5.3 Surface Theory with Differential Forms

Transcription

1 5.3 Surface Theory with Differential Forms 1 Differential forms on R n, Click here to see more details Differential forms provide an approach to multivariable calculus (Click here to see more detailes) that is independent of coordinates. Let U be an open set in R n. A differential 0-form ( zero form ) is defined to be a smooth function f (here smooth means that f is differentiable at any order) on U. If v R n, then f has a directional derivative D v f (see introduction to differentiable function in Chapter one), which is another function on U whose value at a point p U is the rate of change (at p) of f in the v direction: D v f(p) = d dt f(p + tv) t=0. In particular, if v = e j, where {e 1 = (1, 0,..., 0), e 2 = (0, 1,..., 0),..., e n = (0,..., 1)} is the standard basis of R n, then D ej f = f x j, is the partial derivative of f with respect to the jth coordinate function, where x 1, x 2,...x n are the coordinate functions on U. By their very definition, partial derivatives depend upon the choice of coordinates: if new coordinates y 1, y 2,...y n are introduced, then f x j = n i=1 y i x j f y i. The first idea leading to differential forms is the observation that D v f(p) is a linear function of v: D v+w f(p) = D v f(p) + D w f(p), D cv f(p) = cd v f(p) for any vectors v, w and any real number c. This linear map from R n to R is denoted df p and called the (exterior) derivative of f at p. Thus df p (v) = D v f(p). The object df can be viewed as a function on U, whose value at p is not a real number, but the linear map df p, in other wor, df assigns, at every point p U, a linear 1

2 map df p from R n to R. This is just a special case of (general) differential 1-forms. Since any vector v is a linear combination v j e j of its components, df is uniquely determined by df p (e j ) for each j and each p U (i.e. df p (v) = v j df p (e j )) which are just the partial derivatives of f on U. Thus df provides a way of encoding the partial derivatives of f. It can be decoded by noticing that the coordinates x 1, x 2,...x n are themselves functions on U (x j maps each point in U the j-th coordinate), and so define differential 1-forms dx 1, dx 2,...dx n (most time, we also write it as dx 1, dx 2 2,...dx n ), in other wor, (5.1.1) dx j (e i ) = δ ij, ordx j (v) = v j, for v = (v 1,..., v n ) R n, In general, for any 0-form f, (5.1.2) df = n i=1 f x i dxi. The meaning of this expression is given by evaluating both sides at an arbitrary point p: on the right hand side, the sum is defined pointwise, so that df p = n i=1 f x i (p)dxi p. Applying both sides to e j, the result on each side is the j-th partial derivative of f at p (note: dx i p is actually independent of p, i.e. dx i p = dx i ). Example. Let f = e x2 +y, then df = f f dx + x y dy = +y 2xex2 dx + e x2 +y dy. We now give the general definition of 1 forms on U R n : Definition A 1 form φ on U R n ( either n = 2 or n = 3) assigns, for every p U R n, a is a linear map φ p : R n R. Remarks: (1) In linear algebra, given a vector space V, the set of all linear maps f : V R is called the dual space of V. So the alternative definition of 1 form φ is that the 1 form φ on U R n ( either n = 2 or n = 3) assigns, for every p U R n, an element φ p R n, where R n is the dual space of R n. 2

3 (2) In particular, the 1 forms dx 1,..., dx n are defined by the property that for (see (5.1.1)) any vector v = (v 1,..., v n ) R n, dx i p (v) = v i. The dx i form a basis for the space of all 1 forms on R n, so any 1 form φ on U R n may be expressed in the form n (5.1.3) φ = φ i dx i, i=1 where φ 1,..., φ n are functions on U. If v = (v 1,..., v n ) R n, then n φ p (v) = f i (p)v i. i=1 (3) Note that {dx 1,..., dx n } is in fact the standard basis of the dual space R n, it is dual to the standard basis {e 1 = (1, 0,..., 0), e 2 = (0, 1,..., 0),..., e n = (0,..., 1)} of R n. Example. φ = sin xdx + y 2 dy is a 1-form on R 2. You might ask the following question: given a differential 1-form φ on U, when does there exist a function f on U such that φ = df? By comparing (5.1.2) and (5.1.3) reduces this question to the search for a function f whose partial derivatives f/ x i are equal to n given functions φ i. For n > 1, such a function does not always exist: any smooth function f satisfies 2 f x i x j = so it will be impossible to find such an f unless 2 f x j x i, for all i and j. φ j x i φ i x j = 0. Differential 2 forms on R n. For two 1 forms φ, ψ, define the wedge product φ ψ as follows, for p R n, φ p ψ p : R n R n R is given by, for every (v, w) R n R n, (5.1.4) φ p ψ p (v, w) = φ(v)ψ(w) ψ(v)φ(w). 3

4 The wedge product is a way to produce 2 forms from the given two 1 forms. By the definition, we have φ ψ = ψ φ, and φ φ = 0. A basis for the 2-forms on R n is given by the set {dx i 1 dx i 2 : 1 i 1 < i 2 n}. Any 2-forms Ω on U can be expressed in the form Ω = i 1 <i 2 f i1,i 2 dx i 1 dx i 2, where f i1,i 2 are functions on U. In an alternative definition, Definition A 2-form Ω on U R n assigns, at ach point p U, an alternating bilinear mapping Ω p : R n R n R. Here, alternating means that Ω p (v, w) = Ω p (w, v), and bilinear means that Ω p is linear in both v and w. We can also definition the concept of the k-form in a similar way as i 1 <i 2 < <i k f i1,i 2,,i k dx i 1 dx i k. The exterior derivative takes that takes k forms to (k +1) forms. It is defined in a similar way as above. For example, if φ = I =k f I dx i 1 dx i k, then the exterior derivative dφ of φ is the (k + 1) form which is given by (5.1.5) dφ = df I dx i 1 dx i k. I =k If φ is a p form and ψ is a q form, then the Leibniz rule takes the form d(φ ψ) = dφ ψ + ( 1) p φ dψ. Example. Let ω = fdx + gdy + hdx be a 1-form on U R 3, then dω = df dx+dg dy +dh dz = (g x f y )dx dy +(h y g z )dy dz +(f z h x )dz dx. Very Important Theorem: d 2 = 0. Proof. We only check for 0-forms, i.e. for functions. In fact, by definition, n f d(df) = d j=1 x i dxi = 2 f i,j x i x j dxj dx i 4

5 = i<j ( 2 f x j x i 2 f x i x j ) dx j dx i = 0. This verifies the functions case. The general result follows from the similar method. pull-backs: Given a map F = (x 1, x 2, x 3 ) : U R 2 R 3, and ω = fdx + gdy + hdx be a 1-form on U R 3, then we can define F ω, the pull-back of ω by F as F ω = f F dx 1 + g F dx 2 + h F dx 3 ( x1 = f F du du + x ) ( 1 dv dv x2 + g F du du + x ) ( 2 dv dv x3 + h F du du + x ) 3 dv dv, where (u, v) is the coordinate system on R 2. Note that F ω is a differential form on U R 2, and it is easy to check that (do it by yourself) F dω = df ω. 2 Differential forms on surfaces We now define the concept of the differential forms on the surface M in R 3. To do so, we need to look at the tangent space T p (M). Recall that when M is flat, i.e. M = U or M = R 3, then its tangent space T p M is the whole space R 2, i.e. T p U = R 2 for every point p U. So when we look at the definition of 1-forms φ on U R 3, actually we read it as follows: a 1-form φ on U R 3 assgins, at each poitn p U, a linear map ω p : T p (U) = R 2 R, i.e., φ p T p (U) = R 2. We now give the general definition of differential forms on M. Definition Given a surface M in R 3, a (differential) 1 form ω on M assigns, at every p M, a linear map ω p : T p (M) R, i.e., ω p T p (M). A (differential) 2 form Ω on M assigns, at every point p M, a two form Ω p 2 T p (M). Every k form on M is always zero for k 3. For a smooth function f on M, the exterior derivative of f is the 1 form df with the property that for any p M, v p T p (M), (5.2.1) df p (v p ) = D v (f)(p), 5

6 where D v (f)(p) is the directional derivative of f at p with respect to the direction v p (see (3.4.1) for the definition of the directional derivative). So as we indicated before, df provides a way of encoding all the directional derivative of f at p. Let σ : U R 2 R 3 be a parametrization of M. Then the coordinates u, v on R 2 be also regarded as functions on M sending the point σ(u, v) to u and v respectively. Hence, du and dv are well-defined 1-forms on M (more precisely on σ(u)). It can be easily checked by definition that (5.2.2) du p (σ u p ) = 1, du p (σ v p ) = 0, dv p (σ u p ) = 0, dv p (σ v p ) = 1. Hence {du, dv} is dual to {σ u, σ v }. For any smooth function f on M, In terms of parametrization σ : U R 2 R 3 of M, we can write (5.2.3). df = Dσ u fdu + Dσ v fdv We we write f(u, v) = f σ(u, v), since Dσ u f(p) = ( f u)(u 0, v 0 ), Dσ v f(p) = ( f u)(u 0, v 0 ), where σ(u 0, v 0 ) = p, we have (5.2.4) df = f u du + f v dv where f u = f u, f v = f v. More precisely, we should write (5.2.4) as σ df = f u du + f v dv, but we simply write it as in (5.2.4) without the danger of confusion. In other wor, if we work everything on R 2 by the pulling back through σ, then we can regard du, dv as the standard 1-forms on R 2 which is dual to {(1, 0), (0, 1)}. In terms of the local parametrization σ : U R 2 R 3 of M, any 1 form ω on M can be written as ω = adu + bdv where a, b are functions on σ(u). Similarly, every 2 form Ω can be locally (on σ(u)) written as Ω = Adu dv, where A is a function on σ(u). For every 1-form ω on M, the exterior derivative of ω is a 2-form, which is defined in a similar way as in the previous section, by ω = adu + bdv, then dω = da du + db dv. (you can check that it is independent of the choice of the choices of paramertizations). It is easy to check that d 2 = 0. The exterior operator d has the following important property: d 2 = 0, i.e. for every function f on M, d(d(f)) = 0. 6

7 3 The method of moving frames for curves Let x(t) = (x 1 (t), x 2 (t), x 3 (t)) be a space curve. Let e 1 be the unit-tangent vector. Let e 2 such that de 1 / = κe 2. e 2 is called the principal normal and κ is the curvature. Let e 3 = e 1 e 2 is the binormal. Then {e 1, e 2, e 3 } form an orthonormal basis(frenet frame). Write 3 de i = ω ij e j, j=1 where ω ij are 1-forms. Since < e i, e j >= δ ij, form d < e i, e j >= 0, we have ω ij = ω ji. Hence the matrix (ω) ij is a 3 3 skew-symmetric matrix, whose entries are differential 1-forms. From the skew-symmetric, we have ω jj = 0. From the selection of e 1, e 2 and e 3, we have ω 13 = ω 31 = 0. Hence, we have (Frenet formula): de 1 = κe 2 de 2 = κe 1 + τe 3 de 3 = τe 2. 4 The method of moving frames for surfaces 1. Stucture equations I suggest you to read the appendix below before you start this section. Let M be a surface and σ : U R 3 be a local parametrization of M. Recall that the vectors {σ u p, σ v p } is a basis for T p (M). Let e 1 (p), e 2 (p) be an orthonormal basis for T p (M), p U (such orthonormal basis always exists by applying Gram-Schmidt orthonormalization procedure) and let e 3 (p) = n(p) be the unit normal(gauss map). The key point of this section is that we are working on the orthonormal basis, NOT just the basis {σ u p, σ v p }. The basis {e 1 (p), e 2 (p), e 3 (p)} serves as a moving frame for R 3, where e 1 (p), e 2 (p) is an orthonormal basis for T p (M). Let σ : U R 3 be a local parametrization of M. Consider dσ = σ u du + σ v dv. Since {e 1 (p), e 2 (p)} is a basis for T p (M), we write (5.3.1) σ u = λ 1 e 1 + λ 2 e 2, σ v = λ 3 e 1 + λ 4 e 4. 7

8 Hence we can rewrite dσ as (5.3.2) dσ = ω 1 e 1 + ω 2 e 2, where ω 1, ω 2 are differential 1 forms on M, ω 1 = λ 1 du + λ 3 dv, ω 1 = λ 2 du + λ 4 dv. We claim that {ω 1, ω 2 } is dual to {e 1, e 2 }. To prove the claim, we first note that, form (5.3.1) dσ(σ u ) = σ u, and dσ(σ v ) = σ v, so dσ(v) = v for every tangent vector. Then the claim can be derived by using (5.3.2) and dσ(e 1 ) = e 1 and dσ(e 2 ) = e 2. The differential 1 forms ω 1, ω 2 keep track of how our point moves around on M. Next we want to see how the frame itself twists, we will define 1-forms ω ij, i, j = 1, 2, 3. Consider de i, the exterior derivative of e i. Note that de i is a vector-valued (the image is in R 3 ) differential 1 form, and since {e 1, e 2, e 3 } is linearly independent, so it is a basis for R 3. Therefore we can 3 (5.3.4) de i = ω ij e j, j=1 where ω ij, i, j = 1, 2, 3 are differential 1-forms. There are total nine of such 1-forms. First we claim that ω ij = ω ji. In fact, since e i e j = δ ij, by differentiating, we have, de i e j + e i de j = 0. This implies that ω ij = ω ji. Hence, we have only three meaningful 1-forms ω 13, ω 23 and ω 12. Others are just zero. If v p T p (M), then ω ij p (v p ) tells us how fast e i is twisting towar e j at p as we move with velocity v p. Together with the above two 1-forms above, we obtain, in total, five 1-forms: ω 1, ω 2, ω 13, ω 23, ω 12. To summarize, we have the following Equations for moving frame (5.3.5) dσ = ω 1 e 1 + ω 2 e 2, (5.3.6) de 1 = ω 12 e 2 + ω 13 e 3, (5.3.7) de 2 = ω 21 e 1 + ω 23 e 3, (5.3.8) de 3 = ω 31 e 1 + ω 32 e 2, where ω ij = ω ji. Recall that the shape operator (see section 3.3) is S p = dn = de 3, so the shape operator is embodied in the equation de 3 = ω 31 e 1 + ω 32 e 2 = (ω 13 e 1 + ω 23 e 2 ). 8

9 We now calculate the matrix of the shape operator S p respect to the basis {e 1, e 2 }. In fact, S p (e 1 ) = de 3 (e 1 ) = ω 31 (e 1 )e 1 ω 32 (e 1 )e 2 = ω 13 (e 1 )e 1 + ω 23 (e 1 )e 2, S p (e 2 ) = de 3 (e 2 ) = ω 13 (e 2 )e 1 + ω 23 (e 2 )e 2. Thus, by the definition, the matrix of the shape operator S p respect to the basis {e 1, e 2 } is ( ) ω13(e 1 ) ω 13(e 2 ). ω 23 (e 1 ) ω 23 (e 2 ) Recall that the the Gauss curvature (see chapter 4), so we get is the determinant of the matrix of S p K = det(s p ) = ω 13 (e 1 )ω 23 (e 2 ) ω 13 (e 2 )ω 23 (e 1 ) = (ω 13 ω 23 )(e 1, e 2 ). Since ω 13 ω 23 is a 2-form, and the vector space of the two forms has dimension one (as we noted sarlier) we can write So from above, ω 13 ω 23 = λω 1 ω 2. K = (ω 13 ω 23 )(e 1, e 2 ) = λω 1 ω 2 (e 1, e 2 ) = λ. So we have a very important expression for the Gauss curvature (5.3.9). ω 13 ω 23 = Kω 1 ω 2. Most of our results will come from the following: Theorem 5.2.1(Structure Equations): (5.3.10) dω 1 = ω 12 ω 2 ; dω 2 = ω 1 ω 12 ; and (5.3.11) 3 dω ij = ω ik ω kj. Proof: Use the property that d 2 = 0 and using (5.3.5), we have 0 = d(dσ) = d(ω 1 e 1 ) + d(ω 2 e 2 ). 9 k=1

10 This derives the first two equations by using (5.3.6) and (5.3.7). Also, using (5.3.6), (5.2.7), (5.3.8) and d 2 = 0, we get This derives the second equation. The structure equations give: Gauss Equation: dω 12 = ω 13 ω 23, d(de i ) = dω ij ω ij de j = 0. Mainardi-Codazzi Equation: dω 13 = ω 12 ω 23 ; dω 23 = ω 12 ω 13. From (5.3.9), we have Hence we get ω 13 ω 23 = Kω 1 ω 2. dω 12 = Kω 1 ω 2. This provides a new and simple proof of Gauss s theorem egregium (see section 5.1), since from above, we see that K only depen on ω 1 and ω 2, so the Gauss curvature is an intrinsic quantity! 2. Further look of Gauss equation and the Codazzi equations. To see why the equation dω 12 = ω 13 ω 23 is the same as the Gauss equation we derived before (in section 5.1), we consider a orthogonal parametrization, i.e. F = 0. In this case, e 1 = e u / E, e 2 = e v / G. Then, ω 1 = Edu, ω 2 = Gdv (since {ω 1, ω 2 } is the dual basis to e 1, e 2 ). We now calculate ω 12. Write ω 12 = Adu + bdv, we need to determine A and B. From the structure equation, ω 2 ω 12 = dω 1 = ( E) v du dv. Also ω 2 ω 12 = ( Gdv) (Adu + Bdv) = GAdu dv. Hence GA = ( E)v. This implies that Similarily, we get A = ( E) v. G B = ( G) u. E 10

11 Hence ω 12 = ω 21 = ( E) v du + ( G) u dv. G E [( ) ( ) ] ( E)v ( G)u dω 12 = + du dv. G E On the other hand, from a direct computation, we can get ω 13 = (de 1 ) e 3 = ω 23 = (de 2 ) e 3 = v u e du + f dv, E E f du + g dv. G G So ω 13 ω 32 = ω 13 ω 23 = eg f 2 EG du dv. Hence, Gauss equation dω 12 = ω 13 ω 2,3 is equivalent to [( ) ( ) ] ( E)v ( G)u + = eg f 2, G v E u EG which is the same as the Gauss equation we derived before in the case F = 0 (see section 5.1). Similarily, we can verify the Codazzi equations listed above are the same as what we have derived earlier. 3. Normal curvature and geodesic curvature revisited. Let σ : U M be a parametrization, and let C be a curve on M given by α(s) = σ(u(s), v(s)) where s is the arc-lenght parameter. Let T(s) = α (s) be the tangent vector to C, and let ē 1 (s) = T(s), ē 2 (s) = n(s) T(s), ē 3 (s) = n(s), where n is the unit normal to the surface M. Then, {ē 1 (s), ē 2 (s), ē 3 (s)} is an orthnormal moving frame along C (which is called Darboux frame). Then we have dα(s) = ē 1 (s), dē 1 (s) = κ g ē 2 (s) + κ n ē 3 (s), dē 2 (s) = κ g ē 1 (s) + τ g (s)ē 3 (s), 11

12 dē 3 (s) = κ n ē 1 (s) τ g (s)ē 2 (s), where κ g is the geodesic curvature, κ n is the normal curvature and τ g (s) is called the geodesic torsion. Take an orthonormal moving frame (Darboux frame) {ē 1, ē 2, ē 3 } on M with ē 3 = n, such that the restriction of this frame to the curve C is {ē 1 (s), ē 2 (s), ē 3 (s)} We write dē 1 = ω 12 ē 2 + ω 13 ē 3, dē 2 = ω 21 ē 1 + ω 23 ē 3, dē 3 = ω 31 ē 1 + ω 32 ē 2, where ω ij = ω ji. We have the following theorem: Theorem Let C be a curve on M, then κ g = ω 12 (ē 1 ), κ n = ω 13 (ē 1 ), τ g = ω 23 (ē 1 ). Proof: Since α(s) = σ(u(s), v(s)), Hence, for 1 i, j 3, ē 1 = T = dα(s) = du σ u + dv σ v. Hence This finishes the proof. ω ij (T) = < dē i (T), ē j > = < du dē i(σ u ) + dv dē i(σ v ), ē j > = < du ēi,u + dv ēi,v, ē j > = < dē i, ē j >. κ g = ω 12 (T), κ n = ω 13 (T), τ g = ω 23 (T). We now re-derive the formula of κ g in terms of the orthogonal paramerization. Let x : U S be a orthogonal parametrization, i.e F = 0 (where {E, F, G} is its 12

13 first fundamental form). Let C be a curve on S and let {ē 1, ē 2, ē 3 } be the Darboux frame. Then, form above κ g = ω 12 (ē 1 ). On the other hand, consider another (natural) orthonormal frame: i.e. let e 1 = x u / E, e 2 = x v / G, e 3 = e 3 = n. Then the frame {e 1, e 2, e 3 } is also an orthonormal moving frame. Let ω 1, ω 2 be the dual of e 1, e 2 and let ω 12 be the connection form. Then, as we derived before, ω 12 = ω 21 = ( E) v du + ( G) u dv. G E Next, we want to derive a relationship between ω 12 and ω 12. To do so, let θ be the angle from x u to ē 1 = T, then we have ( ) ( ) ( ) ē1 cos θ sin θ e1 =. ē 2 sin θ cos θ e 2 Since {ω 1, ω 2 } is the dual basis of {e 1, e 2 }(resp., { ω 1, ω 2 } is the dual basis of {ē 1, ē 2 }), hence ω 1 = cos θω 1 + sin θω 2 Hence ω 2 = sin θω 1 + cos θω 2 ω 12 = ω 12 + dθ. ω 12 = ω 12 + dθ = ( E) v du + ( G) u dv + dθ. G E Since we have e 1 = T = dα(s) = du x u + dv x v, κ g = ω 12 (ē 1 ) = ( E) v du(t)+ ( G) u dv(t)+dθ(t) = ( E) v G E G Now dθ = θ u du + θ v dv, so du(s) +( G)u dv(s) E +dθ(t). Hence dθ(t) = θ u du(s) κ g = ( E) v G du(s) + θ v dv(s) + ( G) u E = dθ. dv(s) + dθ. 13

14 Therefore we re-proved the following Liouville theorem: Theorem Let x : U S be a orthogonal parametrization, i.e F = 0 (where {E, F, G} is its first fundamental form). Let C be a curve on S. Then κ g = ( E) v G du(s) + ( G) u E dv(s) + dθ. 6. A simple proof of Gauss-Bonnet theorem. The formula dω 12 = Kω 1 ω 2 is the key in apply Green s theorem to prove Gauss-Bonnet theorem. The formula can be re-written as Kdσ = dω 12, where dσ = ω 1 ω 2. From the Gauss equation and Stoke s Theorem, the Gauss-Bonnet formula follows immediately for an oriented surface M with (piecewise smooth) boundary M on which we can globally define a moving frame. That is, we can reprove the local Gauss-Bonnet formula quite effortlessly. Proof: We start with an arbitrary moving frame e 1, e 2, e 3, and take a Darboux frame (i.e. a moving frame for the surface with e 1 tangent to M) ē 1, ē 2, ē 3 along M. We write ē 1 = cos θe 1 +sin θe 2, ē 2 = sin θe 1 +cos θe 2 (where θ is smoothly chosen along the smooth pieces of M and the exterior angle ɛ j at P j gives the jump of theta as we cross P j ). Then, by Stokes theorem, we have M Kdσ = dω 12 = ω 12 = ( ω 12 dθ) = κ g +(2π ɛ j ). M M M M 7. Covariant Derivative, Connection Form The covariant derivativ D v e 1 is the tangential component of de 1 (v) = ω 12 (v)e 2 + ω 13 (v)e 3. Hence, D v e 1 = pr(de 1 (v)) = ω 12 (v)e 2. Similarly, D v e 2 = ω 21 (v)e 1. The form ω 12 is called the connection form and it measures the tangential twist of e 1 and e 2. 14

15 5 Appendix: Review of Surface Theory We review here the theory of the surfaces we have learnt so far. Let M be a surface and σ : U M be an orthogonal parametrization (i.e. F = 0 in the first fundamental form). Recall that the vectors {σ u, σ v } span T p (M). From 0 = F = σ u σ v, we see that σ u and σ v are orthogonal. Let e 1 = σ u, e 2 = σ v. Then {e 1 (p), e 2 (p)} E G forms an orthonormal basis for T p (M), p M (i.e. e 1 (p) = 1, e 2 (p) = 1 and e 1 (p) e 2 (p) = 0). So {e 1, e 2 } is called a moving frame for the tangent spaces of S (since for each point p M, we have that {e 1 (p), e 2 (p)} forms an orthonormal basis for T p (M), here the name moving because p varies from M and the name frame comes from the fact it is a basis). Such orthonormal basis exists becuase we can always apply the Gram-Schmidt orthnormalization procedure if σ is not an orthogonal parametrization. Let e 3 = n be the unit normal(gauss map). Then {e 1, e 2, e 3 } forms an (moving) orthonormal basis for R 3. We have (for an orthogonal parametrization) (do your own calculations using the formulas in section 5.1), in below, E, F, G is first fundamental form with F = 0 and e, f, g are the second fundamental form. (e 1 ) u = E v 2 EG e 2 + e E e 3, (e 1 ) v = G u 2 EG e 2 + f E e 3, (e 2 ) u = E v 2 EG e 1 + f G e 3, (e 2 ) v = G u 2 EG e 1 + g G e 3, (e 3 ) u = e e 1 f e 2, E G (e 3 ) v = f e 1 g e 2. E G Gauss equation: K = 1 ( ) ( ) ( 2 E)v ( G)u +. EG G v E u 15

16 Codazzi equations: ( ) e E v ( ) g G u ( ) f E u ( ) f G v g ( E) v G f ( G)u = 0, EG e ( G) u E f ( E)v = 0. EG 16