
Surface Theory with Differential Forms by Min Ru 1 Differential forms It often becomes simpler if we make use the notion of differential forms (exterior calculus). In this section, we shall introduce the notion of differential forms. We first recall some concept from linear algebra. Definition 1.1 Let T be a real vector space of dimension n.A 1−form ω(which is also called a linear functional) is a linear map ω : T → R. The set of all 1−forms ω is denoted by T ∗, which is called the “dual space” of T . Definition 1.2 Let T be a vector space and T ⊕ T be the direct sum. A 2−form on T is a mapping Ω: T ⊕ T → R which is bilinear and skew-symmetric: (i) Ω(aX + bY, Z) = aΩ(X, Z) + bΩ(Y, Z), (ii) Ω(X, Y ) = −Ω(Y, X). The set V2 T ∗ of all 2−forms on T is a vector space with the following addition and scalar multiplication: (Ω1 + Ω2)(X, Y ) = Ω1(X, Y ) + Ω2(X, Y ), (aΩ(X, Y ) = aΩ(X, Y ). ∗ Definition 1.3 Let ω1, ω2 ∈ T . We define the exterior product, or wedge product of ω1, ω2, denoted by ω1 ∧ ω2 as follows: ω1 ∧ ω2 : T ⊕ T → R, (ω1 ∧ ω2)(X, Y ) = ω1(X)ω2(Y ) − ω1(Y )ω2(X). So, by the definition, ω1 ∧ ω2 is a 2−form. 1 By the definition, we have ω1 ∧ ω2 = −ω2 ∧ ω1, and ω ∧ ω = 0. V2 ∗ For n = 2 we will show that T has dimension = 1. In fact, if e1, e2 is a basis for T , then e1 ∧ e2 spans V2 T ∗. Definition 1.4 For a surface S, a (differential) 1−form ω on S assigns, at every p ∈ S, a ∗ linear map ω |p: TpS → R, i.e., ω |p∈ Tp S. V2 ∗ A 2−form Ω on S assigns, at every point p ∈ S, a two form Ω |p∈ Tp S. We now consider how to express 1−forms and 2−forms on S, in terms of local coordinates. 3 Let x : U → R be a parameterizations for the surface S. Every vector vp ∈ Tp(S) can be expressed as vp = ξ1xu + ξ2xv. Define the differential 1-form du (resp. dv) on U(or x(U)) by du|p(vp) = ξ1 (resp. dv|p(vp) = ξ2), i.e. {du, dv} is the dual base of {xu, xv}. Then every 1−form ω can be locally (on U) written as ω = adu + bdv, where a, b are functions on on U(or x(U)). Every 2−form Ω can be locally (on U) written as Ω = Adu ∧ dv. Every 3−form on S is always zero! Denote by Ak(S) be the set of all (differentiable) k-forms on S, where A0(S) is the set of all (differentiable) functions on S. We define the exterior operator(differential) d : Ak(S) → Ak+1(S) as follows: 2 (a) For every f ∈ A0(S) (i.e. f is a differentiable function on S), we define ∂f ∂f df = du + dv. ∂u ∂v (b) For every 1-form ω ∈ A1(S), if we write ω = adu + bdv, then dω = da ∧ du + db ∧ dv. The exterior operator d has the following important property: d2 = 0, i.e. for every f ∈ A0(S), d(d(f)) = 0 (prove it as an easy exercise!). Finally, we note that, dx = xudu + xudv. For every v ∈ Tp(S), write vp = ξ1xu + ξ2xv, then dx(v) = (xudu + xudv)(ξ1xu + ξ2xv) = ξ1xu + ξ2xv = v. Hence, we have dx(v) = v for every v ∈ Tp(S). 2 The method of moving frames for curves Let x(t) = (x1(t), x2(t), x3(t)) be a space curve. Let e1 be the unit-tangent vector. Let e2 such that de1/ds = κe2. e2 is called the principal normal and κ is the curvature. Let e3 = e1 × e2 is the binormal. Then {e1, e2, e3} form an orthonormal basis(Frenet frame). Write 3 X dei = ωijej, j=1 where ωij are 1-forms. Since < ei, ej >= δij, form d < ei, ej >= 0, we have ωij = −ωji. Hence the matrix (ω)ij is a 3 × 3 skew-symmetric matrix, whose entries are differential 1- forms. From the skew-symmetric, we have ωjj = 0. From the selection of e1, e2 and e3, we have ω13 = ω31 = 0. Hence, we have (Frenet formula): de 1 = κe ds 2 de 2 = −κe + τe ds 1 3 3 de 3 = −τe . ds 2 3 The method of moving frames for surfaces 1. Structure equations ∂x ∂x Let S be a surface and x : U → S be a parametrization. The vectors ∂u , ∂v span TpS. Let e1(p), e2(p) be an orthonormal basis for Tp(S), p ∈ U (such orthonormal basis exists by applying Gram-Schmidt orthonormalization procedure) and e3(p) be the unit normal(Gauss map). Then e1, e2, e3 are smooth vector fields. Write dx = ω1e1 + ω2e2 + ω3e3. We first claim that ω3 = 0 and ωi(ej) = δij, 1 ≤ i, j ≤ 2, where δij = 0 if i 6= j and δii = 1. In fact, by the definition of d, 3 X dx = xudu + xvdv = < xudu + xvdv, ej > ej. j=1 Hence, since < xu, e3 >= 0, < xv, e3 >= 0, ω3 =< xudu + xvdv, e3 >= 0. To see ωi(ej) = δij, 1 ≤ i, j ≤ 2, we just calculate ω1(e1) and ω1(e2). Since dx = ω1e1 +ω2e2, we have ω1(e1) =< dx(e1), e1 >=< e1, e1 >= 1. Similarly, we get ω1(e2) =< e2, e1 >= 0. This proves the claim. From the claim, we can write dx = ω1e1 + ω2e2, 4 where ω1, ω2 is a dual basis to {e1, e2}. Recall that the First Fundamental Form is I = Edu2 + 2F dudv + Gdv2. Since I(v) =< 2 2 2 2 v, v >, if we write v = ξ1e1 +ξ2e2, then I(v) = ω1(ξ1e1 +ξ2e2)+ω2(ξ1e1 +ξ2e2) = ξ1 +ξ2 =< v, v > (notice that ω1, ω2 is a dual basis to {e1, e2}). Hence, we can write 2 2 I = ω1 + ω2. Write 3 X dei = ωijej. j=1 Then ωij are (differential) 1-forms. There are total nine such 1-forms. First we claim that ωij = −ωji. In fact, since ei · ej = δij, by differentiating, we have, dei · ej + ei · dej = 0. This implies that ωij = −ωji. Hence, we have only three “meaningful” 1-forms ω13, ω23 and ω12. Together with the above two 1-forms above, we obtain, in total, five 1-forms: ω1, ω2, ω13, ω23, ω12. To summarize, we have the following Equations for moving frame (compare to the Frenet formula): dx = ω1e1 + ω2e2, where ωi(ej) = δij, 1 ≤ i, j ≤ 2, and de1 = ω12e2 + ω13e3, de2 = ω21e1 + ω23e3, de3 = ω31e1 + ω32e2, where ωij = −ωji. 5 Theorem 3.1 ω13 = aω1 + bω2 ω23 = bω1 + cω2. Proof: We alway can write ω13 = aω1 + b1ω2, ω23 = b2ω1 + cω2. We now claim that b1 = b2. In fact, by the definition, we have de3 = ω31e1 + ω32e2 = −(aω1 + b1ω2)e1 − (b2ω1 + cω2)e2. Hence, b1 = − < de3(e2), e1 >, b2 = − < de3(e1), e2 >. However, e3 = n (the unit normal to the surface) and we have showed that dn is a self-adjoint operator, so < de3(e2), e1 >=< e2, de3(e1) >. Hence b1 = b2. This proves the theorem. Note: From above, we see that a = ω13(e1), b = ω13(e2) = ω23(e1), c = ω23(e2). Hence the matrix a b − b c is actually the matrix of dn = −SP (since e3 = n) with respect to the basis {e1, e2}. So, the Gauss curvature 2 K = ac − b = ω13(e1)ω23(e2) − ω13(e2)ω23(e1) = ω13 ∧ ω23(e1, e2). We can express the first and second fundamental form in terms of these differential forms. Theorem 3.2 I = ω1ω1 + ω2ω2; II = ω1ω13 + ω2ω23, √ 2 ω1 ∧ ω2 = EG − F du ∧ dv = dA. 6 Proof: To show I = ω1ω1 + ω2ω2, we check, by the definition, I(e1) =< e1, e1 >= 1, and (ω1ω1 + ω2ω2)(e1) = ω1(e1)ω1(e1) + ω2(e1)ω2(e1) = 1. Similarly, I(e2) = (ω1ω1 + ω2ω2)(e2). Hence, by linearity, I = ω1ω1 + ω2ω2. The other proofs are similar. Most of our results will come from the following: Theorem 3.3 (Structure Equations): dω1 = −ω2 ∧ ω12; dω2 = −ω1 ∧ ω21; and 3 X dωij = ωik ∧ ωkj. k=1 Proof: Use the property that d2 = 0, we have 0 = d(dx) = d(ω1e1) + d(ω2e2). This derives the first two equations. Also d(dei) = dωij − ωij ∧ dej = 0. This derives the second equation. Remark: The statement of the equations dω1 = −ω2 ∧ ω12; dω2 = −ω1 ∧ ω21 is a very important theorem(Levi-Civita theorem), which tells us that the ω12 can be derived from ω1, ω2 and their differential(so it is an intrinsic property). The form ω12 is called the connection form. The covariant differential(which is called the connection) De1 is the tangential component of de1 = ω12e2 + ω13e3. That is De1 = pr(de1) = ω12e2. Similarly, De2 = ω21e1. This is why the form ω12 is called the connection form. It measures the tangential twist of e1 and e2. It also gives the geometric properties of the surface(parallel transport, geodesics, etc.) 7 The structure equations give: Gauss Equation: dω12 = −ω13 ∧ ω23, Mainardi-Codazzi Equation: dω13 = ω12 ∧ ω23; dω23 = −ω12 ∧ ω13. 2. Calculation of the Gauss curvature Recall K = (ω13 ∧ ω23)(e1, e2). Since ω13 ∧ ω23 is a two-form, so we can write ω13 ∧ ω23 = λω1 ∧ ω2. Hence, K = (ω13 ∧ ω23)(e1, e2) = λω1 ∧ ω2(e1, e2) = λ. Therefore, we have ω13 ∧ ω23 = Kω1 ∧ ω2. This, together with the Gauss equation, gives the following important theorem(formula) about the Gauss curvature K dω12 = −Kω1 ∧ ω2.
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