DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

96 Chapter 3. Surfaces: Further TopicsHow do we know such a moving frame exists? If x: U → M is a parametrized surface, we can startwith our usual vectors x u , x v and apply the Gram-Schmidt process to obtain an orthonormal basis.Or, if U is a region containing no umbilic points, then we can choose e 1 and e 2 to be unit vectorspointing in the principal directions (this approach was tacit in many of our proofs earlier).If x: M → R 3 is the inclusion map (which we may choose, in a computational setting, toconsider as the parametrization mapping U → R 3 ), then we define 1-forms ω 1 ,ω 2 on M bydx = ω 1 e 1 + ω 2 e 2 ;i.e., for any V ∈ T P M,wehave V = ω 1 (V)e 1 + ω 2 (V)e 2 ,soω α (V) =I(V, e α ) for α =1, 2. So far,ω 1 and ω 2 keep track of how our point moves around on M. Next we want to see how the frameitself twists, so we define 1-forms ω ij , i, j =1, 2, 3, byde i =3∑ω ij e j .j=1Note that since e i · e j = const for any i, j =1, 2, 3, we have( 3∑ )0=d(e i · e j )=de i · e j + e i · de j = ω ik e k · e j += ω ij + ω ji ,k=1( 3∑k=1ω jk e k)· e iso ω ji = −ω ij for all i, j =1, 2, 3. (In particular, since e i is always a unit vector, ω ii =0for all i.)If V ∈ T P M, ω ij (V) tells us how fast e i is twisting towards e j at P as we move with velocity V.Note, in particular, that the shape operator is embodied in the equationde 3 = ω 31 e 1 + ω 32 e 2 = − ( ω 13 e 1 + ω 23 e 2).Then for any V ∈ T P M we have ω 13 (V) =II(V, e 1 ) and ω 23 (V) =II(V, e 2 ). Indeed, when wewriteω 13 = h 11 ω 1 + h 12 ω 2ω 23 = h 21 ω 1 + h 22 ω 2for appropriate coefficient functions h αβ ,wesee that the matrix of the shape operator S Prespect to the basis {e 1 , e 2 } for T P M is nothing but [ ]h αβ .Most of our results will come from the followingwithTheorem 3.1 (Structure Equations).dω 1 = ω 12 ∧ ω 2 and dω 2 = ω 1 ∧ ω 12 , and3∑dω ij = ω ik ∧ ω kj for all i, j =1, 2, 3.k=1Proof. From the properties of the exterior derivative, we have( 3∑ ) ( 3∑ )0 = d(dx) =dω 1 e 1 + dω 2 e 2 − ω 1 ∧ ω 1j e j − ω 2 ∧ ω 2j e jj=1j=1= ( dω 1 − ω 2 ∧ ω 21)e1 + ( dω 2 − ω 1 ∧ ω 12)e2 − ( ω 1 ∧ ω 13 + ω 2 ∧ ω 23)e3 ,