DANIEL HUG
ABSTRACT. For a convex bodyK⊂Rn, thekth projection function ofKassigns to any k-dimensional linear subspace ofRnthek-volume of the orthogonal projection ofKto that subspace. LetKandK0be convex bodies inRn, letK0be centrally symmetric and satisfy a weak curvature assumption. Leti, j∈Nbe such that1≤i < j≤n−2with (i, j)6= (1, n−2). Assume thatKandK0have proportionalith projection function and proportionaljth projection function. Then we show thatKandK0are homothetic. IfK0
is a Euclidean ball, we thus obtain characterizations of Euclidean balls as convex bodies having constanti-brightness and constantj-brightness.
1. INTRODUCTION AND STATEMENT OF RESULTS
Nakajima’s problem is concerned with the determination of convex bodies inRnby two projection functions. A convex body in Euclidean spaceRnis a compact convex set with nonempty interior. LetG(n, k)be the Grassmannian ofk-dimensional linear subspaces of Rn,k∈ {0, . . . , n}. Thekth projection functionπk(K)of a convex bodyKis defined by
πk(K) :G(n, k)→R, L7→Vk(K|L),
whereVk(K|L)is thek-volume of the orthogonal projectionK|LofKtoL. For a Eu- clidean ball, all projection functions are constant functions. A converse statement is true for centrally symmetric convex bodies: A centrally symmetric convex bodyKhaving one constant projection functionπk(K), for somek ∈ {1, . . . , n−1}, must be a Euclidean ball.
Examples of non-spherical convex bodies having one constant projection function are well known. Fork = 1these are the bodies of constant width, which have been studied extensively. See the surveys [4], [15]. The casek=n−1of convex bodies of constant brightness has first been studied by Blaschke who constructed smooth bodies of revolu- tion with constant brightness. Further examples, also without rotational symmetry, can be obtained by approximation and convolution arguments, applied to surface area measures of convex bodies (cf. [14]). In the intermediate cases, i.e. for2 ≤ k ≤ n−2, the clas- sical examples of smooth convex bodies with constantkth projection function (bodies of constantk-brightness) are bodies of revolution. The existence of such bodies has been shown by Firey [7] (cf. also [13]). In [12], Goodey and Howard prove the existence of smooth convex bodies with constantk-brightness not having rotational symmetry and ob- tain a parametric description of these bodies, for2≤k≤n−3. They also provide new examples of smooth bodies of revolution having constant brightness, for2≤k≤n−2, by a perturbation argument.
Date: July 12, 2007.
2000Mathematics Subject Classification. 52A20, 52A21, 52A38, 52A40.
Key words and phrases. Constant width, constant brightness, projection function, characterization of Eu- clidean balls.
The author was supported in part by the European Network PHD, FP6 Marie Curie Actions, RTN, Contract MCRN-511953.
1
In its original form, Nakajima’s problem asks whether two constant projection functions are sufficient to characterize balls [19]; see [3], [4], [6], [8], [10], [15]. In dimension three and for smooth convex bodies the affirmative answer has been given by Nakajima himself in the early 20th century. Recently, inR3 the smoothness hypothesis has been removed by Ralph Howard. By a well-known reduction argument, this implies also an affirmative answer for the case of constant width and constant 2-brightness in arbitrary dimensions.
For smooth convex bodies in general dimensions, Nakajima’s problem has been settled in [17] with the exception of two cases. Further results for general convex bodies have subsequently been obained in [18].
In the present paper, we consider a centrally symmetric convex bodyK0 in place of the Euclidean ball. We do not assume any smoothness ofKorK0and settle most of the unresolved cases. The approach developed in [18] and based on [17] is crucial and will be refined further.
Theorem 1.1. LetK, K0be convex bodies inRn, and letK0be centrally symmetric with positive principal radii of curvature on some Borel subset of the unit sphere of positive measure. Let1 ≤i < j ≤n−2be integers with(i, j)6= (1, n−2). Assume that there are positive constantsα, βsuch that
πi(K) =α πi(K0) and πj(K) =β πj(K0).
ThenKandK0are homothetic.
In [17], the cases(1, j)withj < (n+ 1)/2and(1,3)(forn= 5) have been settled.
Thus the cases which remain open are the following:(1, n−2)forn≥6, and(i, n−1) for n ≥ 4, in the general setting. For smooth convex bodies, the cases(1, n−1)and (n−2, n−1)are still unresolved.
The assumption of positive principal radii of curvature on a set of positive measure on the unit sphere is equivalent to requiring that the determinant of the Hessian matrix of the support function ofK0is nonzero on a set of unit vectors of positive measure. The Hessian is defined for almost all unit vectors due to Aleksandrov’s theorem. For instance, the assumption is satisfied, ifK0contains a small smooth piece in its boundary with positive Gauss curvature. However, polytopes do not satisfy the assumption.
The special case whereK0is a Euclidean ball is the following corollary.
Corollary 1.2. LetKbe a convex body inRn. Let1≤i < j ≤n−2be integers with (i, j)6= (1, n−2). Assume thatKhas constanti-brightness and constantj-brightness.
ThenKis a Euclidean ball.
Nakajima’s problem is closely related to the problem of determining a convex body from its projection functions. We say that a convex bodyK0is determined by itskth projection functionπk(K0), if for any convex bodyKwithπk(K) =πk(K0)it follows thatK0is a translate ofKor a translate ofK∗(the reflection ofKin the origin). Continuing work of Schneider [21], Christina Bauer [1] showed that special convex polytopesPare determined by just one projection functionπk(P), where2 ≤ k ≤ n−2. From this she deduced that most convex bodies are determined by theirkth projection function. Fork = 1and k =n−1convex bodies which are determined by theirkth projection function must be centrally symmetric. In striking contrast to these results, Campi [2], Gardner and Volˇciˇc [9], and Goodey, Schneider and Weil [11], [10] constructed examples of non-congruent pairs of convex bodiesK, K0 for whichπk(K) = πk(K0)holds for allk = 1, . . . , n.
The latter authors even exhibit a dense set of convex bodies not determined by all of their projection functions. This clearly shows that in general some additional assumption onK0
such as central symmetry cannot be avoided. A result by Chakerian and Lutwak [5] implies that a centrally symmetric convex bodyK0 is determined by any two of its projection functions. This does not solve Nakajima’s problem, since here the assumption is more restrictive. However, we will use the result from [5] as an important ingredient in our proof.
I am grateful to Paul Goodey and Ralph Howard for showing me their manuscript [12]
prior to publication
2. PREPARATIONS
Leth·,·idenote the scalar product andSn−1the unit ball ofRn. The support function hK of a convex body K ishK(u) = max{hx, ui : x ∈ K},u ∈ Rn. It is positively homogeneous of degree one and convex. By Aleksandrov’s theorem on second order dif- ferentiability of convex functions, the second differentiald2hK(x)in Aleksandrov’s sense exists for almost all (with respect to Lebesgue measure)x ∈Rn. Hered2hK(x)is con- sidered as a linear map ofRn. Homogeneity implies that the second order differential also exists for almost all (with respect to spherical Lebesgue measure) unit vectorsu∈Sn−1. For further explanations and definitions we refer to [20] and [18].
In the following, we writehfor the support function ofKandh0for the support func- tion of a convex bodyK0. Lethbe second order differentiable atu∈Sn−1. The restriction d2h(u)|u⊥of the linear mapd2h(u)to the orthogonal complementu⊥ofuis a selfadjoint linear map with respect to the Euclidean structure. Then−1real and nonnegative eigen- values ofd2h(u)|u⊥are the principal radii of curvature ofK. By Aleksandrov’s theorem, these radii of curvature are defined for almost all unit vectorsu. The product of the radii of curvature ofKin directionuis justdet d2h(u)|u⊥
, the determinant of the Hessian of the support function ofK, whenever the second differential exists.
The determinant of the Hessian of the support function of a convex bodyKis related to the surface area measure Sn−1(K;·)of K of order n−1. The connection can be described as follows. The surface area measure Sn−1(K;·)is a measure on the Borel sets of the unit sphere. It can be defined via a local Steiner formula or as the (n−1)- dimensional Hausdorff measure of the reverse spherical image of Borel sets on the unit sphere (cf. [20]). The Radon-Nikodym derivative of the surface area measure ofKcan be expressed in terms of the support functionhofK, at points of second order differentiability of h. To state a precise result, for a fixed unit vector u ∈ Sn−1 andi ∈ N, we put ωi :=
v∈Sn−1:hv, ui ≥1−(2i2)−1 . Henceωi ↓ {u}, as i → ∞, in the sense of Hausdorff convergence of closed sets. Further we writeσn−1 for spherical Lebesgue measure.
Lemma 2.1. Let K ⊂ Rn be a convex body. Ifu ∈ Sn−1 is a point of second order differentiability of the support functionhofK, then
i→∞lim
Sn−1(K;ωi)
σn−1(ωi) = det d2h(u)|u⊥ .
The connection of the surface area measure Sn−1(K;·)of a convex body K to the projection function of ordern−1ofKis given by the following well-known equation. If uis a unit vector, then
πn−1(K)(u⊥) =1 2
Z
Sn−1
|hv, ui|Sn−1(K;dv),
for all unit vectorsu. We will apply this relation with respect to subspaces ofRnas ambient spaces.
For nonnegative numbersx1, . . . , xn−1andI⊂ {1, . . . , n−1}we define xI :=Y
i∈I
xi
which is defined as 1 ifI=∅.
The following lemma is proved in [12].
Lemma 2.2. Let2≤k≤n−3,b >0, and letx1, . . . , xn−1andy1, . . . , yn−1be positive numbers. Assume thatxI +yI = 2bwheneverI ⊂ {1, . . . , n−1} with|I| =k. Then there is a subsetR⊂ {1, . . . , n−1}with|R|=n−2and there are numbersx, y > 0 such thatxi =xandyi=yfori∈R.
Proof. First, we consider the casek=n−3.
Assume thatx16=x2. Then, fori∈ {3, . . . , n−1},
x1·x3· · ·xˇi· · ·xn−1+y1·y3· · ·yˇi· · ·yn−1 = 2b, x2·x3· · ·xˇi· · ·xn−1+y2·y3· · ·yˇi· · ·yn−1 = 2b.
The notationxˇimeans thatxiis omitted from the product. Subtracting the first equation from the second, we get
(x2−x1)·x3· · ·xˇi· · ·xn−1+ (y2−y1)·y3· · ·yˇi· · ·yn−1= 0.
Hencey16=y2and
x2−x1 y1−y2
= y3· · ·yˇi· · ·yn−1 x3· · ·xˇi· · ·xn−1
,
fori= 3, . . . , n−1. This implies that there is a constantc >0such thatyi =c·xi for i= 3, . . . , n−1. Since
x3· · ·xˇi· · ·xn−1 x1+y1cn−4
= 2b
fori= 3, . . . , n−1andn−1≥4, we conclude thatx:=x3=. . .=xn−1, and therefore alsoy:=y3=. . .=yn−1.
Ifx2=x3, then
xn−3+y2yn−4= 2b.
Moreover, fromx3· · ·xn−1+y3· · ·yn−1= 2bwe obtain xn−3+yn−3= 2b.
We conclude thaty2=y. Thus we can chooseR={2, . . . , n−1}.
Finally, ifx26=x3, then the first part of the proof shows that necessarily x1=x4=· · ·=xn−1 and y1=y4=· · ·=yn−1.
We know thatx3=x4andy3=y4. Hence we can chooseR={1, . . . , n−1}r{2}.
This completes the proof of the special casek=n−3. Now let2≤k≤n−3. Then the cardinality of{x1, . . . , xn−1} is at most 2. In fact, for any numbers1 ≤ i1 < i2 <
i3≤n−1there is a setI⊂ {1, . . . , n−1}with{i1, i2, i3} ⊂Iand|I|=k+ 2. Then xi, yi,i∈I, satisfy the assumptions of the special case, which implies that the cardinality of the set{xi1, xi2, xi3}is at most 2.
Thus (after a permutation of indices) we can assume that there is somel∈ {0, . . . , n− 1}such thatx1=. . .=xl=:x6= ¯x:=xl+1=. . .=xn−1. Assume that2≤l≤n−3.
Thenx1 = x2 6= xn−2 = xn−1. ChoosingI such that{1,2, n−2, n−1} ⊂ I ⊂ {1, . . . , n−1} and|I| =k+ 2 ≥4, we obtain a contradiction by applying the special case already established. Hence there is a setR⊂ {1, . . . , n−1}with|R|=n−2and a numberx >0such thatxi=xfori∈R.
By what we have shown and by the assumption,xk+yI = 2bforI⊂Rwith|I|=k.
HenceyI =yI0for allI, I0 ⊂Rwith|I|=|I0|=k. Since2≤k≤n−3, there is some
y >0such thatyi=yfor alli∈R.
3. PROOF
We can assume thatα= 1. Then by assumption
Vi(K|U) =Vi(K0|U) and Vj(K|L) =β Vj(K0|L)
forU ∈G(n, i)andL∈G(n, j). FixLand considerU ⊂L. ThenK|LandK0|Lbelong to the samei-th projection class with respect toLandK0|Lis centrally symmetric. By a result of Chakerian and Lutwak [5]
Vj(K0|L)≥Vj(K|L)
with equality if and only ifK0|Lis a translate ofK|L. Hence, forL∈G(n, j), Vj(K|L) =β Vj(K0|L)≥β Vj(K|L).
Thusβ≤1with equality if and only ifK0|Lis a translate ofK|Lfor allL∈G(n, j).
The remaining part of the proof is devoted to showing that alsoβ ≥1. Once this has been established, it follows thatK0|Lis a translate ofK|Lfor allL∈G(n, j), and hence K0is a translate ofK.
LetPbe the set of allu ∈ Sn−1such thathandh0(the support functions ofKand K0) are second order differentiable atuand at−uanddet d2h0(u)|u⊥
6= 0. By Alek- sandrov’s theorem and by assumption the setPhas positive spherical Lebesgue measure.
In particular,P6=∅. Hence we can choose a vectoru∈Pwhich will be fixed for the rest of the proof.
Next we chooseW ∈G(n, j+ 1)withu∈Wand putU :=u⊥∩W ∈G(n, j). From Vj((K|W)|U) =β Vj((K0|W)|U)
we get Z
Sn−1∩W
|hv, ui|SjW(K|W;dv) + Z
Sn−1∩W
|hv, ui|SjW(K∗|W;dv)
= 2β Z
Sn−1∩W
|hv, ui|SjW(K0|W;dv),
where the upper indexW indicates that the measure is considered with respect toW as the ambient space. The injectivity of the cosine transform on even measures yields
SjW(K|W;·) +SjW(K∗|W;·) = 2β SjW(K0|W;·).
Sinceu∈P,hK|W,hK∗|W andhK0|W are second order differentiable atuwith respect toW as the ambient space. Hence Lemma 2.1 applied inW yields
det d2hK|W(u)|u⊥∩W
+ det d2hK∗|W(u)|u⊥∩W
= 2β det d2hK0|W(u)|u⊥∩W . To rewrite this relation, we define the linear maps
L(h)(u) :=d2h(u)|u⊥ :u⊥→u⊥, L(h0)(u) :=d2h0(u)|u⊥:u⊥→u⊥. Using the exterior calculus as in [17], [18], we arrive at
∧jL(h)(u) +∧jL(h)(−u) = 2β ∧jL(h0)(u).
SinceL(h0)(u)is an isomorphism, due to our choice ofu∈P, we can define Lh0(h)(u) :=L(h0)(u)−1/2◦L(h)(u)◦L(h0)(u)−1/2. As in [17], [18] we thus obtain
(3.1) ∧jLh0(h)(u) +∧jLh0(h)(−u) = 2β ∧jid, whereidis the identity map onu⊥. In the same way, we also get (3.2) ∧iLh0(h)(u) +∧iLh0(h)(−u) = 2 ∧iid .
In this situation, Lemma 3.4 in [17] implies thatLh0(h)(u)andLh0(h)(−u)have a com- mon orthonormal basis of eigenvectorse1, . . . , en−1, wherex1, . . . , xn−1denote the corre- sponding eigenvalues (relative principal radii of curvature) ofLh0(h)(u)andy1, . . . , yn−1 are the eigenvalues ofLh0(h)(−u). Applying these basis vectors to (3.2) and (3.1), we get the polynomial equations
(xI+yI = 2, |I|=i, xJ+yJ = 2β, |J|=j, (3.3)
whereI, J ⊂ {1, . . . , n−1}.
We distinguish three cases.
Case 1:xi= 0for somei∈ {1, . . . , n−1}. Assume thatx1= 0. Theny1·yJ0 = 2β wheneverJ0⊂ {2, . . . , n−1}with|J0|=j−1. This immediately implies thatyi>0for i= 1, . . . , n−1, and thenyJ0 =yJ00for allJ0, J00⊂ {2, . . . , n−1}with|J0|=|J00|= j−1. Thus we conclude thaty2=. . .=yn−1=:y.
If alsoxi = 0for somei 6= 1, say x2 = 0, then in the same way we gety1 = y3 = . . . =yn−1 =y. Hencey1 =. . . =yn−1 =y. But thenyi = 2andyj = 2βby (3.3), which implies thatβ ≥1.
Ifx2, . . . , xn−1>0, then we infer thatxI+yi = 2wheneverI⊂ {2, . . . , n−1}with
|I|=i. This shows thatxI =xI0 forI, I0 ⊂ {2, . . . , n−1}with1 ≤ |I|=|I0| =i≤ n−3. Clearly, this implies thatx2=. . .=xn−1. Therefore by (3.3) we have
xi+yi= 2, xj+yj= 2β . By Jensen’s inequality,
1 = xi+yi
2 =
xi+yi 2
j/i
≤ (xi)j/i+ (yi)j/i
2 =β ,
i.e.β≥1.
Case 2:yi= 0for somei∈ {1, . . . , n−1}. This is treated as Case 1.
Case 3:x1, . . . , xn−1>0andy1, . . . , yn−1>0. We can apply Lemma 2.2 withk=i ork=j(as appropriate) and combine this information with (3.3) to get
xi+yi= 2, xj+yj= 2β .
Here again we use thatj ≤ n−2 = |R|, whereR is as in Lemma 2.2. By Jensen’s inequality, we deduce thatβ≥1.
As described before, this concludes the proof in all cases that can occur.
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FACHBEREICH MATHEMATIK, UNIVERSITAT¨ DUISBURG-ESSEN, CAMPUS ESSEN, D-45117 ESSEN, GERMANY
E-mail address:daniel.hug@uni-due.de URL:http://www.uni-due.de/∼hm0045
