SOME RESULTS ON THE GEOMETRY OF MINKOWSKI PLANE. Bing Ye Wu

Transcription

1 ARCHIVUM MATHEMATICUM (BRNO Tomus 46 (21, SOME RESULTS ON THE GEOMETRY OF MINKOWSKI PLANE Bing Ye Wu Abstract. In this paper we stuy the geometry of Minkowski plane an obtain some results. We focus on the curve theory in Minkowski plane an prove that the total curvature of any simple close curve equals to the total Lansberg angle. As the result, the sum of oriente exterior Lansberg angles of any polygon is also equal to the total Lansberg angle, an when the Minkowski plane is reversible, the sum of interior Lansberg angles of any n-gon is n 2 times of the total Lansberg angle. Our results generalizes the 2 classical results in plane geometry. We also obtain a new characterizations of Eucliean plane among Minkowski planes. 1. Introuction A Minkowski norm on the real n-space R n is a function F : R n R which is positively homogeneous of egree 1, strictly convex an smooth off the origin. The pair (R n, F is calle a Minkowski space, an it is sai to be reversible if F (y = F ( y for any y R n. When n = 2, (R 2, F is calle a Minkowski plane. In this paper we stuy the geometry of Minkowski plane an obtain some results. The main tool we use is the polar expression of (R 2, F. We first focus on the curve theory an obtain the Frenet type formula for curves in (R 2, F, an prove that the total curvature of any simple close curve equals to the total Lansberg angle of (R 2, F. As the result, the sum of oriente exterior Lansberg angles of any polygon is also equal to the total Lansberg angle, an when F is reversible, the sum of interior Lansberg angles of any n-gon is n 2 2 times of the total Lansberg angle. Our results generalizes the classical results in plane geometry. We also characterize the Eucliean plane among Minkowski planes by a function which is essentially efine by Zhongmin Shen. 2. Polar expression of Minkowski plane Let be the stanar Eucliean norm on R 2, then for any Minkowski norm F on R 2, the function φ(y = F (y y is a positively homogeneous function of egree zero 2 Mathematics Subject Classification: primary 53C6; seconary 53B4. Key wors an phrases: Minkowski plane, polar expression, Lansberg angle, Frenet frame. This project was supporte by the Fun of the Eucation Department of Fujian Province of China (No JA9191. Receive January 6, 21. Eitor J. Slovák.

2 178 B. Y. WU on R 2 \{}. Let (r, θ be the stanar polar coorinates of R 2, then it is clear that φ epens only on θ, an φ = φ(θ is perioical with perio 2π (not necessarily minimal perio. Hence we can view φ as a function on unit circle S. Now we can express the Minkowski norm F as F = r φ(θ. Recall that the funament tensor of F is efine by g ij (y = 1 2 F 2 2 y i y, by a irect computation we obtain the following j polar expression for g ij (y = g ij (θ: (2.1 g 11 = φ 2 2φφ cos θ sin θ + ((φ 2 + φφ sin 2 θ, (2.2 (2.3 g 22 = φ 2 + 2φφ cos θ sin θ + ((φ 2 + φφ cos 2 θ, g 12 = φφ cos 2θ ((φ 2 + φφ cos θ sin θ, (2.4 et(g ij = g 11 g 22 g 2 12 = φ 3 (φ + φ. By these formulas we clearly have the following Proposition 2.1. Let φ: S R be a positive smooth function, then F = r φ(θ is a Minkowski norm on R 2 if an only if φ + φ >. Let S = {y R 2 : F (y = 1} be the inicatrix of R 2. The Minkowski norm F efines a Riemannian metric ĝ = g ij (yy i y j on the puncture plane R 2 \. We view S as the close curve in Riemannian surface (R 2 \, ĝ, its arc length Λ = Λ(F with respect to ĝ is calle the total Lansberg angle of (R 2, F. If F is Eucliean, then we certainly have Λ = 2π, while for general Minkowski norm, Λ may be smaller or bigger than 2π (see e.g., [1, 2]. On the other han, if F is reversible, then the inequality Λ 2π always hols, with the equality hols if an only if F is Eucliean [5]. Notice that the equation of S can be written as y = y(θ = 1 φ(θ (cos θ, sin θ, by (2.4 the arc element is (2.5 Θ = an consequently, (2.6 Λ = g 11 g 22 g 2 12 (y1 y 2 y 2 y 1 = S Θ = 2π φ θ. φ θ, Since φ + φ >, we have Θ θ >, an thus Θ an θ can be mutually expresse, an the parameter Θ is also known as the Lansberg angle. For two nonzero vectors y 1, y 2 R 2, the oriente Lansberg angle (y 1, y 2 an the Lansberg angle (y 1, y 2 between y 1 an y 2 are efine by an (y 1, y 2 = θ2 θ 1 φ θ = Θ 2 Θ 1 (y 1, y 2 = (y 1, y 2,

3 SOME RESULTS ON MINKOWSKI PLANE 179 respectively. Here θ 1 an θ 2 are polar angles of y 1 an y 2, an Θ 1 an Θ 2 the Lansberg angles of y 1 an y 2, respectively. 3. The curve theory in Minkowski plane For a given Minkowski plane (R 2, F, the funamental tensor an the Cartan tensor are efine by g ij (y = 1 2 F 2 2 y i y an C j ijk (y = 1 g ij 2, respectively. For y k X = X i y, Y = Y i i y, Z = Z i i y, write g i y (X, Y = g ij (yx i Y j, C y (X, Y, Z = C ijk (yx i Y j Z k. Let D be the stanar flat connection on (R 2, F, we have (3.1 Z g y (X, Y = g y (D Z X, Y + g y (X, D Z Y + 2C y (X, Y, Z. Recall that C y (y,, =, we call I y = C y (u, u, u the Cartan scalar of F, where u is a vector satisfying g y (y, u =, g y (u, u = 1. Now let c = c(s = ( y 1 (s, y 2 (s be a curve in (R 2, F with arc parameter s, thus the tangent vector fiel T = c s is a unit vector fiel along the curve, namely, F 2 (T = g T (T, T = 1. By (3.1, we get = T g T (T, T = 2g T ( st, T, here T, then c = c(s is a (part of straight line; if s T = D T T. Clearly, when s T, let s an write N = It is easy to see from (3.1 that s T k := s g T, T ( st, st s N = D T N = at + bn. ( ( a = g T s N, T = T g T (N, T g T N, s T = k, ( b = g T s N, N = 1 ( 2 T g T (N, N C T s T, N, N = ki T. Consequently, we get the following Frenet formulas for the curve c in (R 2, F : (3.2 s T = kn, s N = k(t + I T N. Here k is calle the curvature of c, an T, N the Frenet frame of c. If F is Eucliean, then I T =, an (3.2 is reuce to the stanar Frenet formulas for curves in Eucliean plane. Let F = r φ(θ be the polar expression of F. In polar coorinate we can express the unit tangent vector fiel T as T = T (θ = 1 (cos θ, sin θ. φ(θ

4 18 B. Y. WU A irect computation yiels θ T = 1 ( sin θ φ φ(θ φ φ2 (θ + φ = 2 (θ (3.3 φ 2 (θ where θ = θ + θ, an θ is etermine by (3.4 cos θ = Combining (2.1 (2.3 an (3.4 we get g T ( (cos θ, sin θ, (cos θ, sin θ an thus cos θ, cos θ φ φ sin θ (cos θ, sin θ, φ (θ φ2 (θ + φ 2 (θ, sin θ φ(θ = φ2 (θ + φ 2 (θ = φ 2 (θ + (φ 2 + φφ (θ sin 2 (θ θ + φ(θφ (θ sin 2(θ θ = φ4 (θ + φ 3 (θφ (θ φ 2 (θ + φ 2 (θ g T ( θ T, θ T = (θ φ(θ, which together with (2.5 yiels k = g T ( s T, s T = As in the Eucliean case, we call, φ θ = Θ. s s k r = Θ s the relative curvature of curve c. Now let c = c(s be a smooth simple close curve in (R 2, F, an let Ω be the finite omain with bounary c. The positive orientation of c is efine in a way such that Ω is on the left when goes forwar along c. As the curve in Eucliean plane (R 2,, it is well-known that the incremental of the polar angle along c is 2π, namely, θ = 2π. Hence, the total curvature of c is k r s = Θ = namely, we have prove the following c c c 2π φ θ = Λ, Theorem 3.1. The total curvature of any smooth simple close curve in (R 2, F equals to the total Lansberg angle of F.

5 SOME RESULTS ON MINKOWSKI PLANE 181 As in the Eucliean case, Theorem 3.1 can be generalize to the case when c is only piecewise smooth by replacing the curve near the non-smooth points with some smooth curves an then taking the limit. Corollary 3.2. Let c = c(s: [, L] (R 2, F is a piecewise smooth simple close curve with non-smooth points c(s i, i = 1,, n. Let α i = (T (s i, T (s i +, c i = {c(s : s i s s i+1 }, 1 i n 1, an c n = {c(s : s n s L} {c(s : s s 1 }, then n n k r s + α i = Λ. c i i=1 In particular, the sum of oriente exterior Lansberg angles of any polygon is equal to the total Lansberg angle. It is clear that when F is reversible, then the Lansberg angle of any straight angle is Λ 2, an we can prove the following corollary just as in the Eucliean case. Corollary 3.3. The sum of interior Lansberg angles of any n-gon in a reversible Minkowski plane is n 2 2 Λ. i=1 4. A new characterization of Eucliean plane In this last section we shall provie a new characterization of Eucliean plane among Minkowski planes. Let us first introuce a function efine by Professor Zhongmin Shen. Let (R n, F be a Minkowski n-space. For vector y R n \{}, we obtain a hyperplane W y = {u R n : g y (y, u = }. Taking a basis e 1,..., e n of R n such that e 1 = y, an e 2,..., e n is a basis for W y. Let { ( n } B n = (y i R n : F y i e i < 1, { By n 1 = (y a R n 1 : F i=1 ( n a=2 } y a e a < 1. In [5] Shen efine a function ζ = ζ(y on (R n \{}, F as following: ζ(y = vol(bn vol(b n 1 vol(by n 1 F (y vol(b n. Here B k enotes the k-imensional unit ball in Eucliean k-space R k. Clearly, function ζ is inepenent of the choice of basis e 2,..., e n for W y, an ζ = 1 when F is Eucliean. Shen aske that whether or not F is Eucliean when ζ = 1. In the following we shall consier the problem of this type for n = 2. For this purpose let us first introuce the reversibility λ = λ(f of a Minkowski plane (R 2, F as following (see [4]: λ = λ(f = max y F (y F ( y.

6 182 B. Y. WU In term of polar expression, we have φ(θ (4.1 λ = max θ [,2π] φ(θ + π We consier a function ζ = ζ(y on a Minkowski plane (R 2 \{}, F as following: (4.2 ζ(y = Λ 4λ vol(b 1 y F (y vol(b 2. It is clearly that up to a constant, ζ is essentially the same as ζ. We have the following result which provies a new characterization for Eucliean plane among Minkowski planes. Theorem 4.1. Let ζ = ζ(y be a function on Minkowski plane (R 2 \{}, F efine by (4.2. Then F is Eucliean if an only if ζ(y = 1 for any y. Proof. The necessity is trivial, so we nee only to prove the sufficiency. Assume that ζ(y = 1 for any y. Let F = r φ(θ be the polar expression of F, an let y = (cos θ, sin θ. Taking e 1 = y, e 2 = (cos θ, sin θ, where θ = θ + θ, an θ is etermine by (3.4. Then by (2.1 (2.3 it can be verifie that e 2 W y, an consequently, (4.3 vol(b 1 y = 1 φ(θ + 1 φ(θ + π. In the following we nee to compute vol(b 2. Recall that Let then the following transformation B 2 = {(y 1, y 2 R 2 : F (y 1 e 1 + y 2 e 2 < 1}. Ω = {(y 1, y 2 (R 2, F : F (y 1, y 2 < 1}, (y 1, y 2 (y 1, y 2 = (y 1 cos θ + y 2 cos θ, y 1 sin θ + y 2 sin θ maps B 2 onto Ω, an y 1 y 2 = (y 1 cos θ + y 2 cos θ (y 1 sin θ + y 2 sin θ = sin(θ θy 1 y 2 = φ(θ φ2 (θ + φ 2 (θ y1 y 2,

7 SOME RESULTS ON MINKOWSKI PLANE 183 an thus (4.4 vol(b 2 φ2 (θ + φ = y 1 y 2 = 2 (θ B φ(θ 2 φ2 (θ + φ = 2 (θ 2π 1 φ(ξ ξ r r φ(θ φ2 (θ + φ = 2 (θ 2π ξ 2φ(θ φ 2 (ξ. Ω y 1 y 2 From (4.3, (4.4 an the assumption that ζ(y = 1 for any y, we have ( Λ 1 (4.5 λ φ(θ + 1 = 2 2π φ 2 (θ + φ 2 ξ (θ φ(θ + π φ 2 (ξ, which together with (4.1 yiels (4.6 Note that Λ φ(θ Λ 2λ ( 1 φ(θ + 1 φ(θ + π = 2π φ 2 (θ + φ 2 ξ (θ φ 2 (ξ. θ = θ + θ = φ2 (θ + φ(θφ (θ φ 2 (θ + φ 2 θ, (θ an the incremental of θ is 2π when θ goes from to 2π, it is easy to know from (4.6 that 2π Λ 2 θ 2π φ 2 (θ ( φ 2 (θ + φ(θφ (θ θ( 2π θ 2, φ 2 (θ which together with the Cauchy-Schwartz inequality implies that 2π Λ 2 ( φ 2 (θ + φ(θφ (θ 2π θ ( 2π θ φ 2 (θ 2 φ θ = Λ 2. As the result, above inequality becomes an equality, an φ 2 (θ + φ(θφ (θ = C φ 2 (θ for some positive constant C. It is equivalent to et(g ij (y = C by (2.4, an thus the mean Cartan tensor I k = 1 2 log ( et(g y k ij (y =, an by Deiche s theorem [3], F is Eucliean. So we are one. References [1] Bao, D., Lackey, B., Raners surfaces whose Laplacians have completely positive symbol, Nonlinear Anal. 38 (1999, [2] Bao, D., Shen, Z., In the volume of unit tangent spheres in a Finsler manifol, Results Math. 26 (1994, [3] Deiche, A., Über ie Finsler-Räume mit A i =, Arch. Math. (Basel 4 (1953,

8 184 B. Y. WU [4] Raemacher, H. B., A sphere theorem for non-reversible Finsler metrics, Math. Ann. 328 (24, [5] Shen, Z., Lectures on Finsler Geometry, Worl Sci., Singapore, 21. Department of Mathematics, Minjiang University, Fuzhou, Fujiang, 3518, China bingyewu@yahoo.com.cn