Surface Theory with Differential Forms

by Min Ru

1 Differential forms

It often becomes simpler if we make use the notion of differential forms (). In this , we shall introduce the notion of differential forms. We first recall some concept from linear algebra.

Definition 1.1 Let T be a real of n.A 1−form ω(which is also called a linear functional) is a 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 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 (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 (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 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(, , 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.

This provides a new and simple proof of Gauss’s theorem egregium, since from above, we

see that K only depends on ω1 and ω2, so the Gauss curvature is an intrinsic quantity!

3. Further look of Gauss equation and the Codazzi equations.

To see why the equation dω12 = −ω13 ∧ ω23 is the same as the Gauss equation we derived √ before, we consider a orthogonal parametrization, i.e. F = 0. In this case, e1 = eu/ E, e2 = √ √ √ ev/ G. Then, ω1 = Edu, ω2 = Gdv (since {ω1, ω2} is the dual basis to e1, e2). From the structure equation, √ −ω2 ∧ ω12 = dω1 = −( E)vdu ∧ dv,

8 √ ω1 ∧ ω12 = −ω1 ∧ ω21 = dω2 = −( G)udu ∧ dv.

To solve the above two equations for ω12, let

ω12 = Adu + Bdv.

Then, the first equation above gives √ √ −( Gdv) ∧ (Adu + Bdv) = −( E)vdu ∧ dv. √ √ Hence A = −( E)v/ G. Similarily, by solving the second equation we can get B = √ √ ( G)u/ E. Hence √ √ ( E)v ( G)u ω12 = − √ du + √ dv. G E Therefore, " √ ! √ ! # ( E)v ( G)u dω12 = √ + √ du ∧ dv, G v E u On the other hand, from a direct computation, we can get

e f ω13 = (de1) · e3 = √ du + √ dv, E E f g ω23 = (de2) · e3 = √ du + √ dv. G G Hence eg − f 2 ω13 ∧ ω32 = −ω13 ∧ ω23 = − √ du ∧ dv. EG

Hence, Gauss equation dω12 = −ω13 ∧ ω2,3 is equivalent to √ √ " ( E) ! ( G) ! # eg − f 2 − √ v + √ u = √ , G v E u EG which is the same as the Gauss equation we derived before(in the case F = 0).

Similarly, we can verify the Codazzi equations listed above are the same as what we have derived earlier.

9 4. curvature.

Let x : U → S be a parametrization, and let C be a curve on S given by α(s) = x(u(s), v(s)) where s is the arc-length parameter. Let T(s) = α0(s) be the tangent vector to C, and let e¯1(s) = T(s), e¯2(s) = n(s) × T(s), e¯3(s) = n(s), where n is the unit normal to the surface S. Then, {e¯1(s), e¯2(s), e¯3(s)} is an orthonormal moving frame along C (which is called ). Then we have

dα(s) = e¯ (s), ds 1 de¯ (s) 1 = κ e¯ (s) + κ e¯ (s), ds g 2 n 3 de¯ (s) 2 = −κ e¯ (s) + τ (s)e¯ (s), ds g 1 g 3 de¯ (s) 3 = −κ e¯ (s) − τ (s)e¯ (s), ds n 1 g 2 where κg is the geodesic curvature, κn is the normal curvature and τg(s) is called the geodesic torsion.

Take an orthonormal moving frame (Darboux frame) {e¯1, e¯2, e¯3} on S with e¯3 = n, such that the restriction of this frame to the curve C is {e¯1(s), e¯2(s), e¯3(s)} We write

de¯1 =ω ¯12e¯2 +ω ¯13e¯3.

de¯2 =ω ¯21e¯1 +ω ¯23e¯3,

de¯3 =ω ¯31e¯1 +ω ¯32e¯2, whereω ¯ij = −ω¯ji. We have the following theorem:

Theorem 3.4 Let C be a curve on S, then

κg =ω ¯12(e¯1), κn =ω ¯13(e¯1), τg =ω ¯23(e¯1).

10 Proof: Since α(s) = x(u(s), v(s)), dα(s) du dv e¯ = T = = · x + · x . 1 ds ds u ds v Hence, for 1 ≤ i, j ≤ 3,

ω¯ij(T) = < de¯i(T), e¯j > du dv = < de¯ (x ) + de¯ (x ), e¯ > ds i u ds i v j du dv = < e¯ + e¯ , e¯ > ds i,u ds i,v j de¯ = < i , e¯ > . ds j Hence

κg =ω ¯12(T), κn =ω ¯13(T), τg =ω ¯23(T).

This finishes the proof.

We now re-derive the formula of κg in terms of the orthogonal paramerization. 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 and let {e¯1, e¯2, e¯3} be the Darboux frame. Then, form above

κg =ω ¯12(e¯1). √ On the other hand, consider another (natural) orthonormal frame: i.e. let e1 = xu/ E, √ e2 = xv/ G, e3 = e3 = n. Then the frame {e1, e2, e3} is also an orthonormal moving frame.

Let ω1, ω2 be the dual of e1, e2 and let ω12 be the connection form. Then, as we derived

before, √ √ ( E)v ( G)u ω12 = −ω21 = − √ du + √ dv. G E

Next, we want to derive a relationship betweenω ¯12 and ω12. To do so, let θ be the angle from xu to e¯1 = T, then we have       e¯1 cos θ sin θ e1         =     . e¯2 − sin θ cos θ e2

11 Since {ω1, ω2} is the dual basis of {e1, e2}(resp., {ω¯1, ω¯2} is the dual basis of {e¯1, e¯2}), hence

ω¯1 = cos θω1 + sin θω2

ω¯2 = − sin θω1 + cos θω2

ω¯12 = ω12 + dθ.

Hence √ √ ( E)v ( G)u ω¯12 = ω12 + dθ = − √ du + √ dv + dθ. G E Since dα(s) du dv e = T = = x + x , 1 ds ds u ds v we have √ √ √ √ ( E)v ( G)u ( E)v du(s) ( G)u dv(s) κg =ω ¯12(e¯1) = − √ du(T)+ √ dv(T)+dθ(T) = − √ + √ +dθ(T). G E G ds E ds

Now dθ = θudu + θvdv, so du(s) dv(s) dθ dθ(T) = θ + θ = . u ds v ds ds

Hence √ √ ( E)v du(s) ( G)u dv(s) dθ κg = − √ + √ + . G ds E ds ds Therefore we re-proved the following Liouville theorem:

Theorem 3.5 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 √ √ ( E)v du(s) ( G)u dv(s) dθ κg = − √ + √ + . G ds E ds ds

6. A simple proof of Gauss-Bonnet theorem. The formula

dω12 = −Kω1 ∧ ω2

12 is the key in apply Green’s theorem to prove Gauss-Bonnet theorem. The formula can be re-written as

Kdσ = −dω12, where dσ = ω1 ∧ ω2.

From the Gauss equation and Stoke’s Theorem, the Gauss-Bonnet formula follows im- mediately 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 e1, e2, e3, and take a Darboux frame (i.e. a moving frame for the surface with e1 tangent to ∂M) e¯1, e¯2, e¯3 along ∂M. We write e¯1 = cos θe1 + sin θe2, e¯2 = − sin θe1 + cos θe2 (where θ is smoothly chosen along the smooth pieces of ∂M and the exterior angle j at Pj gives the jump of theta as we cross Pj). Then, by Stokes’ theorem, we have

ZZ ZZ Z Z Z X Kdσ = − dω12 = − ω12 = − (¯ω12 − dθ) = − κgds + (2π − j). M M ∂M ∂M ∂M

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