UNIT 2. Curvatures of a Curve

UNIT 2. Curvatures of a Curve

UNIT 2. Curvatures of a Curve ========================================================================================================= ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- Convergence of k-planes, the osculating k-plane, curves of general type in Rn, the osculating flag, vector fields, moving frames and Frenet frames along a curve, orientation of a vector space, the standard orientation of Rn, the distinguished Frenet frame, Gram-Schmidt orthogonalization process, Frenet formulas, curvatures, invariance theorems, curves with prescribed curvatures. ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- One of the most important tools of analysis is linearization, or more generally, the approximation of general objects with easily treatable ones. E.g. the derivative of a function is the best linear approximation, Taylor’s polynomials are the best polynomial approximations of the function around a point. Adapting this idea to the theory of curves, the following questions arise naturally. Given a curve g :[a,b]----------L Rn and a point g(t)----- on it, find the straight line or circle (or conic or polynomial curve of degree < n) that approximates the curve around g(t)----- best or find the k-plane (k-sphere, quadric surface etc.) that is tangent to the curve at g(t)----- with the highest possible order. We shall deal now with the problem of finding the k-plane that fits to a curve at a given point best. The classical approach to this problem is the following. A k-plane is determined uniquely by (k+1) of its points that do not lie in a (k-1)-plane. Let us take k+1 points g(t0 ),g(t 1 ),...,g(t k ) on the curve. If g is a curve of "general type" then these points span a unique k-plane, which will be denoted by A(t ,...,t ). The k-plane we look for is 0 k ----- the limit position of the k-planes A(t0 ,...,t k ) as t 0 ,...,t k tend to t. To properly understand the last sentence, we need a definition of convergence of k-planes. --------------------------------------------------Definition. Let X1 ,X 2 ,...; X be k-planes. We say that the sequence of k-planes X ,X ,... tends to X if one can find points p eX p eX and linearly 1 2 j j j j independent direction vectors v1,...,v kof X j and v 1,...,v k of X such that lim p = p and lim vj = v for i = 1,...,k as j tends to infinity. j* i i ----------------------------------------Exercise. Show that a sequence of k-planes can have at most one limit. ----------------------------------------Solution. Suppose that the sequence X1 ,X 2 ,... has two limits X and Y. Then by the definition, one can find points p , q eX peX qeY and linearly j jj j j j j independent direction vectors {v1,...,v k} and {w 1,...,w k} of X j and {v1,...,v k} of X and {w 1,...,w k} of Y, such that lim p j= p, lim q j = q, 1 j j lim vi= v iand lim w i= w i for i = 1,...,k as j tends to infinity. Since {vj,...,v j} and {w j,...,w j} span the same linear space, which contains the 1 k 1 k j direction vector pj-q j, there exists a unique kxk matrix (ars ) 1<r,s<k and a vector (bj ,...,b j ) such that 1 kk vj= S a jw j for i=1,...,k. (*) is=1 is s k j j pj-q j= S b sw s j s=1 j The components ars of this matrix and the numbers bs can be determined by solving the system (*) of linear equations, thus by Cramer’s rule, they are rational functions (quotients of polynomials) of the components of the vectors vjand w j. The denominator of the quotient expressing aj and b j is a i i j j j rs r non-vanishing kxk minor of the matrix with rows w1,w 2,...,w k. Using the facts that rational functions are continuous and that the denominator of the j j fraction expressing ars or b r tends to a non-vanishing minor of the matrix with rows w ,w ,...,w if we choose the minors mentioned above properly, we 1 2 k j j get that the limits lim ars = a rs , lim b r = b r exist as j-----L 8. Taking j-----L 8 in (*) we obtain k vi= S a isw s for i=1,...,k, s=1 k p -q = S b w , s=1 s s from which follows that the vectors v1,...,v kand w 1,...,w k span the same linear space, i.e. the k-planes X and Y must be parallel and that the point q is a common point of them consequently X = Y. 44444 ------------------------------Remark. The set of all k-dimensional linear/affine subspaces of an n-dimensional linear space has a natural topology, which can easily be described with the help of the factor space topology construction. Assume that a topological space (X,t) is divided into a disjoint union of its subsets. Such a subdivision can always be thought of as a splitting of X into the equivalence classes of an equivalence relation ~ on X. Denoting by Y = X/~ the set of equivalence classes we have a natural mapping p:X----------L Y assigning to an element x e X its equivalence class [x] e Y. -------------------------------------------------------Proposition. The set t’={ U C Y: p-1(U) e t } is a topology on Y. -------------------------Proof. The proof follows from the following set theoretical identities. 2 i) p-1(o) = o, p -1(Y) = X ; ii) p-1(UnV) = p -1(U) n p -1(V); iii) p-1( u U ) = u p -1(U ). 44444 ieIi ieI i --------------------------------------------------Definition. The family t’ is called the ---------------------------------------------------------------------------------------------------------factor space topology on Y. Now consider the set n k V(n,k) = {(x1,...,x k) e (R ): x 1,...,x k are linearly independent}. V(n,k) is an open subset in (Rn) k = R nk hence it inherits a subspace topology from the standard topology of Rnk. If we define two elements of V(n,k) to be equivalent if they span the same k-dimensional linear subspace of Rn, then the set Gr(n,k) = V(n,k)/~ of equivalence classes is essentially the same as the set of all k-dimensional linear subspaces of Rn. This set becomes a topological space with the factor space topology. Topological spaces of the form Gr(n,k) are called----------------------------------------------------------------------------------------------- Grassmann manifolds. ----------------------------------------Exercise. Show that Gr(n,k) is homeomorphic to Gr(n,n-k). We can define affine Grassmann manifolds similarly. We set ~---------------------------------------------------------------------------------------------------------------------------------- n k+1 V(n,k) = {(x0,...,x k) e (R ): x 0,...,x k are not in a (k-1)-plane}, furnish V(n,k)~ with the subspace topology inherited from Rn(k+1) and define an equivalence relation on V(n,k)~ by (x ,...,x )(y ,...,y ) 46 x ,...,x and y ,...,y span the same k-plane. ~ 0 k ~ 0 k 0 k 0 k V(n,k)/~ is the set of affine k-dimensional subspaces that has the factor space topology on it. ----------------------------------------Exercise. Show that convergence of k-planes as defined above is the same as convergence with respect to the topology we have just constructed. Definition. Let g :[a,b]----------L Rn be a curve, t----- e (a,b), 1<k<n. If the -------------------------------------------------- ----- k-planes A(t0 ,...,t k ) are defined for parameters close enough to t and their limit exists as t ,...,t -----L t,----- then the limit is called the osculating 0 k ----- -------------------------------------------------- -----------------------------------k-plane of the curve g at t. The osculating 1-plane of a curve is just the -----------------------------------tangent of the curve. (The word "osculate" comes from the Latin "osculari - to kiss"). The definition of the osculating k-plane is justified by intuition. It is the k-plane that passes through k+1 points "infinitely close" to a given 3 point. However, it would not be easy to determine the osculating k-plane using directly its definition. Fortunately, we have the following theorem. n ----- -----------------------------------Theorem. Let g :[a,b]----------L R be a smooth curve, t e (a,b), 1<k<n. If the derivatives g’(t),-----g’’(t),..., ----- g(k)(t)----- are linearly independent, then the osculating k-plane of g is defined at t----- and it is the k-plane that passes through g(t)----- with direction vectors g’(t), -----g’’(t),..., ----- g(k)(t).----- To prove the theorem we need some preparation. n --------------------------------------------------Definition. Let f :[a,b]----------L R be a vector valued function, t0 ,t 1 ,....e [a,b] are different numbers. The------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- higher order divided differences or ----------------------------------------------------------------------------------------------------difference quotients are defined recursively f0 (t 0 ) := f(t 0 ) f(t ) - f(t ) f (t ,t ) := -----------------------------------------------------------------1 0 1 0 1 t1 - t 0 ... f (t ,...,t ) - f (t ,...,t ) f (t ,t ,...,t ) := -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------k-1

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