Vector fields transverse to the boundary of a manifold

jford1906

I'm trying to work up some examples to help me understand this concept. Would the periodic flow on a solid torus be transverse to it's boundary?

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WWGD

Gold Member
Transverse I think means here not lying in the tangent space to the boundary points.

jford1906

So since it's a periodic flow, say given by $$dx = -y, dy=x \mbox{ and }dz = 0,$$ all trajectories are parallel to the boundary, so they would not lie in the tangent space to any boundary points, and the flow would be transverse?

WWGD

Gold Member
That sounds right.

Rock. Thanks!

lavinia

Gold Member
So since it's a periodic flow, say given by $$dx = -y, dy=x \mbox{ and }dz = 0,$$ all trajectories are parallel to the boundary, so they would not lie in the tangent space to any boundary points, and the flow would be transverse?
Transverse to the boundary means that the part of the flow that is on boundary does not lie in the tangent space to the boundary. So the flow must point either outwards or inwards.

zinq

I'm trying to work up some examples to help me understand this concept. Would the periodic flow on a solid torus be transverse to it's boundary?
First of all, there are a lot of different periodic flows on a solid torus, so no one flow is "the" periodic flow.*

Secondly (as Lavinia has said), a flow being transverse to a manifold's boundary means that the flow is never tangent to the boundary. It follows that on each component of the boundary, the flow is either pointing only inwards, or else pointing only outwards on that component. Since the solid torus has only one boundary component, of course, a flow transverse to its boundary must be everywhere outward or everywhere inward on the boundary.

Clearly, any flow transverse to the boundary of any manifold cannot be periodic, since a periodic flow means that at a certain time T > 0, the flow carries all points back to where they started.
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* For example: If θ is the angular coordinate in the solid torus D2 × S1 (the disk cross the circle = {exp(iθ) : 0 <= θ < 2π}), then the simplest periodic flow is given by the vector field that is d/dθ everywhere. But now we can imagine cutting D2 × S1 along say the disk D2 × {exp(i0)}, giving one exposed disk a rotation by angle (2πp/q) where p/q is any rational number, and the reattaching the exposed disks to make a solid torus again. In this case, the flow will carry all points back to where they started at time 2πq. The core circle of the solid torus will flow back to where it started in less time, namely 2π, than any other trajectory.

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