Carla Peri INTEGRAL GEOMETRY IN MINKOWSKI PLANE 1 ...

REND. SEM. MAT.UNIVERS. POLITECN. TORINOVol. 45\1 (1987)Carla PeriINTEGRAL GEOMETRY IN MINKOWSKI PLANESunto. Alcune classiche formule di geometria integrale nel piano euclideo vengono esteseal piano di Minkowski.1. Introduction.Integral geometry in Minkowski plane has been investigated mainly byChakerian [2] and Guggenheimer [3]. In particular Chakerian introduced akinematic density for polygonal lines and proved Poincare's formula.In the present paper we extend this notion to a wide class of curves andwe define a kinematic density for convex sets. Using such densities we generalizesome formulas of Euclidean integral geometry and derive the isoperimetricinequality and Bonnesen's inequality.Moreover while Chakerian assumed that the metric of the plane wassymmetric, we derive the foregoing results in the unsymmetric case as well.For the basic notions on Minkowski geometry consult, the fundamentalworks of Biberstein [1] and Guggenheimer [3].In the following we shall use the notations introduced by Guggenheimerin [3].2. Preliminaries.Let C be a closed convex curve, called indicatrix, which includes theorigin 0 of the Euclidean plane IR 2 . We assume C is "sufficiently" diffe-Classificazione per soggetio AMS (MOS, 1980): 53C65

108rentiable and with positive finite curvature everywhere.For every v £ IR 2 let A E C be the unique point for which OA isparallel to v, as oriented vectors. The Minkowski length of v is defined by+ \v\(1) H *"-WT'where | • | denotes the Euclidean length.The points of IR 2 with the metric associated to the length (1) shall bereferred to as the Minkowski plane.If || — z>|| = ||?|| for any v G IR 2 (i.e. if C is a centrally symmetriccurve with center at the origin O) we shall refer to the plane as thesymmetric Minkowski plane.Let us denote the vectors from O to the indicatrix C by C. To themare associated vectors T(C) by [T(C), C] = 1, where [ , ] denotes thedeterminant. The vectors T(C), reported from O, describe an oval T calledisoperimetrix. The curve T is easily seen to be the polar reciprocal of C with7Trespect to the Euclidean unit circle, rotated through —r- .Let x{t) be a differentiable vector function, its Minkowski arclengtho x is the parameter defined up to an additive constant by(2)dxdo X= 1.We denote the Minkowski arclength of T by a and the area enclosedby T by II, then 0

109For the properties of these trigonometric functions consult [3].Following Chakerian [2], the angle a> between two directions C(ot)and C(f3), with a

110provided we consider a polygonal line inscribed in T with sufficiently smallsides. Chakerian's proof may be also carried over to the unsymmetric case.Let T 0 , C 0 be the vectors joining the origin 0 of the plane to thepoints corresponding to a = 0 on the isoperimetrix T and on the indicatrixC, respectively.Let T 0 be a fixed curve of finite length L 0 . We assume that T 0 isdefined by the equations x =x(a 0 ), y = y(o 0 ) relative to the arclength o 0as parameter and to the frame (0;T 0 ,C 0 ).Consider a "moving" curve T of length L and arclength a. For"moving" curve T we mean a curve that belongs to the set of the curvescongruent to I\ Let P be a point on T determined by the coordinates(xp,y p ) relative to the frame (0; T 0 , C 0 ) and let C(a) be the unit tangentvector to F at P. We consider the frame (P; T(a), C(a)) "moving" with I\Let X = X(o), Y = Y(o) be the equations of T referred to the arclengtho as parameter and to the frame (P } T(oc), C(ot)).Using the trigonometric relations defined by Guggenheimer in [3] weobtain the following equations for T with respect to the frame (0; T 0 , C 0 )x = x,p + X(o) cm (a, 0) — Y(o) sm (0 ,ai)Then the intersection points of T 0 and r are given by solving the sys­(x(o Q )\ —x p 4- X(o)cm(a,0) — Y(o)sm(0,a)tem!y =y p + X(o)st(0,a) + Y(o) cm (0 ,a),\y(o 0 )=y p +X(o)st(0.,oi)+Y{o)cm(0,a),in the unknowns o 0 , o.Differentiation of these equations yields(dxp =x'do 0 -[X'cm{(x,0)- Y'sm(0,&)]do + [Xsm(0,a)+ Yxcm(a,0)]da\dy. p =y'do 0 -[X'st(0,a) + Y'cm(0,a)]do - [Xcm(0,a)- Yxst(0,a)] da ,where the prime symbol denote differentiation with respect to the arclengthsa 0 or a and x is the unimodular centro-affine curvature of T to the center0. For the differentiation of trigonometric functions see [3].By exterior multiplication we getdV '= dXpAdy p Ada = \-x'X'st(0,a) -x'Y'cm(0,a) ++ y'X'cm(a,0)-y'Y'sm(0,a)\do Q Ado Ada ,

where the absolute value is used because all densities are assumed to bepositive.If we denote by C(j3 0 ), C(j3) the unit tangent vectors to T 0 and T atQ £ T 0 H T, respectively, we have x' = — sm(0,f$ 0 ), y' == cm(0,P 0 ) andX' = -sm(a,P), Y' = cm(a,p).Hence, by using the trigonometric identities, there followsdT = \sm(0,P 0 ) [—sm(a,f3)st(0,

112and OQ is the arclength of T*, then do* 0 = || — C(j3 0 )|| do 0 .Therefore we havedo 0 do + 2l I do*do = 2(L 0 +L%)L ,0 -'O -'O -'0where LQ is the length of T*.Then we have:Minkowskian Poincare formula(4) / ndr = 2(L 0 +L*)L .J{r 0 nr#(}}In the symmetric case L 0 = LQ and (4) becomes(4') I ndr = 4L 0 L,^{r 0 nr^}exactly as in Euclidean plane.4. Kinematic density for convex sets and Santalo-type formulae.From now on we restrict our attention to bounded convex sets withboundary of class C 2 .We shall say that a convex set K 0 is congruent to a convex set K : iftheir boundaries bK 0 , bK x are congruent with respect to the relation definedin the previous section.Clearly two congruent convex sets have the same area and the sameperimeter. Following Guggenheimer [3], we assume as area of a convex setits affine area.Let K be a convex set and let P{x,y) be a fixed point on bK.If C(a) is the unit tangent vector to bK at P, then we define thekinematic density for the set of convex sets congruent to K by(5) dK = dx \dy Ada.Let K 0} K be two convex sets of area S 0 , S and perimeter L 0 , L res-

113pectively. We want to calculate the measure n(K; K 0 K 0 =£0) of the setof convex sets congruent to K that intersect K 0 .Let P 0 EbK 0 , PGdK and let p 0 (P), p((3) be the support functionsof K 0 and K referred to the origin P 0 , P respectively. We denote byp*(j3) the support function of the set K* obtained by reflecting K in P.If we translate K so that it remains externally tangent to K 0 , then Ptraces a convex curve with support function p((3) -po(@) + p*(P)> Hencethe measure of the set of translates of K intersecting K 0 is given by the areaof the convex set with support function p(P).It is easy to verify that this area is 5 0 +5*'+ 25^, where S* is thearea of K*, while S* x = (1/2) / p*(/3)^a 0 with a 0 arclength of bK 0 .JdK 0Of course S* = S, since the area is an unimodular affine invariant.If we integrate 25^ with respect to da and note that da = d(3 fora 0 constant, we obtainwhere L* denotes the perimeter of K*.Therefore we have:LMinkowskian Santalb formula2112S* x da = L 0 L* ,(6) n(K ; K H K 0 * 0) = 2 n(S 0 + S) + L 0 L * .In particular if /C 0 shrinks to a point P, the measure n{K\PE.K)the set of convex sets congruent to K containing P isof(7) fjL(K;PeK) = 2TlS .Subtracting (7) from (6) we see that(8) v(K;Kn.K o ¥=0 t P$K)=2IlS o + L o L* •where ix{K; K H K 0 =£ 0; P $ K) is the measure of the set of convex setscongruent to K intersecting K 0 but which do not contain a fixed pointIn the symmetric case L* = L, whence formulae (6)-(8) become

114(6') n(K;Kr\K o ¥=0) = 2Il(S o +S) +L 0 L(7') n(K,PeK) = 2IlS(8') jJi(K } KnK o =*0,PeK) = 2nS o +L o Land formally coincide with Santalo's formulae for convex sets in Euclideanplane.Let K Q be a convex set of area S 0 and perimeter L 0 and let K lf... ,K n be » convex sets congruent to a convex set K of area 5 and perimeterL.We denote the set K 0 D (K x O ... C\ K n ) by K w and the area of K nby S n .From (7), exactly as in the Euclidean plane [4], we have(9) I S n dK x ... dK n = (2nS) w S 0 .More generally, if S r n denotes the area of the set of those points of K 0that are covered precisely by r sets Kj, from (7) and (8) there follows(10) [ S r n dK x ... dK n = (* ) (2n5) r (2n5 0 + L 0 L*r- r S 0where L* is the perimeter of the convex set K* obtained by reflecting Kin a point.Since the sets Kj are independent,formula (6) yields(11) / dK 1 ...dK n = ( I dK\ =(2n(5 0 +5) + L 0 L*) n .\K 0 n Kj * a} \ {*o n K i **}/Thus the average values of the areas S n and S n are(12) E(S n ) =d3) *(s;> =r(2n5) w S 0(2n(S 0 +S) + L 0 L*)"(")(2nS) r (2n5 0 + L 0 L*) n ' r S,'(2n(S 0 +5) + L 0 L*)"

1155. The isoperimetric inequality.Let K 0 ,K be two convex sets of area S and perimeter L. We denoteby m,- the measure of the set of convex sets congruent to K whose boundarieshave i points in common with dK 0 . Then formulae (6) and (4) becomem 2 + WI4 + m 6 + ... = 4115 + LL* ,2m 2 + 4m A + 6m 6 + ... = 2(L + L*)L .Note that the convex sets with i odd belong to a set of measure zero.Thus the isoperimetric inequality follows(14) L 2 -4IIS>0.This inequality has been already obtained by Guggenheimer [3] in a differentway.Where upon we shall derive a stronger inequality. Following Biberstein[1], we call anticircle of radius r the translate of the image of the isoperimetrixT in a homothety of center O.Consider a convex set K of area S and perimeter L. Let r,- be theradius of a maximal anticircle contained in K and let r e be the radius ofa minimal anticircle that contains K, then r f - < r e . Moreover let r be theradius of an anticircle K 0 such that(15) rj < r < r e .If Wj is the measure of the set of the sets congruent to K whoseboundaries have exactly i points in common with the boundary of K 0 , thenwe have, as above,m 2 + m 4 + ... = 2nCS 0 + S) + L 0 L* ,2m 2 + 4w 4 + ... = 2(L 0 + L*)Lwhere 5 0 and L 0 are the area and the perimeter of K 0 while LQ is theperimeter of the set KQ obtained by reflecting K 0 in a point.From the previous formulae there followsL 0 L 4- L*L -L 0 L*- 2Il(S 0 + S) > 0 .On the other hand, if we fix K and "move" K Q we obtain in thesame wayL 0 L + L 0 L* -L^L -2n(S 0 + 5)> 0 .

116Adding these inequalities we getLL 0 -2T1(S 0 +5)>0.Since an anticircle of radius r has area II r 2previous inequality becomes(16) rL-S-Ilr 2 >0,and perimeter 2IIr thewhere r satisfies the condition (15).From (16), we obtain, exactly as in the Euclidean case 4t, the following(17) L 2 - 4IIS > j [(L - 2llr,) 2 + (2Ur e - L) 2 ] ., LThis inequality shows that L 2 — 4115 = 0 only when r t - = r e = —- ,i.e. when K is an anticircle. Moreover we can statePROPOSITION. If a bounded convex set K is not an anticircle then the setof the convex sets congruent to K whose boundaries have at least fourpoints in common with the boundary of K has measure greater then zero.From (17) by using the inequality 2(x 2 + y 2 ) > (x +y) 2 , we canfinally deriveGeneralized Bonnesen inequality(18) L 2 -4ns>Ii 2 (r e -ri) 2(compare [4] for the Euclidean case).

117REFERENCES[1] O. Biberstein, Elements de geometrie differentielle Minkowshienne, thesis, Universityof Montreal, 1957.[2] G.D. Chakerian, Integral geometry in the Minkowski plane, Duke Math. J., 1962,375-381.[3] H. Guggenheimer, Pseudo-Minkowski differential geometry, Ann. Mat. Pura Appl.(4)70, 1965, 305-370.[4] L.A. Santalo, Integral geometry and geometric probability, Addison-Wesley, 1979.[5] M.I. Stoka, Geometrie Integrale, Mem. Sci. Math., 165, Gauthier-Villars, Paris,1968.CARLA PERI - Istituto di Matematica generale, finanziaria ed economica - UniversitaCattolica del Sacro Cuore - Largo Gemelli, 1 - 20123 MILANOLavoro pervenuto in redazione il 21/1/1987