DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

§3. Surface Theory with Differential Forms 97so from the fact that {e 1 , e 2 , e 3 } is a basis for R 3 we infer thatSimilarly, we obtaindω 1 = ω 2 ∧ ω 21 = −ω 2 ∧ ω 12 = ω 12 ∧ ω 2 and dω 2 = ω 1 ∧ ω 12 .( 3∑ )0 = d(de i )=d ω ik e k ==k=13∑dω ij e j −j=13∑k=1(dω ik e k − ω ik ∧3∑ ( 3∑ω ik ∧ ω kj)e j =j=1∑so dω ij − 3 ω ik ∧ ω kj =0for all i, j.k=1k=1We also have the following additional consequence of the proof:3∑ )ω kj e jj=13∑j=1(dω ij −Proposition 3.2. The shape operator is symmetric, i.e., h 12 = h 21 .□3∑ω ik ∧ ω kj)e j ,Proof. From the e 3 component of the equation d(dx) =0 in the proof of Theorem 3.1 we have0=ω 1 ∧ ω 13 + ω 2 ∧ ω 23 = ω 1 ∧ (h 11 ω 1 + h 12 ω 2 )+ω 2 ∧ (h 21 ω 1 + h 22 ω 2 )=(h 12 − h 21 )ω 1 ∧ ω 2 ,k=1so h 12 − h 21 =0.□Recall that V is a principal direction if de 3 (V) isascalar multiple of V. So e 1 and e 2 areprincipal directions if and only if h 12 =0and we have ω 13 = k 1 ω 1 and ω 23 = k 2 ω 2 , where k 1 andk 2 are, as usual, the principal curvatures.It is important to understand how our battery of forms changes if we change our moving frameby rotating e 1 , e 2 through some angle θ (which may be a function).Lemma 3.3. Suppose e 1 = cos θe 1 + sin θe 2 and e 2 = − sin θe 1 + cos θe 2 for some function θ.Then we haveω 1 = cos θω 1 + sin θω 2ω 2 = − sin θω 1 + cos θω 2ω 12 = ω 12 + dθω 13 = cos θω 13 + sin θω 23ω 23 = − sin θω 13 + cos θω 23Note, in particular, that ω 1 ∧ ω 2 = ω 1 ∧ ω 2 and ω 13 ∧ ω 23 = ω 13 ∧ ω 23 .Proof. We leave this to the reader in Exercise 1.□It is often convenient when we study curves in surfaces (as we did in Sections 3 and 4 of Chapter2) to use the Darboux frame, a moving frame for the surface adapted so that e 1 is tangent to thecurve. (See Exercise 3.) For example, α is a geodesic if and only if in terms of the Darboux framewe have ω 12 =0as a 1-form on α.Let’s now examine the structure equations more carefully.