j=1 ωj k E j. (3.1) j=1 θj E j, (3.2)

Transcription

1 3. Cartan s Structural Equations and the Curvature Form Let E,..., E n be a moving (orthonormal) frame in R n and let ωj k its associated connection forms so that: de k = n ωj k E j. (3.) Recall that ωj k = ωj k and in particular ωk k = 0. Let θj be the basic forms associated with this moving frame given by: dx = n θj E j, (3.2) where x = (x, x 2,..., x n ). Recall that each x j is considered as the jth coordinate function on R n defined by x j (p) = p j for each p = (p, p 2,..., p n ) R n. Now let us differentiate (3.) and (3.2): apply operator d which turns vector-valued -forms to vector-valued 2-forms. But we must be cautious. There is no problem with the left hand sides: from d 2 = 0 we have d(de k ) d 2 E k = 0 and d(dx) = d 2 x = 0. For the right hand sides, naturally we apply the product rule. The product rule for -forms says d(fω) = df ω+fdω, where ω is any -form where f is any smooth function. We can replace f by a vector-valued function, say F, and the same identity holds: d(fω) = df ω + Fdω. However, in (3.) and (3.2), the one forms ωj k and θj are written in front of our vectorvalued functions E j. So we have to figure out a formula for d(ωf). Here it is: d(ωf) = d(fω) = Fdω + df ω = (dω)f ω df. (3.3) The minus sign in front of the last term appears because we have switched the order of the -forms df (vector-valued) and ω. Were we careless we would missed this minus sign and would made a big blunder! Applying d on both sides of (3.2), using d 2 = 0 and (3.3), changing index, and then substituting de l by means of (3.), we get 0 = d 2 x = (dθ j E j θ j de j ) = dθ j E j θ k de k = dθ j E j l= ( n θ l ωj l E j ) = k= (dθ j n l= θl ω j l ) E j. Since at each point E,..., E n form an orthonormal basis, we have dθ j = n l= θl ω j l, (3.4)

2 which are called Cartan s first structural equations. Similarly, (3.) gives 0 = d 2 E k = n (dωj k E j ω j k de j) = n dωj k E j n l= ωl k de l ( = dω j n ) k E j ωk l ωj l E j = (dω jk n ) l= ωl kω j l E j. Thus it follows that l= dω j k = n l= ωl k ω j l, (3.5) which are called Cartan s second structural equations. You must learn to appreciate the beauty of these basic structural equations in differential geometry due to Elie Cartan, who is considered by many as the greatest differential geometer. Example 3.. Verify the structural equations of the forms associated with the Frenet- Serrat frame of a curve in R 3 ; (see 4.). Solution: Recall (.5) from 4. with the identification E = T, E 2 = N, E 3 = B, (the unit tangent vector, the normal vector, and the binormal vector respectively), and ω = ω 2 2 = ω 3 3 = ω 3 = ω 3 = 0, ω 2 = ω 2 = ω, ω 3 2 = ω 2 3 = η. Also dx = θ E = θ T (this is because dx is tangential and the tangent of a curve at any point is the one dimensional space spanned by T) so that θ 2 = θ 3 = 0. A curve has only one parameter when it is described by a parametric equation x = x(t). Thus -forms such as θ, ω and η associated with the curve can be written in the form f(t)dt. But d(f(t)dt) = df(t) dt = f (t)dt dt = 0. So dθ = dω = dη = 0. Thus it remains to check 3 l= θl ω j l = 0 for the first structural equations and 3 l= ωl j ωk l = 0 for the second structural equations. Again, since all -forms here can be written in the form f(t)dt and since f (t)dt f 2 (t)dt = 0, these identities are obvious. Thus, both sides of the structure equations for the Frenet-Serrat frame along a curve are vanishing. This tells us that in some sense the local geometry of a curve is not interesting at all. So, let us turn to surfaces. Let E j (j =, 2, 3) be a Darboux frame for a surface S with E and E 2 tangential and E 3 N normal. (Here we assume that S is oriented so that the pair (E, E 2 ) 2

3 has a compatible orientation at each point. We also assume that the Darboux frame given here has the same orientation as the usual one for R 3 at each point. These assumption will be needed at one subtle point in the future, but let us not worry about this now.) Recall that the basic forms θ j and the connection forms ω j k are determined by the relations: dx = θ E + θ 2 E 2 (that is, θ 3 = 0), and de = ωe ωe 3 3, de 2 = ω2e + ω2e 3 3, de 3 = ω3e + ω3e 2 2. (3.6) The structural equations (3.4) and (3.5) become dθ = θ 2 ω 2, dθ 2 = θ ω 2, θ ω 3 + θ 2 ω 3 2 = 0, (3.7) dω 2 = ω 3 ω 2 3, dω 3 = ω 2 ω 3 2, dω 3 2 = ω 2 ω 3. (3.8) Let F be a vector field on the surface S and tangent to the surface. Consider the vectorvalued differential df. Here we have to be cautious: that F is tangential does not imply that df is also tangential. In fact, in general we expect that the normal component (df N)N of F is nonzero because F has to turn as S bends, and the acceleration has a nonzero normal component when the motion is changing its direction, even when the speed remains constant. For each vector (or vector-valued -form) v attached to a point on S, denote by v the tangential component of v, and v its normal component. Explicitly, v = v + v, where v = (v E )E + (v E 2 )E 2 and v = (v E 3 )E 3. (3.9) For a vector field (or a vector-valued -form) F tangent to our surface S, let d F = (df), d F = (df), (3.0) which will be called the covariant differential and the normal differential of the vector field F respectively. The covariant differential is responsible for the intrinsic geometry of our surface, while the normal differential determines its extrinsic shape. To get some idea about the jargon here, take a piece of paper. You can lay it flat on a table, or roll it up into a cylinder or cone. No matter how you change its shape, the intrinsic geometry is still flat. Now we state the product rules: Rule PC. d (gf) = dg. F + g d F. Rule PC2. d (ωf) = dω. F ω d F. 3

4 Rule PN. d (gf) = g d F. Rule PN2. d (ω F) = ω d F. To verify these identities, write down d(gf) = dg. F + g df and d(ωf) = dω. F ω df, and take the tangential components and normal components; (the detail is left to the reader). Notice that F = F because F is tangential. In the present section we concentrate on covariant differentials for which only Rules PC and PC2 are needed. Unlike d 2 = 0, in general d 2 is nonvanishing. In fact, this is really what makes the covariant differential d interesting. To compute d 2 F for a tangential vector field F, put F = f E + f 2 E 2, where f = F E and f 2 = F E 2 ; (here we use superscript for the components f and f 2 of F instead of subscripts in order to see a pattern). Now d E = (de ) = (ω 2 E 2 + ω 3 E 3 ) = ω 2 E 2. (3.) (Notice that ω = 0, E 2 is tangential and E 3 is normal.) Similarly we have d E 2 = ω 2E, which can be rewritten as ω 2 E if you like. (A systematic way for writing down both identities is d E k = 2 ωj k E j; k =, 2. Insertion of the extra terms ω E and ω 2 2E 2 is harmless because they vanish. This systematic way is needed if we want to study geometric objects of higher dimensions.) Thus d F = d (f E ) + d (f 2 E 2 ) To compute d 2 F d (d F), we begin with = df. E + f d E + df 2. E 2 + f 2 d E 2 = df. E + f ω 2 E 2 + df 2. E 2 + f 2 ω 2E = (df + f 2 ω 2)E + (df 2 + f ω 2 )E 2. d ((df + f 2 ω2)e ) = d(df + f 2 ω2). E (df + f 2 ω2) d E = (df 2 ω2 + f 2 dω2)e (df + f 2 ω2) ωe 2 2 = (df 2 ω2 + f 2 dω2)e (df ω)e 2 2, because ω2 ω 2 = ω2 ( ω2) = 0. Similarly, we have d((df 2 + f ω)e 2 ) = (df ω 2 + f dω)e 2 2 (df 2 ω2)e. 4

5 Thus d 2 F d 2 {f E + f 2 E 2 } = f 2 dω 2 E + f dω 2 E 2. (3.2) So d 2 sends a vector field F = f E + f 2 E 2 (which can be regarded as a vector-valued 0-form) to a vector-valued 2-form, say ϕ E + ϕ 2 E 2, such that [ ] ϕ = ϕ 2 [ 0 dω 2 dω 2 0 ] [ f f 2 ], where dω 2 = ω 3 2 ω 3. We may regard d 2 an operator-valued 2-form and its matrix representation with respect to the basis E, E 2 is given by Ω = [ 0 dω 2 dω 2 0 ]. (3.3) We call this operator-valued 2-form d 2, or the matrix-valued 2-form Ω given above, the curvature form for the surface S. The following tensor property for the curvature form d 2 is important: for all vector fields F tangent to the surface S and for all (nice and smooth) function g on S, we have d 2 (gf) = g d 2 F. (3.4) In contrast, d (gf) g d F (in general); instead we have d (gf) = dg. F+g d F. (Thus, d 2 is a tensor, but d is not.) We are going to give two proofs of (3.4) because of its importance. First proof: start with gf = (gf )E + (gf 2 )E 2. Applying (3.2), we have d 2 (gf) = (gf 2 )dω2e + (gf )dωe 2 2 = g(f 2 dω2e + f dωe 2 2 ) = g d 2 F. Second proof: we have d (gf) = dg. F + g. d F and hence d 2 (gf) = d (dg. F) + d (g d F) = d 2 g F dg d F + dg d F + g d 2 F = g d 2 F. (The second proof applies to more general situations.) From ω 2 = ω2 we see that Ω is a skew symmetric matrix. The entry dω2 of Ω turns out to be independent of the choice of our frame E, E 2. Indeed, suppose that F, F 2 is another frame, having the same orientation; (here we have used the assumption on the orientation of the surface S). Then F = (cos θ)e (sin θ)e 2 and F 2 = (sin θ)e + (cos θ)e 2 for some function θ on S. It 5

6 follows from (0.2) that d 2 E 2 = dω 2 E. To show that dω 2 is independent of frames, it is enough to check d 2 F 2 = dω 2 F. Indeed, by (3.4), d 2 F 2 = d 2 (sin θ E + cos θ E 2 ) = sin θ. d 2 E + cos θ. d 2 E 2 = sin θ. dω 2 E 2 + cos θ. dω2 E = sin θ. d( ω2) E 2 + cos θ. dω2 E = dω2 (cos θ E sin θ E 2 ) = dω2 F. Done. The 2-from dω2 is called the Pfaffian of Ω. (The Pfaffian here has nothing to do with Pfaffian equations. It is a numerical invariant associated with a skew symmetric form on an even dimensional linear space. Strictly speaking, it is a topic in the theory of multilinear algebra.) Now a surface is a two dimensional object. The top forms defined on it are 2-forms and they must be a multiple of the area form θ θ 2. In particular, dω 2 = K θ θ 2 (3.5) for some scalar function K on S, which is called the Gauss curvature for S. The reader is asked to show that K here is independent of the orientation of S; see Exercise (e). Example 3.2. Find the (Gauss) curvature K for the unit sphere S 2 by means of the parametrization given in Example.4 in 4.. Solution: From that example we have θ = cos φ dθ, θ 2 = dφ, ω 2 = sin φ dθ. So dω 2 = d( ω 2 ) = d(sin φ) dθ = cos φ dφ dθ = cos φ dθ dφ and θ θ 2 = cos φ dθ dφ. Thus dω 2 = θ θ 2, telling us that the curvature of S 2 is a positive constant: K =. In the present section we mainly study the intrinsic geometry of a surface S, which only depends on the metric tensor of S, but not on the way how S is embedded in R 3. As a consequence, the Gauss curvature here is defined intrinsically. In the next section we study the extrinsic geometry of S, and from that point of view one can define the total curvature for S. Gauss discovered that the total curvature (an extrinsic concept) coincides with what we call the Gauss curvature (an intrinsic concept). He was so pleased with this discovery that he called it theorem agregium, which means a beautiful theorem. As you will see, using Cartan s structural equations, the proof is only one line long! 6

7 Exercises. (a) Check Rules PC, PC2, PN and PN2. (b) Show that, if X and Y are vector fields tangent to a surface S, then d (X Y) = d X Y + X d Y. Deduce that, if F is a unit vector field tangent to S, then F d F = 0. (c) Show that the Gauss curvature K is independent of the orientation of S. (Hint: there are two simple ways to indicate the change of orientation: first, change the order of E and E 2 ; second, replace one of E or E 2 by its negative. Either way is easy to work with.) 2. In each of the following parts, a parametric surface r (x, y, z) = r(u, v) and a Darboux frame E k (k =, 2, 3) are given. Find (in terms of parameters u and v):. the fundamental forms θ, θ 2 and connection forms ω 2, ω 3, ω 3 2; 2. the metric ds 2 = (θ ) 2 + (θ 2 ) 2, the area form θ θ 2 ; 3. the covariant differentials d E, d E 2 the normal differentials d E, d E 2 ; and 4. the Gaussian curvature K. (a) Let X = X(t), Y = Y (t) be a curve in XY -plane with an arc-length parametrization: X (t) 2 + Y (t) 2 =. The cylinder S based on this curve is given by x(u, v) = X(u), y(u, v) = Y (u), z(u, v) = v. The Darboux frame for S is given by E = r u = (dx/du, dy/du, 0), E 2 = r v = (0, 0, ), E 3 = E E 2. (b) The unit sphere S 2 : r = (cos u cos v, sin u cos v, sin v) with the Darboux frame E = ( sin u, cos u, 0), E 2 = ( cos u sin v, sin u sin v, cos v), E 3 = E E 2 = r. (c) The circular cone with the circle x 2 + y 2 = as the base curve and the point A = (0, 0, ) as the apex, together with the Darboux frame given as follows: r(u, v) = (( v) cos u, ( v) sin u, v), E = r/ u, normalzed = ( sin u, cos u, 0), E 2 = r/ v, normalized = 2 ( cos u, sin u, ), E 3 = E E 2. (d) The unit sphere S 2, parametrized by stereographic projection: r(u, v) = (2uh, 2vh, (u 2 + v 2 )h), where h = /(u 2 + v 2 + ) 7

8 with the Darboux frame is obtaining by normalizing r u, r v, r u r v. (e) The parametric surface r(u, v) = ( u 2 /2, uv/ 2, v 2 /2 ) (see Exercise 6 in 2.8) with the Darboux frame obtained by normalizing r u, (r u r v ) r u, r u r v. (Patience is needed for completing the computation.) 3. Decomposition (3.9) uses the usual inner product u v = u v + u 2 v 2 + u 3 v 3 for two vectors u = (u, u 2, u 3 ) and v = (v, v 2, v 3 ) in R 3. For certain problems (e.g. in theory of relativity) it is more appropriate to use other types of inner products. For example, in deriving the hyperbolic metric (2.8) in 4.2, we consider the hyperboloid H 2 : x 2 y 2 + z 2 = in R 3 with the inner product of two vectors u = (u, u 2, u 3 ) and v = (v, v 2, v 3 ) defined to be u v = u v + u 2 v 2 u 3 v 3 corresponding to the indefinite metric dx 2 + dy 2 dz 2 chosen. Parametrize the lower sheet of H 2 by r(u, v) = (cos u sinh v, sin u sinh v, cosh v). Let E k be vector fields determined by r u = sinh u E, r v = E 2 and r = E 3. (a) Verify that the vector fields E, E 2, E 3 form a Darboux frame by checking: E E =, E 2 E 2 =, E 3 E 3 = and E E 2 = E E 3 = E 2 E 3 = 0, (b) Compute dr, de k to find θ, θ 2 and ω k j ( k, j 3). (c) Find the metric (θ ) 2 + (θ 2 ) 2, the area form θ θ 2, as well as the curvature K. (d) Find the covariant differentials d E, d E 2, normal differentials d E, d E 2, as well as the second covariant differentials d 2 E and d 2 E 2. 8