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.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.)
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?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:
Thanks, that's it, just refreshed my memory ,sorry for my laziness.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.