Math 553x1 σιι− June 4, 1996 Notes on Cartan's Method of Moving Frames Andrejs Treibergs The method of moving frames is a very efficient way to carry out computations on surfaces. Chern's Notesy give an elementary introduction to differential forms. However these are brief and without examples. In this handout, following Chern, I review my version of the notation and develop a couple of explicit examples. Hopefully these remarks will help you see what's going on. Derivation of the equations. We begin by quickly reviewing Cartan's formulation of local differential geometry in terms of moving frames. Our indices shall have the range i; j; k; : : : = 1; 2 and A; B; C; : : : = 1; 2; 3 and we follow the convention that sums are over repeated indices. Let S ⊂ R3 be a surface and let the dot product of R3 be given h· ; ·i. Let a local chart for S be given by the map X : U ! S where U ⊂ R2 is an open set. In a neighborhood of a point we choose a local orthonormal frame, smooth vector fields fe1; e2; e3g such that (1) heA; eBi = δAB; the Kronecker delta. We choose the frame adapted in such a way that e3 is the unit normal vector and e1 A and e2 span the tangent space TpS. The corresponding coframe field of one forms f! g is defined by the differential 1 2 (2) dX = ! e1 + ! e2: In local coordinates (u1; u2) 2 U the one forms are linear functionals of the form 1 2 1 1 2 2 (3) !(·) = P1(u ; u )du + P2(u ; u )du where P and Q are smooth functions in U. dui are the differentials for the coordinate functions ui : U ! R which form a basis for the linear functionals on the vector spaces T(u1;u2)U. Thus, for vectors which in local coordinates have the expression @ @ V = v1(u1; u2) + v2(u1; u2) ; @u1 @u2 the one form (3) acts by 1 1 2 1 2 2 1 2 1 2 !(V ) = v (u ; u ) P1(u ; u ) + v (u ; u ) P2(u ; u ): For simplicity sake we have only described one forms on U. But by the usual identifications between X : U ! X(U) or dX : T(u1;u2)U ! TpS, the one forms can be interpreted as linear functionals on TpS as well. For example, if we choose vector fields Ei in U such that dX(Ei) = ei then we may set! ~(ei) := !(Ei). yS. S. Chern, Euclidean Differential Geometry Notes (Math 140), University of California, Berkeley 2 Andrejs Treibergs i i 2 1 2 2 2 i In particular,! ~ (eA) = δ A. It also means that if the metric takes the form ds = (! ) + (! ) then ! is i i automatically a coframe and the vector fields ei determined by duality! ~ (ej) = δ j are automatically the corresponding orthonormal frame. One forms can be integrated on curves in the usual way. If α : [0;L] ! U is a piecewise smooth curve where α(t) = u1(t); u2(t) then Z Z L Z 0 1 2 ! := !(α (t)) dt = P1du + P2du α([0;L]) 0 α([0;L]) is the usual line integral. Two one forms may be multiplied (wedged) to give a two form, which is a skew symmetric bilinear form on the tangent space. In general terms, if # and ! are one forms then for vector fields Y , Z we have the formula (# !)(Y; Z) := #(Y )!(Z) − #(Z)!(Y ): b In local coordinates this gives 1 2 1 2 1 2 (p1du + p2du ) (q1du + q2du ) := (p1q2 − p2q1)du du : b b Because three vectors are dependent there are no skew symmetric three forms in R2 and the most general two form is β = A(u1; u2)du1 du2: b When evaluated on the vectors @ @ @ @ V = v1(u1; u2) + v2(u1; u2) Z = z1(u1; u2) + z2(u1; u2) @u1 @u2 @u1 @u2 the two form gives β(V; Z) = A(u1; u2) v1(u1; u2)z2(u1; u2) − v2(u1; u2)z1(u1; u2) : A two form, say β, may be integrated over a region R ⊂ U by the formula Z Z β = A(u1; u2) du1du2 R R where du1du2 denotes Lebesgue measure on U. The first fundamental form, the metric, has the expression from (2) ds2 =hdX; dXi 1 2 1 2 (4) =h! e1 + ! e2;! e1 + ! e2i =(!1)2 + (!2)2: 1 2 In this notation, the area form is ! ! . That's because by using (2) to write in terms of the ei basis, the area of the parallelogram spanned bybdX(@=@u) and dX(@=@v) is @ @ @ @ @ @ !1 !2 − !1 !2 = !1 !2 ; : @u @v @v @u b @u @v The Weingarten equations express the rotation of the frame when moved along the surface S X B (5) deA = !A eB: Cartan's Method of Moving Frames 3 B This equation defines the 3 × 3 matrix of one forms !A which is called the matrix of connection forms of 3 E . The fact that the frame is orthonormal implies that when δAB = heA; eBi is differentiated, using (5), 0 =dδAB =dheA; eBi (6) =hd eA; eBi + heA; d eBi X C X C =h !A eC ; eBi + heA; !B eC i B A =!A + !B : B This equation says that the matrix of connection forms is skew so there are only three distinct !A . Geo- metrically it says that the motion of the frame vectors is already determined in large part by the motion of the other vectors in the frame. i The forms !3 determine the motion of the normal vector and hence define the second fundamental form. Thus in local coordinates, the second fundamental form is given using (4),(5) and (6), II(·; ·) = − hd e3; dXi 1 2 1 2 = − h!3 e1 + !3 e2 ;! e1 + ! e2i (7) 1 1 2 2 = − !3 ⊗ ! − !3 ⊗ ! 3 1 3 2 = !1 ⊗ ! + !2 ⊗ ! : We may express the connection forms using the basis 3 1 2 !1 =h11! + h12! ; (8) 3 1 2 !2 =h21! + h22! : Thus inserting into (7), X i j II(·; ·) = hij ! ⊗ ! : In particular, if one searches through all unit tangent vectors Vφ := cos(φ)e1 + sin(φ)e2; for which II~ (Vφ;Vφ) is maximum and minimum, one finds that the extrema occur as eigenvectors of hij and that the principal curvatures ki are the corresponding eigenvalues. In particular, the Gauß and mean curvatures are K = k1 k2 = det(hij) = h11h22 − h12h21; (9) 1 1 1 H = 2 (k1 + k2) = 2 tr(hij) = 2 (h11 + h22): Examples. We now work two examples showing that these forms may be found explicitly using the same amount of work that it would take using other notations. The first is the case of surfaces of revolution 0 f(u) cos v 1 X(u; v) = @ f(u) sin v A g(u) where f and g are functions which describe the generating curve. For simplicity, let us assume f_2 +g _2 = 1; f_f¨+g _g¨ = 0: 4 Andrejs Treibergs Differentiating X we find 0 f_ cos v 1 0 −f sin v 1 dX = @ f_ sin v A du + @ f cos v A dv (10) g_ 0 = du e1 + fdv e2; where we have taken a convenient frame 0 f_ cos v 1 0 − sin v 1 0 −g_ cos v 1 _ e1 = @ f sin v A ; e2 = @ cos v A ; e3 = @ −g_ sin v A : g_ 0 f_ One checks that (1) holds. From (10) it also follows that (11) !1 = du; !2 = f dv: Hence the first fundamental form (4) is ds2 = du2 + f 2 dv2: B Now let us compute the Weingarten equations. Since there are only three independent !A we actually only need to work out half of this: 0 ¨ 1 0 _ 1 f cos v −f sin v g¨ de = f¨sin v du + f_ cos v dv = f_ dv e + du e 1 @ A @ A 2 _ 3 g¨ 0 f 0 − cos v 1 _ de2 = @ − sin v A dv = −f dv e1 +g _ dv e3 0 0 −g¨ cos v 1 0 g_ sin v 1 g¨ de = −g¨ sin v du + −g_ cos v dv = − du e − g_ dv e ; 3 @ A @ A _ 1 2 f¨ 0 f where we have used −g¨g_ = f¨f_. It follows that the matrix of connection forms is 0 1 2 3 1 0 g¨ 1 !1 !1 !1 0 f_ dv du _ B C B f C B 1 2 3 C B _ C B !2 !2 !2 C = B −f dv 0g _ dv C : @ A @ g¨ A 1 2 3 − du −g_ dv 0 !3 !3 !3 f_ Hence we may compute the second fundamental form (8) using (11) 3 g¨ g¨ 1 !1 = du = ! ; f_ f_ g_ ! 3 =g _ dv = !2: 2 f Finally, we obtain the second fundamental form matrix 0 g¨ 1 0 h h _ 11 12 = B f C : h h @ g_ A 21 22 0 f Cartan's Method of Moving Frames 5 Since hij is diagonal, ei are the principal directions and ki = hii are the principal curvatures. Therefore, −f¨ fg¨ + f_g_ K = ;H = : f 2ff_ As a second example consider the Enneper's Surface 0 1 3 2 1 u − 3 u + uv B C X(u; v) = B v − 1 v3 + vu2 C : B 3 C @ A u2 − v2 Differentiating yields 0 1 − u2 + v2 1 0 2uv 1 2 2 2 2 dX = @ 2uv A du + @ 1 − v + u A dv = 1 + u + v (du e1 + dv e2) 2u −2v where we have taken the frame 0 1 − u2 + v2 1 0 2uv 1 0 −2u 1 2 2 2 2 B 1 + u2 + v2 C B 1 + u + v C B 1 + u + v C B C B C B C B 2uv C B 1 − v2 + u2 C B 2v C e1 = B C ; e2 = B C ; e3 = B C : B 1 + u2 + v2 C B 1 + u2 + v2 C B 1 + u2 + v2 C B C B C B C @ 2u A @ −2v A @ 1 − u2 − v2 A 1 + u2 + v2 1 + u2 + v2 1 + u2 + v2 Thus we have !1 = 1 + u2 + v2 du; !2 = 1 + u2 + v2 dv: It follows that 0 −2 1 @ 1 1 de = (1 + u2 + v2) du e + 0 du 3 @u 1 + u2 + v2 3 1 + u2 + v2 @ A −2u 0 0 1 @ 1 1 +(1 + u2 + v2) dv e + 2 dv: @v 1 + u2 + v2 3 1 + u2 + v2 @ A −2v Hence 3 2 du 2 1 !1 = − he1; de3i = = ! ; 1 + u2 + v2 (1 + u2 + v2)2 3 −2 dv −2 2 !2 = − he2; de3i = = ! : 1 + u2 + v2 (1 + u2 + v2)2 Thus 2 h11 = −h22 = ; h12 = 0 (1 + u2 + v2)2 −4 so H = 0 and K = −4 1 + u2 + v2 .