CHAPTER THREE: SURFACES IN R 3

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1 CHAPTER THREE: SURFACES IN R 3 June 15, THE SHAPE OPERATOR Let M be a surface, P M. Define the shape operator S P as the negative (directional) derivative of the unit normal n (as a vector valued function on M), i.e. S P (v) = D v n(p ). If x : U R 3 is a parametrization of M with x(u 0, v 0 ) = P, and if write n(u, v) := (n x)(u, v), then, from the previous discussion (see (3.2.1) and (3.2.2)), S P (v) = ( λn u µn v )(u 0, v 0 ) if v = λx u P + µx v P. (3.4.1) We note that the shape operator S P played an important role in the study of the geometry of surfaces. Recall that in the theory of curves, we used T (t), the derivative (differential) of the unittangent vector T to the curve, to study the geometry of curves (like curvature, torsions) etc. In the surface case, we replace T by n, and replace the derivative by directional derivative. This is the motivation of introducing the shape operator S P. A very important fact about S P is that S P (v) is, in fact, in T P (M) for every v T P (M). Theorem For every P M and v T P (M), S P (v) T P (M). Moreover, the map S P : T P (M) T P (M) is a 1

2 symmetric linear map, where S P is symmetric means that, for any u, v T P (M), we have S P (u) v = u S P (v). S P is called the shape operator at P. Proof. Step 1: Since n n = 1, by taking the differentiation w.r.t. to u, we get n u n = 0. Hence S P (x u ) = n u T P (M). For the same reason, S P (x v ) = n v T P (M). Thus S P (v) T P (M) by linearity. Step 2: Since {x u, x v } forms a basis of T P (M), we only need to verify, by linearity, In fact, S P (x u ) x v = x u S P (x v ). LHD = S P (x u ) x v = D xu n(p ) x v = n u x v = n v x u = D xv n(p ) x u = RHS, where, in above, we used the identity n u x v = n v x u, which can be proved as follows: x v T P (M) = n x v = 0 = 0 = (n x v ) u = n u x v +n x uv, = n u x v = n x uv. Similarly, using n x u = 0, we can obtain that n v x u = n x uv. Hence, n u x v = n v x u. This proves the theorem. Q.E.D. 3.5 THE SECOND FUNDAMENTAL FORM We now use the shape-operator to definition the Second Fundamental Form. 2

3 Definition. Let M be a regular surface. The second fundamental form of M, denoted by II, assigns, for each P M, a map II P : T P (M) T P (M) R defined by II P (u, v) = v S P (u), for u, v T P (M). Similar to the first fundamental form case, from bilinearlty, II P (ax u + bx v, cx u + dx v ) = e(ac) + f(bc + ad) + g(bd). where e = II P (x u, x u ), f = II P (x u, x v ), g = II P (x v, x v ). Hence, we sometimes also call the data {e, f, g} the second fundamental form, if no confusion arises. Note that, by the definition e = x u S P (x u ) = x u n u = n x uu, where, in the last equation, we used the fact that x u n = 0, so (x u n) u = 0, which gives x u n u = n x uu. Similarly, f = n u x v = n x uv ; g = x v n v = n x vv. Example. Consider a surface of revolution (see example in 3.1) where φ 2 + ψ 2 = 1. x(u, v) = (φ(u) cos v, φ(u) sin v, ψ(u)) 3

4 Then x u = (φ (u) cos v, φ (u) sin v, ψ (u)), x v = ( φ(u) sin v, φ(u) cos v, 0). Hence, x u x v = ( φ(u)ψ (u) cos v, φ(u)ψ (u) sin v, φ(u)φ (u)), x u x v = φ(u) (since φ 2 + ψ 2 = 1). Therefore Also, n = x u x v x u x v = ( ψ (u) cos v, ψ (u) sin v, φ (u)). x uu = (φ (u) cos v, φ (u) sin v, ψ (u)), x uv = ( φ (u) sin v, φ (u) cos v, 0), x vv = ( φ(u) cos v, φ(u) sin v, 0). So the Second Fundamental Form is: e = x uu n = φ(u) ψ (u) φ (u)ψ (u), f = x uu n = 0, g = x vv n = φ(u)ψ (u). If we take φ(u) = sin u, ψ(u) = cos u, then the Second Fundamental Form for the unit sphere x(u, v) = (sin u cos v, sin u sin v, cos u), 0 < u < π, 0 < v < 2π, is e = 1, f = 0, g = sin 2 u. 3.6 THE MATRIX OF THE SHAPE OPERATOR S P Let x : U M is a parametrization of M. We find the matrix of S P : T P (M) T P (M) under the basis {x u P, x v P }. The 4

5 matrix is very important later (in Chapter 4) in the calculation of the curvatures. Let S P (x u ) = a 11 x u P + a 21 x v P, S P (x v ) = a 12 x u P + a 22 x v P, then, from the linear algebra, the matrix of S P with respect to the basis {x u P, x v P } is A = a 11 a 12 a 21 a 22 We now calculate a 11, a 12, a 21, a 22. For simplicity (without confusion), we just write x u P, x v P as x u, x v by dropping P. From S P (x u ) = a 11 x u + a 21 x v, we get, by taking dot-product with x u, and S P (x u ) x u = a 11 x u x u + a 21 x v x u S P (x u ) x v = a 11 x u x v + a 21 x v x v. Since x u x u = E, x u x v = F, x v x v = G, S P (x u ) x u = e, S P (x u ) x v = f where {E, F, G} is the First Fundamental Form and {e, f, g} is the Second Fundamental Form, the above equations become Similarly, from we get e = a 11 E + a 21 F, f = a 11 F + a 21 G. S P (x v ) = a 12 x u + a 22 x v, f = a 12 E + a 22 f, g = a 12 F + a 22 G. 5

6 Hence e f f g = E F F G a 11 a 12 a 21 a 22 Therefore, the matrix of the shape operator S P T P (M) with respect to the basis {x u, x v } is where F I = A = E F F G a 11 a 12 a 21 a 22 = FI 1 F II,, F II = e f f g : T P (M) (3.6.1) By the direct calculation, we have, in above, eg ff a = fg gf b = fe ef c = ge ff d = EG F 2. (3.6.2) Example. Consider a surface of revolution x(u, v) = (φ(u) cos v, φ(u) sin v, ψ(u)) where φ 2 +ψ 2 = 1. As we calculated, E = 1, F = 0, G = φ 2 (u) and e = φ (u)ψ (u) φ (u)ψ (u), f = 0, g = φ(u)ψ (u). Hence, and F II = F I = φ 2 (u), φ (u)ψ (u) φ (u)ψ (u) 0 0 φ(u)ψ (u) 6

7 Therefore, the the matrix of the shape operator S P : T P (M) T P (M) with respect to the basis {x u, x v } is A = F 1 I F II = φ (u)ψ (u) φ (u)ψ (u) 0 0 ψ (u)/φ(u) Note that we can also use (3.6.2) to calculate the matrix directory. Finally, to convince you that the shape operator S p indeed describes the geometry (shape) of the surface near the point of p, we prove the following result. Theorem If the shape operator of x : U R 3 is zero everywhere, then x is a part of a plane. Proof: From the condition that the shape operator of x : U R 3 is zero everywhere, we have n u = n v 0, so that n is constant. Since n is perpendicular to both x u and x v, n x u = 0 and n x v = 0. Thus (n x) u = n x v = 0 and (n x) v = n x v = 0. This implies that n x is constant, say d. Then the surface x satisfy the equation n x = d, where n is a constant. This equation is the equation of a plane. In fact, if we write n = (a, b, c), then this equation is ax + by + cz = d. This proves that x is a part of a plane. Q.E.D. RECAP 7

8 The Shape operator S P = D v n(p ) : T P (M) T P (M) is a linear map. The purpose of introducing the shape operator S P is to describe (various) curvatures for the surface M (see Chapter 4 later). The matrix of S P with respect to the basis {x u, x v } is where F I = A = a 11 a 12 a 21 a 22 E F F G = FI 1 F II,, F II = By the direct calculation, we have, a = c = eg ff fe ef e f f g fg gf b = ge ff d = EG F 2. The Second Fundamental Form II assigns, for each P M, a map II P : T P (M) T P (M) R defined by II P (u, v) = v S P (u). For every u, v T P (M), u = ax u + bx v, v = cx u + dx v. II P (u, v) = e(ac) + f(bc + ad) + g(bd), where e = n x uu, f = n x uv, g = n x vv. 8