# Is tangent bundle TM the product manifold of M and T_pM?

• A
mathwonk
Science Advisor
Homework Helper
2020 Award
I thought so, but the word "distinguish" seemed to imply more than is true. My misunderstanding. (I had also mentioned this invariant in post #41.)

lavinia
Science Advisor
Gold Member
I thought so, but the word "distinguish" seemed to imply more than is true. My misunderstanding. (I had also mentioned this invariant in post #41.)
[email protected] I am guessing that what you are trying to say is that different knots can be inequivalent and that this inequivalence can be described topologically. That is true. The difference just isn't through the tangent bundles.

WWGD
Science Advisor
Gold Member
WWGD: I have tried to explain in #27 why all tangent bundles of all embedded circles are isomorphic, namely they are all isomorphic to the intrinsic tangent bundle. I.e. such isomorphisms do not need bobtailed by deformation, i.e. isotopy. In fact the same problem arises for why the knotted circles homeomorphic to the standard embedded circle. I.e. there is no nice deformation from onto the other but they are still homeomorphic via a map that goes back (and forth) to the intrinsic circle. So I am saying that if you believe the knotted circle is diffeomorphic to the standard unknotted one, you should also believe their tangent bundles are isomorphic. I.e. forming the tangent bundle is a "functor", and functors always take isomorphisms of one sort to isomorphisms of another sort.

but perhaps you are subconsciously trying to ask whether there is actually a diffeomorphism of the entire embedding space that restricts to a diffeomorphism of the two differently embedded circles, and carries one bundle to the other. That would be no, I guess. but that is not the definition of bundle isomorphism.

e.g. some informal descriptions of homeomorphism ("rubber sheet geometry")give the impression that one should always be able to deform one object into another homeomorphic one, but that is not the correct definition.

check out the answer to this question on mathoverlow:

https://math.stackexchange.com/questions/469992/tangent-bundles-of-exotic-manifolds
Thanks, Wonk, I was thinking of generalized knots, i.e., non-isotopic embeddings in higher dimensions. I need to review definitions like that of "trivially -embedded". Do you ( or anyone) know it, by chance?

mathwonk
Science Advisor
Homework Helper
2020 Award
well from what i can google it means the manifold is the boundary of an embedded manifold of one higher dimension. so a circle is trivially embedded if it is the boundary of an embedded disc, hence presumably unknotted.

• WWGD
WWGD
Science Advisor
Gold Member
well from what i can google it means the manifold is the boundary of an embedded manifold of one higher dimension. so a circle is trivially embedded if it is the boundary of an embedded disc, hence presumably unknotted.
Thanks, that's it, just refreshed my memory ,sorry for my laziness.

lavinia
Science Advisor
Gold Member
@WWGD @mathwonk

While this whole subject of embeddings is off the original subject I came across an amazing example of an embedding of a flat torus in ##R^3##. The existence of this embedding was known from Nash's first embedding theorem but only recently have computer graphic images of it been produced.

From Hilbert's theorem one knows that there is no smooth embedding of a flat torus in ##R^3##. The theorem says that every closed smooth surface in ##R^3## must have a point of positive Gauss curvature. Gauss curvature is computed as the determinant of the differential of the Gauss mapping - map a point on the surface to the unit sphere by parallel translating the unit normal to the origin - and so requires the embedding to be at least ##C^2##.

The flat torus therefore can not have a continuously differentiable Gauss map even though it has a well defined tangent plane at each of its points - since it is ##C^1##. Along a curve on this torus, the path traced by the unit normal is not differentiable. Some authors on the web call the path a "##C^1## fractal".

The reason that this weird torus is called a flat is the embedding preserves distances and infinitesimal angles..

Here is a link to some computer images.

https://io9.gizmodo.com/5905144/the-bizarre-object-we-thought-it-was-impossible-to-visualize

From the images one sees that the surface is covered with an infinite sequence of wave patterns - called "corrugations" - with each tube like wave being covered by smaller waves and so on ad infinitum. A person walking on the surface would be constantly bumped around as he steps into the unending cluster of divots that the waves make. The "fractal" nature of these waves guarantees that he will be knocked around no matter how small his steps.

- The induced Riemannian metric inherited from the embedding confuses me a little. It would seem to be a continuously differentiable choice of inner products on each tangent plane but not smooth. The differential of the embedding is a bundle isomorphism but the map is not a diffeomorphism since the embedded torus is not a smooth manifold.

- I wonder if one could still define the Gauss curvature as the limit of the ratio of the area traversed by the Gauss map on the sphere divided by the area of its domain on the embedded torus.

- Nash's theorem has another incredible consequence. It says that one can map the unit sphere into an arbitrarily small volume and preserve all of its metrical relations.

Last edited:
• WWGD