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

[lavinia]

Yes, thanks, that is sort of what I was getting at, that one should refer to a specific embedding when talking about the bundle. It may be, e.g., a knotted copy of the circle, like the trefoil, instead of the standard embedding.

Every smooth embedding always gives a bundle isomorphism since it is a diffeomorphism. If it is not smooth then there is no bundle morphism at all. A trefoil knot is a smooth embedding.

 

[Orodruin]

What embedding?

I think this is a very important point that passes above the heads of many students, in particular if manifolds are introduced as embedded spaces. It cannot be emphasised enough. One of the main points of the differential geometry formalism is to only use properties of the manifold itself regardless of any embeddings. Far too many students get stuck in thinking about manifolds in terms of embedded spaces in some flat embedding space.

 

[mathwonk]

now just for fun, note that the grass man of 1 diml subspaces of R^3 is just real projective plane, so any embedding of the circle in R^3 gives a map from the circle to RP^2 (sending a point of the circle to the line through the origin parallel to the tangent line to the circle at the point) and two such are homotopic iff they pull back the same bundle. Now imagine the usual embedding of a circle in a plane in R^3 and I think you can see the induced map to RP^2 wraps twice around the equator hence defines a trivial element of homotopy, hence the tangent bundle is trivial. similarly but harder to visualize is that a trefoil knot seems to induce by its tangent bundle a map that wraps 4 times around the equator but still the trivial homotopy element hence again the trivial tangent bundle. so the classifying space approach seems to show both that a standard and a knotted embedding of the circle define the same tangent bundle and also that this bundle is trivial.

 

[lavinia]

now just for fun, note that the grass man of 1 diml subspaces of R^3 is just real projective plane, so any embedding of the circle in R^3 gives a map from the circle to RP^2 (sending a point of the circle to the line through the origin parallel to the tangent line to the circle at the point) and two such are homotopic iff they pull back the same bundle. Now imagine the usual embedding of a circle in a plane in R^3 and I think you can see the induced map to RP^2 wraps twice around the equator hence defines a trivial element of homotopy, hence the tangent bundle is trivial. similarly but harder to visualize is that a trefoil knot seems to induce by its tangent bundle a map that wraps 4 times around the equator but still the trivial homotopy element hence again the trivial tangent bundle. so the classifying space approach seems to show both that a standard and a knotted embedding of the circle define the same tangent bundle and also that this bundle is trivial.

I was going to say that the classifying map for any embedding of the circle into $R^3$ can be factored through the Gauss mapping into the sphere and the sphere is simply connected. But this is the same as saying that the tangent bundle to the circle has a non-zero section - since otherwise there would be no mapping into the sphere - so one proves it is trivial by assuming it.

 

[WWGD]

I think this is a very important point that passes above the heads of many students, in particular if manifolds are introduced as embedded spaces. It cannot be emphasised enough. One of the main points of the differential geometry formalism is to only use properties of the manifold itself regardless of any embeddings. Far too many students get stuck in thinking about manifolds in terms of embedded spaces in some flat embedding space.

True, but a lot of use deal with other perspectives, like Topological and Differential-Topological ones where the embeddings are relevant, if not the essential issue itself.

 

[Orodruin]

True, but a lot of use deal with other perspectives, like Topological and Differential-Topological ones where the embeddings are relevant, if not the essential issue itself.

I disagree. A manifold is a topological space. Embeddings are definitely not the essential issue of topology. The more essential issue is what is meant by continuity, comvergence, and homeomorphisms.

 

[WWGD]

I disagree. A manifold is a topological space. Embeddings are definitely not the essential issue of topology. The more essential issue is what is meant by continuity, comvergence, and homeomorphisms.

My background is more along the lines of Knot theory where the embedding ( and its isotopy type, knotedness, other characteristics) _is_ the subject matter. EDIT: While I agree that it is an important, if not essential matter to study properties of manifolds that are independent on the embedding, there is some importance too, in other areas outside of D.G, to study the dependence on the choice of embedding.

 

[Orodruin]

My background is more along the lines of Knot theory where the embedding ( and its isotopy type, knotedness, other characteristics) _is_ the subject matter.

I suspected as much, but knot theory is a subfield of topology, not the main matter of study in the field. The main issues in topology itself remains what I quoted in my previous post. I also believe many people have the wrong impression that topology has to do with continuous transformations (who has not seen an animation of the coffee cup transforming into a donut?) and I think it is important to point out that it is not the main issue. Before you get to knot theory, you must pass several ideas that are more fundamental in topology. How points are topologically distinct, what it actually means for a function to be continuous in the first place, etc.

 

[lavinia]

This part of the thread led me ask whether a closed manifold can have two different differentiable structures whose tangent bundles are not isomorphic as bundles. Another way to ask this question is whether the tangent bundle is a topological invariant of the manifold.

Interestingly, Milnor proved that there are topological manifolds with two different differentiable structures whose tangent bundles are not bundle isomorphic. One of the tangent bundles is trivial and the other is not.

If one embeds a topological manifold in a smooth manifold so that it has a well defined tangent plane at each point, then the resulting tangent bundle might depend on the embedding.

 

[mathwonk]

here is a discussion of the related question of how the determinant of the holomorphic cotangent bundle, or just its cohomology class, can vary within the diffeomorphism or homeomorphism class of a given manifold.

https://mathoverflow.net/questions/70429/is-canonical-class-a-topological-invariant

 

[mathwonk]

knots are of course interesting, and were traditionally studied by the fundamental group of the complement, and more recently by the "Jones polynomial" and its variants. I am largely ignorant of this theory, in spite of having heard some nice talks by Vaughn Jones, who draws rather beautiful pictures of knots.

I was hoping to find a link to a video of the talks but only found this announcement with abstracts:

http://www.math.uga.edu/events/10th-annual-cantrell-lectures

 

[lavinia]

here is a discussion of the related question of how the determinant of the holomorphic cotangent bundle, or just its cohomology class, can vary within the diffeomorphism or homeomorphism class of a given manifold.

https://mathoverflow.net/questions/70429/is-canonical-class-a-topological-invariant

Can you explain this?

 

[mathwonk]

not really, just saying that along the lines of milnor's examples, there are theorems that say choosing the complex structure differently on a given smooth manifold cannot greatly change the holomorphic cotangent bundle. But more variation is possible if the new complex structure also defines a different smooth structure on the given topological manifold. Specifically, if one has a diffeomorphism between two complex manifolds, or say just smooth projective complex surfaces, there is an induced isomorphism of cohomology groups. Q: what does this isomorphism do the "canonical class" K, a cohomology class associated to the holomorphic cotangent bundle? The answer seems to be that it can take K to -K, and maybe more variation than that in some cases, but i have not mastered this theory. (some name - omitted)

see last Theorem on page 297 here:

https://books.google.com/books?id=VgG9AwAAQBAJ&pg=PA297&lpg=PA297&dq=canonical+class+of+diffeomorphic+complex+manifolds&source=bl&ots=jnHZyQxxCc&sig=U8aaZECNU5rtTG8DohRqBbuUMy4&hl=en&sa=X&ved=0ahUKEwjAg6ObxtDXAhUnqVQKHRNPB-wQ6AEIVzAH#v=onepage&q=canonical class of diffeomorphic complex manifolds&f=false

 

[WWGD]

This part of the thread led me ask whether a closed manifold can have two different differentiable structures whose tangent bundles are not isomorphic as bundles. Another way to ask this question is whether the tangent bundle is a topological invariant of the manifold.

Interestingly, Milnor proved that there are topological manifolds with two different differentiable structures whose tangent bundles are not bundle isomorphic. One of the tangent bundles is trivial and the other is not.

If one embeds a topological manifold in a smooth manifold so that it has a well defined tangent plane at each point, then the resulting tangent bundle might depend on the embedding.

But you may have non-isotopic embeddings, both smooth , on a large dimensional manifold. EDITThis is the generalization of the iusual $S^1$ knots.

 

[lavinia]

But you may have non-isotopic embeddings, both smooth , on a large dimensional manifold.

Not sure what you are trying to say.

 

[WWGD]

Not sure what you are trying to say.

Would their reapective bundles be isomorphic if the embeddings are non-isotopic or if the differentiable structures are inequivalent?

 

[lavinia]

Would their reapective bundles be isomorphic if the embeddings are non-isotopic or if the differentiable structures are inequivalent?

A trefoil knot would seem not to be smoothly isotopic to a circle. Yet their tangent bundles are isomorphic.

The 7 sphere has different differentiable structures but all exotic 7 spheres have trivial tangent bundles.

I don't know anything about knots but one thing that distinguishes a knot in $S^3$ is the fundamental group of its complement. However all knots have a trivial tangent bundle.

 

[mathwonk]

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

 

[mathwonk]

lavinia: according to wikipedia, the fundamental group of the complement is not sufficient to distinguish between two knots up to "equivalence". not sure what you meant ??

https://en.wikipedia.org/wiki/Knot_group

 

[lavinia]

lavinia: according to wikipedia, the fundamental group of the complement is not sufficient to distinguish between two knots up to "equivalence". not sure what you meant ??

https://en.wikipedia.org/wiki/Knot_group

Right. My point was that this was one method that gives information whereas the tangent bundle does not.

 

[mathwonk]

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]

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.)

Right.@WWGD 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]

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]

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]

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]

@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.