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lavinia

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The question is what flows on a surface can be geodesic flows. Specifically, starting with a smooth vector field on a surface - perhaps with isolated singularities - when is there a Riemannian metric so that the vector field has constant length and is tangent to geodesics on the surface?

Here is an attempt but much more needs to be done.

If I have not made a mistake in computation then here goes.

If the vector field is length 1 it may be viewed as a map into the tangent unit circle bundle. Under this map, the connection 1 form pulls back to a 1 form on the surface. If e1 is the vector field and e2 is the orthogonal vector field with e1,e2 a positively oriented basis for the tangent space at each point, then -[e1,e2], negative the Lie bracket of e1 and e2, is dual to this 1 form - I think. If e1 is tangent to geodesics, then [e1,e2] is a multiple of e2 as can be seen from direct computation - unless I have made an error.

Write [e1,e2] = se2

Let w be the pull back of the connection 1 form. Then dw = -Kvol where K is the Gauss curvature and vol is the volume element. So

dw(e1,e2) = -K = e1.w(e2) - w([e1,e2]) = e1.<e2,-[e1,e2]> - <-[e1,e2],[e2,e2]>

Note that I omitted 1 term on the right because it is zero by assumption.

= -e1.s - s[itex]^{2}[/itex] so one has the differential equation

ds/dx = K + s[itex]^{2}[/itex] where x is the arc length parameter along the geodesic.

From this one gets information right away. For instance suppose the Gauss curvature is positive. Then s is increasing along each curve. Therefore the geodesic can not be closed - although it may actually close off at the singularities. But if K = 0 and s = 0 the geodesic can be closed as is illustrated in the flat cylinder.

Further if K > 0 so that s is increasing, the geodesics can not be dense in any region (Cauchy sequence argument I think) and any spiral must have finite length.

Interestingly, in the case of constant positive curvature this integrates to the tangent - forgetting the constant K[itex]^{1/2}[/itex] - which diverges in finite time. So one also gets a bound on the length of the geodesics. On the sphere it diverges right at the north and south poles.

So I am asking for ideas on how to push this further.

BTW: s can always be solved for in an open region of any point. So this problem is really asking whether it can be solved for globally on the surface. For instance suppose I have a vector field with a singularity of index -1. Cant it be globally tangent to geodesics?

If this is all wrong - please cut it to pieces.

Here is an attempt but much more needs to be done.

If I have not made a mistake in computation then here goes.

If the vector field is length 1 it may be viewed as a map into the tangent unit circle bundle. Under this map, the connection 1 form pulls back to a 1 form on the surface. If e1 is the vector field and e2 is the orthogonal vector field with e1,e2 a positively oriented basis for the tangent space at each point, then -[e1,e2], negative the Lie bracket of e1 and e2, is dual to this 1 form - I think. If e1 is tangent to geodesics, then [e1,e2] is a multiple of e2 as can be seen from direct computation - unless I have made an error.

Write [e1,e2] = se2

Let w be the pull back of the connection 1 form. Then dw = -Kvol where K is the Gauss curvature and vol is the volume element. So

dw(e1,e2) = -K = e1.w(e2) - w([e1,e2]) = e1.<e2,-[e1,e2]> - <-[e1,e2],[e2,e2]>

Note that I omitted 1 term on the right because it is zero by assumption.

= -e1.s - s[itex]^{2}[/itex] so one has the differential equation

ds/dx = K + s[itex]^{2}[/itex] where x is the arc length parameter along the geodesic.

From this one gets information right away. For instance suppose the Gauss curvature is positive. Then s is increasing along each curve. Therefore the geodesic can not be closed - although it may actually close off at the singularities. But if K = 0 and s = 0 the geodesic can be closed as is illustrated in the flat cylinder.

Further if K > 0 so that s is increasing, the geodesics can not be dense in any region (Cauchy sequence argument I think) and any spiral must have finite length.

Interestingly, in the case of constant positive curvature this integrates to the tangent - forgetting the constant K[itex]^{1/2}[/itex] - which diverges in finite time. So one also gets a bound on the length of the geodesics. On the sphere it diverges right at the north and south poles.

So I am asking for ideas on how to push this further.

BTW: s can always be solved for in an open region of any point. So this problem is really asking whether it can be solved for globally on the surface. For instance suppose I have a vector field with a singularity of index -1. Cant it be globally tangent to geodesics?

If this is all wrong - please cut it to pieces.

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