§3. Cartan's Structural Equations and the Curvature Form Let E1, ..., En

§3. Cartan’s Structural Equations and the Curvature Form

n k Let E1,..., En be a moving (orthonormal) frame in R and let ωj its associated connection forms so that: X n j dEk = ω Ej. (3.1) j=1 k

k j k j Recall that ωj = −ωk and in particular ωk = 0. Let θ be the basic forms associated with this moving frame given by: Xn j dx = θ Ej, (3.2) j=1 where x = (x1, x2, . . . , xn). Recall that each xj is considered as the jth coordinate function on Rn deﬁned by

n xj(p) = pj for each p = (p1, p2, . . . , pn) ∈ R .

Now let us diﬀerentiate (3.1) and (3.2): apply operator d which turns vector-valued 1-forms to vector-valued 2-forms. But we must be cautious. There is no problem with the 2 2 2 left hand sides: from d = 0 we have d(dEk) ≡ d Ek = 0 and d(dx) = d x = 0. For the right hand sides, naturally we apply the product rule. The product rule for 1-forms says d(fω) = df∧ω+fdω, where ω is any 1-form where f is any smooth function. We can replace f by a vector-valued function, say F, and the same identity holds: d(Fω) = dF ∧ ω + Fdω. k j However, in (3.1) and (3.2), the one–forms ωj and θ are written in front of our vector- valued functions Ej. So we have to ﬁgure out a formula for d(ωF). Here it is:

d(ωF) = d(Fω) = Fdω + dF ∧ ω = (dω)F − ω ∧ dF. (3.3)

The minus sign in front of the last term appears because we have switched the order of the 1-forms dF (vector-valued) and ω. Were we careless we would missed this minus sign and would made a big blunder! Applying d on both sides of (3.2), using d2 = 0 and (3.3), changing index, and then substituting dE` by means of (3.1), we get Xn Xn Xn 2 j j j k 0 = d x = (dθ Ej − θ ∧ dEj) = dθ Ej − θ ∧ dEk j=1 j=1 k=1 n n n X X ³Xn ´ X ³ Xn ´ j ` j j ` j = dθ Ej − θ ∧ ω Ej = dθ − θ ∧ ω Ej. j=1 ` `=1 ` j=1 `=1 j=1

Since at each point E1,..., En form an orthonormal basis, we have

Xn dθj = θ` ∧ ωj, (3.4) `=1 `

1 which are called Cartan’s ﬁrst structural equations. Similarly, (3.1) gives

Xn Xn Xn 2 j j j ` 0 = d Ek = (dω Ej − ω ∧ dEj) = dω Ej − ωk ∧ dE` j=1 k k j=1 k `=1 n n n X X ³Xn ´ X ³ Xn ´ j ` j j ` j = dω Ej − ωk ∧ ω ∧ Ej = dω − ωkω Ej. k j=1 ` k `=1 ` j=1 `=1 j=1

Thus it follows that

Xn j ` j dω = ωk ∧ ω , (3.5) k `=1 ` which are called Cartan’s second structural equations. You must learn to appreciate the beauty of these basic structural equations in diﬀerential geometry due to Elie Cartan, who is considered by many as the greatest diﬀerential geometer.

Example 3.1. Verify the structural equations of the forms associated with the Frenet- Serrat frame of a curve in R3; (see §4.1).

Solution: Recall (1.5) from §4.1 with the identiﬁcation E1 = T, E2 = N, E3 = B, (the unit tangent vector, the normal vector, and the binormal vector respectively), and

1 2 3 3 1 2 1 3 2 ω1 = ω2 = ω3 = ω1 = ω3 = 0, ω1 = −ω2 = ω, ω2 = −ω3 = η.

1 1 Also dx = θ E1 = θ T (this is because dx is tangential and the tangent of a curve at any point is the one dimensional space spanned by T) so that θ2 = θ3 = 0. A curve has only one parameter when it is described by a parametric equation x = x(t). Thus 1-forms such as θ1, ω and η associated with the curve can be written in the form f(t)dt. But

d(f(t)dt) = df(t) ∧ dt = f 0(t)dt ∧ dt = 0.

1 P3 ` j So dθ = dω = dη = 0. Thus it remains to check `=1 θ ∧ ω` = 0 for the ﬁrst structural P3 ` k equations and `=1 ωj ∧ ω` = 0 for the second structural equations. Again, since all 1-forms here can be written in the form f(t)dt and since f1(t)dt ∧ f2(t)dt = 0, these identities are obvious. Thus, both sides of the structure equations for the Frenet-Serrat frame along a curve are vanishing. This tells us that in some sense the local geometry of a curve is not interesting at all. So, let us turn to surfaces.

Let Ej (j = 1, 2, 3) be a Darboux frame for a surface S with E1 and E2 tangential and E3 ≡ N normal. (Here we assume that S is oriented so that the pair (E1, E2)

2 has a compatible orientation at each point. We also assume that the Darboux frame given here has the same orientation as the usual one for R3 at each point. These assumption will be needed at one subtle point in the future, but let us not worry about this now.) j j Recall that the basic forms θ and the connection forms ωk are determined by the relations: 1 2 3 dx = θ E1 + θ E2 (that is, θ = 0), and

2 3 1 3 1 2 dE1 = ω1E2 + ω1E3, dE2 = ω2E1 + ω2E3, dE3 = ω3E1 + ω3E2. (3.6)

The structural equations (3.4) and (3.5) become

1 2 1 2 1 2 1 3 2 3 dθ = θ ∧ ω2, dθ = θ ∧ ω1, θ ∧ ω1 + θ ∧ ω2 = 0, (3.7)

2 3 2 3 2 3 3 1 3 dω1 = ω1 ∧ ω3, dω1 = ω1 ∧ ω2, dω2 = ω2 ∧ ω1. (3.8)

Let F be a vector ﬁeld on the surface S and tangent to the surface. Consider the vector- valued diﬀerential dF. Here we have to be cautious: that F is tangential does not imply that dF is also tangential. In fact, in general we expect that the normal component (dF · N)N of F is nonzero because F has to turn as S bends, and the acceleration has a nonzero normal component when the motion is changing its direction, even when the speed remains constant. For each vector (or vector-valued 1-form) v attached to a point on S,

denote by v> the tangential component of v, and v⊥ its normal component. Explicitly,

v = v> + v⊥, where v> = (v · E1)E1 + (v · E2)E2 and v⊥ = (v · E3)E3. (3.9)

For a vector ﬁeld (or a vector-valued 1-form) F tangent to our surface S, let

d>F = (dF)>, d⊥F = (dF)⊥, (3.10)

which will be called the covariant differential and the normal differential of the vector field F respectively. The covariant differential is responsible for the intrinsic geometry of our surface, while the normal differential determines its extrinsic shape. To get some idea about the jargon here, take a piece of paper. You can lay it flat on a table, or roll it up into a cylinder or cone. No matter how you change its shape, the intrinsic geometry is still “flat”. Now we state the product rules:

Rule PC1. d>(gF) = dg . F + g d>F.

Rule PC2. d>(ωF) = dω . F − ω ∧ d>F.

3 Rule PN1. d⊥(gF) = g d⊥F.

Rule PN2. d⊥(ω F) = −ω ∧ d⊥F.

To verify these identities, write down

d(gF) = dg . F + g dF and d(ωF) = dω . F − ω ∧ dF, and take the tangential components and normal components; (the detail is left to the reader). Notice that F> = F because F is tangential. In the present section we concentrate on covariant diﬀerentials for which only Rules PC1 and PC2 are needed.

2 2 Unlike d = 0, in general d> is nonvanishing. In fact, this is really what makes the 2 covariant diﬀerential d> interesting. To compute d>F for a tangential vector ﬁeld F, put 1 2 1 2 F = f E1 + f E2, where f = F · E1 and f = F · E2; (here we use superscript for the components f 1 and f 2 of F instead of subscripts in order to see a pattern). Now

2 3 2 d>E1 = (dE1)> = (ω1 E2 + ω1 E3)> = ω1 E2. (3.11)

1 1 (Notice that ω1 = 0, E2 is tangential and E3 is normal.) Similarly we have d>E2 = ω2E1, 2 which can be rewritten as −ω1E1 if you like. (A systematic way for writing down both P2 j 1 2 identities is d>Ek = j=1ωkEj; k = 1, 2. Insertion of the extra terms ω1E1 and ω2E2 is harmless because they vanish. This systematic way is needed if we want to study geometric objects of higher dimensions.) Thus

1 2 d>F = d>(f E1) + d>(f E2) 1 1 2 2 = df . E1 + f d>E1 + df . E2 + f d>E2 1 1 2 2 2 1 = df . E1 + f ω1E2 + df . E2 + f ω2E1 1 2 1 2 1 2 = (df + f ω2)E1 + (df + f ω1)E2.

2 To compute d>F ≡ d>(d>F), we begin with

1 2 1 1 2 1 1 2 1 d>((df + f ω2)E1) = d(df + f ω2) . E1 − (df + f ω2) ∧ d>E1 2 1 2 1 1 2 1 2 = (df ∧ ω2 + f dω2)E1 − (df + f ω2) ∧ ω1E2 2 1 2 1 1 2 = (df ∧ ω2 + f dω2)E1 − (df ∧ ω1)E2,

1 2 1 1 because ω2 ∧ ω1 = ω2 ∧ (−ω2) = 0. Similarly, we have

2 1 2 1 2 1 2 2 1 d((df + f ω1)E1) = (df ∧ ω1 + f dω1)E2 − (df ∧ ω2)E1.

4 Thus 2 2 1 2 2 1 1 2 d>F ≡ d>{f E1 + f E2} = f dω2 E1 + f dω1 E2. (3.12)

2 1 2 So d> sends a vector ﬁeld F = f E1 + f E2 (which can be regarded as a vector-valued 1 2 0-form) to a vector-valued 2-form, say ϕ E1 + ϕ E2, such that · ¸ · ¸ · ¸ 1 1 1 ϕ 0 dω2 f 1 3 1 2 = 2 2 , where dω2 = ω2 ∧ ω3. ϕ dω1 0 f

2 We may regard d> an operator-valued 2-form and its matrix representation with respect to the basis E1, E2 is given by · ¸ 1 0 dω2 Ω = 2 . (3.13) dω1 0

2 We call this operator-valued 2-form d>, or the matrix-valued 2-form Ω given above, the curvature form for the surface S. The following tensor property for the curvature 2 form d> is important: for all vector ﬁelds F tangent to the surface S and for all (nice and smooth) function g on S, we have

2 2 d>(gF) = g d>F. (3.14)

In contrast, d>(gF) 6= g d>F (in general); instead we have d>(gF) = dg. F+g d>F. (Thus, 2 d> is a tensor, but d> is not.) We are going to give two proofs of (3.14) because of its 1 2 importance. First proof: start with gF = (gf )E1 + (gf )E2. Applying (3.12), we have

2 2 1 1 2 2 1 1 2 2 d>(gF) = (gf )dω2E1 + (gf )dω1E2 = g(f dω2E1 + f dω1E2) = g d>F.

Second proof: we have d>(gF) = dg. F + g. d>F and hence

2 d>(gF) = d>(dg. F) + d>(g d>F) 2 2 = d g F − dg ∧ d>F + dg ∧ d>F + g d>F 2 = g d>F.

2 1 (The second proof applies to more general situations.) From ω1 = −ω2 we see that Ω is 1 a skew symmetric matrix. The entry dω2 of Ω turns out to be independent of the choice of our frame E1, E2. Indeed, suppose that F1, F2 is another frame, having the same orientation; (here we have used the assumption on the orientation of the surface S). Then

F1 = (cos θ)E1 − (sin θ)E2 and F2 = (sin θ)E1 + (cos θ)E2 for some function θ on S. It

5 2 1 1 follows from (10.12) that d>E2 = dω2 E1. To show that dω2 is independent of frames, it 2 1 is enough to check d>F2 = dω2 F1. Indeed, by (3.14),

2 2 2 2 d>F2 = d>(sin θ E1 + cos θ E2) = sin θ. d>E1 + cos θ. d>E2 2 1 1 1 = sin θ. dω1 E2 + cos θ. dω2 E1 = sin θ. d(−ω2) E2 + cos θ. dω2 E1 1 1 = dω2 (cos θ E1 − sin θ E2) = dω2 F1.

1 Done. The 2-from dω2 is called the Pfaffian of Ω. (The Pfaffian here has nothing to do with Pfaffian equations. It is a numerical invariant associated with a skew symmetric form on an even dimensional linear space. Strictly speaking, it is a topic in the theory of multilinear algebra.) Now a surface is a two dimensional object. The “top forms” defined on it are 2-forms and they must be a multiple of the area form θ1 ∧ θ2. In particular,

1 1 2 dω2 = K θ ∧ θ (3.15) for some scalar function K on S, which is called the Gauss curvature for S. The reader is asked to show that K here is independent of the orientation of S; see Exercise 1 (e).

Example 3.2. Find the (Gauss) curvature K for the unit sphere S2 by means of the parametrization given in Example 1.4 in §4.1.

1 2 2 Solution: From that example we have θ = cos φ dθ, θ = dφ, ω1 = sin φ dθ. So

1 2 dω2 = d(−ω1) = −d(sin φ) ∧ dθ = − cos φ dφ ∧ dθ = cos φ dθ ∧ dφ

1 2 1 1 2 2 and θ ∧ θ = cos φ dθ ∧ dφ. Thus dω2 = θ ∧ θ , telling us that the curvature of S is a positive constant: K = 1.

In the present section we mainly study the intrinsic geometry of a surface S, which only depends on the metric tensor of S, but not on the way how S is embedded in R3. As a consequence, the Gauss curvature here is deﬁned intrinsically. In the next section we study the extrinsic geometry of S, and from that point of view one can deﬁne the total curvature for S. Gauss discovered that the total curvature (an extrinsic concept) coincides with what we call the Gauss curvature (an intrinsic concept). He was so pleased with this discovery that he called it theorem agregium, which means a beautiful theorem. As you will see, using Cartan’s structural equations, the proof is only one line long!

6 Exercises

1. (a) Check Rules PC1, PC2, PN1 and PN2.

(b) Show that, if X and Y are vector ﬁelds tangent to a surface S, then

d (X · Y) = d>X · Y + X · d>Y.

Deduce that, if F is a unit vector ﬁeld tangent to S, then F · d>F = 0. (c) Show that the Gauss curvature K is independent of the orientation of S. (Hint: there are two simple ways to indicate the change of orientation: ﬁrst, change the order

of E1 and E2; second, replace one of E1 or E2 by its negative. Either way is easy to work with.)

2. In each of the following parts, a parametric surface r ≡ (x, y, z) = r(u, v) and a

Darboux frame Ek (k = 1, 2, 3) are given. Find (in terms of parameters u and v): 1 2 2 3 3 1. the fundamental forms θ , θ and connection forms ω1, ω1, ω2; 2. the metric 2 1 2 2 2 1 2 ds = (θ ) + (θ ) , the area form θ ∧ θ ; 3. the covariant diﬀerentials d>E1, d>E2

the normal diﬀerentials d⊥E1, d⊥E2; and 4. the Gaussian curvature K. (a) Let X = X(t), Y = Y (t) be a curve in XY -plane with an arc-length parametrization: X0(t)2 + Y 0(t)2 = 1. The cylinder S based on this curve is given by x(u, v) = X(u), y(u, v) = Y (u), z(u, v) = v. The Darboux frame for S is given by

∂r ∂r E = = (dX/du, dY/du, 0) , E = = (0, 0, 1), E = E × E . 1 ∂u 2 ∂v 3 1 2 (b) The unit sphere S2: r = (cos u cos v, sin u cos v, sin v) with the Darboux frame

E1 = (− sin u, cos u, 0), E2 = (− cos u sin v, − sin u sin v, cos v), E3 = E1 × E2 = r. (c) The circular cone with the circle x2 + y2 = 1 as the “base curve” and the point A = (0, 0, 1) as the apex, together with the Darboux frame given as follows:

r(u, v) = ((1 − v) cos u, (1 − v) sin u, v),

E1 = ∂r/∂u, normalzed = (− sin u, cos u, 0), 1 E2 = ∂r/∂v, normalized = √ (− cos u, − sin u, 1), E3 = E1 × E2. 2

(d) The unit sphere S2, parametrized by stereographic projection:

r(u, v) = (2uh, 2vh, (u2 + v2 − 1)h), where h = 1/(u2 + v2 + 1)

7 with the Darboux frame is obtaining by normalizing ru, rv, ru × rv. √ (e) The parametric surface r(u, v) = ( u2/2, uv/ 2, v2/2 ) (see Exercise 6 in §2.8)

with the Darboux frame obtained by normalizing ru,(ru ×rv)×ru, ru ×rv. (Patience is needed for completing the computation.)

3. Decomposition (3.9) uses the usual inner product u · v = u1v1 + u2v2 + u3v3 for two 3 vectors u = (u1, u2, u3) and v = (v1, v2, v3) in R . For certain problems (e.g. in theory of relativity) it is more appropriate to use other types of inner products. For example, in deriving the hyperbolic metric (2.8) in §4.2, we consider the hyperboloid 2 2 2 2 3 H : −x − y + z = 1 in R with the inner product of two vectors u = (u1, u2, u3)

and v = (v1, v2, v3) deﬁned to be u · v = u1v1 + u2v2 − u3v3 corresponding to the indeﬁnite metric dx2 + dy2 − dz2 chosen. Parametrize the lower sheet of H2 by

r(u, v) = (cos u sinh v, sin u sinh v, − cosh v). Let Ek be vector ﬁelds determined by

ru = sinh u E1, rv = E2 and r = E3.

(a) Verify that the vector ﬁelds E1, E2, E3 form a Darboux frame by checking:

E1 · E1 = 1, E2 · E2 = 1, E3 · E3 = −1 and E1 · E2 = E1 · E3 = E2 · E3 = 0,

1 2 k (b) Compute dr, dEk to ﬁnd θ , θ and ωj (1 ≤ k, j ≤ 3). (c) Find the metric (θ1)2 + (θ2)2, the area form θ1 ∧ θ2, as well as the curvature K.

(d) Find the covariant differentials d>E1, d>E2, normal differentials d⊥E1, d⊥E2, 2 2 as well as the second covariant differentials d>E1 and d>E2.