\cup V$.
The cycle has $n$ even vertices and contains $0\in E_D$.
Every odd cycle vertex already belongs to $O_n$. Thus
\[
 |A|\le2^{n-1}+|D|+n-1.
\]
The complement of $A$ consists of even vertices and is
independent, so $A$ is a vertex cover. Each added even
vertex has an odd neighbor in $A_0$, so $Q_n[A]$ is connected.

For $u\in A$, let $h(u)$ be its graph distance to $V$ in
$Q_n[A]$. Define
\begin{equation}\label{eq:real-distance-successor}
 \begin{aligned}
 T(v^i)&=v^{i+1} &&(i\bmod 2n),\\
 T(u)&=\min\{w\in A:w\sim u,\ h(w)=h(u)-1\}
       &&(u\in A\setminus V),
 \end{aligned}
\end{equation}
where the minimum uses binary order. Connectedness makes
each choice possible. Distances decrease until the specified
cycle is reached, so it is the only directed cycle of $T$;
it uses every coordinate. Lemmas~\ref{lem:real-cube-assignment}
and~\ref{lem:real-proper-support} therefore give
$\ill(K)\le|A|$ for every body in the definition of $M_n$.
This proves \eqref{eq:real-connected-domination}.

Griggs proves
\[
 \gamma_c(Q_d)=(1+o(1))\frac{2^d}{d}
 \qquad(d\longrightarrow\infty)
\]
in \cite[Theorems~2 and~4]{Griggs}.
Substitute $d=n-1$ and use $n-1=o(2^{n-1}/n)$ to obtain
\eqref{eq:real-asymptotic-error}.
\end{proof}

For $n=3$, the body $P_n$ below is the regular cuboctahedron:
its vertices are the signed permutations of $(1,1,0)$.
Kiss and de Wet give a four-light-source construction for this
body~\cite[Theorem~3.4, pp.~66--67]{KissDeWet}. Together with the
general lower bound four in $\R^3$, this already yields
$\ill(P_3)=4$. Their illumination \emph{parameter} is a different
quantity; here we use the number of sources in their construction.
The following proof treats the whole family directly.

\begin{proposition}\label{prop:real-symmetric-examples}
For $n\ge3$, the real $1$-symmetric body
\[
 P_n=\left\{x\in[-1,1]^n:\sum_{j=1}^n|x_j|\le n-1\right\}
\]
is not an affine cube and satisfies $\ill(P_n)=2^{n-1}$.
\end{proposition}
\begin{proof}
The vertices of $P_n$ are exactly the signed permutations of
$(1,\ldots,1,0)$. To see this, a point with
$\sum_j|x_j|<n-1$ has a coordinate of modulus less than one
that can be perturbed in both directions while preserving both
constraints. If equality holds and two coordinate moduli lie
strictly between zero and one, transfer a small amount from one
modulus to the other, retaining the signs; again the point is
not extreme. There cannot be exactly one nonintegral modulus,
since the sum of the moduli is the integer $n-1$. The remaining
possibility is precisely $n-1$ coordinates of modulus one and
one zero. Conversely, in a convex decomposition of such a point,
the cube constraints fix all its unit coordinates in every term,
and the sum constraint then forces the remaining coordinate to
be zero. Thus each displayed point is a vertex.

Fix a direction $d$ and the index $j$ of the zero coordinate of
one of these vertices. Illumination requires $d_i$ to have sign
opposite to the vertex coordinate for every $i\ne j$, by the
active cube constraints. These requirements determine at most
one vertex with zero coordinate $j$; if a required $d_i$ is
zero there is no such vertex. Consequently $d$ illuminates at
most $n$ vertices. There are $n2^{n-1}$ vertices, so
$\ill(P_n)\ge2^{n-1}$.

For the reverse inequality take
\[
 D=\left\{d\in\{-1,1\}^n:\prod_{j=1}^n d_j=1\right\}.
\]
At a vertex $x$ with $x_j=0$, prescribe $d_i=-x_i$ for $i\ne j$
and choose the unique remaining sign making $d\in D$.
For $0<t<1$,
\[
 \|x+td\|_\infty<1,\qquad
 \sum_{i=1}^n|x_i+td_i|=n-1-(n-2)t<n-1.
\]
Thus $d$ illuminates $x$, and the $2^{n-1}$ members of $D$
illuminate $P_n$. The body $P_n$ is $1$-symmetric directly from
its definition. Its $n2^{n-1}$ vertices differ in number from
the $2^n$ vertices of an affine cube for $n\ge3$.
To justify the passage from vertices to the boundary, express
any boundary point as a convex combination of vertices and
move one vertex of positive weight into the interior along
an illuminating direction. The resulting convex combination
is interior and is a positive step from the original point.
\end{proof}

These examples give the lower bound $2^{n-1}\le M_n$.
Together with \eqref{eq:real-asymptotic-error}, they show that
the upper bound has the correct leading term. They do not
determine the smaller-order difference $M_n-2^{n-1}$.

\section{Fractional lower bounds and equality families}\label{sec:fractional-equality}

The polydisc case of the proposition below is the published
fractional formula~\cite[Theorem~1.3]{RSS}. Its extension to the
specified inner bodies agrees with the Hanner formulas
in~\cite{Families}. We retain a direct proof to establish the
lower bound for every intermediate convex body, including bodies
without coordinate symmetry.

On $\C^2$ put
\begin{equation}\label{eq:planar-equality-envelope}
 B=\{z:|z_1|+|z_2|\le1\},\qquad
 L(a,b)=\{z:|z_1|+a|z_2|\le1,\ b|z_1|+|z_2|\le1\},
\end{equation}
where $0<a,b\le1$. Then $B\subseteq L(a,b)$, and both
bodies meet each coordinate axis in its unit disc.

\begin{proposition}\label{prop:fractional-equality-minkowski}
Fix $p,r\in\mathbb Z_{\ge0}$ with $p+2r=n\ge1$, and choose
$a_j,b_j\in(0,1]$ with $a_jb_j\ge1/2$.
On disjoint coordinate blocks put
\begin{equation}\label{eq:fractional-equality-envelopes}
 H=\mathbb D^p\times B^r,\qquad
 E=\mathbb D^p\times\prod_{j\in[r]}L(a_j,b_j).
\end{equation}
If a convex body $C$ satisfies $H\subseteq C\subseteq E$, then
\[
 \ill^*(C)\ge2^n.
\]
Moreover, for every $0\le t<1$,
\begin{equation}\label{eq:fractional-equality-minkowski}
 \ill^*((1-t)H+tC)=2^n.
\end{equation}
If $C$ is complex $1$-unconditional, all these bodies are complex
$1$-unconditional. Invertible complex linear images preserve the
conclusions.
\end{proposition}
\begin{proof}
We construct the boundary measure giving the lower bound.
On a disc take twice uniform probability measure on its unit
circle. Any fixed nonzero direction illuminates exactly an
open semicircle, so the illuminated mass is one; a zero
direction illuminates no point.

On $L(a,b)$ let $\sigma_1,\sigma_2$ be uniform probability
measures on the two unit axis circles, and put
$\nu_{a,b}=2\sigma_1+2\sigma_2$.
At $\xi e_1$, the first defining inequality in
\eqref{eq:planar-equality-envelope} is active. Hence
illumination by $d$ requires
\[
 \operatorname{Re}(\overline\xi d_1)+a|d_2|<0.
\]
Indeed, if a positive step enters the interior, convexity
makes the one-sided derivative of the active convex function
strictly negative. At $\xi e_2$, illumination similarly
requires $\operatorname{Re}(\overline\xi d_2)+b|d_1|<0$.
For $d_1d_2\ne0$ put $s=|d_2|/|d_1|$. The proportions
illuminated on the two circles are therefore at most
\[
 q_1=\begin{cases}\pi^{-1}\arccos(as),&as<1,\\0,&as\ge1,\end{cases}
 \qquad
 q_2=\begin{cases}\pi^{-1}\arccos(b/s),&b/s<1,\\0,&b/s\ge1.\end{cases}
\]
If one is zero, their sum is at most $1/2$. Otherwise set
$v=as$ and $w=b/s$. Since $v^2+w^2\ge2vw=2ab\ge1$,
we have $w\ge\sqrt{1-v^2}$ and hence
$\arccos v+\arccos w\le\pi/2$.
Thus $q_1+q_2\le1/2$ in all cases. If exactly one coordinate
of $d$ is zero, only half of its own axis circle can be
illuminated, and none of the other one, since $a,b>0$.
The zero vector illuminates neither circle. Every fixed
direction consequently illuminates $\nu_{a,b}$-mass at most one.

Take the product $\nu$ of these measures over the $p$ disc
factors and the $r$ two-dimensional factors of $E$.
Its total mass is $2^p4^r=2^n$, and its support lies in
$H\cap\partial E$. At every point in that support all factor
points are boundary points. A direction illuminating $E$
there must illuminate each factor; the product measure
therefore assigns mass at most one to its illuminated set.
The points supporting $\nu$ also belong to $\partial C$.
Illumination in $C$ implies illumination in $E$, because
$\operatorname{int}C\subseteq\operatorname{int}E$.
For any fractional illuminating measure $\mu$ for $C$,
Tonelli's theorem now gives
\[
 2^n\le\int\mu\{d:d\text{ illuminates }x\}\,d\nu(x)
 =\int\nu\{x:d\text{ illuminates }x\}\,d\mu(d)
 \le\mu(\C^n\setminus\{0\}).
\]
This proves the lower bound. It also applies to
$(1-t)H+tC$, which lies between $H$ and $E$.

For the upper bound we give a fractional illuminating measure
of mass $2^n$ for $H$. On each disc use twice uniform circle
measure on directions. On $B$ use $2\sigma_1+2\sigma_2$,
now as a direction measure. At a boundary point of $B$,
choose any nonzero coordinate; the inward open semicircle
on that axis decreases $|z_1|+|z_2|$ for small positive steps
and has measure one. Thus this direction measure illuminates
every boundary point of $B$. Take the product of the factor
direction measures. A boundary point of $H$ has illuminating
mass at least one: at boundary factors the mass is at least
one, and at interior factors every direction works for a
sufficiently small step. The finitely many steps can be
replaced by their minimum, since a segment from a body point
to an interior point is interior except possibly at its
initial point. This product measure has mass $2^n$.

Finally, a fractional illuminating measure for a
full-dimensional convex body $A$ also illuminates $A+D$
for any compact convex set $D$. Given
$x\in\partial(A+D)$, choose a decomposition $x=y+z$ with
$y\in A$ and $z\in D$. Then $y\in\partial A$, since
an interior $y$ would make $x$ interior. Whenever
$y+sd\in\operatorname{int}A$, one has
$x+sd\in\operatorname{int}A+D\subseteq\operatorname{int}(A+D)$.
Hence the illuminating mass at $x$ is at least the mass at
$y$. This is also the Minkowski-sum fact in
\cite[Fact~3.2]{RSS}.
Apply it to $A=(1-t)H$ and $D=tC$. The restriction $t<1$
ensures that $A$ is full-dimensional, and dilation preserves
the fractional upper bound just proved for $H$.
This gives equality in \eqref{eq:fractional-equality-minkowski}.
Coordinate-rotation invariance is preserved by Minkowski
addition, and invertible complex linear maps preserve
illumination and the corresponding measures.
\end{proof}

\begin{remark}\label{rem:fractional-conjecture}
Taking $p=0$, $r=1$, and $C=H=B$ gives $\ill^*(B)=4$.
This body is not an affine image of $\mathbb D^2$. Writing
$\operatorname{ext}K$ for the set of extreme points of $K$, we have
\[
 \operatorname{ext}B=(\partial\mathbb D\times\{0\})
       \cup(\{0\}\times\partial\mathbb D),\qquad
 \operatorname{ext}(\mathbb D^2)=\partial\mathbb D\times\partial\mathbb D.
\]
For the first identity, a boundary point with both coordinates
nonzero is the convex combination of its two normalized axis
points, with weights $|z_1|$ and $|z_2|$; interior points are not
extreme. Each unit axis point is extreme, since a convex
decomposition forces equality in its coordinate-disc constraint,
fixing that coordinate, and the defining norm then makes the
other coordinate zero. For the second identity, strict convexity
of each disc makes a point with both coordinates of modulus one
extreme, while any coordinate of modulus less than one admits a
nontrivial symmetric perturbation within its disc.
The first extreme-point set has two connected components and the
second is connected. Invertible affine maps preserve extreme
points and connectedness. Thus this example, also discussed
in~\cite{Families,Unconditional}, contradicts the equality clause
of \cite[Conjecture~1.2]{RSS}. It does not contradict the proposed
upper bound $2^n$.
\end{remark}

For an arbitrary endpoint $C$ between $H$ and $E$, the lower
bound is proved, but the upper bound $\ill^*(C)\le2^n$ is
not established here. The equalities for $t<1$ alone do not
justify passing to $t=1$.

\appendix
\section{One family for all normalized bodies}\label{app:simultaneous}

Let $U_n$ be the least cardinality of a finite family of directions
that illuminates every normalized complex $1$-unconditional convex
body in $\C^n$. The family must be chosen independently of the body.
This is a stronger requirement than bounding $\ill(K)$ separately
for each $K$.

\begin{proposition}\label{prop:simultaneous}
Put $m=\lceil2n/3\rceil$ and
\[
 L_n=(2^{m+1}-1)\binom nm,\qquad
 B_n=L_n+\sum_{k=m+1}^n\binom nk.
\]
Put $C_n=B_n-(n-m)$. Then
\begin{equation}\label{eq:universal-sandwich}
 L_n\le U_n\le C_n\le B_n,
\end{equation}
and
\begin{equation}\label{eq:universal-asymptotic}
 U_n\sim\frac{3^{n+1}}{\sqrt{\pi n}}.
\end{equation}
\end{proposition}
\begin{proof}
The directions in the proof of Lemma~\ref{thm:support-bound}
depend only on $n$, the finite matchings and cyclic vectors, and their
common positive margin. The normal estimates use only normalization
of $K$. Hence the same family works for every normalized $K$, proving
$U_n\le C_n$.

For every $m$-set $S\subset[n]$, when $m<n$, consider the normalized body
\[
 K_S=\{z\in\C^n:
       \|z_S\|_\infty+\|z_{S^c}\|_\infty\le1\}.
\]
At a point $x$ of its coordinate torus
\[
 |x_j|=1\ (j\in S),\qquad x_j=0\ (j\notin S),
\]
the directional derivative of its norm is
\[
 \max_{j\in S}\operatorname{Re}(\overline{x_j}d_j)
       +\|d_{S^c}\|_\infty.
\]
Illumination forces this derivative to be negative: for a convex function, an interior point along the ray forces a negative one-sided derivative at its initial point. Consequently
\[
 \operatorname{Re}(\overline{x_j}d_j)
       <-\|d_{S^c}\|_\infty\quad(j\in S),
 \qquad
 \min_{j\in S}|d_j|>\max_{j\notin S}|d_j|.
\]
Thus the $m$ largest coordinate magnitudes of $d$ determine $S$ uniquely. The groups of directions serving these tori for different $m$-sets are disjoint. Within the group for $S$, projection must illuminate the $m$-polydisc torus, and hence the polydisc, requiring at least $2^{m+1}-1$ directions by the exact polydisc illumination theorem~\cite{RSS}. Summation gives $U_n\ge L_n$. If $m=n$, the polydisc itself gives this conclusion.

Finally, $B_n-L_n\le2^n$. Stirling's formula, with $m=2n/3+O(1)$, yields
\[
 2^{m+1}\binom nm\sim\frac{3}{\sqrt{\pi n}}\,3^n.
\]
The same asymptotic holds for $C_n=B_n-(n-m)$; squeezing between the lower and upper bounds proves the claim.


\end{proof}

\subsection{The common upper-bound construction}

We include the coordinate-support argument needed for the
upper bound, so that the simultaneous assertion has its full
proof here. Its use as an individual-body upper bound is
also recorded in the companion manuscript~\cite{Unconditional}.

\begin{lemma}\label{thm:support-bound}
For every $n\ge1$, a single family of at most
\begin{equation}\label{eq:support-bound}
 C_n=(2^{m+1}-1)\binom nm+\sum_{k=m+1}^{n}\binom nk-(n-m),
 \qquad m=\lceil2n/3\rceil,
\end{equation}
directions illuminates every normalized complex
$1$-unconditional body in $\C^n$.
\end{lemma}

We use the cyclic construction in the proof of Proposition~2.7
of~\cite{RSS}. For $b\ge1$, put $q_b=2^{b+1}-1$. The vectors
\begin{equation}\label{eq:cyclic-vectors}
 \left\{\left(\exp\frac{2\pi\ii\,2^{j-1}t}{q_b}\right)_{j\in[b]}:
 t\in\mathbb Z/q_b\mathbb Z\right\}
\end{equation}
meet every product of $b$ open semicircles, and hence illuminate
$\mathbb D^b$. We use this published result as an external input.

The following projection property is already
\cite[Lemma~4.3]{RSSSym}; we include its short proof.
\begin{lemma}\label{lem:cyclic-projections}
Append to \eqref{eq:cyclic-vectors} the column of frequency $2^b$. Every projection of the resulting $(b+1)$-column matrix onto $b$ columns meets every product of open semicircles.
\end{lemma}
\begin{proof}
Since $2^{b+1}\equiv1\pmod{q_b}$, multiplication of the row parameter $t$ by $2$ cyclically permutes the $b+1$ frequencies. It is a permutation of the rows because $q_b$ is odd. Any omitted column can therefore be moved to the last position, reducing the assertion to \eqref{eq:cyclic-vectors}.
\end{proof}


\begin{lemma}\label{lem:binary-support-tree}
Let a finite rooted tree have nonempty coordinate sets as its nodes. Its root is a set $T$ of cardinality $b$. Each node has at most two children, and each child is obtained by deleting one coordinate from its parent. Then $q_b$ directions suffice, simultaneously for every normalized complex $1$-unconditional body, to illuminate all boundary points whose supports occur in the tree.
\end{lemma}
\begin{proof}
Arrange the vectors of \eqref{eq:cyclic-vectors}, with their additional
coordinate from Lemma~\ref{lem:cyclic-projections}, as rows of a matrix.
Label $b$ columns by $T$, leaving one additional column. At a node with support $S$, let $G$ be the set of rows satisfying fixed
open-semicircle conditions on $b-|S|$ columns outside $S$. Exactly one
column outside $S$ remains unrestricted. At the root, $G$ contains every
row, and the additional column is unrestricted.

The active columns $S$ and the constrained columns together comprise exactly $b$ columns. Lemma~\ref{lem:cyclic-projections} implies that $G$, projected onto $S$, meets every product of open semicircles. Indeed, prescribe the desired semicircles on $S$ and retain all the fixed constraints.

Suppose the children are $S\setminus\{i\}$ and $S\setminus\{j\}$. Split $G$ by imposing, in its unrestricted column, respectively the upper and lower open semicircles. These conditions are disjoint. The previously unrestricted column becomes constrained, and the deleted column $i$ or $j$ becomes the unrestricted column of the corresponding child. Thus the invariant persists. For a single child, use just one of the two semicircles. Rows satisfying neither chosen condition stop at the current node; rows at leaves also stop. Every row consequently visits one nested path
\[
 T=S_0\supset S_1\supset\cdots\supset S_r,
 \qquad |S_{\ell-1}\setminus S_\ell|=1.
\]
The group of a node is exactly the set of rows visiting it. In particular, rows that stop early remain available at that node and all its ancestors.

For a node $S$, define on the torus of unit complex coordinates
\[
 g_S(\zeta)=\max_{p\in G}\min_{j\in S}
                   \bigl[-\operatorname{Re}(\overline{\zeta_j}p_j)\bigr].
\]
The covering property makes $g_S$ positive everywhere. It is continuous,
being a finite maximum of finite minima of continuous functions, so its
minimum on the compact torus is positive. Take $\beta>0$ no larger than
these minima for all the finitely many nodes. Thus, for every node $S$
and every $(\zeta_j)_{j\in S}$ with $|\zeta_j|=1$, some row $p\in G$ satisfies
\begin{equation}\label{eq:tree-inward-inequalities}
 \operatorname{Re}(\overline{\zeta_j}p_j)\le-\beta
 \qquad(j\in S).
\end{equation}
Choose $0<\delta<\min\{1,\beta/n\}$.

For a row with path $S_0,\ldots,S_r$, give amplitude $1$ to the coordinates in $S_r$. The coordinate deleted at step $\ell$ receives amplitude $\delta^{r-\ell+1}$, and coordinates outside $T$ receive zero. Multiply these positive amplitudes by the corresponding entries of the row to obtain $d$. At every visited node $S_\ell$,
\begin{equation}\label{eq:tree-amplitude-gap}
 |d_j|\ge\delta^{r-\ell}\quad(j\in S_\ell),
 \qquad
 |d_j|\le\delta^{r-\ell+1}\quad(j\notin S_\ell).
\end{equation}

Now let $x\in\partial K$ have support $S_\ell$. Select a visiting row satisfying \eqref{eq:tree-inward-inequalities} for $\zeta_j=x_j/|x_j|$, and put $a=\delta^{r-\ell}$. For every supporting normal, \eqref{eq:normalized-normal-estimates} and \eqref{eq:tree-amplitude-gap} give
\begin{align*}
 [u,d]
 &\le -\beta a\sum_{j\in S_\ell}|u_j|
       +\delta a\sum_{j\notin S_\ell}|u_j|\\
 &\le a\bigl[-\beta+(n-|S_\ell|)\delta\bigr]
             \sum_{j\in S_\ell}|u_j|<0.
\end{align*}
The last sum is positive since $h_K(u)>0$. The criterion \eqref{eq:normal-criterion} completes the proof.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{thm:support-bound}]
All choices below are made independently of the normalized body $K$.
Write $a_k=\binom nk$ and $a_{n+1}=0$. For $k\in[n-1]$, between the $k$-sets and the $(k+1)$-sets, the Boolean incidence graph has degrees $n-k$ and $k+1$, respectively. We assign a $k$-set to a containing parent, allowing each parent at most two children. Write $N(\mathcal A)$ for the set of neighbors of a vertex family $\mathcal A$.

If $a_k\le2a_{k+1}$, edge counting gives, for every family $\mathcal A$ of $k$-sets,
\[
 |N(\mathcal A)|\ge\frac{n-k}{k+1}|\mathcal A|
 \ge\frac{|\mathcal A|}{2}.
\]
Hall's matching theorem~\cite{Hall}, after replacing each upper vertex by two slots, assigns every $k$-set to a parent. If $a_k\ge2a_{k+1}$, reverse edge counting gives, for every family $\mathcal B$ of upper vertices,
\[
 |N(\mathcal B)|\ge\frac{k+1}{n-k}|\mathcal B|
 \ge2|\mathcal B|.
\]
To apply Hall's theorem to the duplicated upper vertices, take any set
of slots and let $\mathcal B$ be their underlying vertices. There are
at most $2|\mathcal B|$ slots and at least that many lower neighbors by
the displayed estimate. Thus Hall's theorem fills both slots of every
upper vertex. The unused $k$-sets become roots. The top level is handled separately: $[n]$ is a root and has no parent.
Consequently, including $k=n$ with $a_{n+1}=0$, the numbers of roots are
\[
 r_k=\max\{a_k-2a_{k+1},0\}.
\]
The choices between successive pairs of layers are independent: a vertex's parent and its children are governed by different matchings. Together they form a binary forest containing every nonempty subset exactly once.

The positive root counts occur precisely at $k\ge m=\lceil2n/3\rceil$. Applying Lemma~\ref{lem:binary-support-tree} separately to the trees gives a family of at most
\begin{align*}
 &\sum_{k=m}^{n}(a_k-2a_{k+1})(2^{k+1}-1)\\
 &=(2^{m+1}-1)a_m+\sum_{k=m+1}^{n}a_k.
\end{align*}
directions. Here the last equality follows from
\[
 \sum_{k=m}^n(a_k-2a_{k+1})2^{k+1}=2^{m+1}a_m,
 \qquad
 \sum_{k=m}^n(a_k-2a_{k+1})=a_m-\sum_{k=m+1}^n a_k.
\]


For every $k\ge m$ the number $r_k$ of roots at level $k$ is positive.
Prescribe the nested sets $S_k=[k]$, $m\le k\le n$.
The set $S_n$ is already the top root. For $m\le k<n$, the matching
between levels $k$ and $k+1$ leaves $r_k>0$ lower vertices unmatched.
Apply a coordinate permutation to this matching alone so that one
unmatched vertex becomes $S_k$. Containment and all capacities are
preserved. Whether a $k$-set is a root depends only on this matching;
its children are assigned by the separate matching between levels
$k-1$ and $k$. The permutations therefore impose no compatibility
condition across level pairs. We can make every $S_k$ a root at once.

Use the upper and lower open semicircles at every split in the tree construction.
In each root's matrix, the row indexed by $t=0$ has entry $1$ in every column.
It lies on the boundary of both semicircles, so it stops at the root and
is never needed in a descendant's group.
Take one common positive margin $\beta$ for all groups in the forest,
and choose $0<\delta<\min\{1,\beta/n\}$.

Remove the $n-m+1$ rows indexed by $t=0$ associated with the roots
$S_m,S_{m+1},\ldots,S_n$, and replace them by the single direction
\[
 d_j=
 \begin{cases}
  1,&j\le m,\\
  \delta^{j-m},&j>m.
 \end{cases}
\]
At support $S_k$, its smallest active amplitude is $\delta^{k-m}$,
and every inactive amplitude is at most $\delta^{k-m+1}$.
Whenever the deleted row supplies the margin $\beta$ at that root,
the replacement has positive real coordinates and the required amplitude gap.
Writing $a=\delta^{k-m}$, the same normal estimate gives
\[
 [u,d]\le a[-\beta+(n-k)\delta]\sum_{j\in S_k}|u_j|<0
\]
for every nonzero supporting normal at any point for which that row
was selected. The sum is positive by \eqref{eq:normalized-normal-estimates}.
At all other nodes and for all other selected rows the construction is unchanged.
This saves $n-m=\lfloor n/3\rfloor$ directions.
All matchings, rows, margins, and magnitudes depend only on $n$.
Thus the same family works for every normalized body and has
at most $C_n$ directions.
\end{proof}

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\end{document}