Non-vanishing vector field on S1xS2

In summary, the question is asking to define a non-vanishing vector field on S1 x S2 and sketch one on the Klein bottle. The solution involves combining non-vanishing vector fields on both S1 and S2. For the Klein bottle, a constant vector field in the direction parallel to the identified edges is a possible solution. This was the approach taken by the person asking the question, but the correct answer may differ.
  • #1
IAmAZucchini
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Homework Statement



Hi everyone,

This is a question on my tensor analysis/differential geometry homework due tomorrow, and I'm just not sure of the answer. The problem is to define a non-vanishing vector field V on S1 x S2.

Part b of the question is to "sketch a nonvanishing vector field on the Klein bottle".

Homework Equations



A vector field V : M [tex]\rightarrow[/tex] TM is nonvanishing if V(p) is never equal to the zero tangent vector in TpM.

So if I'm not mistaken, S1 x S2 = { x[tex]\in[/tex]R5 : x12 + x22 = 1, x32 + x42 + x52 =1 }

The Attempt at a Solution



So I figure, if I combine nonvanishing vector fields on both S1 andS2, I'll get one for the S1 x S2 manifold. I know that on S1, a nonvanishing vector field is just V(x1,x2) = (-x2, x1 ) because of the parametrization of a circle. However, a 2-sphere cannot be parametrized in terms of one coordinate, and even so, the hairy ball theorem states that it doesn't have a nonvanishing vector field. Will any vector field do, since the S1 vector field I'm using is already nonvanishing? The textbook (The Geometry of Physics by Frankel) doesn't anything about nonvanishing fields. Even if this is the case, I can't think of the parametrization that'll give me a good tangent field, since it has to give 0 when dotted with (x1,x2,x3)...

For part b) of the question (the Kelin bottle), I'm pretty sure that a constant vector field in the direction parallel to the sides identified in the same direction on the rectangle will work, since it's a smooth manifold. However, if someone could confirm that this is right, I'd really appreciate it...

Thanks for any help anyone offers!

_______
Alright, I had to hand it in, but if anyone knows the *right* answer, please do post it! I'll be going to office hours on thursday, probably, to find out anyways. My final answer was: since we can define a nonvanishing vector field on S1 V1 (x1, x2) -> (-x2,x1), and since there exists some vector field V2 tangent to S2, the combination of them V = (V1(x1, x2), V2(x3,x4,x5) ) will be nonvanishing since V1 is nonvanishing. Lame and vague/handwavy, but it's all I got.
 
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  • #2
For part b, I said that a constant vector field in the direction of the edges identified will work, because it would be nonvanishing since it's smooth.
 

FAQ: Non-vanishing vector field on S1xS2

What is a non-vanishing vector field on S1xS2?

A non-vanishing vector field on S1xS2 is a continuous vector field that does not have any points where the vector is zero. In other words, at every point on the surface, there is a non-zero vector that is tangent to the surface.

Why is a non-vanishing vector field on S1xS2 important?

Non-vanishing vector fields on S1xS2 have important applications in geometry and topology. They are used to study the curvature and topology of surfaces, and they can also be used in physical systems to model flows and dynamics.

What does S1xS2 represent in this context?

S1xS2 represents the product space of a circle (S1) and a sphere (S2). This means that the surface is a 3-dimensional space made up of points that can be described by a pair of coordinates (θ, φ) where θ represents a point on the circle and φ represents a point on the sphere.

How is a non-vanishing vector field on S1xS2 different from a regular vector field?

The main difference is that a non-vanishing vector field does not have any points where the vector is zero, while a regular vector field may have points where the vector is zero. This means that a non-vanishing vector field is always tangent to the surface, while a regular vector field may not be.

Can a non-vanishing vector field on S1xS2 exist on any surface?

No, a non-vanishing vector field can only exist on surfaces that have non-zero Euler characteristic. For example, a non-vanishing vector field cannot exist on a torus (doughnut-shaped surface) because it has a zero Euler characteristic.

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