ν(u, v) = N(u, v) G(r(u, v)) E r(u,v) 3.

Transcription

1 5. The Gauss Curvature Beyond doubt, the notion of Gauss curvature is of paramount importance in differential geometry. Recall two lessons we have learned so far about this notion: first, the presence of the Gauss curvature is reflected in the fact that the second covariant differential d 2 in general is not zero, while the usual second differential d2 is; second, the Gauss curvature K defined intrinsically via dω2 = K θ θ 2 is the same as the one defined extrinsically as the determinant of the second fundamental form: II(X, Y) = dx E 3, Y. In the present section we explore two classical aspects of the Gauss curvature: its geometric interpretation via the Gauss map and the Gauss formula for its computation. In the next section we will see more about the Gauss curvature when we discuss geodesics. As usual, let S be a surface and let E k (k =, 2, 3) be a Darboux frame along S. The fundamental forms θ j (j =, 2) and the connection forms ωj k ( j, k 3) are defined in the usual way; see 4.. (The reader is advised to review Cartan s structural equations.) We shall regard the unit E 3 at any point p on S as a point on the unit sphere S 2. To facilitate this point of view, formally we define the Gauss map G for the surface S to be the mapping from S to S 2, which sends a point p on S to the normal vector E 3 at p: G: S S 2, G(p) = E 3 p. (5.) Let r = r(u, v) be a parametrization of the surface S. For clarity, we denote by σ the mapping from a domnain D in uv-plane to the surface S which sends (u, v) to r(u, v). (We need σ here for forming pullbacks by σ.) Consider the map ν from D to the unit sphere S 2, which sends (u, v) to the normal vector E 3 at r(u, v): ν : D S 2, ν(u, v) = N(u, v) G(r(u, v)) E r(u,v) 3. Here N is considered to be a parametrization of the sphere S 2 ; (basically, N is just E 3 considered as a function of u, v). Now both N(u, v) and (r u r v )/ r u r v are

2 unit normal vectors at r(u, v). We assume that they coincide; (this really means that the surface S is properly oriented, and our parametrization and Darboux frame both agree with this orientation). Recall the area form for the surface α S σ ds, and σ ds = 2 dr dr = r u r v du dv = r u r v du dv N = α S N; (5.2) (see Example 2.4 in 2.2). The area form for S 2 is α S 2 = ν ds, with ν ds = 2 dn dn = 2 de 3 de 3 = 2 (ω 3E + ω 2 3E 2 ) (ω 3E + ω 2 3E 2 ) = ω 3 ω 2 3 E 3 ; (5.3) (the reader should have no trouble to fill in the missing detail in this computation). So ν ds = ω 3 ω 2 3 N = dω 2 N = dω 2N = K α S N = K σ ds in view of the first identity in (3.8) and (3.5). We conclude ν ds = K σ ds, in particular, α S 2 = K α S. (5.4) Let P is a small patch of the surface S around a point p on S, which comes from a small region U in D: σ(u) = P. The image of P in S 2 via the Gauss map is G(P ) = G(σ(U)) = ν(u). The area of P is a(g(p )) = U α S 2 a(p ) = U α S, while the area of its image G(P ) is = U K α S K(p) U α S = K(p) a(p ). So the absolute value of the Gaussian curvature K(p) is the limit of the ratio a(g(p ))/a(p ), when the region P shrinks to the point p: K(p) = lim P p area of G(P ) area of P. (5.5) This identity helps us to visualize the Gaussian curvature. a plane, G is constant and hence G(P ) is always a single point. a(g(p ))/a(p ) is always zero and consequently K 0. For example, when S is Thus, for a plane, Similarly, when S is a cylinder, G(P ) is always lying in a great circle and again we have K 0. When S is the unit sphere S 2, G is the identity map, telling us G(P ) = P. Hence K for the unit sphere. (The standard classical proof of (5.5) uses computation involving the second fundamental form in a complicated way. We avoid this and shorten the proof via a simple aspect of the super Lie algebra of vector-valued differential forms. Neat!) 2

3 The computation of the curvature K for a given surface S in general is very messy. However, when the metric tensor g jk of S satisfies g 2 = g 2 = 0, we can derive a useful formula for computing K. Putting the parametric equation for S as r = r(u, v), this means r u r v = g 2 = 0. Also, g = r u r u = r u 2 and g 22 = r v r v = r v 2. We can take our moving frame to be E = r u r u and E 2 = r v r v. Recall the identity dr = θ E + θ 2 E 2. On the other hand, dr = r u du + r v dv = r u E. du + r v E 2. dv = g du. E + g 22 dv. E 2. Hence θ = g du and θ 2 = g 22 dv. Thus the area form is θ θ 2 = g g22 du dv = g du dv, where g is the determinant of the 2 2 matrix [g jk ] associated with the metric tensor of S. Recall that dω 2 = K θ θ 2. Hence our aim is to find ω 2 and then differentiate it. Since ω 2 is a -form with two varables u and v, we can put it as ω 2 = hdu + kdv, where h, k are functions of u, v. The first structural equations give Similarly, we have dθ = θ 2 ω 2 = g 22 dv (hdu + kdv) = h g 22 du dv. dθ 2 = θ ω 2 = θ ( ω 2) = g du ( hdu kdv) = k g du dv. On the other hand, dθ = d( g du) = d( g ) du ( g = u du + ) ( g g dv du = v v ) du dv. Similarly we have dθ 2 = (d g 22 ) dv = ( ) u g22 du dv. Thus we have h = g = ( ) g22 v g22 2 g v g = 2 g v g. Similarly we have k = 2 g u g 22. Hence dω 2 = d(hdu + kdv) = dh du + dk dv = (h u du + h v dv) du + (k u du + k v dv) dv = ( h v + k u )du dv = g (k u h v ) θ θ 2 = K θ θ 2, 3

4 where K = ( { } 2 g g u u g 22 + { }) g v v g (5.6) which is the formula we are looking for, called the Gauss formula. In case g = g 22 and g 2 = g 2 = 0, that is, the metric tensor of S can be written as ds 2 = G(du 2 + dv 2 ) (with G = g = g 22 = g), this formula can be simplified. In this case, (u, v) is called a conformal coordinate system, and (5.6) becomes K = ( 2G = 2G { u G ( u } u G { u log G + { }) u G v G } + { }) v v log G perhaps we should keep this formula in mind and treat it as a rule: = log G. 2G Rule GF. If g 2 = g 2 = 0 and G = g = g 22, then K = log G. 2G A deep theorem in differential geometry says that every (smooth) surface has local conformal coordinates. This theorem seems to be valuable only for theoretical purposes. Example 5.. The surface S obtained by revolving z = f(x) about the z-axis can be written in parametric equations as x = r cos θ, y = r sin θ and z = f(r). Find the curvature K for S in terms of the function f. Solution: The same computation as that of Example 2. in 4.2 leads to the identity dx 2 + dy 2 = dr 2 + r 2 dθ 2. On the other hand, from z = f(r) we have dz = f (r)dr and hence dz 2 = f (r) 2 dr 2. So the metric tensor for S is ds 2 = dx 2 + dy 2 + dz 2 = ( + f 2 )dr 2 + r 2 dθ. Thus we have g = + f 2, g 22 = r 2 and g 2 = g 2 = 0. Also g g g 22 = r + f 2. Notice that g is independent of θ. Therefore recipe (5.6) gives K = ( ) 2 g. 2r g r = r + f 2 in view of the trivial identity yy = 2 (y2 ). 4 d dr = d, (5.7) + f 2 2r dr + f 2

5 Example 5.2. Find a surface of revolution with curvature K. Solution: We look for a function f(r) satisfies d 2r dr ( + f 2 ) =, that is d dr ( + f 2 ) = 2r. Thus ( + f 2 ) = r 2 + c for some constant c. Let us choose c = 0. Then f = ± r 2 /r. The meaning of the last identity is indicated in the following graph for f, which is called a tractrix. Identity (5.6) tells us that, in the special case of g 2 = 0, the curvature K is determined by the metric tensor g jk. In fact, this is true without the condition g 2 = 0. Thus, so long as a metric is given, it makes sense to talk about the curvature K. In other words, the curvature is an intrinsic quantity, not depending on the shape of the given surface. Example 5.3. Find the Gauss curvature for the hyperbolic (upper half) plane H 2 = {(x, y) R 2 y > 0} with the metric given by ds 2 = (dx 2 + dy 2 )/y 2. Solution. We may apply Rule GF with G = /y 2 to obtain K = ( ) 2 2G log G = y2 log y y x y 2 log y = y 2 y y =. Hence the Gaussian curvature of H is a negative constant: K =. That the curvature of hyperbolic plane H 2 is a constant actually follows from the homogeneity of H: everywhere looks like the same. In fact, H 2 has a Lie group structure and its metric (dx 2 + dy 2 )/y 2 is derived from an invariant -form, namely the Maurer- Cartan form for this Lie group. So the fact that H 2 has a constant curvature is not surprising, provided you realize this Lie group structure.) More generally, if the group of isometries of a manifold acts transitively, then this manifold has a constant total curvature. Such a manifold is usually called a homogeneous space. 5

6 Exercises. (a) Give some detail of the computation in (5.3). (b) Show that identity (5.4) can be rewritten as N u N v = K r u r v. 2. In each of the following parts, find the Gauss curvature K by the Gauss formula. (a) The sphere in sterographic coordinates, with ds 2 = (du 2 + dv 2 )/( + u 2 + v 2 ) 2. (b) The disk model D for hyperbolic geometry, with ds 2 = (du 2 + dv 2 )/( u 2 v 2 ) 2. (c) The helicoid r(u, v) = (u cos v, u sin v, bv). 3. Show that the answers given in Example 5. and in Example 4.3 agree. Hint: A careful book-keeping tells us how to identify various symbols: r = Y (s) and Z(s) = f(y (s)). Thus, by the chain rule, we have dz/ds = (df/dr)(dy/ds). So /[ + (df/dr) 2 ] = (dy/ds) 2 /[(dy/ds) 2 + (dx/ds) 2 ] = (dy/ds) 2. Thus 2r d dr + (df/dr) 2 = 2Y (s) because ds/dr = (dr/ds) = Y (s). d dr Y (s) 2 = 2Y (s) 2Y (s)y (s) ds dr = Y (s) Y (s), 4. Let S be a surface with a Darboux frame E k ( k 3). As usual we denote by (j =, 2) its fundamental forms (let us recall that dr = θ E + θe 2 ) and ω k j ( j, k 3) its connection forms. Fix a constant λ > 0. Consider the parallel θ j surface S λ consisting of r λ = r + λn, where r is a general point on S and N E 3 the corresponding unit normal vector at that point. (a) Verify: dr λ dr λ = dr dr + λ 2 dn dn = ( + λ 2 K) dr dr. Hint: dr de 3 = (θ E + θ 2 E 2 ) (ω 3E + ω 2 3E 2 ) = (θ ω 3 + θ 2 ω 3 2) E 3 = 0. (b) Show that, by translating E j from their base point r on S to r λ r + λn for each point on S, we get a Darboux frame for S λ. Also show that the connection forms of S λ with respect to this Darboux frame is still ωj k, but the fundamental forms become Θ j = θ j + λω j 3 (j =, 2). Hint: Part (a) tells us that E 3 is also normal to S λ. (c) Show that, if K is the total curvature and H is the mean curvature of S, then the total curvature K λ and the mean curvature H λ of S λ are given by K λ = K λh + λ 2 K and Hλ = 6 H λk λh + λ 2 K.

7 Hint: Use (5.7) and part (b) above. In the next two exercises we study minimal surfaces, that is, surfaces with mean curvatures equal to zero (everywhere). 5. (a) Show that the total curvature K of a minimal surface is 0. Hint: Recall that H and K are just the trace and the determinant of a symmetric 2 2 matrix [h ij ]; see (5.). So this is just a simple problem in algebra. (b) Show that the Gauss map G: S S 2 of a minimal surface S is conformal. The meaning of this statement is explained as follows. Recall that the metric tensor for S is (θ ) 2 + (θ) 2, where the -forms θ j come from dr = θ E + θ 2 E 2, where r is a general point on S. Since E 3 is normal to both S and S 2, we can also regard E j as a Darboux frame for the sphere S 2. Now the Gauss map sends r to E 3 at r and de 3 = ω 3E + ω 2 3E 2. So the metric tensor for S 2 can be expressed as (ω 3) 2 + (ω 2 3) 2. That G is conformal means that there is a scalar function λ such that (ω 3) 2 + (ω 2 3) 2 = λ { (θ ) 2 + (θ 2 ) 2}. Hint: Use (4.8), h 2 = h 2 and H = h + h 22 = (a) Let r = r(u, v) be a conformal parametrization of a minimal surface S, that is, its metric tensor has the form ds 2 (θ ) 2 + (θ 2 ) 2 = G(u, v)(du 2 + dv 2 ). Show that r = r(u, v) is harmonic: r ( x, y, z) = 0, in other words, its components x = x(u, v), y = y(u, v), z = z(u, v) are harmonic as functions of u, v. Hint: We have g = g 22 = G while g 2 = g 2 = 0. So, by part (c) of Exercise 5 in 4.4, we have H = (L uu + L vv )/G 2 = 0, giving us (r uu + r vv ) N = 0. It remains to check (r uu + r vv ) r u = 0 and (r uu + r vv ) r v = 0, which can be verified by using the assumption that u, v are conformal coordinates. Remark. By our previous exercise, we see that if the Gauss map of a minimal surface S is locally invertible, then we can use any conformal coordinate system for S 2 (such as the sterographic coordinate system) to give (local) coordinate patches for S. Thus our assumption that (u, v) is a conformal coordinate system is not stringent. The rest of this exercise depends heavily on complex analysis. (b) Consider u, v as the real and the imaginary parts of a complex variable w: w = u + iv so that x, y, z (the components of r = r(u, v) becomes functions of w. Let x = x/w, ỹ = y/w and z = z/w (recall that x/w = (/2)((x/u) (x/v)); 7

8 see (.5) in 3.) and let Φ = ( x, ỹ, z). Check that x, ỹ and z are holomorphic functions satisfying x 2 + ỹ 2 + z 2 = 0. (c) Show that if f and g are meromorphic functions defined by f = x + iỹ and g = z/( x + iỹ), then we have ( f( g 2 ) Φ =, 2 f( + g 2 ) ), fg, 2i called Weierstrass s representation for the minimal surface S. (d) Following the notation from the previous part, show that the function G in the metric tensor G(du 2 +dv 2 ) is given by G = 4 f 2 (+ g 2 ) 2. Compute the unit normal field N = r u r v / r u r v, L uu = r uu N, L uv = r uv N and L vv = r vv N, with answers in terms f and g and their derivatives. Then use the formula for total curvature K given in Exercise 5 of 4.4 to derive K = 6 g 2 / f 2 ( + g 2 ) 4. Hint: Let Φ be the complex vector obtained by taking complex conjugates of components of Φ. Then r u = Φ + Φ and r v = i(φ Φ). Notice that Φ Φ = 0 and Φ Φ = 0. An easy computation shows r u r u = r v r v = r u r v = Φ Φ and r n r v = i Φ Φ. Note that, for any holomorphic function φ, φ u φ/u = φ and φ v φ/v = iφ. (e) Show that, if X, Y, Z are functions such that X = 2 F ( G2 ), Y = 2i F ( + G2 ) and Z = F G for some functions F and G, then X 2 + Y 2 + Z 2 = 0. (f) Show that, if X(w), Y (w), Z(w) (w = u + iv) are holomorphic functions with X 2 + Y 2 + Y 2 = 0, if P X, P Y and P Z are primitives of X, Y and Z respectively (that is, P X is a holomorphic and dp X /dw = X, etc.), and if x, y, z are real parts of X, Y and Z respectively, then the parametric equations x = x(u, v), y = y(u, v) and z = z(u, v) give a minimal surface in conformal coordinates u, v. 8