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Advances in Mathematics
www.elsevier.com/locate/aim
Sharp affine isoperimetric inequalities for the volume decomposition functionals of polytopes
✩Yude Liu, Qiang Sun,Ge Xiong∗
SchoolofMathematicalSciences,TongjiUniversity,Shanghai,200092,PRChina
a r t i c l e i n f o a b s t r a c t
Articlehistory:
Received26January2021
Receivedinrevisedform4May2021 Accepted28June2021
Availableonlinexxxx
CommunicatedbyErwinLutwak
MSC:
52A40 52B60
Keywords:
Polytope Cone-volume
Affineisoperimetricinequality Subspaceconcentrationcondition
The nth power of the volume functional Vn of polytopes P in Rn, according to dimensions of the spaces spanned by any n unit outer normal vectors of P, is decomposed into n homogeneous polynomials of degree n. A set of newsharp affineisoperimetric inequalitiesfor these volume decomposition functionals in R3 are established, which essentiallycharacterizethegeometricandalgebraicstructures ofpolytopes.
©2021ElsevierInc.Allrightsreserved.
1. Introduction
Thesetting for this paper isthe n-dimensionalEuclidean space, Rn. A convex body isacompact convexset thathas anonempty interior. Denote byKno theset of convex
✩ ResearchoftheauthorswassupportedbyNSFCNo.11871373.
* Correspondingauthor.
E-mailaddresses:[email protected](Y. Liu),[email protected](Q. Sun), [email protected](G. Xiong).
https://doi.org/10.1016/j.aim.2021.107902 0001-8708/©2021ElsevierInc. Allrightsreserved.
bodies inRn with the origin o in their interiors. A polytope in Rn is the convex hull of a finite set of points in Rn provided it has positive volume Vn (i.e., n-dimensional Lebesgue measure). The convex hullof asubset of these points is called a face of the polytopeifitliesentirelyontheboundaryofthepolytopeandifithaspositive(n−1)- dimensionalLebesguemeasure.WritePon fortheset ofpolytopesinRn withtheorigin intheirinteriors.
Let P ∈ Pon and ubeaunitouternormal vectortoaface F ofP.Thecone-volume VP({u}) of P associated with u is the volume of the convex hull of the origin o and face F.Thesimplest formofcone-volumeisreducedto theareaformulaoftrianglesin ancient geometry.
However, it is strikingthatcombining the notionsof cone-volume of polytopes, lin- ear independence of vectors and dimension of spaces, a new geometry of polytopes is emerged: Thenth powerof volumeof polytopes inRn is naturallydecomposedinto n terms, and each term is a homogeneous polynomial of degree n. See Theorem 3.1 for details. Inthefollowing,weintroducetheso-called volumedecomposition functional of polytopes.
Definition1.1. SupposeP ∈ Pon,andu1,u2,. . . ,uN aretheunitouternormalvectorsof itsfaces.WedefinethekthvolumedecompositionfunctionalXk(P),k= 1,2,. . . ,n−1,n, by
Xk(P)n =
dim(span{ui1,...,uin})=k
VP({ui1})VP({ui2})· · ·VP({uin}).
As usual, span{ui1,. . . ,uin} in the above definition denotes the linear subspace spanned by n normal vectors ui1,. . . ,uin of the polytope P. Obviously Xk(P)n is a homogeneous polynomialincone-volumesofdegree n, k= 1,2,. . . ,n;Xk(P) iscentro- affine invariant, i.e., Xk(T P) = Xk(P) for T ∈ SL(n); and Xk(λP) = λnXk(P) for λ>0.
It isinterestingthatifk=n,then Xn(P) = (
ui1∧···∧uin=0
VP({ui1})VP({ui2})· · ·VP({uin}))n1,
whichis identical tothefunctionalU introducedbyE.Lutwak, D.Yang andG.Zhang (LYZ) [22] toattackthe longstandingunsolvedSchneider projection probleminconvex geometry;Fork=n,allthefunctionalsXk(P) aretotallynew.
It is interestingthatthe nth powerof thevolumefunctional Vn(P) satisfiesthefol- lowing tidyidentity
Vn(P)n =X1(P)n+X2(P)n+· · ·+Xn(P)n, (1.1) which says that the nth power of volume Vn(P) of a polytope P is decomposed into n homogeneous polynomials Xk(P)n, k = 1,2,. . . ,n. It seems that the investigation
ofthese polynomials are helpfultofind outthedistributionof cone-volumesand outer normalvectorsofthepolytopeP.Therefore,thegeometricoralgebraicstructureofthe polytopeP ismoreclearlyvisualized.
Moreover,theidentity(1.1) suggeststhatthesenewfunctionalsXk,k= 1,2,. . . ,n−1, arecomplementary totheLutwak-Yang-ZhangU functional.So,insomesensewetrace theoriginofLYZ’sU functionalforthefirsttime.
Fortheproofofidentity(1.1),please referto Theorem3.1inSection3.
Atthis moment, we would dwell onthe story about thenth volume decomposition functionalXn,whatcertainlymirrorsthesignificanceof thesetofnewfunctionals Xk, k= 1,2,. . . ,n−1.
In2001, LYZ[22] conjectured thatifP is apolytopeinRn with its centroidat the origin,then
Xn(P)
Vn(P) (n!)n1
n (1.2)
withequalityifandonlyifP isaparallelotope.
It took more than one dozen years to completely settle this conjecture. The LYZ conjecture for origin-symmetric polytopes was firstly solved by He, Leng and Li [12], with an alternate proof given by Xiong [28]. The LYZ conjecture in R2 and R3 was alsoconfirmed in[28].A final solutionwasattributedto Henkand Linke[13].In 2016, BöröczkyandHenk[2] provedthatLYZ’sconjectureisalsoaffirmativeforconvexbodies.
ItisworthmentioningthatintheprocessofsolvingtheLYZconjecture,anessential
“concentration phenomenon” of cone-volumes of polytopes was discovered: If P is a polytope in Rn with its centroid at the origin and the unit outer normals of P are u1,u2,. . . ,uN, thenforeachsubspaceξ⊆Rn,
ui∈ξ
VP({ui})≤ dimξ
n Vn(P) (1.3)
withequalityforasubspaceξifandonlyifthereexistsasubspaceξ complementaryto ξinRn,suchthat{uj:uj ∈/ξ}⊆ξ.
In2013,Böröczkyand LYZ[4] originallyposedthe notionof subspaceconcentration condition (seeSection 2for itsdefinition) forfinite Borelmeasures on the unitsphere Sn−1,andprovedthatthisconditionissufficientandnecessarytoguaranteetheexistence ofsolutionstoevenlogarithmicMinkowskiproblems.
For more investigations and applications on subspace concentration condition, we referto[3,7,5,8,17,19,27,31].In[15],theauthorspointedoutthatsubspaceconcentration conditionisalsoconnectedtotheYau-Tian-Donaldsonconjectureinalgebraicgeometry.
BacktothenewvolumedecompositionfunctionalsXk,k= 1,2,. . . ,n−1,inlightof theidentity(1.1),wearetemptedto raisethefollowingproblem.
Problem X.LetP be apolytopeinRn withitscentroidat theorigin.Doesthere exist aconstantc(n,k) dependingonnandk,k∈ {1,2,. . . ,n−1},suchthat
Xk(P)
Vn(P) ≤c(n, k)?
Inthisarticle,theProblem XinR3is satisfactorilysolved.
Theorem 1.1.If P isapolytope inR3 withitscentroid attheorigin,then X1(P)
V3(P) ≤1 3
23
, X2(P) V3(P) ≤2
3 13
, and X3(P) V3(P) ≥ 213
323, and equalityholds ineach inequalityifand onlyif P isaparallelepiped.
Theorem 1.2.If P isapolytope inRn with itscentroidattheorigin, then X1(P)
Vn(P) ≤n1n−1 with equalityifand onlyif P aparallelotope.
RestrictedtoP3n,i.e.,thesetofpolytopesinRnwhoseanythreeouternormalvectors (uptotheirantipodalnormalvectors)arelinearindependent,weprovethefollowing.
Theorem 1.3.If P isapolytope inP3n with itscentroidattheoriginandn≥3,then X2(P)
Vn(P) ≤nn1−1[(2n−1−1)(n−1)]n1 with equalityifand onlyif P aparallelotope.
Thisarticleisorganizedasfollows.InSection3,weinvestigatebasicpropertiesofthe functionals Xk anddemonstratewhypolytopesshouldbe“positioned”withcentroidat theoriginintheProblemX.ForpolytopesPin(k+ 1)-generalposition,weestablishthe inversion formulaforXk(P),whichresemblestheclassicalinclusion-exclusionprinciple inspirit.Inthethirdpart,weverifythatX3 doesnot attain itsmaximumat parallelo- topesinR4,whichsuggeststhecomplexity oftheProblemXinhigherdimensions.
In Section4, we provetheTheorem 1.2 byusing themaximum principle forconvex functions. By solvingtwo constrained optimizationproblems, we present theproofs of Theorem 1.1 and Theorem 1.3 in Section 5 and Section 6, respectively. Meanwhile, delicateclassificationsviageometryofsubspacesareemployed.
Intheappendix,weshowthatthesetPkn,k= 1,2,. . . ,n,isdenseinKno inthesense of Hausdorffmetric.
2. Preliminaries
WritePcn and Psn forthe set ofpolytopesinRn with centroidat theoriginand the set of polytopes inRn symmetricwith respect to the origin,respectively. LetGn,k be thesetofk-dimensionalsubspacesofRn.
The standardinner product of the vectors x,y ∈ Rn is denotedby x·y. We write
|x|2=x·x,andSn−1={x∈Rn :|x|= 1}fortheboundaryoftheEuclideanunitball B inRn.TheletterμwillbeusedexclusivelytodenoteafiniteBorelmeasureonSn−1. Forsuchameasure μ,wedenotebysuppμitssupportset.
ThesupportfunctionhK :Rn →Rn ofaconvexbodyK isdefined,forx∈Rn,by hK(x) = max{x·y :y∈K}.
Observethatsupportfunctionsare positivelyhomogeneous ofdegreeoneandsubaddi- tive.
ThesetKon isoftenequippedwiththeHausdorffmetric δ.ForK,L∈ Kno, δ(K, L) = sup
u∈Sn−1|hK(u)−hL(u)|. AhyperplaneofRn canbe writtenintheform
Hu,α={x∈Rn:x·u=α}
withu∈Rn\{o}andα∈R.ThehyperplaneHu,α boundsaclosed halfspace Hu,α− ={x∈Rn:x·u≤α}.
Cone-volumemeasureisanaturalgeneralizationofcone-volumeofpolytopestocon- vexbodies.ForK∈ Kno,itscone-volumemeasure VK isafiniteBorelmeasureonSn−1, definedforeachBorelset ω⊆Sn−1by
VK(ω) = 1 n
x∈ν−1K(ω)
x·νK(x)dHn−1(x), (2.1)
where νK : ∂K → Sn−1 is the Gauss map of K, defined on ∂K, theset of points of
∂K thathave auniqueouter unitnormal, and Hn−1 is (n−1)-dimensionalHausdorff measure.
Cone-volume measures have appeared in [1,9–11,20,21,23–26,29,30], and have been intensivelyinvestigatedinrecentyears.See,e.g.,[6,2,3,7,5,8,13,14,16,28].
FollowingBöröczkyandLYZ[4], wepresentthedefinitionofsubspaceconcentration condition and the celebratedBöröczky-LYZ existence theorem of solutionsto the even logarithmicMinkowskiproblem.
Definition 2.1.A finiteBorelmeasure μonSn−1 issaidto satisfythesubspaceconcen- tration inequalityif,foreverysubspaceξ ofRn,suchthat0<dimξ < n,
μ(ξ∩Sn−1)≤ 1
nμ(Sn−1)dimξ. (2.2)
The measure is said to satisfy the subspace concentration condition if in addition to satisfyingthesubspaceconcentrationinequality(2.2),whenever
μ(ξ∩Sn−1) = 1
nμ(Sn−1)dimξ,
forsomesubspaceξ,thenthereexistsasubspaceξ,whichiscomplementarytoξinRn, so thatalso
μ(ξ∩Sn−1) = 1
nμ(Sn−1)dimξ, or equivalentlyso thatμisconcentratedonSn−1∩(ξ∪ξ).
Lemma 2.1.(Böröczky-Lutwak-Yang-Zhang, [4]) A non-zero finite even Borel measure on theunitsphereSn−1 isthecone-volumemeasureofanorigin-symmetricconvexbody in Rn ifand onlyif itsatisfiesthesubspaceconcentrationcondition.
3. FundamentalpropertiesofXk andk-generalposition 3.1. FundamentalpropertiesofXk
Theorem 3.1.If P isapolytopein Rn with theoriginin itsinterior,andtheunitouter normalvectorsofP areu1,u2,. . . ,uN,thenVn(P)n =X1(P)n+X2(P)n+· · ·+Xn(P)n. Proof. SinceP containstheorigininitsinterior, byDefinition1.1,itfollows that
Vn(P)n= (VP({u1}) +VP({u2}) +· · ·+VP({uN}))n
= n k=1
dim(span{ui1,...,uin})=k
VP({ui1})VP({ui2})· · ·VP({uin})
= n k=1
Xk(P)n, as desired.
Remark1.Thefunctionals Xk,k= 1,2,. . . ,n,arenotcontinuousinPon.
Wesee thisbyconstructinganexample.LetPi= conv{Ai,Bi,Ci,Di},i∈N,where Ai = (1i −1,1),Bi = (−1i + 1,1),Ci = (1i + 1,−1),Di = (−1i −1,−1).That is, Pi is
Fig. 1.Pi.
anisoscelestrapezoidwithsymmetric tothey-axis inR2, whosesizesof upperbottom and lowerbottom are(2−2i) and (2+2i), respectively; thedistance from theorigin o toAiBi andCiDi is1,respectively.SeeFig.1.
Then,suppVPi ={e1,−e1,ui,vi}andui=−vi.Bycalculatingdirectly,wehave V2(Pi) = 1
2·(2−2
i + 2 + 2
i)·2 = 4, VPi({e1}) =1
2·(2−2
i)·1 = 1−1 i, VPi({−e1}) = 1
2·(2 + 2
i)·1 = 1 +1 i,
VPi({ui}) =VPi({vi}) = V2(Pi)−VPi({e1})−VPi({−e1})
2 = 1.
AccordingtothedefinitionofX1,itfollowsthat X1(Pi)2=
u∧v=0
VPi({u})VPi({v})
=(1−1
i)·(1−1
i + 1 +1
i) + (1 +1
i)·(1 + 1
i + 1−1
i) + 1·1 + 1·1
=6.
Note thatlimi→∞Pi=C,C= [−1,1]2, andX1(C)2= 4·1·2= 8,itfollowsthat
i→∞lim X1(Pi) =√ 6=√
8 =X1(C) =X1( lim
i→∞Pi),
which impliesthatX1 isnotcontinuous. Withthecontinuity oftheareafunctional V2
and thefactthatV2(Pi)2=X1(Pi)2+X2(Pi)2,itfollowsthatX2 isnotcontinuous.
Now, we explainthe reasonswhypolytopesshouldbe “positioned” with centroidat the origin, when we consider Problem X. Specifically, restricted to the set of origin- symmetric polytopes inRn, weprovethe following theorem by essentially makinguse of Lemma2.1.
Theorem 3.2.supP∈Pn s
Xn(P)
Vn(P) = 1, infP∈Psn
Xk(P)
Vn(P) = 0, k= 1,2,. . . ,n−1.
Proof. ForanaturalN,N > n,let
βi = (1, i, i2, . . . , in−1), i= 1,2, . . . , N.
BythepropertyofVandermondedeterminant,itfollowsthat
det(βtl1, βlt2, . . . , βltn)= 0, ∀ {l1, l2, . . . , ln} ⊆ {1,2, . . . , N},
whichimpliesthatanynelements oftheset {β1,β2,. . . ,βN}arelinearindependent.
Constructing afiniteeven discretemeasureμN onSn−1 suchthat suppμN={±β1, . . . ,±βN}, and
μN({βi}) =μN({−βi}) = 1
2N, i= 1,2, . . . , N, where βi= |ββi
i|.
For any j dimensional subspace ξ of Rn, 0 < j < n, since any n elements of {β1,β2,. . . ,βN} are linear independent, it follows that any (j + 1) elements of {β1,β2,. . . ,βN}arealsolinearindependent,andthereforeξcontainsatmostjelements of {β1,β2,. . . ,βN}.Thus,
μN(ξ∩Sn−1)≤2j· 1 2N < j
n,
whichimpliesthatμN satisfiesthesubspaceconcentrationcondition.
ByLemma2.1, thereexists anorigin-symmetricpolytope PN,suchthatVPN =μN. So,
suppVPN ={±β1, . . . ,±βN}, and VPN({βi}) =VPN({−βi}) = 1
2N, i= 1,2, . . . , N.
FromthedefinitionofXn andthefactthatanynelementsoftheset{β1,β2,. . . ,βN} arelinearindependent,itfollowsthat
Xn(PN)n Vn(PN)n =
u1,u2,...,un∈suppVP
u1∧u2∧···∧un=0
VPN({u1})VPN({u2})· · ·VPN({un}) 1
=2N·2(N−1)· · ·2(N−n+ 1)( 1 2N)n
=N ·(N−1)· · ·(N−n+ 1)
Nn →1, N→ ∞.
So,
0≤ Xk(PN)n
Vn(PN)n ≤Vn(PN)n−Xn(PN)n
Vn(PN)n →0, N → ∞, k= 1,2, . . . , n−1.
Thus,
P∈Psupsn
Xn(P)
Vn(P) = 1, inf
P∈Pns
Xk(P)
Vn(P) = 0, k= 1,2, . . . , n−1, asdesired.
By Theorem 3.2, it yields that supP∈Pn c
Xn(P)
Vn(P) = 1,infP∈PcnXVk(P)
n(P) = 0, k = 1,2,. . . ,n−1.
Theorem3.3. supP∈Pn o
X1(P) Vn(P) = 1.
Proof. Assume thatP is aregular simplex inRn with Vn(P)= 1,and the unit outer normalsofP areu1,u2,. . . ,un+1.MoveP togetasequenceofsimplices{Pi}i∈N sothat VPi({u1})= 1−2i1,and VPi({uj})= 2ni1 ,j = 2,3,. . . ,n+ 1.Then
X1(Pi)n=
dim(span{uj1,uj2,...,ujn})=1
VPi({uj1})VPi({uj2})· · ·VPi({uin})
=
n+1
j=1
VPi({uj})n > VPi({u1})n = 1− 1
2i n
→1, i→ ∞.
So,limi→∞X1(Pi)
Vn(Pi) = 1,asdesired.
Proposition 3.4.If P is a polytope in Rn with the origin in its interior, then the two assertionsare equivalent.
(1) suppVP ⊆ {±u1,. . . ,±uN}, and VP({±ui})>0, i= 1,2,. . . ,N. (2) suppVP ∪suppV−P ={±u1,. . . ,±uN}.
Proof. Assume assertion (1) holds. In light of suppV−P = −suppVP, it yields that suppV−P ⊆ {±u1,. . . ,±uN}.Hence,suppVP∪suppV−P ⊆ {±u1,. . . ,±uN}.Meanwhile, sinceVP({±u})>0 foru∈ {±u1,. . . ,±uN},itfollowsthatu∈suppVPoru∈suppV−P. Thatis,u∈suppVP∪suppV−P.Thus,suppVP∪suppV−P ={±u1,. . . ,±uN}.
Assume assertion (2) holds. Let ui ∈ suppVP ∪suppV−P. Then, ui ∈ suppVP or
−ui ∈suppVP. Thus, VP({±ui})=VP({ui})+VP({−ui})>0.AddedthatsuppVP ⊆ {±u1,. . . ,±uN}, thenassertion(1) holds.
Theorem 3.5.If P ∈ Pon andsuppVP∪suppV−P ={±u1,±u2,. . . ,±uN},then Xk(P)n=
ξ∈Gn,k
i1,i2,...,in∈{1,2,...,N} span{ui1,ui2,...,uin}=ξ
VP({±ui1})VP({±ui2})· · ·VP({±uin}),
k= 1,2, . . . , n.
Proof. From the definition of Xk and the fact that span{v1,. . . ,vn−1,−vn} = span{v1,. . . ,vn−1,vn},byinterchangetheorderinthesummationswehave
Xk(P)n=
v1,...,vn∈suppVP
span{v1,...,vn}∈Gn,k
VP({v1})· · ·VP({vn})
=
ξ∈Gn,k
v1,...,vn∈suppVP
span{v1,...,vn}=ξ
VP({v1})· · ·VP({vn})
=
ξ∈Gn,k
v1,...,vn−1∈suppVP
VP({v1})· · ·VP({vn−1})
vn∈suppVP
span{v1,...,vn}=ξ
VP({vn})
=
ξ∈Gn,k
v1,...,vn−1∈suppVP
VP({v1})· · ·VP({vn−1})
vn∈{u1,u2,...,uN} span{v1,...,vn}=ξ
VP({±vn})
=
ξ∈Gn,k
vn∈{u1,u2,...,uN}
VP({±vn})
v1,...,vn−1∈suppVP
span{v1,...,vn}=ξ
VP({v1})· · ·VP({vn−1})
· · ·
=
ξ∈Gn,k
v1,...,vn∈{u1,u2,...,uN} span{v1,...,vn}=ξ
VP({±v1})VP({±v2})· · ·VP({±vn})
=
ξ∈Gn,k
i1,i2,...,in∈{1,2,...,N} span{ui1,ui2,...,uin}=ξ
VP({±ui1})VP({±ui2})· · ·VP({±uin}),
as desired.
Theorem 3.6. Let P and Q be polytopes in Rn with the origin in their interiors. If VP({±u})=VQ({±u})foranyu∈Sn−1,then Xk(P)=Xk(Q),k= 1,2,. . . ,n.
Proof. Forany u∈suppVP ∪suppV−P,itfollows thatVP{±u}>0 by Proposition3.4.
From the assumption that VQ({±u}) = VP({±u}), it yields that VQ{±u} > 0, and thereforeu∈suppVQ∪suppV−Q.Hence,suppVP∪suppV−P ⊆suppVQ∪suppV−Q.
Similarly,suppVQ∪suppV−Q⊆suppVP∪suppV−P.So,suppVP∪suppV−P = suppVQ∪ suppV−Q.FromTheorem3.5togetherwiththeassumptionthatVP({±u})=VQ({±u}) foranyu∈Sn−1,itfollows thatXk(P)=Xk(Q),k= 1,2,. . . ,n.
Remark2.FromtheequationsXk(P)=Xk(Q),k= 1,2,. . . ,n,wecannot concludethat P =Q. Forinstance,letP be an equilateraltriangleinR2 with centroidat theorigin andunitouternormalsu1,u2,u3;letQbearegularhexagonwithcentroidattheorigin andunitouter normals±u1,±u2,±u3. LetP andQhavethesamearea.Itisevidently thatP andQsatisfytheconditionsinTheorem3.6.
Theorem 3.7.If P is a polytope in Rn with its centroid at theorigin, then there exists anorigin-symmetricpolytope QinRn suchthat Xk(Q)=Xk(P),k= 1,2,. . . ,n.
Proof. Letμbeafinitediscretemeasure onSn−1 suchthatforu∈Sn−1, μ({u}) =μ({−u}) =1
2VP({±u}).
Thenμisanevenmeasure.ForanysubspaceξofRn, μ(ξ∩Sn−1) =
u∈ξ∩Sn−1
μ({u}) =
u∈ξ∩Sn−1
1
2μ({±u})
=
u∈ξ∩Sn−1
1
2VP({±u}) =
u∈ξ∩Sn−1
VP({u}) =VP(ξ∩Sn−1).
Specifically,μ(Sn−1)=VP(Sn−1).Hence,foreverysubspaceξofRnwith0<dimξ < n, μ(ξ∩Sn−1)
μ(Sn−1) = VP(ξ∩Sn−1) VP(Sn−1) .
Since the centroid of P is at the origin, its cone volume measure VP satisfies the subspaceconcentrationcondition.Sodoesthemeasureμ.
AccordingtoLemma2.1,thereexistsanorigin-symmetricpolytopeQsothatVQ =μ.
Thus, VQ({±u}) = μ({±u}) = VP({±u}). By Theorem 3.6, it yields that Xk(Q) = Xk(P),k= 1,2,. . . ,n, asdesired.
ByTheorem 3.7,ityieldsthatsupP∈Pn c
Xk(P)
Vn(P) = supP∈Pn s
Xk(P)
Vn(P),k= 1,2,. . . ,n−1.
3.2. k-generalposition
Inthispart,weposethenotionofk-generalposition,k∈ {1,2,. . . ,n},whichgeneral- izes theessentialgeometricnotion:generalposition.Restrictedtotheclassofpolytopes in(k+ 1)-generalpositioninRn,weestablishtheinversion formula ofthekth volume decomposition Xk. Thefactthatpolytopes ink-general position inRn is densein the set Kon isprovedinAppendixA.
Definition 3.1.Let Z be a finite set of unit vectors in Rn, and Z ∪ (−Z) = {±u1,±u2,. . . ,±uN}. Z is said to be in k-general position, k ∈ {1,2,. . . ,n}, if Z is notcontainedinaclosedhemisphereofSn−1andanykelementsof{u1,u2,. . . ,uN}are linearlyindependent.
ApolytopeP inRnissaidtobeink-generalposition,ifthesetofunitouternormals of P isink-generalposition.
Write Pkn fortheset ofpolytopes inRn whichareink-generalposition andcontain theoriginintheirinteriors.Theinclusionrelation
Pnn⊆ Pn−1n ⊆ · · · ⊆ P3n⊆ P2n=P1n=Pon
is evident.
RecallthatKárolyiandLovász[18] firstposedthenotionofgeneralposition:Afinite set Z of unitvectors inRn is said to be ingeneralposition, ifZ is notcontained ina closed hemisphere of Sn−1 and any n elements of Z are linearly independent. So, the n-generalposition isindeedthegeneralpositioninthesenseof KárolyiandLovász,up to antipodalunitouternormals.
WriteLn⊆ PonforthesetofpolytopesinRnwhoseunitouternormalsareingeneral position. Then,Ln ⊆ Pnn.However,acubeC= [−1,1]n∈ Pnn,butC /∈ Ln.
RestrictedinPk+1n ,k∈ {1,2,. . . ,n−1},weestablishtheinversionformulaofthekth volumedecompositionXk.
Theorem 3.8.If P ∈ Pk+1n andsuppVP ∪suppV−P ={±u1,. . . ,±uN},then
Xk(P)n=
1≤i1<i2<···<ik≤N
k l=1
(−1)k−l
{j1,...,jl}⊆{i1,...,ik}
(aj1+· · ·+ajl)n,
where aj=VP({±uj}),j= 1,. . . ,N.
Proof. From Theorem 3.5, and the assumption that P is in (k+ 1)-general position (therefore, any k-dimensional subspace spanned by unit outer normals of P precisely containsk normalsof P,uptotheirantipodalnormals),itfollowsthat
Xk(P)n=
ξ∈Gn,k
j1,...,jn∈{1,...,N} span{uj1,...,ujn}=ξ
aj1· · ·ajn
=
1≤i1<i2<···<ik≤N
∪nm=1{jm}={i1,...,ik}
aj1· · ·ajn
=
1≤i1<···<ik≤N
l1,...,lk∈N+ l1+···+lk=n
ali11ali22· · ·alikk n l1
× n−l1
l2
· · · n−l1−l2− · · · −lk−1 lk
=
1≤i1<···<ik≤N
l1,...,lk∈N+ l1+···+lk=n
ali11· · ·alikk l1!· · ·lk!n!.
Fixingi1,. . . ,ik, let k
l=1
(−1)k−l
{j1,...,jl}⊆{i1,...,ik}
(aj1+· · ·+ajl)n =
l1,...,lk∈N l1+···+lk=n
c(l1, . . . , lk)ali11· · ·alikk, (3.1)
wherec(l1,. . . ,lk) isthecoefficientofali11· · ·alikk.
Observethatwhen l1,. . . ,lk ∈ N+,ali11· · ·alikk appearsonly ifl =k onthe leftside.
So,
c(l1, . . . , lk) = n l1
n−l1
l2
· · · n−l1−l2− · · · −lk−1 lk
= n!
l1!· · ·lk!.
Whenl1·l2· · ·lk = 0, assumelp1,. . . ,lpm >0 and lp1 +· · ·+lpm =n,m ≤k−1.By comparingthecoefficientofalipp1
1· · ·alipm
pm ofbothsidesoftheequation (3.1),wehave c(l1, . . . , lk) =
k l=m
(−1)k−l
{p1,...,pm}⊆{j1,...,jl}⊆{i1,...,ik}
¯ c
= ¯c k l=m
(−1)k−l k−m l−m
= ¯c
k−m
l=0
(−1)k−m−l k−m l
= ¯c(1−1)k−m= 0,
wherec¯isthecoefficientofalpp11· · ·alppmm in(ap1+· · ·+apm)n.So, k
l=1
(−1)k−l
{j1,...,jl}⊆{i1,...,ik}
(aj1+· · ·+ajl)n =
l1,...,lk∈N+ l1+···+lk=n
ali11· · ·alik
k
l1!· · ·lk!n!.
Hence,
Xk(P)n=
1≤i1<i2<···<ik≤N
k l=1
(−1)k−l
{j1,...,jl}⊆{i1,...,ik}
(aj1+· · ·+ajl)n.
3.3. Thecomplexity of theProblemX
Inthispart,weconstructtwoexamplestoshowthatinR4,thevolumedecomposition functional X3(P) does not attain its maximum at parallelotopes, which indicates the complexity oftheProblemXinhigherdimensions.
Example 1.LetQbe anorigin-symmetricparallelotopeinRn.SupposethatsuppVQ = {±u1,. . . ,±un}andVQ({±ui})= 1,i= 1,2,. . . ,n.Then, Q∈ Pnn andVn(Q)=n.
ByTheorem3.8,itfollowsthat Xk(Q)n= n
k k
l=1
k l
(−1)k−lln. Inparticular,
X1(Q)n= n 1
=n, X2(Q)n= n
2
2n− 2 1
=n(n−1)(2n−1−1), X3(Q)n= n
3
3n− 3 2
·2n+ 1 1
·1n
= n(n−1)(n−2)
2 (3n−1−2n+ 1).
Example 2.Let P be a simplex in R4 with its centroid at the origin. Suppose that suppVP ={u1,u2,u3,u4,u5}andVP({ui})= 1,i= 1,2,3,4,5.Then, P ∈ P44, V4(P)= 5.
ByTheorem3.8,itfollowsthat X3(P)4= 5
3 3
1
14− 3 2
24+ 3 3
34
= 360.
Thus,
X3(P)4 V4(P)4 =360
54 = 72 125. However,takingn= 4 inExample 1,wehave X3(Q)4
V4(Q)4 = 4(4−1)(4−2)
2 (34−1−24+ 1)1 44 = 9
16= 1125
2000 < 1152 2000 = 72
125 =X3(P)4 V4(P)4,
