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Added description of Jacobi equation.

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edited Sep 22 2010 at 17:29

Bill Thurston
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Additional: how to visualize the Jacobi equation.The behavior of geodesics is described by the Jacobi equation, which says that the second derivative of signed distance from a geodesic to infinitesimally nearby geodesics equals-(Gaussian curvature) times distance. Integrating this equation amounts to taking the limit of composition of sequence of elements of $SL(2,\mathbb R)$, acting on (distance, derivative of distance).

This can be visualized by looking at the action of $SL(2,\mathbb R)$ on the hyperbolic plane, which we can coordinatize as the upper half plane $im(z) > 0$. The boundary of the upper half-plane corresponds to the set of slopes of lines in the plane, with 0 corresponding to parallel variations of a geodesic and infinity corresponding to variations of a geodesic that keep distance 0 and turn at an angle. In general, each points on the circle at infinity tell us the curvature of an advancing wave-front. If the point is on the right, the wave front is convex; on the left, concave.

If curvature is 0, the point at infinity is fixed, and the action is a unit speed translation to the right in Euclidean terms, a parabolic transformation in hypebolic terms. Curvature creates an additional effect that moves the point at infinity: positive curvature moves it counterclockwise by adding a parabolic vector field fixing 0, and negative curvature moves it clockwise.

The instantaneous sum of these two motions is rotation around a point in the positive curvature case, and translation along a geodesic in the negative curvature case. Thus for positive curvature, the wave front shapes make complete circuits around the circle at infinity, alternating between local convexity and local concavity. In the negative curvature case, once convex they always remain convex.

I don't want to make this answer even overlier long, but it should be clear from this concrete picture of the Jacobi equation that we can put in smooth little blips of negative curvature without changing the qualitative nature of the wave fronts at the time of the collision that forms the cut locus, and thus not destroying convexity of the precut locus.

Retracted false portion, concerning precut locus for S^2.

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edited Sep 22 2010 at 15:49

Bill Thurston
17.8k●8●62●100

Correction If a tip of the cut locus is not a conjugate point, then the exponential map is a local diffeomeorphism near its preimage, in which case the precut locus is obviously nonconvex. At an isolated tip, the only way to have local convexity is for a family of equal-length geodesics converge to the tip from directions ranging over a 180 degrees. The total curvature of the bigon swept out is the sum of its two angles, which is greater than $\pi$. In a positively curved metric, there can be at most 3 tips of the tree that are focal points in this way, since the total curvature is $4 \pi$, so for any positively curved metric on $S^2$, if any point has cut locus more with more than 3 tips, then the precut locus at that point is not convex. The generic behavior is for the tips of branches to be instantaneous focal points only, so there would be no positive angle of equal geodesics. Now I realize, after further thought upon seeing Ludovic's comment: Even though the conjugate locus in the manifold has cusps, it is the image of a smooth curve in the unit tangent bundle whose preimage as the boundary of the precut locus is smooth, so it changes smoothly with sufficiently smooth changes of the metric and remains convex under perturbation.

The same trick can work to get metrics with convex precut locus on $S^2$ even though they have patches of negative curvature: a very localized $C^3$-bounded change of the metric can make curvature have localized areas of negativity, but it changes the exponential map near the cut locus by a $C^3$ small amounts.The precut locus will be perturbed by only small $C^2$ amount and remain convex.

True but not relevant to the question: I don't think the large angles of focusing on all tips of a cut locus tree can happen for all points in an open set, although I haven't thought it through. If all trees collapse to points, then the sphere is swept out by equal length geodesics between those two points. I think it should be known that if this happens for any starting point, the metric is a constant curvature metric. (For small perturbations this is related to the Radon transform variant

Added conjugate locus picture

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edited Sep 22 2010 at 14:49

Bill Thurston
17.8k●8●62●100

Added For the case in the figure above, we can see the conjugate locus by drawing rays perpendicular to the curve, showing the geodesics; the fold lines in the picture give the conjugate locus. Each tip of the cut locus is a cusp of the conjugate locus; waves coming from one side of the point focus behind the cut locus. In this case, the conjugate locus is visible as an 8-pointed star with 4 long points and 4 short points ending at the ends of the cut locus tree. This figure shows 1400 very thin lines, and shows a larger region than above.

Added picture of cut locus

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edited Sep 22 2010 at 14:17

Bill Thurston
17.8k●8●62●100

Clarified analysis for S^2

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edited Sep 22 2010 at 13:44

Bill Thurston
17.8k●8●62●100

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answered Sep 22 2010 at 12:47

Bill Thurston
17.8k●8●62●100

For a non-injective map like $\exp$, there are many alternative choices of subsets of the domain where the restriction is injective. For a flat 2-torus like $\mathbb R^2/\mathbb Z^2$, one domain for $\exp$ is a square centered at the origin, but there are many others --- any shape that tiles the plane by translations in $\mathbb Z^2$ will do. For $\exp$, there is a canonical for vectors whose geodesics don't reach the cut locus. I'll take this to be your meaning. Is there a standard terminology, perhaps even "injectivity domain" as you used? Even if so, I think it is misleading, so I'll refer to it as the pre-cut locus.

Consider the case $M^2 = \mathbb {RP}^2 = S^2 / \pm 1$. For the standard round metric, the pre-cut locus is the disk of radius $\pi$ in every tangent space. The cut locus remains smooth and convex under small perturbations of the metric, since it's far away from the conjugate locus. But there are also modifications of the metric that make the curvature negative in places without changing the geodesic flow very much.

A positively curved metric tends to focus a family of geodesics, making the wave fronts (curves perpendicular to the geodesics) bend toward the concave sense so that the total geodesic curvature of the wave front decreases (by an amount equal to the total Gaussian curvature they sweep through). Negative curvature bends in the opposite way. But just as camera lenses are often designed with both convex and concave elements, you can put some negative curvature into the mix and still make geodesics focus. The cut locus is determined by the shapes of wavefronts at the time they collide, and wherever he net geodesic curvature of any segment is less at the time of collision than the corresponding angle in the tangent space, the precut locus will be locally convex. (But note that to get the full local convexity description depends on geodesics from both sides).

No: there are metrics on $\mathbb {RP}^2$ with small areas of negative curvature whose precut locus is still convex.

On $S^2$, another phenomenon takes place: the cut locus for the round metric is also the conjugate locus, where geodesics actually focus. In small perturbation of the metric, the cut locus typically becomes a planar tree, which can have arbitrary combinatorial complexity, and can even have infinitely many branches. This is closely connected to the possible cut loci for smooth curves in the plane, that is, the set of centers of disks with interiors inside the curve and boundary circles that touch the curve in two or more points.

The part of the precut locus that maps to the tip of a branch has a 360 degree angle, so it is never convex. The only way to avoid this on $S^2$ is to have the cut locus always be a point. I think this should be known to imply the metric has constant curvature.