Download Appendix Plank Problems

Appendix
Plank Problems
Gergely Ambrus
This appendix gives a brief survey of the functional analytic side of the famous plank
problem, together with some new results. We do not discuss the wide variety of discrete
geometric generalisations that have arisen particularly in the last 20 years; they can
be found, for example, in K. Bezdek’s recent survey paper [8]. The main goal here
is to present the so-called non-commutative plank problem that serves as a common
generalisation of the affine and the complex plank problems. En route, Section 2
provides a combinatorial interpretation of Bang’s Lemma expressed in Proposition 8,
while Proposition 9 gives a first step towards generalising the Lemma for different point
sets. In Section 3 we shall see the different approaches to the affine plank problem,
and show that it suffices to prove it for simplices. Finally, Section 4 explains the noncommutative plank problem. We establish the connections with the affine and complex
versions, present methods to transfer the complex non-commutative problem to a real
setting, and discuss the links with relative width measures. The main result of the
section is the proof of the 2-dimensional case provided by Theorem 14.
1
The original plank problem
In 1932, A. Tarski posed the following problem [14], which has become known as
the “plank problem”:
Conjecture 1 (Original plank problem, Tarski, 1932). If a convex body K ⊂ Rd is
covered by a finite number of planks, then the sum of their widths is not less than the
minimal width of K.
Here a plank means the closed region of Rd between two parallel hyperplanes,
whose distance apart is called the width of the plank. For any unit vector u ∈ S d−1 , the
width of K in direction u, denoted by wu (K), is the length of the orthogonal projection
1
of K onto a line of direction u. The minimal width of K, usually denoted by w(K),
is the minimum of these lengths. Therefore w(K) is the minimal width of a plank
containing K, and it immediately follows that a covering can be achieved with the sum
of the widths being w(K).
Although the formulation of Conjecture 1 concerns convex bodies only, we can
pose the question for a larger class of closed subsets of Rd as well; in particular, the
d-dimensional sphere S d−1 plays an important role. Tarski himself gave a proof of
Conjecture 1 for the cases when K is B 2 or B 3 . Both of these results are implied by
the plank problem for S 2 , whose proof relies on a fact that was already familiar to the
ancient Greeks. The proof below is attributed to Archimedes.
Proposition 2. If both boundary planes of a plank L intersect S 2 , then the surface
area measure of L ∩ S 2 depends only on the width of L.
Proof. By cutting the plank into thin “sheets”, we can suppose that the width of it,
say δ, is arbitrarily small. Assuming that the plank has the form L = {x ∈ R3 : a ≤
hx, ui ≤ a + δ} for some unit vector u ∈ S 2 and real number −1 ≤ a ≤ 1 − δ, the surface
area measure of the intersection L ∩ S 2 is approximately
2
{x ∈ S : hx, ui = a} È
δ
1 − hu, vi2
where v is the normal direction to S 2 at a point of the circle {x ∈ S 2 : hx, ui = a}.
√
Since the radius of this circle is 1 − a2 and hu, vi = a, the above formula simplifies to
2πδ, which does not depend on a.
It follows that the normalised surface area measure of the part of S 2 covered by
any plank is at most the half-width of the plank. This fact immediately solves the
plank problem for S 2 and B 3 . For the case of the unit disc B 2 on the plane, embed R2
into R3 and take a unit ball B in R3 such that its central section is B 2 . Denote the
orthogonal projection from R3 onto the plane of B 2 by ϕ. The inverse image of a plank
in this plane under ϕ is a plank in R3 of the same width. If a collection of planks cover
B 2 , then the inverse images of these planks under ϕ cover S 2 , and hence the sum of
their widths is at least 2. Altogether we obtain the following statement.
Corollary 3 (Tarski [14]). If S 2 , B 2 or B 3 is covered by a collection of planks, then
the sum of the widths of these planks is at least 2, and equality can hold only if all the
planks are parallel to each other.
2
Almost 20 years had passed until Th. Bang gave a solution to the original problem
in 1950 ([5], and a somewhat simplified form in [6]). His approach is based on the
following two lemmas.
Lemma 4 (Bang [6]). If K is a convex body in Rd of minimal width m, and the vector
v ∈ Rd has length h < m/2, then (K − v) ∩ (K + v) is a convex body of minimal width
at least m − 2h.
The second lemma below is usually referred to as Bang’s Lemma. The original
formulation by Bang himself was quite complicated. W. Fenchel [11] gave a simplified
and more transparent interpretation to it (see also K. Ball [3]).
Lemma 5 (Bang’s Lemma). Let (ui )n1 be a sequence of unit vectors in Rd and (wi )n1 a
sequence of positive numbers. Then for any sequence (mi )n1 of reals, there exists a point
of the set
B=
¦X
©
εi ui wi : εi = ±1
that is not covered by the open planks
Li = {x ∈ Rd : |hx, ui i − mi | < wi }.
Note that (mi ) describe the shifts of the planks of normal vector ui and width 2wi .
The system B will be referred to the “Bang system” generated by the planks (Li );
plainly, it is an affine image of the discrete cube {−1, 1}n . For any sequence of signs
ε = (ε1 , . . . , εn ) we write p (ε) for the point
P
εi ui wi .
The proof of Lemma 5 is presented in Section 2 with alternative proofs and generalisations.
Once Lemmas 4 and 5 are established, the proof of the original plank problem is
straightforward.
Proof of Conjecture 1. Suppose that the convex body K has minimal width 1
and the system (Li ) of planks covers K with the sum of their widths being less than 1.
Üi ) with
Let wi and ui be the half-width and the normal direction of Li , and choose (w
P
Üi < 1/2 such that wi < w
Üi for every i. Define B = {
w
P
Üi : εi = ±1} as
εi ui w
above. Then the successive application of Lemma 4 guarantees that the intersection
T
{(K − x) : x ∈ B} has minimal width at least 1 − 2
P
Üi > 0, and therefore it is
w
not empty. This in turn implies that if we translate K so that the origin lies in the
3
intersection, then B ⊂ K, and hence Lemma 5 yields the existence of a point of K that
is not covered by the original planks Li .
It is possible to extend the original plank problem to finite or infinite covers by
planks of the unit ball of an arbitrary Hilbert space H . Note that in a complex Hilbert
space H , a plank is defined as the set {x ∈ H : |hx, vi − m| 6 w} for some v ∈ H ,
m ∈ R and w > 0. The proof of Lemma 5 remains valid if the unit vectors (ui )
are chosen from H . It is also possible to drop the finiteness condition: since H is
reflexive, its unit ball is compact in the weak topology, and therefore any cover with the
interiors of planks contains a finite subcover. Moreover, if
P
wi < 1, then the induced
Bang system B is contained in the unit ball of H . Therefore we obtain the following
extension.
Theorem 6. If the unit ball of a Hilbert space is covered by a sequence of planks, then
the sum of their half-widths is at least 1.
We shall see several generalisations of this theorem in different directions. Let us
present the one which is perhaps most relevant here. In real Hilbert spaces, Theorem 6
can be shown to be essentially sharp. However, in complex Hilbert spaces, in the special
case when all the planks are centred, K. Ball proved a much stronger result in 2001,
whose proof is based upon a complex analogue of Bang’s Lemma.
Theorem 7 (Complex plank problem; Ball [4]). Let (vj )n1 be a sequence of norm 1
vectors in a complex Hilbert space and (tj )n1 be a sequence of non-negative numbers
satisfying
X
t2j = 1.
Then there is a unit vector z for which for every k,
|hvk , zi| > tk .
2
Bang’s Lemma
However innocent it may seem, Lemma 5 served as the starting point for wide-
spread generalisations of the plank theorem. Here come three proofs of it; the first is
the most standard one, based on the ideas of Bang [6], and first published in this form
by Fenchel [11].
4
Proof of Lemma 5. Let n be the number of planks. Denote ε = (ε1 , . . . , εn ) and
introduce the function f on the discrete cube {−1, 1}n as follows:
f (ε) =
X
εi εj hui , uj iwi wj − 2
X
i,j
εi mi wi .
(1)
i
Note that the first term in the definition of f is kp(ε)k2 , while the second term is related
to the shifts of the planks.
Choose ε such that f (ε) is maximal; we claim that p(ε) is not covered by the
planks. Indeed, fix k and define ε̃k as
(
εi
ε̃k =
if i 6= k
−εi if i = k.
Then p(ε̃k ) = p(ε) − 2 εk uk wk . Since f (ε̃k ) 6 f (ε), we have
0 6 f (ε) − f (ε̃k ) = 4
X
i
εi εk hui , uk iwi wk − 4wk2 − 4 εk mk wk
€
€
Š
Š
= 4wk εk hp(ε), uk i − mk − wk .
Therefore, using wk > 0, we obtain that
|hp(ε), uk i − mk | > wk
for every k.
In 1959, inspired by the above proof, M. Bognár noticed that the general case
follows from that in which all planks are centred via a nice geometric idea.
Proof of Lemma 5 (Bognár [9]). First we show that if the intersection of all the
central hyperplanes of the planks Li is not empty, then the assertion is true. This
is equivalent to the statement that if the origin lies in the intersection of the central
hyperplanes, i.e. mi = 0 for every i, then in any translate B + y, y ∈ Rd of the Bang
system, there exists a point which is not covered by the interiors of the planks. Choose
the point x ∈ B + y for which kxk is maximal; if x = p(ε) + y, then the inequality
kxk > kx − 2 εk uk wk k
5
holds for every k, and it immediately implies that
|hx, uk i| > wk .
Now for the general case, suppose that Rd is embedded into Rd+1 as a hyperplane
H, and let ` be a line orthogonal to H with direction vector v. Place the origin at
the intersection of H and `, and let Pt = t v ∈ `, where t > 0. Assume that the
system of planks (Li ) ⊂ H covers the convex body K ∈ H. For any t > 0, define the
(d + 1)-dimensional planks Lti ⊂ Rd+1 so that Pt lies on the central hyperplane of Lti ,
and Lti ∩ H = Li . If Bt denotes the Bang system generated by the planks (Lti ), then it
is easy to see that Bt → B as t → ∞. On the other hand, according to the first part,
for each t there exists a point xt ∈ Bt that is not covered by the interiors of the planks
(Lti ). There exists an accumulation point of (xt ) as t → ∞, which is necessarily a point
in B that is not covered by the interior of any of the original planks.
We now prove that the limit process in the above proof naturally generates the
function (1) used in the original proof of Bang’s Lemma. For any t, the point of Bt
that is guaranteed not to be covered by the planks (Lti ) is the one of maximal distance
from Pt . For a fixed sequence of signs ε, let p(ε) and pt (ε) be the corresponding points
in B and Bt respectively. Let pt (ε) = qt (ε) + dt (ε)v, where qt (ε) ∈ H and dt (ε) ∈ R
(remember that v ⊥ H). Then qt (ε) → p(ε) and dt (ε) → 0 as t → ∞. A simple
calculation yields that
dt (ε) =
X
 ‹
εi
mi wi
1
+o
.
t
t
Therefore,
kpt (ε) − Pt k2 = kqt (ε)k2 + (t − dt (ε))2
= kqt (ε)k2 + t2 − 2
X
εi mi wi + o(1).
Since t2 is a constant term for a fixed t, maximising the function kpt (ε) − Pt k2 is
equivalent to maximise the term kqt (ε)k2 − 2
P
εi mi wi + o(1), and taking the limit
when t → ∞ we obtain (1).
Next, we give a new, combinatorial interpretation for Bang’s Lemma. Suppose
that we have a fixed system of n planks L = (L1 , . . . , Ln ) in a Hilbert space H , where
Lk has normal vector uk and half-width wk as usual. Introduce a directed graph G(L )
on the discrete cube {−1, 1}n as vertex set as follows. Let −
ε→
σ be a directed edge iff
6
ε and σ differ only at the kth coordinate, and p(ε) is covered by the interior of the
plank Lk . Note that this condition ensures that p(σ) is not covered by Lk , and that
p(σ) = p(ε) − 2 εk uk wk .
A walk on the graph G(L ) can be interpreted as a series of “jumps”. Start from
an arbitrary point p(ε) ∈ B. If p(ε) is not covered by any of the open planks, then we
stop at p(ε). Otherwise, suppose that p(ε) is covered by Lk , in which case we move to
the point p(σ), where σ differs from ε only at the kth coordinate; in other words, we
“jump out” from the plank Lk . Now, this move is equivalent to a move along the edge
−
ε→
σ in G(L ). If this series of jumps, or what is the same, the walk in G(L ) terminates
at some point, then Bang’s Lemma is proved for this arrangement of the planks. On the
other hand, if the walk never terminates, then G(L ) must contain a cycle. Therefore
the following statement implies Lemma 5.
Proposition 8. For any arrangement L = (Li )n1 of planks in Rd , the induced graph
G(L ) on {−1, 1}n does not contain a directed cycle.
ε→
σ in G(L )
Proof. First, suppose that all the planks are centred. Then for any edge −
we have kp(ε)k < kp(σ)k, and therefore the norms of the image points of a walk in
G(L ) form a strictly increasing sequence; hence, G(L ) does not contain a cycle.
Now in the general case, we use the idea of Bognár. Embed Rd into Rd+1 as
before. Let Gt be the graph induced by the system of planks (Lti ). For any t, Gt does
not contain a directed cycle. There exists an infinite subset (tj ) ⊂ N for which the
graphs Gtj are the same, say G. Since tj → ∞, we have that Btj → B, (Lti ) → (Li ),
and therefore E(G(L )) ⊂ E(G) (note that strict inclusion can occur). Therefore the
graph G(L ) does not contain a cycle either.
An alternative proof to Proposition 8 can be given by using the function (1). A
simple calculation reveals that whenever −
ε→
σ is an edge in G(L ), then f (ε) < f (σ), and
therefore the f -values of the vertices of a path form a strictly increasing sequence.
The natural question arises here: for what point sets is the assertion of Lemma 5
true? Such examples that we know so far are translates of B and scaled copies of B
with factor at least 1. The following statement provides a further generalisation.
Proposition 9. Let (Li ) be a sequence of planks in Rd with normal vectors (ui ) and
half-widths (wi ). Let B be the Bang system generated by (Li ), and suppose that the set
B is the image of B under the map ϕ : Rd → Rd . If for any x, y ∈ B and for any k,
7
|hx − y, uk i| > 2wk implies |hϕ(x) − ϕ(y), uk i| > 2wk , then there is a point of B that is
not covered by the interiors of the planks (Li ).
Proof. Suppose that B is covered by the interiors of (Li ). For any k, let Xk ⊂ B
be the set of points covered by int Lk . The diameter of the projection of Xk onto a
line parallel to uk is less than 2wk . The condition ensures that the diameter of the
projection of ϕ−1 Xk is less than 2wk as well; therefore ϕ−1 Xk can be covered by the
interior of a translate of Lk . Hence B can be covered by the interiors of translates of
(Li ), which is in contradiction with Lemma 5.
Unfortunately, the number of inequalities to be satisfied by B is very large, and
therefore the limits of the applicability of Proposition 9 are uncertain.
3
The affine plank problem
Having solved the original plank problem, Bang also proposed a more natural
variant of it at the end of his 1951 article [6]. The advantage of the new problem is that,
unlike the original version, it is invariant under affine transformations. To formulate
it, we need a definition: for a convex body K and a plank L with normal direction u,
the relative width of L with respect to K is w(L)/wu (K). The affine invariant version
goes as follows.
Conjecture 10 (Affine plank problem, Bang, 1951). If a convex body K ⊂ Rn is
covered by a finite number of planks, then the sum of their relative widths is not less
than 1.
There had been very few affirmative results about the affine plank problem for
about 40 years. Bang proved the conjecture in the special case of two planks in [7], and
later proofs were given by Moser [13] in 1958, Alexander [1] in 1968 and Gardner [12] in
1988. Alexander [1] proved that the conjecture is equivalent to the following statement,
which in turn is a generalisation of an observation of Davenport [10] connected to the
theory of Diophantine approximation.
Conjecture 11. Given a convex body K in an Euclidean space and n hyperplanes,
there exists a translate of
1
n+1 K
inside K, whose interior does not meet any of the
hyperplanes.
8
For a long time, the most likely way of proving Conjecture 10 was believed to
be similar to the proof of Corollary 3 by Tarski. This method would involve finding a
probability measure on K, such that the measure of the intersection of the convex body
K with a plank is the relative width of the plank, as long as both boundary hyperplanes
of the plank intersect K. Such a measure µ is called a relative width measure with
respect to K, using the terminology of Gardner [12]. If the normal directions of the
covering planks are restricted to belong to a set Θ, then µ is a relative width measure
on K for Θ. By using affine invariance and a simple approximation technique, it can be
assumed that the normal directions of the planks are pairwise orthogonal. Therefore
it would suffice to construct relative width measures on K for the set of coordinate
directions in Rn .
Building a relative width measure for one direction is simple: Let ` be a line of
this direction, ϕ be the projection onto ` (hence ϕ(K) is a segment), and define the
measure µ on K as µ(A) = |ϕ(A)|/|ϕ(K)|. On the other hand, Proposition 2 implies
that relative width measures for all directions exist on ellipsoids in R3 and on ellipses
in R2 and can be obtained as the affine transformation of the normalised surface area
measure on S 2 or its normalised projection to B 2 – the resulting measures are called
canonical on the ellipsoid and the ellipse.
The quest for relative width measures had been partially ended by Gardner [12].
On the one hand, he proved that for any convex body K and any set of at most
two allowed directions, there exists a relative width measure. On the other hand, he
showed that such measures in general do not necessarily exist. In particular, if the set
of possible directions of planks is “ infinite”, that is, every analytic function vanishing
on each line through the origin parallel to the normal directions is identically zero, then
the only possible relative width measures are essentially the one dimensional Lebesgue
measure restricted to a segment, or the canonical relative width measures on the ellipse
and the ellipsoid.
This negative result still leaves open the possibility of constructing relative width
measures for a finite set of directions, that would suffice for proving Conjecture 10.
However, as one of Gardner’s examples shows, if K is the triangle in R2 with vertices
(0, 0), (1, 0) and (0, 1), then there is no relative width measure on K for planks that are
parallel to one of the sides of K. Nevertheless, the plank theorem in this setting is true,
because a suitable translate of the Bang system B sits inside K. This indicates that
9
relative width measures cannot provide the way for tackling the affine plank problem
in general, a fact that will have consequences in Section 4.
In 1991, K. Ball [2] managed to prove Conjecture 10 in the case when K is symmetric, and thus he solved the most important part of this long-open problem. His
approach was to extend Theorem 6 for real Banach spaces; this is sufficient, since every
symmetric convex body K can be seen as the unit ball of the Banach space given by
the norm generated by K. Planks in a Banach space X can be described by linear
functionals: if φ is a unit functional in X ∗ , m is a real number and w is a positive
number, then the set
{x ∈ X : |φ(x) − m| 6 w}
is a plank in X of half-width w. The statement then goes as follows.
Theorem 12 (Ball [2]). If the unit ball of a real Banach space is covered by a (countable)
collection of planks, then the sum of the half-widths of these is at least 1.
For reflexive spaces, the infinite-dimensional result follows immediately from the
finite-dimensional one; in the general case, a careful argument [2] can be used to establish the link between the two.
The basic idea of the proof of Theorem 12 is the following. Suppose that the
number of planks is finite, say n. Let the planks be defined by the unit functionals
(φi )n1 , and let xi ∈ BX such that φi (xi ) = 1, where BX is the unit ball of X. Define
the n × n matrix A = (aij ) by
aij = φi (xj ).
The definition yields that aii = 1.
When X is a real Hilbert space, and φi = hui , .i, then xi = ui , aij = hui , uj i,
hence A is symmetric. If x =
P
i aki ci
P
ci xi for a sequence of scalars (ci )n1 = c, then φk (x) =
= (Ac)k . Therefore Bang’s Lemma can be reformulated as follows:
If A = (aij ) is a real, symmetric matrix with 1’s on the diagonal, (mi ) is a sequence
of reals and (wi ) a sequence of positive numbers, then there exists a choice of signs
ε = (εi ) such that
X
aki εi wi − mi > wk
i
for every k.
10
In the Banach space setting, the matrix A is not symmetric. Nevertheless, an
ingenious method is used to modify and symmetrize the matrix A, hence translating
the problem to a setting where Bang’s Lemma can be used.
Now we return to Conjecture 10. When K is non-symmetric, the Banach space
method cannot be used. As mentioned earlier, it can be assumed that the planks are
pairwise orthogonal, and therefore the set Θ of their normal directions forms a basis
of Rn . We can also suppose that K ⊂ Rn , and that the width of K is 1 in directions
belonging to Θ, hence the relative and the euclidean widths of each plank are the same.
Next we show that it suffices to prove Conjecture 10 for simplices. Assume that
K ⊂ Rd is covered by a collection of planks (Li )d1 where Li is given by
Li = {x ∈ Rd : |hx, ui i − mi | 6 wi },
and the width of K in each direction ui is 1. The goal is to prove
P
wi > 1/2. For
each i, let x2i and x2i+1 be points of K such that
hx2i − x2i+1 , ui i = 1.
We search a point of the form x =
P
ci xi with
P
ci = 1 and ci > 0 for every i, such
that |hx, uk i − mk | > wk for every k. Because of convexity, x ∈ K. The condition is
equivalent to
X
ci hxi , uk i − mk > wk
(2)
i
2d and ũ = (hx , u i)2d ∈ R2d . Then (2) is equivalent
for every k. Define c = (ci )2d
i k 1
k
1 ∈R
to
|hc, ũk i − mk | > wk ,
where c runs over a (2d − 1)-dimensional simplex T embedded in R2d . Define the
planks L̃k by L̃k = {x ∈ R2d : |hx, ũk i − mk | 6 wk }. It is easy to check that the set
{hc, ũk i : c ∈ T } has length 1, and therefore the relative width of L̃i with respect to T is
2wi ; hence the original covering is equivalent to a covering of T with the collection (L̃i ).
11
4
The non-commutative plank problem
The two main directions of generalising the original plank problem (Theorem 6) are
the affine plank problem (Conjecture 10) and the complex plank problem (Theorem 7).
The following statement, proposed by K. Ball, would serve as a common extension of
both results. Recall that the numerical range of an n × n complex matrix H is the set
{z ∗ Hz : z ∈ Cn , kzk = 1}.
If H is Hermitian, then this set is a segment of the real axis bounded by the smallest
and largest eigenvalues of H, and the length of this segment is called the numerical
range of H.
Conjecture 13 (Non-commutative plank problem, Ball). Let (Hk )n1 be a sequence of
d × d Hermitian matrices with numerical range 2 and (tk )n1 a sequence of positive reals
with
P
tk = 1. Then there exists a unit vector z ∈ Cd for which
|z ∗ Hk z| > tk
for every k.
Let us first see how the complex and affine plank problems follow from the noncommutative version.
In order to derive Theorem 7, suppose that (vi )n1 is a sequence of unit vectors in
Cd , (si )n1 a sequence positive numbers, and the centred planks
Li = {z ∈ Cd : |hz, vi i| 6 si }
cover the unit ball of Cd (and hence the unit sphere as well). The goal is to show that
P 2
si > 1.
For each k, define the d × d Hermitian matrix H̃k by
(H̃k )ij = (vk )i (vk )j = h(vk )i , (vk )j i
and let
Hk = H̃k − Id s2k /2,
12
where Id is the d-dimensional identity matrix. For any unit vector z = (zi )d1 in Cd ,
z ∗ Hk z =
X
z¯i (Hk )ij zj −
i,j
X
2
s2
s2k
s2
=
zj (vk )j − k = |hz, vk i|2 − k ,
2
2
2
j
and therefore the numerical range of each Hk is 1.
If
P 2
si < 1, then applying Conjecture 13 with ti = s2i /2 implies that there exists
a unit vector z ∈ Cd , such that
|z ∗ Hk z| >
s2k
2
for every k. Since z ∗ Hk z > −s2k /2, this yields that |hz, vk i| > sk for every k, and thus
the system (Li ) does not cover the unit ball of Cd , a contradiction.
Next, we show the link to the affine plank problem. Let K be the n-dimensional
simplex covered by the system of planks (Lk ), and embed K into Rn+1 :
K = {x ∈ Rn+1 : x1 + . . . xn+1 = 1 , xj > 0 ∀j}.
We can assume that K is covered by the system (Lk ) of (n + 1)-dimensional planks in
Rn+1 , where the normal of each Lk is parallel to the affine subspace of K. Hence Lk
can be defined by
Lk = {x ∈ Rn+1 : |hx, uk i − mk | 6 wk },
where mk ∈ R, wk > 0, and uk ∈ Rn+1 , uk ⊥ (1, . . . , 1), and the magnitude of uk is
chosen such that the length of {hx, uk i : x ∈ K} is 2. This means that the difference
between the largest and smallest coordinates of uk is 2.
The goal is to show that
P
wk > 1. Suppose on the contrary that it is not true,
and for every k define the (n + 1) × (n + 1) diagonal matrices Hk by
(Hk )jj = (uk )j − mk .
The numerical range of Hk in this case is just the difference between the smallest
and largest diagonal entries, which is 2. Therefore we can apply Conjecture 13 yielding
a complex unit vector z ∈ Cd with |z ∗ Hk z| > tk . Let z = (zk )d1 and xk = |zk |2 . Since
P
|zk |2 = 1, we have x = (xk )d1 ∈ K, and for every k,
X
wk = tk 6 |z ∗ Hk z| = |zj |2 ((uk )j − mk ) = |hx, uk i − mk |.
13
Therefore, the d-dimensional affine plank problem follows from the 2d-dimensional noncommutative plank problem.
Next, we establish a connection that proves Conjecture 13 in the 2-dimensional
case via Corollary 3.
Theorem 14. The d = 2 case of the non-commutative plank problem (Conjecture 13)
is equivalent to the original plank problem on S 2 .
Proof. Suppose that the 2 × 2 Hermitian matrix H has the form
„
H=
a+b
u
ū
a−b
with a, b ∈ R. The eigenvalues of H are a ±
Ž
È
b2 + |u|2 , and therefore the numerical
range of H is 2 iff
b2 + |u|2 = 1.
(3)
Furthermore, if z = (z1 , z2 ) ∈ C2 , then
z ∗ Hz = |z1 |2 (a + b) + |z2 |2 (a − b) + 2 <(z¯1 uz2 ).
Therefore the 2-dimensional case of Conjecture 13 can be formulated as follows:
Let (ak )n1 and (bk )n1 be sequences of reals, (tk )n1 be a sequence of positive reals with
P
tk = 1, and (uk )n1 be a sequence of complex numbers such that b2k + |uk |2 = 1 holds
for every k. Then there exists a 2-dimensional complex unit vector z = (z1 , z2 ), such
that for every k,
2
2
|z1 | (ak + bk ) + |z2 | (ak − bk ) + 2 <(z¯1 z2 uk ) > tk .
(4)
Notice that the since the value of (4) is independent of rotations of z (that is, multiplication by a complex scalar of modulus 1), it can be assumed that z1 is a positive real
14
number, say, r. Then |z2 | =
√
1 − r2 , and
|z1 |2 (ak + bk ) + |z2 |2 (ak − bk ) + 2 <(z̄1 z2 uk )
= ak + (2r2 − 1)bk + 2r <(z2 )<(uk ) − 2r =(z2 )=(uk )
¬
¶
= ak + (2r2 − 1, 2r <(z2 ), −2r=(z2 )), (b, <(uk ), =(uk ))
= ak + hz, uk i.
Here both z and uk are unit vectors on S 2 , as it follows from the equations (3) and
kzk2 = (2r2 − 1)2 + 4r2 |z2 |2 = 1.
Moreover, the set of possible z’s cover S 2 . Therefore the problem reduces to finding a
unit vector z on S 2 with
| ak + hz, uk i| > tk
for every k; and this is a case of the original plank problem on S 2 .
The approach of Theorem 14 of transforming the complex problem into a real
setting can be carried out in general as follows.
Suppose that z = (zk )d1 is a unit vector in Cd . Let H be a d × d Hermitian matrix,
and write hij = (H)ij . Then
z ∗ Hz =
X
hkk |zk |2 + 2
k
=
X
X
<(hjk z̄j zk )
j<k
hkk |zk |2 + 2
k
−2
X
X
€
Š
<(hjk ) <(zj )<(zk ) + =(zj )=(zk )
j<k
€
Š
=(hjk ) <(zj )=(zk ) − =(zj )<(zk )
j<k
= hh, zi
where
€
Š
√
√
h = h11 , . . . , hdd , . . . , 2 <(hjk ), − 2 =(hjk ), . . .
and
€
Š
√
√
z = |z1 |2 , . . . , |zd |2 , . . . , 2(<zj <zk + =zj =zk ), 2(<zj =zk − =zj <zk ), . . .
with 1 6 j < k 6 d.
15
(5)
2
Both h and z are points in Rd . Furthermore, z is a unit vector, since
2
kzk =
X
k
4
|zk | + 2
X
2
2
|zj | |zk | =
X
j<k
!
2
|zk |
= 1.
k
The set of z’s arising from unit vectors z ∈ Cd by this transformation hence forms a
subset of S d
2 −1
. This subset is a (2d − 1)-dimensional manifold, because the real and
imaginary parts of the complex numbers z1 , . . . , zd give a parametrization for it, and
the sum of these coordinates equals 1. Therefore the non-commutative plank problem
2 −1
is equivalent to the original plank problem on a subset of S d
.
Since z ∗ Hz is invariant under rotations of z, it can be assumed that z1 is a positive
real r. Then |z2 |2 + · · · + |zd |2 = 1 − r2 , and (5) can also be written as
z ∗ Hz = hh, zi,
where
€
h = h11 − h22 − · · · − hdd , <(h12 ), −=(h12 ), . . ., <(hjk ), −=(hjk ), . . .
Š
(1 6 j < k 6 d)
and
€
z = 2r2 − 1, 2r<(z2 ), 2r=(z2 ), . . . , 2r<(zd ), 2r=(zd ),
. . . , 2(<zj <zk + =zj =zk ), 2(<zj =zk − =zj <(zk )), . . .
with (2 6 j < k 6 d); here h, z ∈ Rd
2 −d+1
Š
. The interesting property of this transfor2 −d+1
mation is the following. Let ϕ denote the orthogonal projection of Rd
onto the
subspace generated by the first 2d − 1 coordinate vectors. Then
kϕ(z)k2 = (2r2 − 1)2 + 4r2 (|z2 |2 + · · · + |zd |2 ) = 1,
moreover, it is easy to see that the set of ϕ(z)’s arising from the unit vectors z ∈ Cd is
exactly S 2d−2 . Since the dependence between the first 2d − 1 and the other coordinates
of z is not described by linear relations, we obtain that if d > 2, then the d-dimensional
non-commutative plank problem is not equivalent to the original plank problem on
16
S 2d−2 by means of this transformation (and probably in general, there is no equivalence
between the two).
The proof of the non-commutative plank problem in the 2-dimensional case provided by Theorem 14 can be understood as constructing a relative width measure on
the unit sphere of C2 . The question immediately arises: is this a possible way of proving the conjecture in the general case? The results of Gardner show that relative width
measures do not exist for convex bodies in Rd , but this does not refer to the sets obtained by transforming the non-commutative problem to the real setting, since these
are not necessarily convex.
Nevertheless, we have seen that the 2d-dimensional non-commutative problem implies the d-dimensional affine plank problem. If such a measure exists, then it would
transfer to the real setting as well. In the case of the 2-dimensional complex problem,
the constructed measure transfers to the Lebesgue measure on a segment, which indeed
proves the 1-dimensional affine plank problem. However, as has been noted in Section 3,
there is an example with 3 directions when there is no relative width measure. This
implies that the non-commutative problem cannot be solved by constructing a suitable
measure, if the number of dimensions is at least 6.
17
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[3] K. M. Ball, Convex Geometry and Functional Analysis, in: Handbook of the
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18