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1. Vector Differentials and Moving Frames, Darboux Frames in Particular

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1. Vector Differentials and Moving Frames, Darboux Frames in Particular CHAPTER 4. GEOMETRY OF SURFACES §1. Vector Differentials and Moving Frames, Darboux Frames in Particular The differential of a vector field in a Euclidean space is straightforward to define. If F n is a vector field in R with components F1,F2,...,Fn, that is, F = (F1,F2,...,Fn), then dF is just the vector-valued differential form with components dF1, dF2, . , dFn, that is, dF = (dF1, dF2, . , dFn). For example, writing r = (x, y, z), we have dr = (dx, dy, dz). Another example: let 2 2 2 2 F = (x − y , 2xy). The components of this vector field are F1 = x − y and F2 = 2xy with dF1 = 2xdx − 2ydy and dF2 = 2xdy + 2ydx and hence dF = 2(xdx − ydy, xdy + ydx), or dF = 2(xdx, ydx) + 2(−ydy, xdy) = G1dx + G2dy, where the “coefficients” G1 and G2 of dF are themselves vector fields given by G1 = 2(x, y) and G2 = 2(−y, x). In general, a vector-valued differential in the Euclidean space Rn can be written as Pn k=1 Vkdxk, where Vk are vector fields. Another description is that a vector-valued differential is an expression like a vector with differential forms as its components. In actual computation of differentials of vector fields, we often use various versions of the product rule, depending on what kind of products we are dealing with. In what follows, F and G are vector fields and f is a scalar field in Rn: Rule VD1 (Product Rule). d(fG) = df. G + fdG, d(F · G) = dF · G + F · dG, and when n = 3, d(F × G) = dF × G + F × dG. (F · G is the dot product and F × G is the cross product of F and G.) Often a vector field or a vector-valued differential spreads over a smaller set called a submanifold, such as a curve or a surface, so that it depends on a smaller number of parameters, as shown in the following example. Example 1.1. Find the differentials dT and dN of the unit tangent field T and the unit normal field N respectively along the parabola y = x2/2. Solution: Put the parabola y = x2/2 into parametric equations as x = t, y = t2/2. Then the velocity vector is given by v = (dx/dt, dy/dt) = (1, t). The unit tangent field T, which is a field of unit vectors tangent to the parabola, can be obtained by normalizing v: T = |v|−1v = (1 + t2)−1/2(1, t) ≡ (1 + x2)−1/2(1, x). 1 To find the normal direction at each point on the parabola, we treat the parabola as the level curve f = 0 of the scalar field f(x, y) = y − x2/2 and compute the gradient of f: ∇f = (∂f/∂x, ∂f/∂y) = (−x, 1). Normalizing ∇f, we get the unit normal field N = |∇f|−1∇f = (1 + x2)−1/2(−x, 1). Alternatively, we can get N by turning T 90o: turning a vector v = (a, b) in 90o anticlockwise, we get v⊥ = (−b, a). By the product rule, we have dT = d((1 + x2)−1/2). (1, x) + (1 + x2)−1/2 d(1, x) = (−x(1 + x2)−3/2). (1, x) dx + (1 + x2)−1/2 (0, 1) dx = (1 + x2)−3/2 {−x(1, x) + (1 + x2)(0, 1)} dx = (1 + x2)−3/2 (−x, 1) dx. Similarly, we have dN = (1 + x2)−3/2 (−1, −x) dx. Putting ω = (1 + x2)−1 dx, we can rewrite the answer as dT = ω N, dN = −ω T. In the above example, the pair of vector fields T, N form an orthonormal frame field, or a moving frame, along the parabola. By an orthonormal frame in the Euclidean space Rn we mean an orthonormal basis in this linear space, which consists of n unit vectors orthogonal to each other, say e1, e2,..., en, with ej · ek = δjk (the Kronecker delta, which is 1 when j = k and is 0 when j 6= k), that is, ej · ej = 1 and ej · ek = 0 for j 6= k, 1 ≤ j, k ≤ n. By a moving frame along a manifold M (such as a curve, or a surface, or a hypersurface) in Rn we mean at each point p of the manifold M, a frame E1(p), E2(p),..., En(p) attaching to p is assigned. Thus each Ek (1 ≤ k ≤ n) individually is a vector field along M, and together they form an orthonormal basis at each point of M. The elegant method of moving frames is credited to Elie Cartan for studying differential geometry, although Darboux used them before him. Suppose that a moving frame E1, E2,..., En is given. Then, for each k (1 ≤ k ≤ n), the differential dEk can be written as a linear combination of E1, E2,..., En with some differential forms as coefficients, say X n j dEk = ω Ej; (1.1) j=1 k j j The differential forms ωk are called the connection forms. We can write out ωk explicitly n by using the orthonormality of the frame {Ek}k=1, that is, Ej · Ek = δjk. Indeed, Xn Xn ` ` j dEk · Ej = ωk E` · Ej = ωkδ`j = ω . `=1 `=1 k 2 j and hence we have ωk = dEk · Ej. Applying “d” to both sides of the orthonormal relation Ej · Ek = δjk and using the product rule, we have dEj · Ek + Ej · dEk = 0, that is, k j k k k ωj + ωk = 0. In particular, when j = k, we have ωk + ωk = 0 and hence ωk = 0. Thus we conclude k j k ωk = 0 and ωk = −ωj for all j, k with 1 ≤ j, k ≤ n. (1.2) In Example 1.1, we may take E1 = T and E2 = N. Relation (1.2) reduces four connection j 2 forms ωk (1 ≤ j, k ≤ 2) to one. In that example, (1.1) becomes dE1 = ω1E2 = ω E2 and 1 2 −1 dE2 = ω2E1 = −ωE1, where ω = (1 + x ) dx. As usual, we denote the Cartesian coordinate functions by x1, x2, . , xn. (For any n point p = (p1, p2, . , pn) in R , xk(p) is the kth component of p, namely, xk(p) = pk. n So xk is indeed a function defined on R .) Put x = (x1, x2, . , xn), which is considered as a vector field. (When n = 2 or 3, we often use the letter r instead. Thus r = (x, y) or r = (x, y, z).) The vector-valued differential form dx = (dx1, dx2, . , dxn) is a linear combination of E1,..., En with some differential forms as coefficients, say Xn k dx = θ Ek. (1.3) k=1 k k As before, we can write out θ explicitly as: θ = dx · Ek. (In the next chapter we shall Pn k 2 see that the “metric tensor” of a submanifold is given by k=1 (θ ) .) 1 2 Example 1.2. Find the covector fields θ and θ associated with the frame E1 = T and E2 = N in Example 1.1. Solution: We have r = (x, y) and dr = (dx, dy) = (dx, d(x2/2)) = (dx, xdx). Since 2 −1/2 2 −1/2 E1 = T = (1 + x ) (1, x) and E2 = N = (1 + x ) (−x, 1), 1 2 −1/2 2 1/2 θ = dr · E1 = (1 + x ) {(dx, xdx) · (1, x)} = (1 + x ) dx, and similarly θ2 = (1 + x2)−1/2{(dx, xdx) · (−x, 1)} = 0. (As we shall see, θ2 = 0 is expectable because E1 and E2 form a so-called Darboux frame for the parabola.) A moving frame E1, E2,..., En along a submanifold M, such as a curve or a surface, is called a Darboux frame for M if each Ek is either tangent to M or normal to M. The moving frame T, N in Example 1.1 is a Darboux frame along the parabola y = x2/2. For a general curve in R2 described by parametric equations x = x(t), y = y(t), we can get a Darboux frame as follows. First compute the velocity v ≡ (dx/dt, dy/dt). Let p v = |v| ≡ (dx/dt)2 + (dy/dt)2 and ds = v dt; (1.4) 3 (|v| is the magnitude of the velocity v, called the speed, and ds, called a line element of p the curve, is often written as ds = dx2 + dy2, which is rather problematic but good for mnemonic purpose). We normalize v to get a unit tangent field T = v/v; (here we tacitly assume that v 6= 0). Turning T 90o anticlockwisely, we get a unit normal vector N. In this way we obtain a Darboux frame along the curve consisting of E1 = T and E2 = N. Now equations (1.1) become dT = ωN and dN = −ωT. Here the connection form ω depends only on one variable, namely the parameter t. Hence it can be written in the form f dt, where f is a function with a single variable t. Recall the relation ds = vdt from (1.4). Let us assume that v 6= 0 so that we may rewrite this relation as dt = v−1ds. Thus we have ω = f dt = κ ds, where κ = v−1f. We call κ = κ(t) the curvature of the parametric curve at point (x(t), y(t)). In Example 1.1, we have v = (1, t) and ω = (1 + t2)−1dt (notice that x = t). Hence v = |v| = (1 + t2)1/2, which gives ds = vdt = (1 + t2)1/2dt. Consequently ω = (1 + t2)−3/2(1 + t2)1/2dt = (1 + t2)−3/2ds. Therefore the curvature of the parabola in that example is given by κ(t) = (1+t2)−3/2 at the point (t, t2/2).
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