LECTURE 3: CURVES IN EUCLIDEAN AND AFFINE SPACES

1. Introduction

When studying submanifolds Σ of a homogenous space G/H, one question that arises frequently is the notion of equivalence: given two submanifolds Σ_{1}, Σ_{2}, when can one be transformed into the other via a symmetry of the underlying space G/H? For instance, in the case of submanifolds of Euclidean space 𝔼^{n} we say that two submanifolds are equivalent if one can be transformed into the other by a rigid motion.

We will approach this general question by considering the restriction of certain frames on the underlying space G/H to the submanifold Σ in question. If f : Σ → G/H is an immersion, then “restriction” actually means pullback: the restriction of a frame to Σ is really a section of the bundle f^{−1}(G) lying over Σ. We will find it useful to consider frames that are adapted to Σ. This means that instead of considering arbitrary frames, we will use the geometry of Σ to choose “nice” frames. This is somewhat analogous to choosing “nice” coordinates at a point on a surface to study the geometry at that point. The beauty of the method of moving frames is that we can do this at all points simultaneously.

Once we have chosen a nice frame along Σ, we can consider the restriction of the Maurer-Cartan form ω and its structure equations to this frame. Again, by “restriction” we actually mean the pullback f^{∗}ω. The structure equations will generally contain quantities which are invariants of Σ; this means that if we change Σ by a symmetry of the ambient space G/H, these quantities remain unchanged. Typical examples of invariants are quantities such as arc length, curvature, etc.

Given an immersion f : Σ → G/H, the choice of a frame amounts to a lifting of f to a map f̃: Σ → G. In order for the frame to contain useful information about the invariants of Σ, the choice of this lifting should be completely determined in some canonical way by the geometry of Σ. Moreover, the lifting itself should be invariant in the sense that

{ \atop (g · f)} = g · f̃

for any g ∈ G. If such an invariant lifting exists, then the question of equivalence is completely answered by the following lemma:

Important Lemma 1: Let f̃_{1}, f̃_{2} : Σ → G be two smooth immersions. Then there exists an element g ∈ G such that

f̃_{1}(x) = gf̃_{2}(x)

for all x ∈ G if and only if f̃_{1}^{∗}ω = f̃_{2}^{∗}ω, where ω is the Maurer-Cartan form of G.

This is not too surprising; if the restriction f̃^{∗}ω contains information about the invariants of Σ, then this information should remain the same if Σ is transformed by an element of G. The converse says that, in fact, f̃^{∗}ω contains enough information about Σ to determine it completely up to a symmetry of the ambient space.

2. Curves in 𝔼^{3}

Consider a smooth parametrized curve x : I → 𝔼^{3} which maps some open interval I ⊂ ℝ into Euclidean space. A frame along x is a choice, for each t ∈ I, of an orthonormal frame {e_{1}(t), e_{2}(t), e_{3}(t)} for the tangent space T_{x(t)}𝔼^{3}. If the curve is “nice enough” (the precise meaning of this will become clear shortly), we can choose such a frame in a way that reflects the geometry of the curve.

Recall that x is regular if x′(t) ≠ 0 for every t ∈ I. The first condition we will require in order that x be “nice enough” is that x must be a regular curve. With this assumption, we can make our first frame adaptation by setting

e_{1}(t) = \frac{x′(t)}{|x′(t)|},

i.e., we require that e_{1}(t) be the unit tangent vector to the curve at x(t). This choice is clearly invariant under the action of E(3). The vector e_{1} is now uniquely determined, but we may still vary e_{2}, e_{3} by an arbitrary rotation in O(2).

Here we make an observation that will simplify the remainder of our computations. Fix t_{0} ∈ I and define the arc length of x at t ∈ I to be

s(t) = ∫_{t_{0}}^{t} |x′(τ)| dτ.

The arc length is clearly invariant under the action of E(3), and since x′(t) ≠ 0 for all t ∈ I, s(t) has an inverse function t(s). By setting x(s) = x(t(s)), we can assume that x is parametrized by arc length, and so e_{1}(s) = x′(s).

In order to make the next adaptation, we need to make another assumption about the curve. We will say that x is nondegenerate if x is regular and, in addition, e′_{1}(s) ≠ 0 for all s ∈ I. In this case, differentiating the equation

⟨e_{1}(s), e_{1}(s)⟩ = 1

with respect to s yields

⟨e′_{1}(s), e_{1}(s)⟩ = 0.

Thus e′_{1}(s) is orthogonal to e_{1}(s), and we can make our second adaptation by setting

e_{2}(s) = \frac{e′_{1}(s)}{| e′_{1}(s) |}.

This vector is called the unit normal vector to the curve at x(s). It has the property that the osculating plane to the curve at x(s) is spanned by e_{1}(s) and e_{2}(s). This choice is also invariant under the action of E(3).

The frame is now essentially unique; because the frame must be orthonormal, e_{3}(s) is determined up to multiplication by ±1. (Indeed, if we consider oriented orthonormal frames, e_{3}(s) is uniquely determined.) e_{3}(s) is called the binormal vector to the curve at x(s). This adapted frame {e_{1}(s), e_{2}(s), e_{3}(s)} is called the Frenet frame of the curve x(s) and defines a lifting of the curve x̃ : I → E(3).

Now we want to consider the Maurer-Cartan forms of E(3) restricted to the Frenet frame. When we pull these forms back via the map x̃ : I → E(3), they become forms on the 1-dimensional space I, and so they must all be multiples of ds. For instance, x̃^{∗}(dx) = x′(s) ds. Recall that the structure equations on E(3) are

dx = ∑_{i=1}^{3} e_{i} ω^{i}

de_{i} = ∑_{j=1}^{3} e_{j} ω_{i}^{j}

and that ω_{j}^{i} = −ω_{i}^{j}. Let \frac{∂}{∂s} denote the vector field on I which satisfies ds(\frac{∂}{∂s}) = 1. Pulling back the structure equations via x̃ and evaluating all the 1-forms on the vector field \frac{∂}{∂s} yields

x′(s) = ∑_{i=1}^{3} e_{i}(s) ω^{i}(\frac{∂}{∂s})

e′_{i}(s) = ∑_{j=1}^{3} e_{j}(s) ω_{i}^{j}(\frac{∂}{∂s}).

But we have chosen our frame so that x′(s) = e_{1}(s) and e′_{1}(s) is a multiple of e_{2}(s), say e′_{1}(s) = κ(s) e_{2}(s). The function κ(s) is called the curvature of x at s. Note that x is nondegenerate if and only if κ(s) ≠ 0 for all s ∈ I.

The structure equations imply that

x′(s) = e_{1}(s) = ∑_{i=1}^{3} e_{i}(s) ω^{i}(\frac{∂}{∂s})

e′_{1}(s) = κ(s) e_{2}(s) = ∑_{j=1}^{3} e_{j}(s) ω_{1}^{j}(\frac{∂}{∂s}).

Therefore,

ω^{1}(\frac{∂}{∂s}) = 1,  ω^{2}(\frac{∂}{∂s}) = 0,  ω^{3}(\frac{∂}{∂s}) = 0,
ω_{1}^{2}(\frac{∂}{∂s}) = κ(s), ω_{1}^{3}(\frac{∂}{∂s}) = 0.

Now define the function τ(s) by

τ(s) = ω_{2}^{3}(\frac{∂}{∂s}).

τ(s) is called the torsion of x at s. Then the structure equations for de_{2}, de_{3} become

e′_{2}(s) = ∑_{j=1}^{3} e_{j}(s) ω_{2}^{j}(\frac{∂}{∂s}) = −κ(s) e_{1} + τ(s) e_{3}

e′_{3}(s) = ∑_{j=1}^{3} e_{j}(s) ω_{3}^{j}(\frac{∂}{∂s}) = −τ(s) e_{2}.

Thus we have the familiar Frenet equations:

[e′_{1}(s) e′_{2}(s) e′_{3}(s) x′(s)] = [e_{1}(s) e_{2}(s) e_{3}(s) x(s)] [\begin{matrix}0 & −κ(s) & 0 & 1 \\ κ(s) & 0 & −τ(s) & 0 \\ 0 & τ(s) & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}]

Note that the matrix on the right multiplied by the 1-form ds is equal to x̃(s)^{−1}d(x̃(s)), and so is exactly the pullback of the Maurer-Cartan form on E(3) via x̃.

Applying Important Lemma 1 yields the following theorem:

Theorem: Two nondegenerate curves x_{1}, x_{2} : I → 𝔼^{3} parametrized by arc length differ by a rigid motion if and only if they have the same curvature κ(s) and torsion τ(s).

3. Curves in 𝔸^{3}

Now consider a smooth parametrized curve x : I → 𝔸^{3} which maps some open interval I ⊂ ℝ into affine space. A frame along x is a choice, for each t ∈ I, of a unimodular basis {e_{1}(t), e_{2}(t), e_{3}(t)} for the tangent space T_{x(t)}𝔸^{3}. The situation is quite different from that of Euclidean space; for instance, the Euclidean notion of arc length of a curve is not invariant under the action of A(3). Moreover, we have much greater freedom in choosing our frame; the only requirement is that det[e_{1} e_{2} e_{3}] = 1.

In the Euclidean case we used the first derivative of x to choose e_{1} and the second derivative to choose e_{2}, stopping along the way to normalize so that the frame would be orthonormal. This determined e_{3} essentially uniquely, but from the structure equations it is clear that e_{3} depends on the third derivative of x. In order for this procedure to work, we had to assume that x was nondegenerate, i.e., that the vectors x′(s), x′′(s) were linearly independent for each s ∈ I.

Since orthonormality is not required in the affine case, our first attempt at an adapted frame might be

e_{1}(t) = x′(t)
e_{2}(t) = x′′(t)
e_{3}(t) = x′′′(t).

In order for this to work, we must assume that the vectors x′(t), x′′(t), x′′′(t) are linearly independent for each t ∈ I; such a curve will be called nondegenerate. For nondegenerate curves the only problem with this choice of frame is that it is not necessarily unimodular. But we can fix this by defining the adapted frame to be

e_{1}(t) = \frac{x′(t)}{\sqrt[3]{det[x′(t) x′′(t) x′′′(t)]}}
e_{2}(t) = \frac{x′′(t)}{\sqrt[3]{det[x′(t) x′′(t) x′′′(t)]}}
e_{3}(t) = \frac{x′′′(t)}{\sqrt[3]{det[x′(t) x′′(t) x′′′(t)]}}.

(Exercise: why is this frame invariant under the action of A(3)?) Now wouldn’t it be nice to get rid of that ugly denominator? Suppose we reparametrize the curve by setting x(s) = x(t(s)) for some invertible function t(s). Then

\frac{dx}{ds} = t′(s)\frac{dx}{dt}

\frac{d^{2}x}{ds^{2}} ≡ t′(s)^{2} \frac{d^{2}x}{dt^{2}} mod \frac{dx}{ds}

\frac{d^{3}x}{ds^{3}} ≡ t′(s)^{3} \frac{d^{3}x}{dt^{3}} mod \{\frac{dx}{ds}, \frac{d^{2}x}{ds^{2}}\}

so det[x′(s) x′′(s) x′′′(s)] = t′(s)^{6} det[x′(t) x′′(t) x′′′(t)]. This suggests that we define

s(t) = ∫_{t_{0}}^{t} \sqrt[6]{det[x′(τ) x′′(τ) x′′′(τ)]} dτ.

This quantity is the affine arc length, and it is invariant under the action of A(3). Note that unlike Euclidean arc length, which depends only on the first derivative of x, the affine arc length depends on the third derivative of x. In fact, this is dependent on the dimension of the ambient affine space; the affine arc length of curves in 𝔸^{n} depends on the nth derivative of x.

Assuming that the curve is parametrized by affine arc length s, we have

e_{1}(s) = x′(s)
e_{2}(s) = x′′(s)
e_{3}(s) = x′′′(s).

Therefore

x′(s) = e_{1}(s)
e′_{1}(s) = e_{2}(s)
e′_{2}(s) = e_{3}(s)
e′_{3}(s) = κ_{1}(s) e_{1}(s) + κ_{2}(s) e_{2}(s)

for some functions κ_{1}(s), κ_{2}(s), called the affine curvatures of x at s. (Exercise: Why is there no e_{3} term in e′_{3}(s)?) Thus the affine analog of the Frenet equations is

[e′_{1}(s) e′_{2}(s) e′_{3}(s) x′(s)] = [e_{1}(s) e_{2}(s) e_{3}(s) x(s)] [\begin{matrix}0 & 0 & κ_{1}(s) & 1 \\ 1 & 0 & κ_{2}(s) & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}] .

Applying Important Lemma 1 yields the following theorem:

Theorem: Two nondegenerate affine curves x_{1}, x_{2} : I → 𝔸^{3} parametrized by affine arc length differ by an affine transformation if and only if they have the same affine curvatures κ_{1}(s), κ_{2}(s).

Exercises

1. Prove Important Lemma 1. (Hint: Define h : Σ → G by

h(x) = f̃_{2}(x)f̃_{1}(x)^{−1}

and show that dh = 0. Therefore, h(x) must be constant .)

2. Prove that the choice of affine frame is invariant under the action of A(3).

3. Use the structure equations of A(3) to compute the pullbacks of the Maurer-Cartan forms ω^{i}, ω_{j}^{i} on A(3) via the map x̃ : I → A(3) for a nondegenerate curve x. Why is there no e_{3} term in the expression for e′_{3}(s)?

4. Consider the case of a curve x : I → 𝔸^{2}. How would you define an adapted frame x̃ : I → A(2) along x? How would you define nondegeneracy for x? How would you define affine arc length for x? Use the structure equations of the Maurer-Cartan forms on A(2) to find a complete set of invariants for affine curves x : I → 𝔸^{2} parametrized by affine arc length.

5. Hopefully you discovered a single invariant κ(s), called the affine curvature of x, in Exercise 4. Suppose that κ = κ(s) is constant. Show that

a) If κ = 0, then x is a straight line or a parabola.

b) If κ > 0, then x is a hyperbola.

c) If κ < 0, then x is a circle or an ellipse.

(Hint: In each case, you should be able to show that x(s) satisfies a differential equation whose general solution is not difficult to find. Since each of these conditions is preserved under affine transformations, you can perform an affine transformation to eliminate most of the arbitrary constants in the general solution. Finally, eliminate the parameter s to show that x lies on the appropriate conic section.)