44 Chapter 2. <strong>Surfaces</strong>: Local TheoryFigure 2.2yet aga<strong>in</strong>, we have:(†)±κ = κN · n = T ′ (0) · n = −T · (n◦α) ′ (0) = −D V n(P ) · V.This leads us to study the directional derivative D V n(P ) more carefully.Proposition 2.1. For any V ∈ T P M, the directional derivative D V n(P ) ∈ T P M. Moreover,the l<strong>in</strong>ear map S P : T P M → T P M def<strong>in</strong>ed byS P (V) =−D V n(P )is a symmetric l<strong>in</strong>ear map; i.e., for any U, V ∈ T P M,wehave(∗)S P (U) · V = U · S P (V)S P is called the shape operator at P .Proof. Forany curve α: (−ε, ε) → M with α(0) = P <strong>and</strong> α ′ (0) = V, weobserve that n◦α hasconstant length 1. Thus, by Lemma 2.1 of Chapter 1, D V n(P )·n(P )=(n◦α) ′ (0)·(n◦α)(0) = 0, soD V n(P )is<strong>in</strong>the tangent plane to M at P . That S P is a l<strong>in</strong>ear map is an immediate consequenceof Proposition 2.3 of the Appendix.Symmetry is our first important application of the equality of mixed partial derivatives. <strong>First</strong>we verify (∗) when U = x u , V = x v . Note that n · x v =0,so0= ( n · x v)u = n u · x v + n · x vu .(Remember that we’re writ<strong>in</strong>g n u for D xu n.) Thus,S P (x u ) · x v = −D xu n(P ) · x v = −n u · x v = n · x vu= n · x uv = −n v · x u = −D xv n(P ) · x u = S P (x v ) · x u .
§2. The Gauss Map <strong>and</strong> the Second Fundamental Form 45Next, know<strong>in</strong>g this, we just write out general vectors U <strong>and</strong> V as l<strong>in</strong>ear comb<strong>in</strong>ations of x u <strong>and</strong>x v :IfU = ax u + bx v <strong>and</strong> V = cx u + dx v , thenS P (U) · V = S P (ax u + bx v ) · (cx u + dx v )= ( aS P (x u )+bS P (x v ) ) · (cx u + dx v )= acS P (x u ) · x u + adS P (x u ) · x v + bcS P (x v ) · x u + bdS P (x v ) · x v= acS P (x u ) · x u + adS P (x v ) · x u + bcS P (x u ) · x v + bdS P (x v ) · x v=(ax u + bx v ) · (cSP (x u )+dS P (x v ) ) = U · S P (V),as required.□Proposition 2.2. If the shape operator S P is O for all P ∈ M, then M is a subset of a plane.Proof. S<strong>in</strong>ce the directional derivative of the unit normal n is 0 <strong>in</strong> every direction at everypo<strong>in</strong>t P ,wehaven u = n v = 0 for any (local) parametrization x(u, v) ofM. ByProposition 2.4 ofthe Appendix, it follows that n is constant. (This is why we assume our surfaces are connected.) □Example 2. Let M be a sphere of radius a centered at the orig<strong>in</strong>. Then n = 1 x(u, v), so foraany P ,wehaveS P (x u )=−n u = − 1 a x u <strong>and</strong> S P (x v )=−n v = − 1 a x v,soS Pidentity map on each tangent plane. ▽is −1/a times theIt does not seem an easy task to give the matrix of the shape operator with respect to thebasis {x u , x v }. But, <strong>in</strong> general, the proof of Proposition 2.1 suggests that we def<strong>in</strong>e the secondfundamental form, as follows. If U, V ∈ T P M,wesetII P (U, V) =S P (U) · V.Note that the formula (†) onp.44shows that the curvature of the normal slice <strong>in</strong> direction V (with‖V‖ =1)is, <strong>in</strong> our new notation, given by±κ = −D V n(P ) · V = S P (V) · V =II P (V, V).As we did at the end of the previous section, we wish to give a matrix representation whenwe’re work<strong>in</strong>g with a parametrized surface. As we saw <strong>in</strong> the proof of Proposition 2.1, we havel =II P (x u , x u )= −D xu n · x u = x uu · nm =II P (x u , x v )= −D xu n · x v = x vu · n = x uv · n =II P (x v , x u )n =II P (x v , x v )=−D xv n · x v = x vv · n.(By the way, this expla<strong>in</strong>s the presence of the m<strong>in</strong>us sign <strong>in</strong> the orig<strong>in</strong>al def<strong>in</strong>ition of the shapeoperator.) We then write[ ] []l m x uu · n x uv · nII P = =.m n x uv · n x vv · nIf, as before, U = ax u + bx v <strong>and</strong> V = cx u + dx v , thenII P (U, V) =II P (ax u + bx v ,cx u + dx v )= acII P (x u , x u )+adII P (x u , x v )+bcII P (x v , x u )+bdII P (x v , x v )