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lavinia

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- Does the topology of the boundary determine the topology of the tube?

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- Thread starter lavinia
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lavinia

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- Does the topology of the boundary determine the topology of the tube?

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quasar987

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homeo: a tube T is a tubular nbhd of the manifold M, so you can think of it as a (closed) nbhd of the zero sectioin of the normal bundle N that intersects each fiber in a closed disk.

lavinia, how about this: Pick a fiber metric on TN. Use this to define a splitting of TN into a vertical bundle V (isomorphic to N) and a horizontal bundle H (isomorphic to TM). This induces a similar splitting of T(T). Now, an orientation of T as a manifold would induce an orientation of N (as a vector bundle), but we know that the normal bundle is not always orientable, so the tube cannot always be orientable. In fact, it is so iff M and its normal bundle are thus.

lavinia, how about this: Pick a fiber metric on TN. Use this to define a splitting of TN into a vertical bundle V (isomorphic to N) and a horizontal bundle H (isomorphic to TM). This induces a similar splitting of T(T). Now, an orientation of T as a manifold would induce an orientation of N (as a vector bundle), but we know that the normal bundle is not always orientable, so the tube cannot always be orientable. In fact, it is so iff M and its normal bundle are thus.

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lavinia

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ok.

The Tubular Neighborhood Theorem shows that around any compact manifold without boundary embedded in another smooth manifold, there is an open neighborhood of it that is diffeomorphic to the total space of the normal bundle to the embedding. The tube is obtained first locally from the exponential mapping in the normal direction to the embedded manifold and then pieced together as the union of these local tubes. Since the embedded manifold is compact the union of finite many of these neighborhoods gives is the tube. The boundary of this tube is diffeomorphic to the unit sphere bundle in the normal bundle.

My question is whether a tube in Euclidean space is always orientable. I would think so since the boundary is a hypersurface. For example a tube around the Klein bottle in R^4 is orientable and its boundary is the the total space of a circle bundle over the Klein bottle.

The geometry of tubes was studies by Weyl and there is a cool book that derives characteristic classes from tube geometry rather than intrinsic geometry.

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lavinia

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I am confused.

You mean just the restriction of the Euclidean volume form to the tube?

My thought was that the boundary as a hypersurface must be orientable but how do I extend this to an orientation of the whole tube? But what you are saying is the the volume form on Euclidean space just does this. OK.

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quasar987

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Can anyone tell me what is wrong with my argument then? Thx.

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lavinia

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Can anyone tell me what is wrong with my argument then? Thx.

I think that the Whitney sum of two non-orientable bundles can be orientable. They just need to be non-orientable for the same reason - their first Stiefel-Whitney classes must be equal.

In Eudlidean space the normal and tangent bundles sum to the trivial bundle which is orientable.

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Oh- are you talking about the boundary of the tubular neighbourhood being orientable? (actually, just looked up orientation for a manifold with boundary, we say a manifold with boundary is orientable if the manifold obtained by glueing together two lots of the manifold with boundary across their boundaries is orientable, wasn't aware of this definition).

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