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Chapter 8 Gauss Map I 8.1 “Curvature” of a Surface We’ve already discussed the curvature of a curve. We’d like to come up with an analogous concept for the “curvature” of a regular parametrized surface S parametrized by x: U Rn. This can’t be just a number — we need at the → very least to talk about the “curvature of S at p in the direction v T (S)”. p ∈ So given v T (S), we can take a curve α: I S such that α(0) = p p ∈ → and α′(0) = v. (This exists by the definition of the tangent plane.) The curvature of α itself as a curve in Rn is d2α (note that this is with respect ds2 to arc length). However, this depends on the choice of α — for example, if you have the cylinder over the unit circle, and let v be in the tangential direction, both a curve that just goes around the cylinder and a curve that looks more like a parabola that happens to be going purely tangentially at p have the same α′, but they do not have the same curvature. But if we choose a field of normal vectors N on the surface, then d2α N is independent of the ds2 · p choice of α (as long as α(0) = p and α′(0) = v). It’s even independent of the magnitude of v — it only depends on its direction vˆ. We call this curvature k (N,vˆ). For the example, we can see that the first curve’s α′′ is 0, and that p the second one’s α′′ points in the negative zˆ direction, whereas N points in 49 the radial direction, so k (N,vˆ) is zero no matter which α you choose. p (In 3-space with a parametrized surface, we can always choose N to be N = xu∧xv .) |xu∧xv| To prove this, we see that α(s) = x(u (s),u (s)), so that dα = du1x + 1 2 ds ds u1 du2x and d2α = d2u1x +du1 du1x + du2x +d2u2x +du2 du1x + du2x . ds u2 ds2 ds u1 ds ds u1u1 s u1u2 ds u2 ds ds u1u2 s u2u2 But by normality, N x = N x = 0, so d2α N = 2 b (N)duiduj, · u1 ¡ · u2 ds2 ·¢ i,j=1 ij¡ ds ds ¢ where b (N) = x N. ij uiuj · P We can put the values b into a matrix B(N) = [b (N)]. It is symmetric, ij ij and so it defines a symmetric quadratic form B = II: T (S) R. If we use p → {xu1,xu2}asabasisforTp(S),thenII(cxu1+dxu2) = (c d) bb1211((NN)) bb1222((NN)) dc . We call II the Second Fundamental Form. ³ ´ ¡ ¢ II is independent of α, since it depends only on the surface (not on α). To show that k (N,vˆ) is independent of choice of α, we see that p d2α du du b (N)duiduj k (N,Vˆ) = N = b (N) i j = i,j ij dt dt p ds2 · ij ds ds ds 2 P ij dt X ¡ ¢ Now, s(t) = t α′(t) dt, so that ds 2 = α′(t) 2 = du1x + du2x 2 = t0| | dt | | | dt u1 dt u2| i,j dduti ddutjR gij, where gij come¡s fr¢om the First Fundamental Form. So ³ ´ P ¡ ¢ b (N)duiduj k (N,vˆ) = i,j ij dt dt p P i,jgijddutiddutj P The numerator is just the First Fundamental Form of v, which is to say its length. So the only property of α that this depends on are the derivatives of its components at p, which are just the components of the given vector v. And in fact if we multiply v by a scalar λ, we multiply both the numerator and the denominator by λ2, so that k (N,vˆ) doesn’t change. So k (N,vˆ) p p depends only on the direction of v, not its magnitude. If we now let k (N) be the maximum value of k (N,vˆ). This exists 1 p p because vˆ is chosen from the compact set S1 T (S). Similarly, we let p ⊂ 50 k (N) be the minimal value of k (N,vˆ). These are called the principle 2 p p curvatures of S at p with regards to N. The directions e and e yielding 1 2 thesecurvaturesarecalled theprincipal directions. Itwill turnoutthatthese are the eigenvectors and eigenvalues of a linear operator defined by the Gauss map. 8.2 Gauss Map Recall that for a surface x: U S in R3, we can define the Gauss map → N: S S2 which sends p to N = xu1∧xu2 , the unit normal vector at p. → p |xu1∧xu2| Then dN : T (S) T (S2); but T (S) and T (S2) are the same plane p p → Np p Np (they have the same normal vector), so we can see this as a linear operator T (S) T (S). p p → For example, if S = S2, then N = p, so N is a linear transform so it is p p its own derivative, so dN is also the identity. p For example, if S is a plane, then N is constant, so its derivative is the p zero map. For example, if S is a right cylinder defined by (θ,z) (cosθ,sinθ,z), 7→ then N(x,y,z) = (x,y,0). (We can see this because the cylinder is defined by x2+y2 = 1, so 2xx′+2yy′ = 0, which means that (x,y,0) (x′,y′,z′) = 0, · so that (x,y,0) is normal to the velocity of any vector through (x,y,z).) Let us consider the curve α with α(t) = (cost,sint,z(t)), then α′(t) = ( sint,cost,z′(t)). So (N α)(t) = (x(t),y(t),0), and so (N α)′(t) = − ◦ ◦ ( sint,cost,0). So dN (x ) = x . So in the basis x ,x , the matrix is p θ θ θ z − { } 1 0 . It has determinant 0 and 1 trace equal to 1. It turns out that the 0 0 2 2 determinant of this matrix only depends on the First Fundamental Form, ¡ ¢ and not how it sits in space — this is why the determinant is the same for the cylinder and the plane. A zero eigenvalue can’t go away no matter how you curve the surface, as long as you don’t stretch it. 51