Riemannian
structures
and free curve-straightening
Anders Linnér
Submitted January 15, 2005 and, in revised form, August 24, 2005.
Abstract. The choice of Riemannian structure on a space of curves affects computational speed, smoothness, and the preservation of symmetries along steepest descent. It is in general impossible to address all of these concerns at once. For instance, to preserve symmetry of reflection it is necessary to sacrifice either computational speed, or smoothness. Another peculiarity is the failure of some Riemannian structures to commute with the derivative operator as the flow projects onto the space of velocities of the curves. By choosing a structure that avoids this behavior, a flow-invariant is discovered for free curve-straightening. Using this invariant, it is possible to predict the limit-length along steepest descent in terms of the initial elastic energy. This yields examples where all negative gradient trajectories converge despite the fact that the Palais-Smale condition is violated. The length of the curves grows monotonically along steepest descent. In Euclidean space this behavior is not shared by the curve-straightening flow applied to curves with both endpoints fixed.
1991 Mathematics Subject Classification. Primary 58F25; Secondary 58E10, 53C21.
Key words and phrases. Curve-straightening, Palais-Smale condition, gradient trajectories.
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(1.1) |
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extends indefinitely so that . Due to the presence of the gradient, the trajectories are affected by the choice of Riemannian structure. In particular, it is possible that smoothness may be lost, and initial symmetries may be destroyed along the trajectory. When corresponds to a space of curves, it is of interest to compare the flow in to the flow in the space of velocities. This relationship is tied to the choice of Riemannian structure. Perhaps surprisingly, the two flows and the derivative operator do not always combine to form a commutative diagram.
Section 4 examines the free curve-straightening flow. The flow-invariant is the main topic here. Free curve-straightening is illustrated by considering circular arcs in Euclidean, spherical, and hyperbolic geometry, subject to either variable or fixed length. The more routine proofs are available in the Appendix.
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Assume , and sufficiently smooth so that is well defined. There are two quantities that simplify the presentation of the gradients:
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and
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Here denotes the partial derivatives of with respect to the second/third variable. The super and subscripts on , and are there to remind the reader that these quantities depend on and . This formalism anticipates the presence of isoperimetric constraints. The quantity introduced here is the negative of the anti-derivative of the so-called Euler operator; see page 16 in [2]. In Mechanics the quantity corresponds to the work.
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Here , , , and . To save space, write , , and later .
Theorem 2.1. For functions such that ,
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Proof. See the Appendix. ÿ
The inner product
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utilized in [8], [9], [10], [11], and [12], is a standard Sobolev inner product. Note that with and , leads to the much simpler formula
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If the integrand of the functional has no explicit dependence on so that , then
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because in this case. Moreover, , so the initial point is fixed. The endpoints are not in general expected to remain fixed under gradient flows. For instance, if and , then .
Theorem 2.2. If is twice continuously differentiable and a critical point, then for all and . The statement
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with equality everywhere, is the Euler-Lagrange equation. The statements , correspond to the natural boundary conditions .
Proof. If the gradient vanishes in one metric, then it vanishes in all metrics. In the present context, the most efficient argument is available using the metric . Set in the formula for the gradient and conclude that . The derivative of the gradient formula reveals that for all . Next note that . Finally, since (2.13) implies
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it is also true that
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and therefore . ÿ
Observe that the statement incorporates the Euler-Lagrange equation as well as the necessary conditions given by the natural boundary conditions. It follows that the equation is more selective than the Euler-Lagrange equation.
Theorem 2.3. The gradients in the three metrics are given by
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and depending on if or
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Proof. See the Appendix. ÿ
Note that the gradient vector fields do not depend on so these are constant vector fields on . If , then
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If , then
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When the metric is or , the vectors in the vector field are smooth ( ) elements in . Only if or is the same true in the metric .
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where is equipped with the metric , , and , respectively.
Theorem 2.4. The projected gradients are given by
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Proof. See the Appendix.
ÿ
Note the absence of in the expression for the projected gradients. It
is also obvious that . It is seen
that, if only the initial point is kept fixed and , then the projected gradient is given by
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provided . In general,
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When only is kept fixed, the general case is given by
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It is somewhat surprising that both and are absent in this last formula. Another peculiarity is that if and , then
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and
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On the other hand if and , then
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and
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The formulas (2.33) and (2.36) are possible to interchange in problems where it is irrelevant which of the two endpoints is fixed. The same remark applies to (2.34) and (2.35). Formulas (2.36) and (2.34) have theoretical and numerical advantages.
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where satisfies the same assumptions as . The constraint is classically known as an isoperimetric constraint. Assume regularity so that implies is onto, and hence . With this assumption, is a closed submanifold of co-dimension 1. Specifically, at each , splits into a one-dimensional normal space and the tangent space
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Note that depends on the metric and that it spans the normal space. The projection of a gradient onto the tangent space is here denoted by . Up to this point the tangent space has played only a minor role since and
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so in both cases the tangent space is independent of . When the regularity assumption is satisfied, then for each metric there is a unique scalar field defined on such that
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The condition determines . In fact, the formula is given by
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To use this formula, recall that
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For instance, when the metric is , use
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and
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The special case with and is given by
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Lemma 2.5. Let be a Hilbert space. Consider two closed linear
subspaces and . Let , , , , and denote the various projections. The following is
always true .
Proof. See the Appendix.
ÿ
With the help of this lemma, the formula for may be applied to gradients that are themselves
the result of projections. This is particularly useful in the presence of
constrained endpoints. If all the
constraints are dealt with at once, or at least all of the isoperimetric
constraints, then is a vector field with the same number of
components as the number of constraints. The formula for this time involves the inverse of a matrix. For
theoretical purposes this may be the preferred approach, but for computations
consideration should be given the iterative approach.
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and let be given by
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where satisfies the same assumptions as before. Define by
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Note that if , then . Let
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With , let
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The gradient vector field in is given by
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The flow along the negative gradient trajectories of is here referred to as the velocity flow. As seen in (3.6), the gradients are free of derivatives of the function . This fact is important when numerical algorithms are developed. Specifically, the flow along the gradient trajectories of is rarely known explicitly, and hence only discrete approximations of are available. This means is also only available as a discrete approximation. In contrast, the flow of yields discrete approximations of and hence discrete approximations of the anti-derivative . The operation is better behaved numerically than the operation .
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given by
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If and , then the derivative operator projects onto . Conversely, lifts to when . Look at the following diagram
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The upper half depends on the metric in . It is natural to ask if the derivative operator along the ‘missing edge’ turns this into a commutative diagram.
Theorem 3.1. Denote the metric in by and let have the induced metric. Suppose the form of is given by
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Consider all and such that both and exist. Now
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if and only if is a positive constant times for all .
Proof. Start with an arbitrary so that . Compute the two directional derivatives and note that they are related by . It follows that
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If is a positive constant times , then and
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because , and is an arbitrary element in . Conversely, if is not a positive constant times , then there is a pair such that and but . Let so that , and observe that
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for this , and hence . ÿ
It follows from this that behaves better than the other metrics when the
velocity flow is treated together with the original flow. The positive constant
alluded to is 1 in this case.
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in the form
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An application of the duBois-Reymond’s lemma shows that
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for some . The difference must be twice differentiable and
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From this it is seen that , and in particular
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If and are twice differentiable, then . When is known, this is a non-homogenous linear ordinary differential equation. The homogenous solution may be expressed in terms of hyperbolic functions as seen in the formula for .
Example 3.2. Suppose in the metric with and . Consider the smooth unconstrained functional
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Consider the initial function given by . The global minimum value of is , and this value is attained at . Click the following link and press Ctrl+A followed by Ctrl+Y, adjust animation speed via Edit|Preferences …|Graphics Options|Animation. The animation simulates the negative gradient trajectory with initial point and limit . The gradient flow slows down as the limit is approached. Smoothness is lost instantly during the descent. The symmetry of reflection is retained.
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In the context of curve-straightening, the traditional Riemannian structure on is
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The gradient of relates to the directional derivative via . Using the previous notation, it is straightforward to verify that the two components of the gradient are given by and .
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The gradient has only one component, . Along the negative gradient trajectory with flow parameter , the turning angles satisfy
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From the gradient formula, it is clear that that is constant. The initial represents the turning angle between the parallel transport of and . It follows that , and hence . One expects a different choice of Riemannian structure on to yield a different formula for the gradient . There are Sobolev metrics that produce gradients of that do not vanish at the initial point; see equation (2.11). With the current metric, the general solution through some initial is the explicit .

Figure 1
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Figure 2 shows snapshots of the flow at the same as before. The equatorial geodesic is also indicated. Figure 2 animation.

Figure 2
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The initial hyperbolic radius is
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According to Mathematica, the initial geodesic curvature, which by the Gauss-Bonnet theorem is the hyperbolic area enclosed plus , is the bewildering
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The turning angle is since the curve has unit length. Figure 3 shows snapshots of the flow at the same . The tangential geodesic is also indicated. Figure 3 animation.

Figure 3
Theorem 4.1. Consider the negative gradient flow trajectory emanating from the initial point defined for by
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where
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It is always true that for all ,
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Proof.
Let
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The derivative is
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The negative disappears since the flow is in the negative gradient direction. The derivative with respect to eliminates the constant in the gradient. The gradient and the operator do not automatically satisfy a commutative law. There are Riemannian structures on based on other selections of Sobolev inner products that violate the commutative law. The structure of the present article is not one of them; see Theorem 3.1. The next step is to use the -component of the gradient. Observe that
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Integrate and shuffle the terms to show that this is equivalent to , where is a constant of integration that only depends on the initial . ÿ
Corollary 4.2. The free curve-straightening flow increases the length of the curves in a strictly monotone fashion. Moreover, each initial curve , such that and with turning angle , converges to a geodesic segment of length with initial point and initial direction .
Proof. From the previous proof, it is clear that is positive. Since and , it follows that is bounded from above. Observe that
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so if , it must be that converges to a constant function. All that remains is to determine at the initial curve and use the consequence that . ÿ
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The free flow invariant has the simpler form here, and this leads to the separable equation
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According to Mathematica, the inverse of the solution is given by
(4.18)
It is of course possible to approximate the solution numerically by solving the ordinary differential equation. Compare the free curve-straightening flow of fixed length in the Euclidean plane with its variable length counterpart. The ‘snapshots’ are taken at the same flow times.

Figure 4
Here is the free curve-straightening flow in the Euclidean plane followed much longer ( ). Figure 5 animation.

Figure 5
Figure 6 shows the variable length free curve-straightening flow in the hyperbolic disc. The initial curve is the unit length circle centered at the origin. Figure 6 animation.

Figure 6
Figure 7 shows the variable length free curve-straightening flow in the standard sphere. The initial unit length curve is at ‘the bottom’. The sphere is here invisible. The limit curve is part of the great circle that is tangent to the initial curve at its initial point. Click and hold the right mouse button while moving the pointer to examine Figure 7 three-dimensional.

Figure 7
In arbitrary two-dimensional Riemannian manifolds, the turning angle has its -plot in figure 8.

Figure 8
The length is plotted in figure 9.

Figure 9
As suggested by this and the different examples, the length ‘converges faster’ than the turning angle. ‘The free curve-straightening flow prefers to spend the available elastic energy towards lengthening rather than flattening the turning angle.’ The limit of the length is here .
Acknowledgement: Many thanks to Franz Pedit for the invitation to visit GANG and for our many useful and interesting discussions.
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Department of Mathematical Sciences, Northern Illinois University, DeKalb, Illinois 60115, USA
E-mail address: alinner@math.niu.edu