Chern/Banchoff: Differential Geometry

2. Curvature and Fenchel's Theorem

If X is an immersed curve, with X'(t) 0 for all t in the domain, then we may define the unit tangent vector T(t) to be X'(t)/|X'(t)|. If the parameter is arclength, then the unit tangent vector T(s) is given simply by X'(s). The line through X(t₀) in the direction of T(t₀) is called the tangent line at X(t₀). We can write this line as Y(u) = X(t₀) + uT(t₀), where u is a parameter that can take on all real values.

Since T(t)T(t) = 1 for all t, we can differentiate both sides of this expression, and we obtain 2T'(t)T(t) = 0. Therefore T'(t) is orthogonal to T(t). The curvature of the space curve X(t) is defined by the condition k(t) = |T'(t)|/|X'(t)|, so k(t)s'(t) = |T'(t)|. If the parameter is arclength, then X'(s) = T(s) and k(s) = |T'(s)| = |X"(s)|.

Proposition 1: If k(t) = 0 for all t, then the curve lies along a straight line.

Proof: Since k(t) = 0, we have T'(t) = 0 and T(t) = A, a constant unit vector. Then X'(t) = T(t)s'(t) = As'(t), so by integrating both sides of the equation, we obtain X(t) = As(t) + B for some constant B. Thus X(t) lies on the line through B in the direction of A.

Curvature is one of the simplest and at the same time one of the most important properties of a curve. We may obtain insight into curvature by considering the second derivative vector X"(t), often called the acceleration vector when we think of X(t) as representing the path of a particle at time t. If the curve is parametrized by arclength, then X'(s)X'(s) = 1 so X"(s)X'(t) = 0 and k(s) = |X"(s)|. For a general parameter t, we have X'(t) = s'(t)T(t) so X"(t) = s"(t)T(t) + s'(t)T'(t). If we take the cross product of both sides with X'(t) then the first term on the right is zero since X'(t) is parallel to T(t). Moreover X'(t) is perpendicular to T'(t) so |T'(t) x X'(t)| = |T'(t)| |X'(t)| = s'(t)² k(t). Thus X"(t) x X'(t) = s'(t)T'(t) x X'(t) and | X"(t) x X'(t)| = s'(t)³k(t). This gives a convenient way of finding the curvature when the curve is defined with respect to an arbitrary parameter. We can write this simply as k(t) = |X"(t) x X'(t)|/| X'(t) X'(t)|^3/2.

Note that the curvature k(t) of a space curve is non-negative for all t. The curvature can be zero, for example at every point of a curve lying along a straight line, or at an isolated point like t = 0 for the curve X(t) = ( t, t³, 0 ). A curve for which k(t) > 0 for all t is called non-inflectional.

The unit tangent vectors emanating from the origin form a curve T(t) on the unit sphere called the tangential indicatrix of the curve X. To calculate the length of the tangent indicatrix, we form the integral of |T'(t)| = k(t)s'(t) with respect to t, so the length is k(t)s'(t)dt = k(s)ds. This significant integral is called the total curvature of the curve X.

Up to this time, we have concentrated primarily on local properties of curves, determined at each point by the nature of the curve in an arbitrarily small neighborhood of the point. We are now in a position to prove our first result in global differential geometry or differential geometry in the large.

By a closed curve X(t), a t b, we mean a curve such that X(b) = X(a). We will assume moreover that the derivative vectors match at the endpoints of the interval, so X'(b) = X'(a).

Fenchel's Theorem: The total curvature of a closed space curve X is greater than or equal to 2, i.e. k(s)ds 2.

The first proof of this result was found independently by B. Segre in 1934 and later independently by H. Rutishauser and H. Samelson in 1948. The following proof depends on a lemma by R. Horn in 1971:

Lemma: Let G be a closed curve on the unit sphere with length L < 2. Then there is a point M on the sphere that is the north pole of a hemisphere containing G. To see this, consider two points P and Q on the curve that break G up into two pieces G₁ and G₂ of equal length, therefore both less than . Then the distance from P to Q along the sphere is less than so there is a unique minor arc from P to Q. Let M be the midpoint of this arc. We wish to show that no point of G hits the equatorial great circle with M as north pole. If a point on one of the curves, say G₁, hits the equator at a point R, then we may construct another curve G₁' by rotating G₁ one-half turn about the axis through M, so that P goes to Q and Q to P while R goes to the antipodal point R'. The curve formed by G₁ and G₁' has the same length as the original curve G, but it contains a pair of antipodal points so it must have length at least 2, contradicting the hypothesis that the length of G was less than 2.

From this lemma, it follows that any curve on the sphere with length less than 2 is contained in a hemisphere centered at a point M. However if X(t) is a closed curve, we may consider the differentiable function f(t) = X(t)M. At the maximum and minimum values of f on the closed curve X, we have 0 = f'(t) = X'(t)M = s'(t)T(t)M, so there are at least two points on the curve such that the tangential image is perpendicular to M. Therefore the tangential indicatrix of the closed curve X is not contained in a hemisphere, so by the lemma, the length of any such indicatrix is greater than 2. Therefore the total curvature of the closed curve X is also greater than 2.

Corollary: If, for a closed curve X, we have k(t) 1/R for all t, then the curve has length L 2R.

Proof: L = ds Rk(s)ds = Rk(s)ds 2R.

Fenchel also proved the stronger result that the total curvature of a closed curve equals 2 if and only if the curve is a convex plane curve.

I. Fáry and J. Milnor proved independently that the total curvature must be greater than 4 for any non-self-intersecting space curve that is knotted (not deformable to a circle without self-intersecting during the process.)

Exercises

Let X be a curve with X'(t₀) 0. Show that the tangent line at X(t₀) can be written as Y(u) = X(t₀) + uX'(t₀) where u is a parameter that can take on all real values.

The plane through a point X(t₀) perpendicular to the tangent line is called the normal plane at the point. Show that a point Y is on the normal plane at X(t₀) if and only if X'(t₀)Y = X'(t₀) X(t₀)).

Show that the curvature k of a circular helix X(t) = ( r cos(t), r sin(t), pt ) is equal to the constant value k = |r|/(r² + p²). Are there any other curves with constant curvature? Give a plausible argument for your answer.

Assuming that the level surfaces of two functions F(x₁, x₂, x₃) = 0 and G(x₁, x₂, x₃) = 0 meet in a curve, find an expression for the tangent vector to the curve at a point in terms of the gradient vectors of F and G (where we assume that these two gradient vectors are linearly independent at any intersection point.) Show that the two level surfaces x₂ - x₁² = 0 and x₃x₁- x₂² = 0 consists of a line and a "twisted cubic" x₁(t) = t, x₂(t) = t², x₃(t) = t³. What is the line?

What is the geometric meaning of the function f(t) = X(t)M used in the proof of Fenchel's theorem?

Let M be a unit vector and let X be a space curve. Show that the projection of this curve into the plane perpendicular to M is given by Y(t) = X(t) - (X(t)M)M. Under what conditions will there be a t₀ with Y'(t₀) = 0?

2. Curvature and Fenchel's Theorem If X is an immersed curve, with X'(t) 0 for all t in the domain, then we may define the unit tangent vector T(t) to be X'(t)/\|X'(t)\|. If the parameter is arclength, then the unit tangent vector T(s) is given simply by X'(s). The line through X(t₀) in the direction of T(t₀) is called the tangent line at X(t₀). We can write this line as Y(u) = X(t₀) + uT(t₀), where u is a parameter that can take on all real values.
Since T(t)T(t) = 1 for all t, we can differentiate both sides of this expression, and we obtain 2T'(t)T(t) = 0. Therefore T'(t) is orthogonal to T(t). The curvature of the space curve X(t) is defined by the condition k(t) = \|T'(t)\|/\|X'(t)\|, so k(t)s'(t) = \|T'(t)\|. If the parameter is arclength, then X'(s) = T(s) and k(s) = \|T'(s)\| = \|X"(s)\|.
Proposition 1: If k(t) = 0 for all t, then the curve lies along a straight line.
Proof: Since k(t) = 0, we have T'(t) = 0 and T(t) = A, a constant unit vector. Then X'(t) = T(t)s'(t) = As'(t), so by integrating both sides of the equation, we obtain X(t) = As(t) + B for some constant B. Thus X(t) lies on the line through B in the direction of A.
Curvature is one of the simplest and at the same time one of the most important properties of a curve. We may obtain insight into curvature by considering the second derivative vector X"(t), often called the acceleration vector when we think of X(t) as representing the path of a particle at time t. If the curve is parametrized by arclength, then X'(s)X'(s) = 1 so X"(s)X'(t) = 0 and k(s) = \|X"(s)\|. For a general parameter t, we have X'(t) = s'(t)T(t) so X"(t) = s"(t)T(t) + s'(t)T'(t). If we take the cross product of both sides with X'(t) then the first term on the right is zero since X'(t) is parallel to T(t). Moreover X'(t) is perpendicular to T'(t) so \|T'(t) x X'(t)\| = \|T'(t)\| \|X'(t)\| = s'(t)² k(t). Thus X"(t) x X'(t) = s'(t)T'(t) x X'(t) and \| X"(t) x X'(t)\| = s'(t)³k(t). This gives a convenient way of finding the curvature when the curve is defined with respect to an arbitrary parameter. We can write this simply as k(t) = \|X"(t) x X'(t)\|/\| X'(t) X'(t)\|^3/2.
Note that the curvature k(t) of a space curve is non-negative for all t. The curvature can be zero, for example at every point of a curve lying along a straight line, or at an isolated point like t = 0 for the curve X(t) = ( t, t³, 0 ). A curve for which k(t) > 0 for all t is called non-inflectional.
The unit tangent vectors emanating from the origin form a curve T(t) on the unit sphere called the tangential indicatrix of the curve X. To calculate the length of the tangent indicatrix, we form the integral of \|T'(t)\| = k(t)s'(t) with respect to t, so the length is k(t)s'(t)dt = k(s)ds. This significant integral is called the total curvature of the curve X.
Up to this time, we have concentrated primarily on local properties of curves, determined at each point by the nature of the curve in an arbitrarily small neighborhood of the point. We are now in a position to prove our first result in global differential geometry or differential geometry in the large.
By a closed curve X(t), a t b, we mean a curve such that X(b) = X(a). We will assume moreover that the derivative vectors match at the endpoints of the interval, so X'(b) = X'(a).
Fenchel's Theorem: The total curvature of a closed space curve X is greater than or equal to 2, i.e. k(s)ds 2.
The first proof of this result was found independently by B. Segre in 1934 and later independently by H. Rutishauser and H. Samelson in 1948. The following proof depends on a lemma by R. Horn in 1971:
Lemma: Let G be a closed curve on the unit sphere with length L < 2. Then there is a point M on the sphere that is the north pole of a hemisphere containing G. To see this, consider two points P and Q on the curve that break G up into two pieces G₁ and G₂ of equal length, therefore both less than . Then the distance from P to Q along the sphere is less than so there is a unique minor arc from P to Q. Let M be the midpoint of this arc. We wish to show that no point of G hits the equatorial great circle with M as north pole. If a point on one of the curves, say G₁, hits the equator at a point R, then we may construct another curve G₁' by rotating G₁ one-half turn about the axis through M, so that P goes to Q and Q to P while R goes to the antipodal point R'. The curve formed by G₁ and G₁' has the same length as the original curve G, but it contains a pair of antipodal points so it must have length at least 2, contradicting the hypothesis that the length of G was less than 2.
From this lemma, it follows that any curve on the sphere with length less than 2 is contained in a hemisphere centered at a point M. However if X(t) is a closed curve, we may consider the differentiable function f(t) = X(t)M. At the maximum and minimum values of f on the closed curve X, we have 0 = f'(t) = X'(t)M = s'(t)T(t)M, so there are at least two points on the curve such that the tangential image is perpendicular to M. Therefore the tangential indicatrix of the closed curve X is not contained in a hemisphere, so by the lemma, the length of any such indicatrix is greater than 2. Therefore the total curvature of the closed curve X is also greater than 2.
Corollary: If, for a closed curve X, we have k(t) 1/R for all t, then the curve has length L 2R.
Proof: L = ds Rk(s)ds = Rk(s)ds 2R.
Fenchel also proved the stronger result that the total curvature of a closed curve equals 2 if and only if the curve is a convex plane curve.
I. Fáry and J. Milnor proved independently that the total curvature must be greater than 4 for any non-self-intersecting space curve that is knotted (not deformable to a circle without self-intersecting during the process.)
Exercises Let X be a curve with X'(t₀) 0. Show that the tangent line at X(t₀) can be written as Y(u) = X(t₀) + uX'(t₀) where u is a parameter that can take on all real values.
The plane through a point X(t₀) perpendicular to the tangent line is called the normal plane at the point. Show that a point Y is on the normal plane at X(t₀) if and only if X'(t₀)Y = X'(t₀) X(t₀)).
Show that the curvature k of a circular helix X(t) = ( r cos(t), r sin(t), pt ) is equal to the constant value k = \|r\|/(r² + p²). Are there any other curves with constant curvature? Give a plausible argument for your answer.
Assuming that the level surfaces of two functions F(x₁, x₂, x₃) = 0 and G(x₁, x₂, x₃) = 0 meet in a curve, find an expression for the tangent vector to the curve at a point in terms of the gradient vectors of F and G (where we assume that these two gradient vectors are linearly independent at any intersection point.) Show that the two level surfaces x₂ - x₁² = 0 and x₃x₁- x₂² = 0 consists of a line and a "twisted cubic" x₁(t) = t, x₂(t) = t², x₃(t) = t³. What is the line?
What is the geometric meaning of the function f(t) = X(t)M used in the proof of Fenchel's theorem?
Let M be a unit vector and let X be a space curve. Show that the projection of this curve into the plane perpendicular to M is given by Y(t) = X(t) - (X(t)M)M. Under what conditions will there be a t₀ with Y'(t₀) = 0?