# An overview of complex ellipsoids

Jorge Luis Arocha  ,  Javier Bracho  Instituto de matemáticas, UNAM, Mexico
and  Luis Montejano  Instituto de Matematicas, UNAM, Sede Juriquilla,
Queretaro, Mexico

###### Contents

  1. 1 Introduction
  2. 2 Complex symmetry
    1. 2.1 Complex symmetry and projections
    2. 2.2 Complex symmetry and complex ellipsoids
  3. 3 Extremal inscribed and circumscribed complex ellipsoids
    1. 3.1 Auxiliary lemmas
    2. 3.2 Maximal inscribed complex ellipsoids.
    3. 3.3 Minimal circumscribed complex ellipsoids
  4. 4  A characterization of complex ellipsoids
  5. 5 Complex ellipsoids by means of sections and of projections
  6. 6 Convex bodies all whose sections are complex affinely equivalent

##  1\. Introduction

An ellipsoid is the image of a ball under an affine transformation. If this
affine transformation is over the complex numbers, we refer to it as a
_complex ellipsoid_. Characterizations of real ellipsoids have received much
attention over the years, see for example de surveys [26], [28], [17] or
Section 1.12 of [25]; however, characterizations of complex ellipsoids have
been studied very little. The recent interest in them begins with Gromov’s
proof of the Isometric Banach Conjecture over the complex numbers in even
dimensions [16]. Gromov proved that a complex symmetric convex body in complex
space $\mathbb{C}^{2n+1}$ all whose complex hyperplane sections through the
origin are complex linearly equivalent is a complex ellipsoid. For that
purpose, he began by proving that the group of complex linear symmetries of a
compact complex symmetric set of $\mathbb{C}^{n}$ is conjugate to a compact
subgroup of $SU_{n}$ using the uniqueness of the complex ellipsoid of minimal
volume containing a complex symmetric set.

This paper is a review of what is known about complex ellipsoids from the
point of view of convex geometry. In particular, the proof of the Complex
Banach Conjecture for dimensions $n\equiv 0,1,2$, $n\geq 5$, in [9] is
outlined in Section 6. This conjecture is basically a characterization of
complex ellipsoids. Its proof revealed that some ideas from the real case
could be carried on to the complex case, but also that many new intriguing
questions naturally arise. One of them was that if the analog of segments in
the real line is discs in the complex line, the analog of convex bodies in
$\mathbb{R}^{n}$ should be what we called _bombons_ in $\mathbb{C}^{n}$:
bodies whose non-void sections with complex lines are disks. They don’t give
any new examples, because they turned out to be complex ellipsoids, Section 4
is devoted to that proof which first appeared in [2]. One of the main tools
used in many characterizations of complex ellipsoids is the uniqueness of
minimal circumscribed complex ellipsoids of convex bodies; which is reviewed
in Section 3.

The study of complex ellipsoids is naturally related to the study of complex
symmetry. So, characterizing and understanding complex symmetry is vital to
characterizing complex ellipsoids; the next section is devoted to it.

To ease the reading, we omit the term _complex_ from such concepts as lines,
hyperplanes, subspaces, $k$-planes, affine and linear maps, bodies of
revolution and ellipsoids when they are clearly meant to be in the context of
a complex vector space. On the other hand, we will always use the term _real_
when we work with their counterparts over the real numbers. However, we still
use the term complex in the titles of sections and for emphasis in
definitions.

##  2\. Complex symmetry

Let $\mathbb{S}^{1}$ be the multiplicative group of unit complex numbers
$\mathbb{C}$. Let $A\subset\mathbb{C}^{n}$ be a set. We say that $A$ is
_complex symmetric_ , or simply _symmetric_ , if there is a translated copy
$A^{\prime}$ of $A$ such that $\xi A^{\prime}=A^{\prime}$, for every
$\xi\in\mathbb{S}^{1}$. In this case, if $A^{\prime}=A-x_{0}$, we say that
$x_{0}$ is the center of symmetry of $A$. If $-A$ is a translated copy of $A$,
we just say that $A$ is _real-symmetric_. It will be useful to consider the
empty set as a symmetric set. Note that a _a convex body
$A\subset\mathbb{C}^{n}$ is symmetric with center at $x_{0}$ if and only if
for every line $L$ through $x_{0}$, the section $L\cap A$ is a disk centered
at $x_{0}$._ Of course, the unit ball of a finite dimensional Banach space
over the complex numbers $\mathbb{C}$ is symmetric.

The following characterization of symmetry will be very useful.

######  Theorem 1.

A convex body $K\subset\mathbb{C}^{n}$ is symmetric if and only if for every
$\xi\in\mathbb{S}^{1}$, $\xi K$ is a translated copy of $K$.

###### Proof.

Suppose $K$ is symmetric, then there exists $x_{0}\in\mathbb{C}^{n}$ such that
for every $\xi\in\mathbb{S}^{1}$, $\xi(K-x_{0})=K-x_{0}$. This implies that
$\xi K-\xi x_{0}=K-x_{0}$ and hence that

$\xi K=K+(\xi-1)x_{0}\,.$

Thus, $\xi K$ is a translated copy of $K$.

Suppose now that for every $\xi\in\mathbb{S}^{1}$, $\xi K$ is a translated
copy of $K$. Hence

$\xi K=K+\kappa_{\xi}\,,$

where $\kappa_{\xi}\in\mathbb{C}^{n}$ depends on $\xi$.

For $\xi=-1$, we get that $-K=K+\kappa_{-1}$. This implies that $K$ is real-
symmetric with center $-\frac{\kappa_{-1}}{2}$. Define
$x_{0}=-\frac{\kappa_{-1}}{2}$; it is the center of real-symmetry of $K$.

We then have that for any given $\xi\in\mathbb{S}^{1}$, $\xi K$ is real-
symmetric with center $\xi x_{0}$. On the other hand, since $\xi
K=K+\kappa_{\xi}$, the center of real-symmetry of $\xi K$ is
$x_{0}+\kappa_{\xi}$. Since there is a unique center of real-symmetry of $\xi
K$, we have that

$\xi x_{0}=x_{0}+\kappa_{\xi}\,.$

Therefore, $\kappa_{\xi}=(\xi-1)x_{0}.$

To conclude the proof, note that $\xi K=K+(\xi-1)x_{0},$ implies that
$\xi(K-x_{0})=K-x_{0}$ for every $\xi\in\mathbb{S}^{1}$ and therefore, $K$ is
symmetric by definition. ∎

The analog of the following theorem but over the real numbers was originally
proved by Rogers [27] in 1964. Using it, it is possible to prove that if all
real hyperplane sections of a convex body through a point $O$ are real-
symmetric, it is because the body is itself real-symmetric. Indeed, if the
point ${O}$ is not the center of symmetry of the body, then Larman [23]
proved, in the real case, that the body must be a real-ellipsoid. This result
is known as the _false center theorem_. We still don’t know if the false
center theorem is true over the complex numbers.

######  Theorem 2 (Complex Rogers’ Theorem).

Let $K_{1},K_{2}\subset\mathbb{C}^{n}$, $n\geq 3$, be convex bodies with
${p_{1}\in int(K_{1})}$ and $p_{2}\in int(K_{2})$. Suppose that for every
hyperplane $H$ through the origin, the section $(H+p_{1})\cap K_{1}$ is
positively homothetic to $(H+p_{2})\cap K_{2}$. Then, $K_{1}$ is positively
homothetic of $K_{2}.$

###### Proof.

Asume first that $K_{1}$ and $K_{2}$ are strictly convex bodies. Note that if
$K_{2}^{\prime}$ is positively homothetic to $K_{2}$, then $K_{2}^{\prime}$
also satisfies the conditions of the lemma. So that using a positive
homothecy, we may assume $p_{1}=p_{2}$ and ${K_{2}\subset int(K_{1})}$. Now,
dilate $K_{2}$ from $p_{1}$ until it touches the boundary of $K_{1}$.
Therefore, without loss of generality we may assume that:

  1. (1)

$K_{2}\subset K_{1},$

  2. (2)

$p_{1}=p_{2}$, and

  3. (3)

$K_{1}$ and $K_{2}$ share a common support real-hyperplane $\Delta$ at
$x_{0}$.

Let $L$ be the line through $x_{0}$ and $p_{1}=p_{2}$. Consider a hyperplane
$H$ that contains $L$. By hypothesis, there is a positive homothecy $h$ that
sends $H\cap K_{1}$ onto $H\cap K_{2}$. We claim that $h$ has center $x_{0}$.
Since such a homothecy sends support real-hyperplanes (of $H\cap K_{1}$) to
parallel support real-hyperplanes (of $H\cap K_{2}$), it has to send
$H\cap\Delta$ to itself because both $K_{1}$ and $K_{2}$ are strictly convex
and the homothecy is positive. So, $h$ has center
$x_{0}=K_{1}\cap\Delta=K_{2}\cap\Delta$; let $r_{H}>0$ be its ratio. Since
$x_{0}\in L$, $h$ restricts to a positive homothecy that sends $L\cap K_{1}$
to $L\cap K_{2}$. Because $L\cap intK_{1}\neq\emptyset$, the ratio of
homothecy $r_{H}$ must be the same $r$ for all $H$ that contain $L$. Because
$n\geq 3$, every point lies in a hyperplane through $L$, so that the homothecy
with center at $x_{0}$ and ratio $r$ sends $K_{1}$ to $K_{2}$ as we wished to
prove.

The proof of the non-strictly convex case requires one more step of a slightly
more technical nature that follows straightforwardly from the real case given
in [27]. ∎

Note that in Theorem 2 all homotecies are real homotecies. So, it is
interesting to ask if a similar result exists for complex homotecies. Namely:
_if $K_{1}$ and $K_{2}$ are convex bodies in $\mathbb{C}^{n}$, $n\geq 3$,
containing the origin in their interior and for every hyperspace $H$, there
exists a complex number $\xi$ (depending on $H$) such that_

$\xi(H\cap K_{1})=H\cap K_{2}\,,$

_is it true that $\,\xi_{0}K_{1}=K_{2}\,$, for some $\xi_{0}\in\mathcal{C}$?_

As a corollary to Rogers’ Theorem we have:

######  Corollary 1.

Let $K_{1},K_{2}\subset\mathbb{C}^{n}$, $n\geq 3$, be convex bodies with
$p_{1}\in int(K_{1})$ and $p_{2}\in int(K_{2})$. Suppose that for every
hyperspace $H$ the section $(H+p_{1})\cap K_{1}$ is a translated copy of
${(H+p_{2})\cap K_{2}}$. Then, $K_{1}$ is a translated copy of $K_{2}.$

###### Proof.

Translations are homothecies of ratio 1. So, by Theorem 2, we conclude that
there is a positive homothecy $h$ sending $K_{1}$ to $K_{2}$, with ratio of
homothecy $r>0$. To prove that $h$ is a translation we must show that $r=1$.
Let $p_{3}=h(p_{1})$. Then, for every hyperplane $H$ through the origin the
section $(H+p_{3})\cap K_{2}$ is positively homothetic to $(H+p_{2})\cap
K_{2}$, with ratio of homothecy $r$. Let $H^{\prime}$ be a hyperplane through
the origin with the property that $(H^{\prime}+p_{3})=(H^{\prime}+p_{2})$.
Hence, $(H^{\prime}+p_{3})\cap K_{2}=(H^{\prime}+p_{2})\cap K_{2}$, and being
positively homothetic with ratio of homothecy $r>0$, we conclude that $r=1$. ∎

######  Theorem 3.

In dimension $n\geq 3$, a convex body $K\subset\mathbb{C}^{n}$ all of whose
hyperplane sections through an interior point $x_{0}$ are symmetric, is itself
symmetric.

###### Proof.

To prove that $K$ is symmetric, by Theorem 1 it is enough to prove that for
every $\xi\in\mathbb{S}^{1}$, $\xi K$ is a translated copy of $K$.

Fix $\xi\in\mathbb{S}^{1}$, thus $\xi x_{0}$ is an interior point of $\xi K$.
It is enough to prove that for every hyperspace $H$, $(H+x_{0})\cap K$ is a
translated copy of $(H+\xi x_{0})\cap\xi K$; because this implies, by
Corollary 1, that $\xi K$ is a translated copy of $K$.

By hypothesis, $(H+x_{0})\cap K$ is symmetric, hence by Theorem 1,
$(H+x_{0})\cap K$ is a translated copy of $\xi\big{(}(H+x_{0})\cap K\big{)}$,
but

$\xi\big{(}(H+x_{0})\cap K\big{)}=(\xi H+\xi x_{0})\cap\xi K=(H+\xi
x_{0})\cap\xi K\,.$

∎

We now turn our attention to characterizing complex symmetry by means of
projections.

###  2.1. Complex symmetry and projections

The affine image of a symmetric set is a symmetric set. Essentially this is so
because, if $\xi K=K$, for every $\xi\in\mathbb{S}^{1}$, and $f$ is a linear
map then $\xi f(K)=f(\xi K)=f(K)$. In particular, the orthogonal projection of
a symmetric set onto a line is always a disk. The following simple lemma will
be useful in the sequel.

######  Lemma 1.

A convex body $K\subset\mathbb{C}^{n}$ is symmetric if and only if it has a
translated copy all whose orthogonal projections onto $1$-dimensional
subspaces are disks centered at the origin.

###### Proof.

The necessity is clear. For sufficiency, assume $\pi(K)$ is a disk centered at
the origin for every orthogonal projection $\pi$ onto a $1$-dimensional
subspace.

We shall prove that for every $\xi\in\mathbb{S}^{1}$, $\xi K=K$.

Suppose not. Then there is $x\in K$ and $\xi\in\mathbb{S}^{1}$ such that $\xi
x\notin K$. By convexity there is a real-hyperplane $\Delta$ in
$\mathbb{R}^{2n}=\mathbb{C}^{n}$ that separates $\xi x$ from $K$. Let $H$ by
the unique hyperplane contained in $\Delta$. Let $\pi:\mathbb{C}^{n}\to L$ be
the orthogonal projection to the $1$-dimensional subspace $L$ orthogonal to
$H$. Then the real line $\pi(\Delta)$ separates $\pi(K)$ from $\pi(\xi
x)=\xi(\pi(x))$ in the line $L$. On the other hand, $\pi(x)$ lies in the disk
$\pi(K)$ centered at the origin, therefore $\xi(\pi(x))$ lies in $\pi(K)$,
which is a contradiction. ∎

######  Theorem 4.

For $k\geq 2$, a convex body all whose orthogonal projections onto $k$-planes
are symmetric is symmetric.

###### Proof.

Consider a convex body $K\subset\mathbb{C}^{n}$, $n\geq 3$, all whose
orthogonal projections onto hyperplanes are symmetric. For every line $L$
through the origin, let $c_{L}$ be the center of the orthogonal projection of
$K$ to $L^{\perp}$, the orthogonal hyperplane to $L$, and let $L^{\prime}$ be
the parallel line to $L$ through $c_{L}$. We claim that

$\bigcap_{L}L^{\prime}\text{ is a single point }\\{x_{0}\\}.$

Let $L_{1}$ and $L_{2}$ be two different lines through the origin. The
orthogonal projection of $K$ to $L_{1}^{\perp}\cap L_{2}^{\perp}$ is symmetric
because it is the affine image of symmetric convex bodies (in $L_{1}^{\perp}$
and $L_{2}^{\perp}$); let $a$ be its center of symmetry and let $\Gamma$ be
the orthogonal plane to $L_{1}^{\perp}\cap L_{2}^{\perp}$ passing through $a$.
The lines $L_{1}^{\prime}$ and $L_{2}^{\prime}$ are both in $\Gamma$ because
the symmetry centers $c_{L_{1}}$ and $c_{L_{2}}$ are orthogonally projected to
$a$. Therefore, since $L_{1}^{\prime}$ and $L_{2}^{\prime}$ are not parallel,
they intersect in a point $\\{x_{0}\\}=L_{1}^{\prime}\cap L_{2}^{\prime}$.

Consider a third line $L_{3}$ through the origin, linearly independent of
$L_{1}$ and $L_{2}$; it exists because $n\geq 3$. By the above argument,
$L_{3}^{\prime}$ intersects both $L_{1}^{\prime}$ and $L_{2}^{\prime}$. But by
linear independence, $L_{3}^{\prime}$ intersects $\Gamma$ in at most one
point, so we must have that $x_{0}\in L_{3}^{\prime}$.

Finally, for the general line $L$ through the origin, $L$ is linearly
independent of at least one pair of the lines $L_{1},L_{2},L_{3}$. So that the
preceding argument yields that $x_{0}\in L^{\prime}$. This proves that
$\bigcap_{L}L^{\prime}=\\{x_{0}\\}$ as we claimed.

Suppose without loss of generality that $x_{0}$ is the origin. Then, for every
$(n-1)$-plane $H$ through the origin, the orthogonal projection of $K$ onto
$H$ is symmetric with center at the origin. This immediately implies that
orthogonal projections of $K$ onto $1$-dimensional subspaces are disks
centered at the origin and therefore, by Lemma 1, $K$ is symmetric.

The theorem now follows by induction, because a convex body all whose
orthogonal projections onto $k$-planes are symmetric has the property that all
orthogonal projections onto $(k+1)$-planes are symmetric, and so on. ∎

At this point the following two questions arise. _Is a convex body all whose
orthogonal projections onto lines are disks, symmetric?_ The projection of an
ellipsoid from an $(n-2)$-affine plane on a line is a disk. Is the converse
true? That is, _if all projections of a convex body on lines are disks, is the
body an ellipsoid?_

###  2.2. Complex symmetry and complex ellipsoids

######  Lemma 2.

A symmetric real ellipsoid is an ellipsoid.

###### Proof.

We need to recall some facts about real ellipsoids, within the real context.
Let $E\subset\mathbb{R}^{n}$ be a $n$-dimensional real ellipsoid centred at
the origin. For every $k$-dimensional real subspace, $H\subset\mathbb{R}^{n}$
with $1\leq k<n$, there exists a complementary $(n-k)$-dimensional real
subspace, $L$ of $\mathbb{R}^{n}$, called its polar subspace with respect to
$E$, such that

$\partial E\cap L=\\{x\in\mathbb{R}^{n}\mid H+x\mbox{ is a k-dimensional real
plane tangent to }\partial E\mbox{ at }x\\}$

(this set is called the shadow boundary of $E$ in the direction $H$).
Moreover, $H$ is the polar subspace of $L$ with respect to $E$, and the
section $L\cap E$ is a $(n-k)$-dimensional real ellipsoid with the following
property: for every $(n-k)$-plane $L^{\prime}$, parallel to $L$, the
corresponding section $L^{\prime}\cap E$ is either the empty set, a point in
$H$ or a real ellipsoid homothetic to $L\cap E$ and centred at $H$. For more
about shadow boundaries see Section 1.12.2 of [25].

Clearly, every ellipsoid is a real ellipsoid which is symmetric. Let
${K\subset\mathbb{C}^{n}=\mathbb{R}^{2n}}$ be a symmetric real ellipsoid
centered at the origin. By induction on the complex dimension $n$, we will
prove that there is a linear isomorphism $g\in GL(n,\mathbb{C})$ such that
$g(K)$ is a ball. The statement is true for $n=1$. Suppose it is true for
dimension $n-1$, we shall prove it for dimension $n$.

Assume the diameter of $K$ is $h$, and let $[-u,u]$ be a diameter of $K$; let
$L$ be the unique line containing the vector $u$. By hypothesis $D=L\cap K$ is
a disk centred at the origin all of whose diameters are also diameters of $K$.
This implies that the polar to $L$ with respect to $E$ is the hyperplane, $H$,
orthogonal to $L$. Then, for every affine line $L^{\prime}$ orthogonal to $H$
and touching $int(K)$, the section $L^{\prime}\cap K$ is a disk with center at
$H$.

By induction we have that $H\cap K$ is an ellipsoid. Therefore, using a linear
isomorphism, we may assume that $H\cap K$ is a $(2n-2)$-dimensional ball of
diameter $h$. To conclude the proof of the lemma, we prove that $K$ is a ball.

Let $\ell$ be a real line subspace contained in $H$ and let $\Delta$ be the
$3$-dimensional real subspace generated by $\ell$ and $L$. Since $(L+x)\cap K$
is a disk with centre at $\ell$ for every $x\in\ell\cap int(K)$, $\Delta\cap
K$ is a real ellipsoid of revolution with axis the line $\ell$. Since the
three axes of this ellipsoid are equal, this implies, that $\Delta\cap K$ is a
$3$-dimensional ball with centre at the origin. Since this holds for every
real $3$-dimensional real subspace containing $L$, we have that $K$ is a ball,
as we wished. ∎

In 1955, in his book; The Geometry of Geodesics [13], Busemann proved the
following Kubota’s Theorem: _if all real planes passing through a point $O$
intersect a convex body (in $\mathbb{R}^{3}$) in real ellipses then the body
is a real ellipsoid_. We will start proving the analog of this theorem for
ellipsoids with the additional hypothesis that the point $O$ is the center of
symmetry of the body. Later, in Theorem 9, we will get rid of this auxiliary
hypothesis.

######  Proposition 1.

If $K\subset\mathbb{C}^{n+1}$, $n\geq 2$, is a symmetric convex body all whose
hyperplane sections through the center are ellipsoids, then $K$ is an
ellipsoid.

###### Proof.

By Lemma 2, it is enough to prove that $K$ is a real ellipsoid. Consider
$\mathbb{C}^{n}=\mathbb{R}^{2n}$. By Theorem 2.12.4 of [25], it is enough to
prove that every real two dimensional subspace intersects $K$ in an ellipse.
Let $\Pi$ be a two dimensional real plane generated by $\\{v_{1},v_{2}\\}$. If
$\Pi$ is a line, $\Pi\cap K$ is a ball, so assume it is not. Let $L_{i}$ be
the line containing $v_{i}$, $i=1,2$. Consequently, $\Pi$ is contained in the
plane $P$ generated by $\\{L_{1},L_{2}\\}$. By hypothesis, $P\cap K$ is a
section of an ellipsoid and hence is itself an ellipsoid. This implies that
$\Pi\cap K$ is an ellipse. Therefore, $K$ is a real ellipsoid. ∎

##  3\. Extremal inscribed and circumscribed complex ellipsoids

Let $A$ be a non-flat, compact subset of $\mathbb{C}^{n}$. An ellipsoid
$\mathfrak{E}$ is called _circumscribed_ if $A\subset\mathfrak{E}$, and it is
_inscribed_ if $\mathfrak{E}\subset\widehat{A}$, where $\widehat{A}$ denotes
the convex closure of $A$. We say that $\mathfrak{E}$ is a _minimal
circumscribed ellipsoid_ (MiCE) if it has the minimal volume among all
circumscribed ellipsoids. On the other hand, $\mathfrak{E}$ is a _maximal
inscribed ellipsoid_ (MaIE) if $\mathfrak{E}$ has maximal volume among all
inscribed ellipsoids.

The purpose of this section is to prove that if a convex set in
$\mathbb{C}^{n}$ contains two inscribed ellipsoids of maximal volume, then one
is a translate of the other. On the other hand, the circumscribed ellipsoid of
minimal volume is unique. The first proofs of uniqueness of MiCE and MaIE for
real ellipsoids in its full generality seem to have appeared independently in
[15] and [30] and for complex ellipsoids in [1]. For real ellipsoids MiCE and
MaIE are known as Löwner–John Ellipsoids and have many applications in several
areas of mathematics (see, e.g., [18], [26], Section 2.12 of [25] and [19]
with the references therein).

The existence of these ellipsoids follows from standard arguments. There is a
ball big enough to contain $A$ and a non-zero sphere contained in
$\widehat{A}$. All ellipsoids are easily parametrized by a matrix and a
vector; among them, we can consider only those that are contained in the big
ball. This set is compact in the parameter space. Moreover, the volume
function is continuous so that the existence of a MiCE and a MaIE follows.

For $x=\left(x_{1},...,x_{n}\right)\in\mathbb{C}^{n}$, let
$\overline{x}=\left(\bar{x}_{1},...,\bar{x}_{n}\right)$, where the bar above
denotes the complex conjugate. Also, we denote by $x\odot y$ the Hadamar
product of $x$ and $y$; that is, the coordinatewise product (see for example
[20] Chapter 5).

A _scalar product_ in $\mathbb{C}^{n}$ is a sesquilinear, Hermitian, positive-
definite functional $\mathbb{C}^{n}\times\mathbb{C}^{n}\rightarrow\mathbb{C}$
denoted by $\left\langle x\cdot y\right\rangle$ for any $x$ and $y$ in
$\mathbb{C}^{n}$. For each scalar product there is a matrix $M$ (Hermitian,
positive-definite) such that $\left\langle x\cdot
y\right\rangle=x^{T}M\overline{y}$. In the case that $M$ is the identity
matrix, the scalar product is the usual Hermite’s product in $\mathbb{C}^{n}$.
It is well known that the eigenvalues of $M$ are real positive numbers and so
is its determinant. It is also known that $M$ is diagonalizable by a unitary
transformation.

An _ellipsoid_ (centered at the origin) is a set

(1) $\left\\{x\in\mathbb{C}^{n}\mid x^{T}M\overline{x}\leq 1\right\\}\,,$

where $M$ is Hermitian and positive-definite. If $M$ is the identity matrix,
this ellipsoid is the _unit ball_ $\mathfrak{B}$.

The set of unitary transformations is the subgroup of
$GL\left(\mathbb{C}^{n}\right)$ that preserves the unit sphere. The modulus of
a scalar $\xi\in\mathbb{C}$ will be denoted by $\left|\xi\right|$. We will
denote by $\left\|\cdot\right\|$ the usual norm in $\mathbb{C}^{n}$, that is,
the norm defined by Hermite’s product.

Let $B\overset{\text{def}}{=}\sqrt{M^{T}}$. We have $M^{T}=BB$ and hence,
$M=B^{T}B^{T}=B^{T}\overline{B}$. From this, we obtain

$x^{T}M\overline{x}=x^{T}B^{T}\overline{B}\overline{x}=\left(Bx\right)^{T}\overline{\left(Bx\right)}=\left\|Bx\right\|^{2}\,.$

Therefore, the ellipsoids in $\mathbb{C}^{n}$ can also be written in the form

$\\{B^{-1}u\mid u\in\mathfrak{B}\\}.$

If we use a unitary transformation to bring $M$ to the diagonal form, then
this can be rewritten as

(2)
$\mathfrak{El}\left(\lambda\right)\overset{\text{def}}{=}\left\\{\lambda\odot
u\mid u\in\mathfrak{B}\right\\}\,,$

where $\mathbf{\lambda}$ is a vector in $\mathbb{R}_{+}^{n}$. Since
$M^{{\frac{1}{2}}}=B$, the two forms 1 and 2 are related by the fact that
$\lambda$ is the diagonal of $M^{-{\frac{1}{2}}}$. The map
$x\mapsto\lambda\odot x=M^{-{\frac{1}{2}}}x$ is an invertible linear map in
$GL\left(\mathbb{C}^{n}\right)$ which maps the unit ball $\mathfrak{B}$ into
the ellipsoid $\mathfrak{El}\left(\lambda\right)$. Therefore

(3) $Vol\mathfrak{El}\left(\lambda\right)=\det
M^{-{\frac{1}{2}}}Vol\mathfrak{B}\,.$

The results that follow do not depend on the translation-invariant measure
chosen to define the volume. We just need the validity of equation 3.

Let $\det\lambda$ denote the product of coordinates of $\lambda$. Of course,
we have ${\det\lambda=\det A^{-{\frac{1}{2}}}}$ and

$Vol\mathfrak{El}\left(\lambda\right)=\det\lambda\,Vol\mathfrak{B}.$

Translates of ellipsoids centered at the origin are also _ellipsoids_ , and
translations do not change volume.

###  3.1. Auxiliary lemmas

For the proofs of Theorems 5 and 7, we need first two lemmas whose proofs are
straightforward computations. Let $\mathbf{1}$ be the vector which has all
coordinates equal to $1$.

######  Lemma 3.

If $\mathbf{\lambda}\in\mathbb{R}_{+}^{n}$ is such that
$\det\mathbf{\lambda}=1$ and $\mathbf{\lambda}\neq\mathbf{1}$, then

$\det\left(\frac{\mathbf{\lambda}+\mathbf{1}}{2}\right)>1\text{.}$

###### Proof.

We have $\left(\lambda_{i}-1\right)^{2}\geq 0$ and therefore
$\left(\lambda_{i}+1\right)^{2}\geq 4\lambda_{i}$. With equality only when
$\lambda_{i}=1$. Taking the product, we get

$\det\left(\mathbf{\lambda}+\mathbf{1}\right)^{2}>4^{n}\det\mathbf{\lambda}$

which is the same as

$\det\left(\mathbf{\lambda}+\mathbf{1}\right)>2^{n}\sqrt{\det\mathbf{\lambda}}=2^{n}\,.$

The lemma follows. ∎

######  Lemma 4.

Let $c,x$ be complex numbers and $\lambda$ be a real number. Then,

$\lambda\left|x\right|^{2}+\left|x-c\right|^{2}=\left(\lambda+1\right)\left|x-\frac{c}{\left(\lambda+1\right)}\right|^{2}+\left(\frac{\lambda}{\lambda+1}\right)\left|c\right|^{2}\,.$

###### Proof.

We have

$\begin{array}[]{c}\lambda\left|x\right|^{2}+\left|x-c\right|^{2}=\lambda\left|x\right|^{2}+\left(x-c\right)\overline{\left(x-c\right)}\\\
=\left(\lambda+1\right)\left|x\right|^{2}-\left(x\overline{c}+c\overline{x}\right)+\left|c\right|^{2}\end{array}$

and completing the square, we obtain

$\begin{array}[]{l}=\left(\lambda+1\right)\left(\left|x\right|^{2}-\dfrac{\left(x\overline{c}+c\overline{x}\right)}{\left(\lambda+1\right)}+\left|\dfrac{c}{\left(\lambda+1\right)}\right|^{2}-\left|\dfrac{c}{\left(\lambda+1\right)}\right|^{2}\right)+\left|c\right|^{2}\\\
=\left(\lambda+1\right)\left|x-\dfrac{c}{\left(\lambda+1\right)}\right|^{2}+\left(\dfrac{\lambda}{\lambda+1}\right)\left|c\right|^{2}.\end{array}$

∎

###  3.2. Maximal inscribed complex ellipsoids.

######  Theorem 5.

Let $A$ be a non-flat compact in $\mathbb{C}^{n}$. Let $\mathfrak{E}_{1}$ and
$\mathfrak{E}_{2}$ be two MaIE contained in $\widehat{A}$. Then, there is a
vector $c\in\mathbb{C}^{n}$ such that $\mathfrak{E}_{2}=\mathfrak{E}_{1}+c$.

###### Proof.

Using a suitable affine transformation we can assume that $\mathfrak{E}_{1}$
is the unit ball $\mathfrak{El}\left(\mathbf{1}\right)$. Let
$c\in\mathbb{C}^{n}$ be the center of $\mathfrak{E}_{2}$. We can use a unitary
transformation to diagonalize the matrix of $\mathfrak{E}_{2}-c$. And
therefore, $\mathfrak{E}_{2}=\mathfrak{El}\left(\lambda\right)+c$ for some
$\lambda\in\mathbb{R}_{+}^{n}$.

Let us first prove that the ellipsoid
$\mathfrak{E}_{3}=\mathfrak{El}\left({\frac{1}{2}}\left(\lambda+\mathbf{1}\right)\right)+{\frac{1}{2}}c$
is contained in the convex closure of $\mathfrak{E}_{1}\cup\mathfrak{E}_{2}$,
that is,

$\mathfrak{E}_{3}=\mathfrak{El}\left(\frac{\lambda+\mathbf{1}}{2}\right)+\frac{c}{2}\subset\widehat{\mathfrak{El}\left(\mathbf{1}\right)\cup\left(\mathfrak{El}\left(\lambda\right)+c\right)}\,.$

For any $x\in\mathfrak{E}_{3}$ there is
$u\in\mathfrak{B}=\mathfrak{El}\left(\mathbf{1}\right)=\mathfrak{E}_{1}$, such
that

$x=\frac{\lambda+\mathbf{1}}{2}\odot u+\frac{c}{2}\,.$

The point $y=\lambda\odot u+c$ is in $\mathfrak{E}_{2}$ and

$\frac{y+u}{2}=\frac{\lambda\odot u+c+u}{2}=x\,.$

This means that $x$ is the middle point of the segment joining $y$ and $u$,
and proves that
$\mathfrak{E}_{3}\subset\widehat{\mathfrak{E}_{1}\cup\mathfrak{E}_{2}}$, as we
wished.

Since $\mathfrak{E}_{1}\cup\mathfrak{E}_{2}\subset\widehat{A}$, we have
$\mathfrak{E}_{3}\subset\widehat{\mathfrak{E}_{1}\cup\mathfrak{E}_{2}}\subset\widehat{A}$.

We know that
$Vol\left(\mathfrak{E}_{1}\right)=Vol\left(\mathfrak{E}_{2}\right)=\det\lambda
Vol\left(\mathfrak{E}_{1}\right)$. Therefore, $\det\lambda=1$. On the other
hand,
$Vol\left(\mathfrak{E}_{3}\right)=\det\left({\frac{1}{2}}\left(\lambda+\mathbf{1}\right)\right)Vol\left(\mathfrak{E}_{1}\right)$.
If $\lambda\neq\mathbf{1}$, Lemma 3 implies that
$\det\left({\frac{1}{2}}\left(\lambda+\mathbf{1}\right)\right)>1$, and hence
$Vol\left(\mathfrak{E}_{3}\right)>Vol\left(\mathfrak{E}_{1}\right)$; but this
contradicts that $\mathfrak{E}_{1}$ is a MaIE. So, we have
$\lambda=\mathbf{1}$ and therefore, $\mathfrak{E}_{2}=\mathfrak{E}_{1}+c$,
which proves our theorem. ∎

Now, we shall see that symmetry guarantees uniqueness of the MaIE.

######  Theorem 6.

If $A$ is a non-flat, compact symmetric set in $\mathbb{C}^{n}$, then its MaIE
is unique.

###### Proof.

We can assume that the center of $A$ is the origin. Let $\mathfrak{E}$ be the
unique real MaIE in $A$. Let $\zeta$ be a scalar of modulus $1$. Since
${\zeta\mathfrak{E}\subset\zeta A=A}$ and
$Vol\left(\zeta\mathfrak{E}\right)=Vol\left(\mathfrak{E}\right)$ then, the
uniqueness of $\mathfrak{E}$ implies that $\mathfrak{E}$ is symmetric. Using
Lemma 2, we get that $\mathfrak{E}$ is not only a real ellipsoid but a complex
one. There is not another MaIE because any ellipsoid is also a real ellipsoid.
∎

###  3.3. Minimal circumscribed complex ellipsoids

The uniqueness of a minimal ellipsoid containing a symmetric set was first
proved by Gromov [16], when he was proving that the group of symmetries of a
compact symmetric set of $\mathbb{C}^{n}$ is conjugate to a compact subgroup
of $SU_{n}$. Here we prove the uniqueness without the additional hypothesis of
symmetry.

######  Theorem 7.

Let $A$ be a non-flat, compact set in $\mathbb{C}^{n}$. Let $\mathfrak{E}_{1}$
and $\mathfrak{E}_{2}$ be two MiCE containing $A$. Then,
$\mathfrak{E}_{1}=\mathfrak{E}_{2}$.

###### Proof.

Using an affine transformation, we can make $\mathfrak{E}_{2}$ the unit ball.
Then, using a unitary transformation, we can diagonalize the matrix of
$\mathfrak{E}_{1}$. Finally, we translate so that the center of
$\mathfrak{E}_{1}$ is the origin and the center of $\mathfrak{E}_{2}$ is some
vector $c\in\mathbb{C}^{n}$. Then,

$\begin{array}[]{l}\mathfrak{E}_{1}=\left\\{x\in\mathbb{C}^{n}\mid\sum\lambda_{i}\left|x_{i}\right|^{2}\leq
1\right\\}=\mathfrak{El}\left(\beta\right)\\\
\mathfrak{E}_{2}=\left\\{x\in\mathbb{C}^{n}\mid\sum\left|x_{i}-c_{i}\right|^{2}\leq
1\right\\}=\mathfrak{El}\left(\mathbf{1}\right)+c\,,\end{array}$

where $\mathbf{\beta}=\mathbf{\lambda}^{-{\frac{1}{2}}}$. Therefore,

$A\subset\mathfrak{E}_{1}\cap\mathfrak{E}_{2}\subset\mathfrak{E}_{3}\overset{\text{def}}{=}\left\\{x\in\mathbb{C}^{n}\mid\sum\lambda_{i}\left|x_{i}\right|^{2}+\left|x_{i}-c_{i}\right|^{2}\leq
2\right\\}.$

Using Lemma 4, the inequality is transformed into

(4)
$\sum\left(\lambda_{i}+1\right)\left|\left(x_{i}-\frac{c_{i}}{\left(\lambda_{i}+1\right)}\right)\right|^{2}\leq
2-\sum\left(\frac{\lambda_{i}}{\lambda_{i}+1}\right)\left|c_{i}\right|^{2}\leq
2\,.$

Let

$\mathfrak{E}_{4}\overset{\text{def}}{=}\left\\{x\in\mathbb{C}^{n}\mid\sum\left(\lambda_{i}+1\right)\left|\left(x_{i}-\frac{c_{i}}{\left(\lambda_{i}+1\right)}\right)\right|^{2}\leq
2\right\\}\,,$

which is an ellipsoid. From inequality 4, we obtain that

$A\subset\mathfrak{E}_{1}\cap\mathfrak{E}_{2}\subset\mathfrak{E}_{3}\subset\mathfrak{E}_{4}\,.$

Let $\mathbf{z}\in\mathbb{C}^{n}$ be the vector with coordinates
$c_{i}/\left(\lambda_{i}+1\right)$. We have

$\mathfrak{E}_{4}=\mathbf{z}+\left\\{x\in\mathbb{C}^{n}\mid\sum\left(\lambda_{i}+1\right)\left|x_{i}\right|^{2}\leq
2\right\\}=\mathfrak{El}\left(\left(\frac{\mathbf{\lambda}+1}{2}\right)^{-{\frac{1}{2}}}\right)+\mathbf{z}\,.$

Since $\det\mathbf{\beta}=1$, we obtain $\det\mathbf{\lambda}=1$. If
$\mathbf{\lambda}\neq\mathbf{1}$, the hypothesis of Lemma 3 are satisfied,
then

$\Delta\overset{\text{def}}{=}\det\left(\frac{\mathbf{\lambda}+1}{2}\right)>1\,.$

Therefore
$Vol\mathfrak{E}_{4}=\Delta^{-{\frac{1}{2}}}Vol\mathfrak{E}_{2}<Vol\mathfrak{E}_{2}$,
which contradicts the minimality of $\mathfrak{E}_{2}$.

So, we conclude that $\lambda=\mathbf{1}$, $\mathfrak{E}_{1}$ is the unit ball
and $\mathfrak{E}_{2}=\mathfrak{E}_{1}+c$. In this case, the inequality (4)
transforms to the following

$\sum\left|\left(x_{i}-\frac{c_{i}}{2}\right)\right|^{2}\leq
1-\sum\left|\frac{c_{i}}{2}\right|^{2}.$

Therefore, $\mathfrak{E}_{3}$ is a ball with center in $c/2$, and radius

$r\overset{\text{def}}{=}\sqrt{1-\sum\left|\frac{c_{i}}{2}\right|^{2}}\,.$

If $c$ is not the origin then $r<1$ and $\mathfrak{E}_{3}$ has volume strictly
less than that of the unit ball $\mathfrak{E}_{1}$. Therefore, $c=\mathbf{0}$
and $\mathfrak{E}_{2}=\mathfrak{E}_{1}.$ ∎

##  4\.  A characterization of complex ellipsoids

The following characterization of the ellipsoids has no analogue over the real
numbers. It was unexpected for us, and was first proved in [2].

######  Theorem 8.

A convex body $K\subset\mathbb{C}^{n}$ is an ellipsoid if and only if for
every line $L\subset\mathbb{C}^{n}$, $L\cap K$ is either empty, a point or a
disk.

For expository reasons, it is convenient to define that a _bombon_ is a convex
body $K\subset\mathbb{C}^{n}$ such that for every line
$L\subset\mathbb{C}^{n}$, $L\cap K$ is either empty or a disk; where points
are regarded as disks of radius zero. Then, the theorem simply says that
bombons and ellipsoids are one and the same.

Note that $1$-dimensional ellipsoids and $1$-dimensional bombons are disks.
Furthermore, ellipsoids are bombons because balls are bombons and the image of
a bombon under an affine map is a bombon. The basic reason for this fact is
that affine maps between lines send disks to disks.

The proof of Theorem 8 is by induction. The general case will follow from
Proposition 1, and the first case, $n=2$, takes most of the work using ideas
of its own.

An _abstract linear space_ consists of a set $X$ together with a distinguished
family of subsets, called _lines_ , satisfying the following property: given
different $x,y\in X$, there is a unique line $L$ containing $x$ and $y$.
Typical examples of abstract linear spaces are the euclidean $n$-space, the
complex $n$-space and their corresponding projective spaces; with their
classic lines as lines.

Our interest in these objects lies in the abstract linear space $LS^{3}$,
where $X=\mathbb{S}^{3}$, the unit sphere of $\mathbb{C}^{2}$, and a line is
the intersection with $\mathbb{S}^{3}$ of a line of $\mathbb{C}^{2}$, when it
is has more than a single point; which is a flat circle in $\mathbb{S}^{3}$.
It is an abstract linear space because two different points in
$\mathbb{S}^{3}$ lie in a unique line.

A subset $A$ of an abstract linear space $X$ is _linearly closed_ if for any
different $x,y\in A$, the line through $x$ and $y$ is contained in $A$. Since
the intersection of linearly closed subsets is linearly closed, given any
$Y\subset X$, there exists a unique minimal linearly closed subset containing
$Y$, called its _linear closure_.

######  Lemma 5.

Suppose $K\subset\mathbb{C}^{2}$ is a bombon with the property that the
ellipsoid of minimal volume containing it is the unit ball. Then
$K\cap\mathbb{S}^{3}$ is a linearly closed subset of the abstract linear space
$LS^{3}$.

###### Proof.

Let $x,y\in K\cap\mathbb{S}^{3}$, $x\neq y$, and let $L$ be the line through
$x$ and $y$. We have to prove that $L\cap\mathbb{S}^{3}\subset K$.

Recall that $\mathfrak{B}$ is the unit ball in $\mathbb{C}^{2}$, whose
boundary is $\mathbb{S}^{3}$. By hypothesis, $L\cap K$ and $L\cap\mathfrak{B}$
are two disks in $L$ which share two different points, $x$ and $y$ on their
boundary. Then, $K\subset\mathfrak{B}$ implies that $L\cap
K=L\cap\mathfrak{B}$, and therefore that $L\cap\mathbb{S}^{3}\subset K$. ∎

######  Lemma 6.

The linearly closed proper subsets of the abstract linear space $LS^{3}$ are
single points and lines.

The proof of this lemma follows immediately from the next two lemmas.

We need to make precise some standard definitions. A topological space $U$ can
be topologically embedded in $\mathbb{C}$, if there is a continuous injective
map $f:U\to\mathbb{C}$. A topological space $A$ is locally embedded in
$\mathbb{C}$ if for every point $x\in A$, there is a neighborhood $U$ of $x$
in $A$ which can be topologically embedded in $\mathbb{C}$.

######  Lemma 7.

Let $A$ be a linearly closed proper subset of the abstract linear space
$LS^{3}$. Then $A$ is embedded in $\mathbb{C}$.

###### Proof.

Since $A$ is proper, there exists a point $w\in\mathbb{S}^{3}$ such that
$w\notin A$. Let $L$ be the line through $w$ and the origin and let
$L^{\perp}$ be the orthogonal line to $L$ at the origin, which we identify
with $\mathbb{C}$.

Let $\pi:\mathbb{S}^{3}\setminus\\{w\\}\to L^{\perp}$ be the geometric
projection from $w$, that is, given $x\in\mathbb{S}^{3}$ different from $w$,
$\pi(x)$ is the intersection with $L^{\perp}$ of the line through $x$ and $w$;
well defined because this line is not parallel to $L^{\perp}$.

We claim that $\pi|_{A}$ is an embedding. Indeed, if $x,y\in A$ are such that
$\pi(x)=\pi(y)$ then their lines through $w$ coincide. If $x$ and $y$ were
different points, it would imply that $w\in A$ because $A$ is linearly closed,
which contradicts the choice of $w$. ∎

######  Lemma 8.

Let $A$ be a linearly closed non-empty subset of the abstract linear space
$LS^{3}$ which is not a point or an line. Then $A$ is not locally embedded in
$\mathbb{C}$.

###### Proof.

$A$ has at least two points, so it contains an line $\ell\subset
A\subset\mathbb{S}^{3}$. But since $A$ is not an line, there is yet another
point $a\in A\setminus\ell$.

For every $x\in\ell$ let $\ell_{x}$ be the line through $a$ and $x$. Note that
$\ell_{x}\cap\ell=\\{x\\}$, and if $x\not=x^{\prime}\in\ell$, then
$\ell_{x}\cap\ell_{x^{\prime}}=\\{a\\}$, because two lines intersect at most
in one point. Let

$\Omega=\bigcup_{x\in\ell}\ell_{x}\subset A\subset\mathbb{S}^{3}\,.$

Since every abstract line of $LS^{3}$ is actually a circle, then
$\Omega\setminus\\{a\\}=\bigcup_{x\in\ell}(\ell_{x}\setminus\\{a\\})$ is an
$\mathbb{R}$-bundle over $\mathbb{S}^{1}$. This implies that, either $\Omega$
is homeomorphic to the real projective plane or $\Omega$ is homeomorphic to a
closed cylinder modulo its boundary,
$\frac{\mathbb{S}^{1}\times[0,1]}{\mathbb{S}^{1}\times\\{0,1\\}}$, where the
point $a\in\Omega$ is precisely
$\frac{\mathbb{S}^{1}\times\\{0,1\\}}{\mathbb{S}^{1}\times\\{0,1\\}}$.

Since $\Omega\subset\mathbb{S}^{3}$, the space $\Omega$ cannot be homemorphic
to the real projective plane. So $\Omega$ is homeomorphic to
$\frac{\mathbb{S}^{1}\times[0,1]}{\mathbb{S}^{1}\times\\{0,1\\}}$ and
consequently, any neighborhood of $a$ in $\Omega$ cannot be topologically
embedded in $\mathbb{C}$, because it contains the cone of two disjoint
circles. Since $\Omega\subset A$, then $A$ is not locally embedded in
$\mathbb{C}$. ∎

_Proof of Theorem 8._ As we have said, the proof is by induction on the
dimension $n$.

_Case_ $n=2$. Suppose $K\subset\mathbb{C}^{2}$ is a bombon, and let
$\mathfrak{E}$ be the ellipsoid of minimal volume containing $K$ (see Theorem
7). We may assume without loss of generality that $\mathfrak{E}$ is the unit
ball of $\mathbb{C}^{2}$; and we must prove that $K=\mathfrak{E}$, which is
equivalent to

$K\cap\partial\mathfrak{E}=K\cap\mathbb{S}^{3}=\mathbb{S}^{3}\,.$

By Lemma 5, $K\cap\mathbb{S}^{3}$ is a linearly closed subset of $LS^{3}$ and
by Lemma 6, we have to prove that $K\cap\mathbb{S}^{3}$ is neither a point nor
an line. It clearly cannot be a point by the minimality condition. So, we are
left to prove that this condition also rules out the line. We give an _ad hoc_
proof of this fact, without using John’s theory on minimal ellipsoids (see for
example [21] or [4]).

Suppose there is a line $L$, such that
$K\cap\mathbb{S}^{3}=L\cap\mathbb{S}^{3}$. First, we consider the case in
which $L$ passes through the origin; without loss of generality, suppose
$L=\mathbb{C}\times\\{0\\}$.

Consider the following family of ellipsoids for small $\epsilon\geq 0$, (see
Fig. 1.a)

$\mathfrak{E}_{\epsilon}=\\{(x,y)\in\mathbb{C}^{2}\mid\frac{|x|^{2}}{(1+\epsilon)^{2}}+\frac{|y|^{2}}{(1-\epsilon)^{2}}\leq
1\\}\,.$

For $\epsilon>0$, the volume of $\mathfrak{E}_{\epsilon}$ is less than the
volume of the unit ball $\mathfrak{E}=\mathfrak{E}_{0}$; in fact, their volume
ratio is $(1-\epsilon^{2})^{2}$. Then, all the sets
$X_{\epsilon}=\mathfrak{E}\setminus\mathfrak{E}_{\epsilon}$ intersect $K$
because of the minimality condition of $\mathfrak{E}$. Furthermore, the
intersection and limit of $X_{\epsilon}$, when $\epsilon\to 0$, is the solid
torus $T=\\{\,(x,y)\in\mathbb{S}^{3}\,|\,|y|\geq 1/\sqrt{2}\,\\}$. So that the
compacity of $K$ implies that there exists a point in $K\cap T\subset
K\cap\mathbb{S}^{3}$, which is far from $L$.

Figure 1. a) Section of the family of ellipsoids $\mathfrak{E}_{\epsilon}$,
with a real plane having real lines in the coordinate axis. b) Section of the
family of balls $\mathfrak{E}+\epsilon c$ with a real plane containing $c$ and
the origin.

If $L$ does not contain the origin, let $c$ be its closest point to the origin
and consider the family of balls $\mathfrak{E}+\epsilon c$, $\epsilon\geq 0$
(see Fig.1.b). As before, the sets
$X_{\epsilon}=\mathfrak{E}\setminus(\mathfrak{E}+\epsilon c)$ intersect $K$
for all $\epsilon>0$, because of the uniqueness of the minimal volume
ellipsoid containing $K$ (Theorem 7). Therefore, $K$ has a point in the limit
of $X_{\epsilon}$, as $\epsilon\to 0$, which is the closed half-hemisphere of
$\mathbb{S}^{3}$ opposite $c$. This contradicts that
$K\cap\mathbb{S}^{3}=L\cap\mathbb{S}^{3}$, and completes the proof for $n=2$.

_General case._ Suppose the theorem is true for $n-1$, we shall prove it for
$n\geq 3$. By induction, for every hyperplane $H$, the section $H\cap K$ is
either empty, a single point or an ellipsoid. This implies that every section
of $K$ is symmetric and consequently, by Theorem 3, $K$ is symmetric.

We may assume without loss of generality that the center of $K$ is the origin.
Now note that every hyperplane section of $K$ through the origin is an
ellipsoid. By Proposition 1, this implies that $K$ is an ellipsoid. ∎

##  5\. Complex ellipsoids by means of sections and of projections

The main purpose of this section is to prove that all sections through a point
$p_{0}$ of a convex body are ellipsoids only when the body is an ellipsoid.
The corresponding $3$-dimensional theorem for sections of real ellipsoids, as
we already mentioned, is due to Kubota in [22] and was independently
discovered by Auerbach, Mazur and Ulam, [3]. In its $n$-dimensional real
version the result was originally proved by Busemann for $p_{0}\in int(K)$
[13], and was extended to any $p_{0}\in\mathbb{E}^{n}$ by Burton, [11]. We
also prove the corresponding theorem for orthogonal projections. All
orthogonal projections of a convex body onto affine $k$-planes are ellipsoids
only when the body is an ellipsoid. For real ellipsoids this result was
observed by Blaschke and Hessenberg [7] without proof and proved by Süss [Su0]
in dimension $3$. Later, Chakerian in [14], gave a proof for the case of $n$
dimensions.

######  Theorem 9.

Let $k$ and $n$ be such that $2\leq k<n$. A convex body in $\mathbb{C}^{n}$
all whose $k$-plane sections through a point $p_{0}$ are ellipsoids is an
ellipsoid.

###### Proof.

It is easy to see that the general case $2\leq k<n$ of this theorem follows
from the case $k=n-1$. Let $K\subset\mathbb{C}^{n}$ be a convex body, $n\geq
3$, and suppose that all hyperplane sections of $K$ through $p_{0}$ are
ellipsoids. Let us first prove that $K$ is a real ellipsoid. Using Theorem
1.12.4 of [25], it is enough to prove that all real plane sections of $K$ are
real ellipses. Let $P$ be a real $2$-dimensional plane through $p_{0}$ and let
$H$ be a hyperplane through $p_{0}$ such that $P\subset
H\subset\mathbb{C}^{n}$. This is possible because $n\geq 3$. By hypothesis,
$H\cap K$ is an ellipsoid. Then, $P\cap K$ is a real ellipse. This implies
that $K$ is a real ellipsoid.

Since every hyperplane section of $K$ through $p_{0}$ is an ellipsoid, we have
that all line sections of $K$ through $p_{0}$ are disks. Remember that any two
parallel sections of a real ellipsoid are homothetic. Hence, every complex
line section, not only the ones through $p_{0}$, are disks and therefore $K$
is a bombon. By Theorem 8, $K$ is an ellipsoid. ∎

An equivalent result is true for orthogonal projections.

######  Theorem 10.

Let $k$ and $n$ be such that $2\leq k<n$. A convex body in $\mathbb{C}^{n}$
all whose orthogonal projections onto $k$-dimensional planes are ellipsoids is
an ellipsoid.

###### Proof.

Suppose the orthogonal projection of the convex set $K\subset\mathbb{C}^{n}$
onto every $k$-dimensional subspace is an ellipsoid, with $2\leq k<n$. Since
ellipsoids are symmetric, by Theorem 4, $K$ is symmetric. By Lemma 2, it now
suffices to prove that $K$ is a real ellipsoid.

Let $P$ be a $2$-dimensional real plane and let $H$ be a complex $k$-subspace
such that $P\subset H\subset\mathbb{C}^{n}$; this is possible because $k\geq
2$. By hypothesis, the orthogonal projection of $K$ onto $H$ is an ellipsoid.
Then, the orthogonal projection of $K$ onto $P$ is a real ellipse. By Theorem
2.12.5 of [25], $K$ is a real ellipsoid. ∎

Our next purpose is to prove that ellipsoids are the most symmetric bodies in
complex space, in the sense that only ellipsoids have all their hyperplane
sections symmetric. For real ellipsoids, this result was first proved by H.
Brunn [10] under the hypothesis of regularity and in general by G. R. Burton
[12]. It is interesting to note the use of topology in the proof of the
corresponding complex result.

######  Theorem 11.

A convex body all whose hyperplane sections are symmetric is an ellipsoid.

###### Proof.

Let $K$ be a convex body all whose hyperplane sections are symmetric and let
$L\subset\mathbb{C}^{n-1}$ be a line such that $L\cap K\not=\emptyset$. We
shall prove that $L\cap K$ is a disk and the theorem follows from Theorem 8.
For that purpose, it will be enough to prove that there exists a hyperplane
$H$ containing $L$, such that the center of $H\cap K$ lies in $L$.

Suppose, without loss of generality, that $L=\\{(0,\dots,0,z)\mid
z\in\mathbb{C}\\}$, and that the origin is in $L\cap K$. For every
$(n-2)$-plane through the origin $\Gamma\subset\mathbb{C}^{n-1}$, we have that
$\Gamma\cap K\not=\emptyset$. Let $\Gamma^{\prime}$ be the hyperplane of
$\mathbb{C}^{n}$ generated by $\Gamma$ and $L$, and let $x_{\Gamma}$ be the
center of symmetry of $\Gamma^{\prime}\cap K$. If $x_{\Gamma}\in L$, there is
nothing to prove; so we may assume, without loss of generality, that for every
$(n-2)$-plane of $\mathbb{C}^{n-1}$ through the origin $\Gamma$, we have that
$x_{\Gamma}\notin L$.

Let $\pi:\mathbb{C}^{n}\to\mathbb{C}^{n-1}$ be the projection onto the first
$n-1$ coordinates. The choice $\Gamma\to\pi(x_{\Gamma})\in\Gamma$ is a
continuous assignment of a non-zero vector in $\Gamma$, for each hyperplane
$\Gamma$ of $\mathbb{C}^{n-1}$. This is a contradiction to the well known fact
that the canonical vector bundle of hyperplanes through the origin of
$\mathbb{C}^{n-1}$ does not admit a non-zero section. See Steenrod’s Book
[29].

The continuity of $\Gamma\to\pi(x_{\Gamma})$ follows from the following fact:
suppose $H_{i},i\geq 1$, is a sequence of hyperplanes in $\mathbb{C}^{n}$ with
the property that $H_{0}=\lim(H_{i})$ (where limits are taken as
${i\to\infty}$). If $H_{i}\cap K$ is symmetric with centre at $c_{i}$ for
$i\geq 1$, then $H_{0}\cap K$ is symmetric with center at ${c_{0}=\lim
c_{i}}$. To see this, let $\Gamma_{i}=H_{i}-c_{i}$ and $K_{i}=(H_{i}\cap
K)-c_{i}$, for $i\geq 0$. Then, $\Gamma_{0}=\lim\Gamma_{i}$, $K_{0}=\lim
K_{i}$ (using the Hausdorff metric) and $K_{i}$ is symmetric with center at
the origin for $i\geq 1$. This implies for every $\xi\in\mathbb{S}^{1}$ and
$i\geq 1$, that $\xi K_{i}=K_{i}$. Therefore,

$\xi K_{0}=\xi(\lim K_{i})=\lim\xi K_{i}=\lim K_{i}=K_{0}$

which implies that $H_{0}\cap K$ is symmetric with center at $c_{0}$. ∎

##  6\. Convex bodies all whose sections are complex affinely equivalent

In 1932, Stephan Banach conjectured that if all $n$-dimensional subspaces of a
Banach Space $V$ are isometric, $n>1$, then $V$ must be a Hilbert space. Of
course, the Banach Space $V$ can be a linear space over the reals or the
complex numbers. If $V$ is a complex Banach space, this conjecture is
equivalent to the following characterization of the ellipsoid. _A symmetric
convex body $B\subset\mathbb{C}^{n+1}$ all whose hyperplane sections through
the origin are linearly equivalent is an ellipsoid._ This problem has a
positive answer when $n$ is even (Gromov [16]) and when $n\equiv 1$ mod $4$
(Bracho and Montejano [9]). The purpose of this section is to give a brief
summary of the ideas and techniques used in the proof of the following
theorem.

######  Theorem 12.

If all hyperplane sections through the origin of a symmetric convex body
$B\subset\mathbb{C}^{n+1}$ are linearly equivalent, $n\equiv 0,1,2$ mod $4$,
then the convex body $B$ is an ellipsoid.

Given a symmetric convex body $K\subset\mathbb{C}^{n}$, let

$G_{K}:=\\{\,g\in GL_{n}(\mathbb{C})\,|\,g(K)=K,\det g>0\,\\}$

be the group of linear isomorphisms of $K$ with positive real determinant. By
Theorem 7, there exists an ellipsoid of minimal volume containing $K$ centered
at the origin. Suppose now that this minimal ellipsoid is the
$(2n-1)$-dimensional unit ball, then every $g\in G_{K}$ is actually an element
of $SU_{n}$, because it fixes the unit ball, so in this case, $G_{K}=\\{\,g\in
SU_{n}\mid g(K)=K\,\\}\subset SU_{n}$.

Let $B\subset\mathbb{C}^{n+1}$, $n\geq 2$, be a symmetric convex body with
center at the origin all whose hyperplane sections through the origin are
linearly equivalent. By the above, it is not difficult to prove that there
exists a symmetric convex body $K\subset\mathbb{C}^{n}$ with center at the
origin and with the property that every hyperplane section of $B$ is linearly
equivalent to $K$ and moreover $G_{K}\subset SU_{n}$. Indeed, using Topology
of compact Lie groups, in particular, the notion of reduction of the structure
group of a fiber bundle (see [29] and [24]), it is possible to prove that if
$n$ is even, then $G_{K}=SU_{n}$ and if $n\equiv 1$ mod $4$, then
$SU_{n-1}\subset G_{K}$. For the details see [9]. Note that if $n$ is even,
due to the fact that $SU(n)$ is transitive, $G_{K}=SU_{n}$ implies that $K$ is
a ball and hence that all hyperplane sections of $B$ are ellipsoids.
Consequently, by Theorem 11, $B$ must be an ellipsoid. But to understand the
convex geometry of the consequences of the fact that $SU_{n-1}\subset G_{K}$,
when $n\equiv 1$ mod $4$, we need the following definition:

A body of revolution is a symmetric convex body $K\subset\mathbb{C}^{n}$ for
which there exists a $1$-dimensional subspace $L$ of $\mathbb{C}^{n}$, called
its axis of revolution, such that for every affine hyperplane $H$ orthogonal
to $L$, we have that $H\cap K$ is either empty, a single point or a
$(2n-2)$-dimensional ball centered at $H\cap L$. Of course, $K$ is a convex
body of revolution if and only if $SU_{n-1}\subset G_{K}$.

With this in mind, using topology of compact Lie groups, it is possible to
prove the following:

######  Proposition 2.

Let $B\subset\mathbb{C}^{n+1}$, $n\equiv 1$ mod 4, $n\geq 5$, be a symmetric
convex body with center at the origin all whose hyperplane sections through
the origin are linearly equivalent. Then, there exists a body of revolution
$K\subset\mathbb{C}^{n}$ with center at the origin and with the property that
every hyperplane section of $B$ through the origin is linearly equivalent to
$K$.

The next step is to use convex geometry to prove that if all hyperplane
sections of a symmetric body are linearly equivalent to a body of revolution
then at least one of the sections must be an ellipsoid. As we will see, the
proof of this result is an interesting combination of ideas of algebraic
topology and convex geometry. We need first to understand the geometry of
those symmetric convex bodies which are linearly equivalent to a body of
revolution. From now on, let us call those bodies _linear bodies of
revolution_. Thus, a linear body of revolution comes equipped with an axis of
revolution, $L$, which is a line, and a hyperplane of revolution, $H$, which
is a complementary hyperplane (but not necessarily orthogonal) to $L$, and it
satisfies that all its sections with affine hyperplanes $H^{\prime}$ parallel
to $H$ are either empty, a point or an ellipsoid centered at $L$ and
homothetic to the ellipsoid $H\cap K$.

A non-spherical real body of revolution has only one real axis of revolution.
The reason is that between $SO_{n-1}$ and $SO_{n}$ there is no connected
subgroup. The analogy over the complex numbers is false, that is, between
$SU_{n-1}$ and $SU_{n}$ there are several connected subgroups, however, as we
will see next, a non-spherical body of revolution has also a single axis of
revolution.

######  Lemma 9.

A linear body of revolution $K\subset\mathbb{C}^{n}$, $n\geq 3$, admitting two
different hyperplanes of revolution, is an ellipsoid.

###### Proof.

By Theorem 7, let $\mathfrak{E}$ be the unique ellipsoid of minimal volume
centered at the origin containing $K$ and we may suppose, without loss of
generality, that $\mathfrak{E}$ is the unit ball. Since every symmetry of $K$
is a symmetry of the unit ball, our hypothesis now implies that $K$ is a body
of revolution with two different axes of revolution. Let $L_{1}$ and $L_{2}$
be two different lines and let $G_{1}$ and $G_{2}$ be the complex rotation
groups around the axis $L_{1}$ and $L_{2}$, respectively; they are both
conjugate to $SU(n-1)$. Suppose $G$ is a compact subgroup of $SU_{n}$ that
contains both $G_{1}$ and $G_{2}$. We shall prove that the action of $G$ in
$\mathbb{S}^{2n-1}$ is transitive. If this is so, and both $L_{1}$ and $L_{2}$
are axes of revolution of $K$, then $G^{0}_{K}=\\{g\in SU_{n}\mid g(K)=K\\}$,
which is compact because $K$ is a compact convex body, would act transitively
on $\partial K$ and $K$ would be a ball.

Let $P$ be the plane generated by $L_{1}$ and $L_{2}$ and let
$\pi_{1},\pi_{2}$ and $\pi_{0}$ be the orthogonal projections onto
$L_{1},L_{2}$ and $P$, respectively. Furthermore, let $D=P\cap
int(\mathfrak{B})$, where $\mathfrak{B}$ is the unit ball of $\mathbb{C}^{n}$.
Consider the set

$U=\pi^{-1}_{0}(D)\cap\mathbb{S}^{2n-1}\,.$

Note that $U$ is an open connected dense subset of $\mathbb{S}^{2n-1}$ because
$\mathbb{S}^{2n-1}\setminus U=P\cap\mathbb{S}^{2n-1}$ is a $3$-sphere
contained in $\mathbb{S}^{2n-1}$, and since $n\geq 3$, its (topological)
codimension is at least 2.

Let $x\in U$. Our purpose is to construct an open neighborhood $W$ of $x$ in
$U$ such that $W$ is contained in the orbit $G\cdot x$ of $x$ under the action
of $G$ in $\mathbb{S}^{2n-1}$. This will be enough to prove the lemma because
$U$ is a connected open dense subset of $\mathbb{S}^{2n-1}$.

Let $H_{1}=\pi_{1}^{-1}(\pi_{1}(x))$. It is the affine hyperplane orthogonal
to $L_{1}$ and passing through $x$, so that $G_{1}\cdot
x=H_{1}\cap\mathbb{S}^{2n-1}$. Let $W_{1}=H_{1}\cap D$. It is an open disk in
an affine line parallel to the line $L_{1}^{\perp}\cap P$, and observe that
restricted to this affine line ($H_{1}\cap P$), the map $\pi_{2}$ is an affine
isomorphism onto $L_{2}$ because $L_{1}\not=L_{2}$. So that
$W_{2}=\pi_{2}(W_{1})$ is an open subset of $L_{2}\cap D$ that contains
$\pi_{2}(x)$.

Let $W=\pi_{2}^{-1}(W_{2})\cap U$. It is an open neighborhood of $x$ in $U$.
We are left to prove that $W$ is contained in the orbit $G\cdot x$.

Given $y\in W$, let $H_{2}=\pi_{2}^{-1}(\pi_{2}(y))$, so that $G_{2}\cdot
y=H_{2}\cap\mathbb{S}^{2n-1}$. Consider the affine subspace $\Gamma=H_{1}\cap
H_{2}$ of dimension $n-2>0$. By construction, $H_{2}$ intersects $W_{1}$ in a
point, so that $\Gamma$ touches the interior of the unit ball $\mathfrak{B}$.
Therefore, $\Gamma\cap\mathbb{S}^{2n-1}=(G_{1}\cdot x)\cap(G_{2}\cdot y)$ is
not empty. This implies that $G\cdot x=G\cdot y$, so that $y\in G\cdot x$, and
hence $W\subset G\cdot x$. ∎

######  Lemma 10.

Every hyperplane section $\Gamma\cap K$ of a linear body of revolution
${K\subset\mathbb{C}^{n}}$, $n\geq 3$, is a linear body of revolution.
Furthermore, if $H$ is the hyperplane of revolution of $K$, then either
$\Gamma=H$ or $\Gamma\cap H$ is a hyperplane of revolution of $\Gamma\cap K$.

###### Proof.

Without loss of generality, we may assume that $K$ is a body of revolution;
that is, if its axis of revolution is the line $L$, then $H=L^{\perp}$ is the
corresponding hyperplane of revolution and we have that $H\cap K$ is a ball
centered at the origin.

Assume $\Gamma\not=H$. We will prove that $K_{1}=\Gamma\cap K$ is a body of
revolution in $\Gamma$ with hyperplane of revolution $H_{1}=\Gamma\cap H$. Let
$L_{1}$ be the line orthogonal to $H_{1}$ in $\Gamma$; it will be the axis of
$K_{1}$.

Given $H_{1}^{\prime}\subset\Gamma$ parallel to $H_{1}$, we have to consider
the intersection of $H_{1}^{\prime}\cap K_{1}=H_{1}^{\prime}\cap K$. Let
$H^{\prime}$ be the affine hyperplane of $\mathbb{C}^{n}$ parallel to $H$ that
contains $H_{1}^{\prime}$. By hypothesis, we have that $H^{\prime}\cap K$ is
either empty, a point or a ball (in $H^{\prime}$) centered at $L$. Therefore,
its intersection with $H_{1}^{\prime}$ (a hyperplane of $H^{\prime}$) is
either empty, a point or a ball. By construction, in the two last cases the
point or the center of the ball lies in $L_{1}$; indeed, the plane generated
by $L$ and $L_{1}$ is orthogonal to $H_{1}$. Therefore, $K_{1}\subset\Gamma$
is a body of revolution as we wished. ∎

######  Lemma 11.

Let $K\subset\mathbb{C}^{n}$ be a linear body of revolution with axis of
revolution $L$, $n\geq 3$. Suppose $\Gamma\subset\mathbb{C}^{n}$ is a
hyperplane containing $L$ for which $\Gamma\cap K$ is an ellipsoid. Then $K$
is an ellipsoid.

###### Proof.

First, we may assume that $K$ is a body of revolution with axis of revolution
$L$, hyperplane of revolution $H=L^{\perp}$ and such that $H\cap K$ is the
unit ball in $H$. By hypothesis and Lemma 10, $\Gamma\cap K$ is an ellipsoid
and a body of revolution with axis of revolution $L$. Using a linear map which
is the identity on $H$ and a dilatation on $L$, we may assume $\Gamma\cap K$
is a unit ball centered at the origin; so that $\Gamma\cap
K=\Gamma\cap\mathfrak{B}$, where $\mathfrak{B}\subset\mathbb{C}^{n}$ is the
unit ball. Our purpose is to prove that $K=\mathfrak{B}$ to conclude the
proof.

For every affine hyperplane $H^{\prime}$ parallel to $H$ that touches the
interior of $K$, we have that both $H^{\prime}\cap K$ and
$H^{\prime}\cap\mathfrak{B}$ are concentric balls. Furthermore, they have the
same radius because their boundaries have non empty intersection (in
$\Gamma$). Consequently, $H^{\prime}\cap K=H^{\prime}\cap\mathfrak{B}$ and
hence $K=\mathfrak{B}$, as we wished. ∎

From now on, let $B\subset\mathbb{C}^{n+1}$ be a symmetric convex body, $n$
odd, $n\geq 5$, all of whose hyperplane sections are non elliptical, linear
bodies of revolution. First of all note that the assignment of the axis of
revolution in every hyperplane is continuous.

For every line $\ell\subset\mathbb{C}^{n+1}$ denote by $\ell^{\perp}$ the
hyperplane subspace of $\mathbb{C}^{n+1}$ orthogonal to $\ell$. Furthermore,
by Lemma 9, denote by $L_{\ell}$ the unique axis of revolution of
$\ell^{\perp}\cap B$ and by $H_{\ell}$ the corresponding $(n-1)$-dimensional
subspace of revolution of $\ell^{\perp}\cap B$. Note that the line $L_{\ell}$
contains the origin because the center of the ellipsoid $H_{\ell}\cap B$ is
the origin.

######  Lemma 12.

Let $B\subset\mathbb{C}^{n+1}$, $n$ odd, $n\geq 5$, be a symmetric convex
body, all whose hyperplane sections are non elliptical, linear bodies of
revolution. Suppose $\ell_{1}$ and $\ell_{2}$ are two different lines with the
property that $L_{\ell_{2}}\subset\ell_{1}^{\perp}$. Then

$H_{\ell_{2}}\cap\ell_{1}^{\perp}=H_{\ell_{1}}\cap\ell_{2}^{\perp}=H_{\ell_{2}}\cap
H_{\ell_{1}}.$

###### Proof.

Consider $\ell_{1}^{\perp}\cap\ell_{2}^{\perp}$, the $(n-1)$-dimensional
subspace of $\ell_{2}^{\perp}$. By hypothesis, $\ell_{2}^{\perp}\cap B$ is a
non elliptical, linear body of revolution with axis of revolution
$L_{\ell_{2}}$. Therefore, since
$L_{\ell_{2}}\subset\ell_{1}^{\perp}\cap\ell_{2}^{\perp}$, we have that
$\ell_{1}^{\perp}\cap\ell_{2}^{\perp}\cap B$ is a linear body of revolution
with corresponding axis of revolution $L_{\ell_{2}}.$ Furthermore, by Lemma
11, $\ell_{2}^{\perp}\cap B$ is not an ellipsoid. Moreover, the unique
hyperplane of revolution of $\ell_{1}^{\perp}\cap\ell_{2}^{\perp}\cap B$ is
$\ell_{1}^{\perp}\cap H_{\ell_{2}}.$

On the other hand, $\ell_{1}^{\perp}\cap\ell_{2}^{\perp}$ is an
$(n-1)$-dimensional subspace of $\ell_{1}^{\perp}.$ Note that
$\ell_{1}^{\perp}\cap\ell_{2}^{\perp}\not=H_{\ell_{1}}$, otherwise
$\ell_{1}^{\perp}\cap\ell_{2}^{\perp}\cap B=H_{\ell_{1}}\cap B$ would be an
ellipsoid contradicting our previous assumption. Since $\ell_{1}^{\perp}\cap
B$ is a non elliptical, linear body of revolution and
$\ell_{1}^{\perp}\cap\ell_{2}^{\perp}\not=H_{\ell_{1}}$, then by Lemma 10,
$\ell_{1}^{\perp}\cap\ell_{2}^{\perp}\cap B$ is a non elliptical, linear body
of revolution with corresponding subspace of revolution
$H_{\ell_{1}}\cap\ell_{2}^{\perp}.$ Consequently, by Lemma 9, since $n-1\geq
3$, we have that
$H_{\ell_{2}}\cap\ell_{1}^{\perp}=H_{\ell_{1}}\cap\ell_{2}^{\perp}$. ∎

We are ready to prove that

######  Theorem 13.

Let $B\subset\mathbb{C}^{n+1}$ be a symmetric convex body, $n$ odd, $n\geq 5$.
Suppose all hyperplane sections of $B$ are linear bodies of revolution. Then,
one of the hyperplane sections is an ellipsoid.

###### Proof.

The assignment $\ell\to L_{\ell}\subset\ell^{\perp}$ is a continuous function
on $\ell$. We start proving that there are two different lines $\ell_{1}$ and
$\ell_{2}$ such that

$L_{\ell_{2}}\subset H_{\ell_{1}}.$

For that purpose, let $L$ be a line in $H_{\ell_{1}}$ and denote by
$L^{\perp}$ the hyperplane of $\mathbb{C}^{n+1}$ orthogonal to $L$. Denote by
$\Pi:\mathbb{C}^{n+1}\to L^{\perp}$ the orthogonal projection along $L$. Of
course, $\Pi$ is a linear map over the complex numbers $\mathbb{C}$.

For every line $\ell\subset L^{\perp}$, consider
$L_{\ell}\subset\ell^{\perp}$. Suppose $L_{\ell}\not=L$, for every
$\ell\subset L^{\perp}$. If this is so, $\Pi(L_{\ell})$ is a line through the
origin in $L^{\perp}$, ortogonal to $\ell$. Furthermore, the assignment
$\ell\to\Pi(L_{\ell})$ is a continuous function on $\ell$. This is a
contradiction to the well known fact that the canonical vector bundle of
hyperplanes through the origin of $\mathbb{C}^{n-1}$ does not admit a line
subbundle, when $(n-1)$ is even. See Steenrod’s Book [29]. This implies the
existence of $\ell_{2}$ such that $L=L_{\ell_{2}}\subset H_{\ell_{1}}$.

Note now that this is a contradiction to Lemma 12 because clearly
$L_{\ell_{2}}\subset\ell_{1}^{\perp}$, hence

$L_{\ell_{2}}\subset
H_{\ell_{1}}\cap\ell_{2}^{\perp}=H_{\ell_{2}}\cap\ell_{1}^{\perp}\subset
H_{\ell_{2}}$

which is impossible. ∎

Once we know at least one of the sections of our body $B$ has a hyperplane
section through the origin which is an ellipsoid, all hyperplane sections
through the origin are ellipsoids and hence, by Theorem 11, $B$ is an
ellipsoid. This concludes the proof of Theorem 12.

The following, and final, result is again a characterization of ellipsoids by
means of their line sections and following the spirit of the above results. As
the proof is shorter and simpler than the proof of Proposition 2, we include
it in its entirety so that the reader has a more complete idea of the
topological ideas developed in this section.

######  Theorem 14.

A convex body $K$ is an ellipsoid if and only if any two non-trivial line
sections of $K$ are affinely equivalent.

###### Proof.

In view of Theorem 8, it will suffice to prove that every non-trivial line
section of $K$ is a disk.

By hypothesis, there is a fixed convex set $C\subset\mathbb{C}$, such that
every non-trivial line section of $K$ is affinely equivalent to $C$. Without
loss of generality, we may assume that $\mathbb{S}^{1}$ is the circumcircle of
$C$, so that $G_{C}$, _the compact group of affine equivalences of
$\mathbb{C}$ that fix $C$_, is a subgroup of $U(1)$. To see that every non-
trivial line section of $K$ is a disk, we will prove that $C$ is the unit
disk, which is clearly equivalent to proving that $G_{C}=U(1)$.

Consider a point in the interior of $K$ and a 2-dimensional affine subspace
through it. To fix ideas, we may assume that the origin is in the interior of
$K$, and think of $\mathbb{C}^{2}\subset\mathbb{C}^{n}$. The space of all
lines through the origin in $\mathbb{C}^{2}$ is the Riemann sphere
$\mathbb{S}^{2}$. Let

$\xi^{2}:\mathbb{\it E}^{2}\to\mathbb{S}^{2}$

be the canonical vector bundle of $1$-dimensional subspaces of
$\mathbb{C}^{2}$. In view of the continuity of the line sections of $K$, our
hypothesis imply that the structure group of $\xi^{2}$ can be taken to be
$G_{C}$. That is, if $\xi^{2}$ is described by a characteristic map
$\tau:\mathbb{S}^{1}\to U(1)$, it factors through $G_{C}$; (or if it is
described by transition functions and an open cover, the transition functions
can be chosen to go to $G_{C}$). See, e.g., [29].

Since the only proper compact subgroups of $U(1)$ (which can be identified
with $\mathbb{S}^{1}$) are finite, but the canonical line bundle is non
trivial, then $G_{C}=U(1)$. This completes the proof. ∎

Acknowledgments. Luis Montejano acknowledges support from CONACyT under
project 166306 and from PAPIIT-UNAM under project IN112614.

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