CHARACTERIZATIONS OF INCLINED CURVES WHICH IS CONCERNED WITH OSCULATING SPHERE IN L FOR SPACE-LIKE CURVES

In this study, we rst calculate the higher ordered curvatures in terms of higher ordered harmonic curvatures for the space-like curves in Ln . Thus, we give characterizations of inclined curves in Ln for space-like curves. Furthermore, we obtain the coordinates of the central point of the space-like curves osculating sphere. Then we calculate the higher ordered curvatures of the space-like curves in terms of the coordinates of the central point of its osculating sphere. At the end, we give another characterization for inclined curves in Ln, in terms of these coordinates.


Introduction
The characterizations of inclined curves in E n is given in [3] and [4] that 1) is an inclined curve in E n , 2) is an inclined curve in E n 1 , det V 0 1 ; V 0 2 ; :::; V 0 n = 0.In Section 3 and Section 5, we show that inclined curves which is space-like in L n have the following characterizations: 2) is an inclined curve in L n 1 , det V 0 1 ; V 0 2 ; :::; V 0 n = 0.In Section 4, we obtain the relation between the higher ordered curvatures and the higher ordered harmonic curvatures.
In addition, characterizations of inclined curves which is concerned with osculating sphere in E n are given in [7].In Section 6, we show that if M is a space-like curve in L n then the coordinates of the central point of the osculating sphere of M are m i : I !R "1 k1(s) ; i = 2 fm 0 i 1 (s) + k i 2 (s)m i 2 (s)g "i 1"i 2 ki 1(s) ; 2 < i n: In Section 7, we give the relation between the functions of m i and k i .In Section 8, we write the relationship between m 0 i and m i values in the matrix form.Let M be a space-like curve in L n+1 , n 4, and let n be an even number.If we denote the coe¢ cient matrix as B n ; we get the following characterizations: 2) det B n = 0 , M L n+1 is an inclined curve in L n ; where m 0 i = dmi ds and s denotes the arc length parameter of M .

Preliminaries
2.1.Symmetric bilinear forms.Let V be a real vector space.A bilinear form on V is an r-bilinear function: We also consider only the symmetric case.A symmetric bilinear form h ; i on V is : (a) positive [negative] de…nite providing that v 6 = 0 implies hv; vi > 0 [ < 0 ] , (b) positive [negative] semide…nite providing that hv; vi 0 [ 0 ] for all v 2 V. (c) nondegenerate providing that hv; wi = 0 for all w 2 V implies v 6 = 0: If h ; i is a symmetric bilinear form on V, then, for any subspace W of V, the restriction h ; i j WXW denoted merely by h ; i j W is also symmetric and bilinear.
The index q of a symmetric bilinear form h ; i on V is the largest integer that is the dimension of a subspace W V on which h ; i j W is negative de…nite.
Thus, 0 q dim V and q =0 if and only if h ; i is positive semide…nite [5].
2.2.Scalar product.A scalar product h ; i on a vector space V is a nondegenerate symmetric bilinear form on V [5].
Lemma 2.1.A scalar product space V 6 =0 has an orthonormal basis.The matrix of h ; i relative to an orthonormal basis e 1 ; e 2 ; :::; e n for V is diagonal.In fact, < e i ; e j >= i j " j where " j =< e j ; e j >= 1 [5]: Lemma 2.2.Let e 1 ; e 2 ; :::; e n be an orthonormal basis for V, with " j = < e j ; e j >.Then, each v 2 V has a unique expression Lemma 2.3.For any orthonormal basis e 1 ; e 2 ; :::; e n for V, the number of negative signs in the signature (" 1 ; " 2 ; :: Lemma 2.4.For any orthonormal basis e 1 ; e 2 ; :::; e n for V, the number of an integer q with 0 q n , changing the …rst q plus signs above the minus, gives tensor The resulting semi-Euclidean space R n q reduces to R n if q = 0.For n 2 , R n 1 is called Minkowski n-space; if n = 4, it is the simplest example of a relativistic spacetime.
Fix the notation as follows: A Lorentz vector space is a scalar product space of index 1 and dimension 2 [5]: 2.3.Lorentzian space.Let M be a smooth connected paracompact Hausdor¤ manifold, and let : TM !M denote the tangent bundle of M. A Lorentzian metric <; > for M is a smooth symmetric tensor …eld of type (0,2) on M such that, for each p 2 M, the tensor <; > p : T P M xT P M !R is a nondegenerate inner product of signature (-, +, . . .,+).In other words, a matrix representation of <; > at p will have one negative eigenvalue, and all other eigenvalues will be positive.
A Lorentzian manifold (M, <; >) is a manifold M together with a Lorentzian metric <; > for M. All noncompact manifolds admit Lorentzian metrics.However, a compact manifold admits a Lorentzian metric if its Euler characteristic vanishes [6].
Lorentzian space is the manifold M =R n together with the metric This space-time is time oriented by the vector …eld @ = @x 1 [6]: 2.4.Curves and curvatures.A curve in a Lorentzian space, L n , is a smooth mapping: where I is open interval in the real line R: The interval I has a coordinate system consisting of the identity map u of I: The velocity vector of at t 2 I is A curve is said to be regular if 0 (t) does not vanish for all t in I: 2 L n is space-like if its velocity vectors 0 are space-like for all t 2 I; similar to time-like and null.
If is a space-like or time-like curve, we can reparametrize it such that < 0 (t); 0 (t) >= " 0 (where " 0 = +1 if is a space-like and " 0 =-1 if is time-like, respectively ).In this case, is said to be unit speed, or it has arc length parameterization.Here and in the sequel, we assume that has arc length parametrization [5].

De…nition 2.6. Let M
L n be the curve with coordinate neighborhood (I , ); and letfV 1 ; V 2 ; :::; V r gbe the Frenet r-frame at (s) with s2 I.Then, the function is called i.th curvature function of the curve M, and for s2 I, k i (s) is called i.th curvature of M at (s): De…nition 2.7.Let M be curve in L n , parametrized by its own arc length.Let us denote the Frenet vector …elds of this curve fV 1 ; V 2 ; :::; V r g.Then, the equality Theorem 2.8.Let M L n be a regular curve with coordinate neighborhood (I , ), and let fV 1 ; V 2 ; :::; V r gbe the Frenet r-frame at (s) with s2 I.Then,

A New Characterization for Inclined Curves in Lorentzian Spaces
for Sace-Like Curves De…nition 3.1.Let be a space-like curve in L n ; and let V 1 be the …rst Frenet vector …eld of .X 2 (L n ) is a constant unit vector …eld.If then is called a general helix (inclined curve) in L n : ' is called slope angle, and the space SpfXg is called slope axis [1].
De…nition 3.2.Assume that is space-like or time-like curve in L n : If the higher ordered curvatures of are k r , 1 < r n 1, then the higher ordered harmonic curvatures H r , 1 r n 2, are Theorem 3.3.Let : I !L n be a general helix (inclined curve), parametrized by its arc length.Let X be a unit and constant vector …eld of L n ; and let fV 1 ; V 2 ; :::; V r g be Frenet r-frame at the point of (s) of .If we consider the angle between 0 and X as ', we have H j : I !R; < V j+2 ; X >= H j cosh ': Then, the value of the H j function at the point of (s) is called as the j-th harmonic curvature according to X at the point of (s) of [1].
Theorem 3.4.Let be space-like curve in L n .Let the Frenet frame of be F = fV 1 ; V 2 ; :::; V n g ; and the higher ordered harmonic curvatures be H 1 ; H 2 ; :::; H n 2 .Then, Proof.()) Let be inclined curve in L n .We denote slope angle of with '; and we also denote slope axis of with SpfXg.From De…nition3.1 we can write < V 1 ; X > = cosh '; and from Theorem 3.1, we can write If we take derivative of the equation Since X 2 Sp fV 1 ; V 2 ; :::; V n g we can write Thus, we have Since X is a space-like and unit vector …eld, we can write 1 = kXk 2 =< X; X > : Thus, using the equation (2) ; we obtain Since < X; X >= 1 and < V j+2 ; V j+2 >= " j+1 , we can write or since " 0 = 1, we can write Hence, we obtain (() Let us assume that , and let us show that is an inclined curve.We know that and X is a space-like vector …eld.Hence, we show that X is a constant unit vector …eld.If we take the derivative of (4) ; we will get Using the value of D V1 H j in De…nition 3.2 and the value of D V1 V j+2 at Theorem 2.5, we obtain Hence, we also obtain D V1 X = 0.As a result, X is a constant vector …eld.Furthermore, < X; X >= +1, and this means that X is a unit vector …eld.Finally, for space-like curve, and the constant unit vector …eld (space-like), X we can write Hence, we obtain < V 1 ; X >= cosh ': This completes the proof of theorem.

Higher Ordered Curvatures in Terms of Higher Ordered Harmonic Curvatures
Theorem 4.1.Let be an inclined curve (space-like) in L n .The relation between the higher ordered curvatures k r , 2 < r n 2, and the higher ordered harmonic curvatures H r , 1 r n 2, is Proof.We will prove the theorem by induction method.From De…nition 3.2; we have For i = 2, (6) gives us Since we assume that H 0 = 0, ( 5) is satis…ed for i = 2.If we extend (7) by 2H 1 ; then we have . This proves that theorem is true for r = 3. Now,let us assume that the theorem is true for r = p 1; and then let us prove that the theorem is also true for r = p.
As our assumption, we have that In the equation (6) for i = p 1; we have and if we replace here the value of k p 1 , from (8), we will obtain Thus, the theorem is also true for r = p.This completes the proof of the theorem.

Another Characterization for Inclined Curvatures in Lorentzian
Spaces for Space-Like Curves Theorem 5.1.Let be space-like curve in L n , n = 2k 4. Let us assume that the Frenet frame of be F = fV 1 ; V 2 ; :::; V n g.Then, is an inclined curve in L n 1 , det V 0 1 ; V 0 2 ; :::; V 0 n = 0: Proof.In this proof, we use the induction method.
()) Let be an inclined curve in L n 1 .Then, we show that det V 0 1 ; V 0 2 ; :::; V 0 n = 0: From Theorem 2.5, we know that Then, we can write From the Theorem 4.1, we may write If we replace here the value of k 3 , from (9) ; we will obtain 2

:
According to hypothesis, if is an inclined curve in L 3 , then we have 4 ) = 0.As a result, the theorem is true for n = 4. Now, let us assume that the theorem is true for n = p:We show that the theorem is also true for n = p + 2.
As our assumption, we have det (V 0 1 ; :::; Now, we show that the theorem is true for n = p + 2.
We have If we replace the value of k p+1 of Theorem 4.1, we will have 2

:
According to the hypothesis, we may write Thus, the last equation becomes det(V 0 1 ; :::; V 0 p+2 ) = 0; and it proves the necessity of the theorem.
(()Let us assume that det(V 0 1 ; :::; V 0 n ) = 0.Then, we will show that is an inclined curve.For the case of n = 4; we can write According to the hypothesis, det(V Since k 1 6 = 0; we have Thus, the theorem is true for n = 4. Now, let us assume that the theorem is true for n = p:We show that the theorem is also true for n = p + 2.
In the case of n = p; we can write det(V Since we assume that k 1 6 = 0; k 3 6 = 0; :::; k p 1 6 = 0; then we obtain that k p+1 = 0.This means that Thus, is an inclined curve in L p+1 .As a result, the theorem is also true for n = p + 2. This means that the theorem is true.

Curves Osculating Sphere
Let M L n be a space-like curve with coordinate neighborhood (I; ); and let fV 1 (s); :::V n (s)g be the Frenet n-frame at (s) with s 2 I. Let b be the center of osculating sphere.Then, we can write b = (s) + m 1 (s)V 1 (s) + ::: + m n (s)V n (s): ) where m i denotes the coordinate functions of the centers of osculating spheres of M .
Let r be the radius of osculating sphere of M .Then, we can write If we take the derivative of (13) with respect to V 1 ; we will obtain and from equation (12) ; we have De…nition 6.1.Let M be a space-like curve in L n+1 ; and let the coordinate functions of osculating sphere of M be m 1 ; :::; m n+1 :Then where k i , 1 i n, are the higher ordered curvatures of the space-like curve.
7. The Relations Between the functions of m i and k i Theorem 7.1.Let M be a space-like curve in L n+1 .Then, the relation between the functions of m i and k i is where m i and k p denote the coordinate functions of the centers of osculating spheres of M and the curvature functions of M L n+1 , respectively.
Proof.We will use the induction method.From De…nition 6.1 we have or for the case of i = j + 1, we have If we get j = 2 in the equation (24) ; we will have On the other hand, if we write p = 2 in the equation (23) ; we will obtain Therefore, the theorem is true for p = 2. Now, we assume that the theorem is true for p = r; and let us prove it for p = r + 1.For p = r; we have Now, we get j = r + 1 in the equation (24).Then, we obtain If we replace k r from (25) into (26), we will obtain This completes the theorem.

Characterization of Inclined Curves Which is Concerned With
Osculating Sphere in L n for Space-Like Curves Theorem 8.1.Let M be a space-like curve in L n+1 , n 4, and let n be an even number.Then, the relationship between m 0 i and m i values in the matrix form as follows: 2  If we denote the coe¢ cient matrix as B n ; we will get the following characterizations: Theorem 8.2.
( For n =4, we know that from ( 27) or On the other hand, from the hypothesis, we know that det B 4 = 0. Therefore, we can write k 2 k 4 = 0: Since we assume that k 2 6 = 0, from Theorem 7.1 we have that is, This proves the theorem for n = 4. Now, let us assume that the theorem is true for n = p; and let us show that the theorem is also true for n = p + 2.Then, from (27) we can write Now, we prove that the theorem is also true for n = p + 2. From (27), we can write Since det B p+2 = 0; we have k 2 k 4 :::k p k p+2 = 0.Here since k 2 6 = 0,.., k p 6 = 0, we obtain k p+2 = 0: Thus, from Theorem 7.1 we can write This proves the necessity of the theorem.
(() Let us assume that n P i=2 " i 1 m 2 i =constant, and we show that det B n = 0.For n = 4; the theorem is true.if we replace the value of k 4 in the equation (29) into the equation (30), Thus, we get det B 4 = 0.This proves that the theorem is true for n = 4. Now, let us assume that theorem is true for n = p; and let us prove that the theorem is true for n = p + 2.
2: We will use the induction method: (() Let us assume that M is an inclined curve in L n .Then, from (1), we know that We show that det B n = 0. Theorem is true for n = 4. From (5), the value of k 4 is Using this value in equation ( 29), we get According to the hypothesis, since M is an inclined curve in L 3 ; can be written, and thus we have det B = 0.This proves that the theorem is true for n = 4. Now, let us assume that the theorem is true for n = p; and let us also prove it for n = p + 2: From ( 5), (31), (34) and (35) the following equation can be written According to the hypothesis, since M is an inclined curve in L n p P i=2 " i+1 H 2 i =constant ) p P i=2 " i+1 (H 2 i ) 0 = 0, thus, det B p+2 = 0.This proves the su¢ ciency of the theorem.
()) Let us assume that det B n = 0; and let us show that the curve M is an inclined curve.Theorem is true for n = 4. Indeed, since we know that and according to the hypothesis, we may write det B 4 = 0: Let us assume that k 2 6 = 0. Thus, we obtain This means that M is an inclined curve in L 3 : Now, let us assume that the theorem is true for n = p; and let us show that the theorem is true for n = p + 2.

2 )
det B n = 0 , M L n+1 is an inclined curve in L n ; where m 0 i = dmi ds and s denotes the arc length parameter of M .Proof.1: ())Let us assume that det B n = 0: We show that n P i=2 " i 1 m 2 i =constant.For n = 4; the theorem is true.