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

• A
Joker93
Hello.
I was trying to prove that the tangent bundle TM is a smooth manifold with a differentiable structure and I wanted to do it in a different way than the one used by my professor.
I used that TM=M x TpM. So, the question is:

Can the tangent bundle TM be considered as the product manifold of a smooth manifold M and its tangent planes TpM with p∈M?.
If this is the case, then can we conclude that it is a smooth manifold because it is a product of smooth manifolds and we can find a differentiable structure that comes from the maps of M and the maps of TpM?

Thank you.

## Answers and Replies

Gold Member
Hello.
I was trying to prove that the tangent bundle TM is a smooth manifold with a differentiable structure and I wanted to do it in a different way than the one used by my professor.
I used that TM=M x TpM. So, the question is:

Can the tangent bundle TM be considered as the product manifold of a smooth manifold M and its tangent planes TpM with p∈M?.
If this is the case, then can we conclude that it is a smooth manifold because it is a product of smooth manifolds and we can find a differentiable structure that comes from the maps of M and the maps of TpM?

Thank you.

In general, the tangent bundle is not a product. For instance, the tangent bundle of the 2 sphere is not the product of a sphere with a 2 dimensional plane. In fact, the only closed surface with a product tangent bundle is the torus.

However, the tangent bundle is always locally a product. It is a product over any smooth coordinate chart.

Joker93
In general, the tangent bundle is not a product. For instance, the tangent bundle of the 2 sphere is not the product of a sphere with a 2 dimensional plane. In fact, the only closed surface with a product tangent bundle is the torus.

However, the tangent bundle is always locally a product. It is a product over any smooth coordinate chart.
Aah, I wrote a proof in a test that depended on the last paragraph you wrote, but I did not think about it not being expandable globally.
Shame..
Thanks though!

Gold Member
Hello.
I was trying to prove that the tangent bundle TM is a smooth manifold with a differentiable structure and I wanted to do it in a different way than the one used by my professor.
I used that TM=M x TpM. So, the question is:

Can the tangent bundle TM be considered as the product manifold of a smooth manifold M and its tangent planes TpM with p∈M?.
If this is the case, then can we conclude that it is a smooth manifold because it is a product of smooth manifolds and we can find a differentiable structure that comes from the maps of M and the maps of TpM?

Thank you.
Adding to what Lavinia said, this is equivalent to the Tangent Bundle being (globally) trivial. EDIT: This is equivalent to the existence of a continuous nowhere-zero vector field.

Joker93
Adding to what Lavinia said, this is equivalent to the Tangent Bundle being (globally) trivial. EDIT: This is equivalent to the existence of a continuous nowhere-zero vector field.
Could you explain why this is please?

Gold Member
Could you explain why this is please?
If you could give me some background re your knowledge, I will try to produce an argument.

Gold Member
This is equivalent to the existence of a continuous nowhere-zero vector field.

This isn't true. Every odd-dimensional sphere admits an everywhere non-vanishing vector field, but $S^n$ only has trivial tangent bundle for $n=1,3,7$. You might mean that having ##n## everywhere-independent vector fields on an ##n##-manifold is equivalent to having trivial tangent bundle (this follows simply by identifying an element ##(x,v)## of the tangent bundle with ##(x,w)\in M\times\mathbb{R}^n## where the coordinates of ##w## are projections of ##v## onto the vector fields at ##x##). For oriented ##2##-manifolds, it's enough to find a single such vector field ##X## because we may produce a second by fixing a metric ##g## and defining ##Y## to be everywhere unit length and orthogonal to ##X## and so that ##(X_p,Y_p)## is positively oriented for all ##p##.

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• lavinia
Gold Member
This isn't true. Every odd-dimensional sphere admits an everywhere non-vanishing vector field, but $S^n$ only has trivial tangent bundle for $n=1,3,7$. You might mean that having ##n## everywhere-independent vector fields on an ##n##-manifold is equivalent to having trivial tangent bundle (this follows simply by identifying an element ##(x,v)## of the tangent bundle with ##(x,w)\in M\times\mathbb{R}^n## where the coordinates of ##w## are projections of ##v## onto the vector fields at ##x##). For oriented ##2##-manifolds, it's enough to find a single such vector field ##X## because we may produce a second by fixing a metric ##g## and defining ##Y## to be everywhere unit length and orthogonal to ##X## and so that ##(X_p,Y_p)## is positively oriented for all ##p##.
And no non-orientable closed manifold has a product tangent bundle but many have zero Euler characteristic e.g. the Klein bottle.

Joker93
If you could give me some background re your knowledge, I will try to produce an argument.
I just know elementary manifold theory(just enough to start studying Riemannian Geometry) and I am currently learning Riemannian geometry.
I have only been learning higher mathematics for the past 3/2 months.

Gold Member
This isn't true. Every odd-dimensional sphere admits an everywhere non-vanishing vector field, but $S^n$ only has trivial tangent bundle for $n=1,3,7$. You might mean that having ##n## everywhere-independent vector fields on an ##n##-manifold is equivalent to having trivial tangent bundle (this follows simply by identifying an element ##(x,v)## of the tangent bundle with ##(x,w)\in M\times\mathbb{R}^n## where the coordinates of ##w## are projections of ##v## onto the vector fields at ##x##). For oriented ##2##-manifolds, it's enough to find a single such vector field ##X## because we may produce a second by fixing a metric ##g## and defining ##Y## to be everywhere unit length and orthogonal to ##X## and so that ##(X_p,Y_p)## is positively oriented for all ##p##.
Sorry, I was trying to say that there exist n independent sections.

You should think of fibre bundles in general as generalized combs, at not as products. Locally you can write each bundle as the product of a (small) open subset of the manifold and the fibre (= local trivialization) but in general there is no way to 'glue' these local trivializations in a consistent way to a global trivialization. Heuristically, you can think of the curvature of the manifold being the obstacle to do so (not always true, but in most cases this picture fits).

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Gold Member
You should think of fibre bundles in general as generalized combs, at not as products. Locally you can write each bundle as the product of a (small) open subset of the manifold and the fibre (= local trivialization) but in general there is no way to 'glue' these local trivializations in a consistent way to a global trivialization.

A fiber bundle need not have a manifold as its base space.

Heuristically, you can think of the curvature of the manifold being the obstacle to do so (not always true, but in most cases this picture fits).

A product bundle over a manifold often has non-zero curvature. Around any point where the curvature tensor is not identically zero, there is a neighborhood where the bundle is a product.

A fiber bundle need not have a manifold as its base space.
Depends on the definition. I usually include it into the definition, but of course this depends on what you want to do.

Around any point where the curvature tensor is not identically zero, there is a neighborhood where the bundle is a product.
Yes, locally.

Gold Member
Are there any examples of n-manifolds; n=2 or higher with non-zero curvature tensor and trivial tangent bundle? For n=1 , we have the circle as a counter, but 1-manifolds are relatively straightforward.

Gold Member
Depends on the definition. I usually include it into the definition, but of course this depends on what you want to do.

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

Yes, locally.

So curvature is not an obstruction to a product bundle.

Also many closed Riemannian manifolds with non-zero curvature have product tangent bundles. E.g. any torus embedded in ##R^3## or the standard 3 sphere in ##R^4##.

• Joker93
Gold Member
Are there any examples of n-manifolds; n=2 or higher with non-zero curvature tensor and trivial tangent bundle? For n=1 , we have the circle as a counter, but 1-manifolds are relatively straightforward.
see post #15

• WWGD
Gold Member
see post #15
How does the tangent bundle distribute with respect to the product, e.g., is there a nice result for ##TM(A\times B) ##? Maybe ##TM_A \times TM_B ##?

Gold Member
How does the tangent bundle distribute with respect to the product, e.g., is there a nice result for ##TM(A\times B) ##? Maybe ##TM_A \times TM_B ##?
yes

Gold Member
yes
Doesn't that then explain the result for Tori, i.e., the triviality of the tangent bundle?

Gold Member
Doesn't that then explain the result for Tori?

yes.

Also the torus is a Lie group - as is the 3 sphere. All Lie groups have trivial tangent bundle.

Every closed orientable 3 manifold has a trivial tangent bundle.

Gold Member
Important examples of fiber bundles where the base is not a finite dimensional manifold are the universal classifying spaces for bundles with structure group a given Lie group.

For instance, the base of the universal classifying space for the discrete group ##Z_2## is the direct limit under inclusion of the finite dimensional projective spaces of all dimensions. Its topology ( I think) is the smallest topology such that intersection of open sets with the finite dimensional projective spaces is open.

Covering spaces are examples of fiber bundles with discrete fiber. The base can be any topological space. The interesting cases are mostly path connected locally path connected spaces.

Here are some references: Steenrod The Topology of Fiber Bundles; Milnor Characteristic Classes

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This thread is based in the subforum of differential geometry and in many books on differential geometry or gauge theory they start right away by defining bundles over (differentiable) manifolds. Of course the most general definition incorporates topological spaces , but dependent on your goal you may restrict definitions, makes life simpler. Examples: Morita - Geometry of Differential Forms; Baez/Muniain - Gauge Fields, Knots and Gravity, etc. etc.

Also many closed Riemannian manifolds with non-zero curvature have product tangent bundles. E.g. any torus embedded in R3R^3 or the standard 3 sphere in R4R^4.

Okay, I didn't know that. Thanks for the information ;)

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Gold Member
This thread is based in the subforum of differential geometry and in many books on differential geometry or gauge theory they start right away by defining bundles over (differentiable) manifolds. Of course the most general definition incorporates topological spaces , but dependent on your goal you may restrict definitions, makes life simpler. Examples: Morita - Geometry of Differential Forms; Baez/Muniain - Gauge Fields, Knots and Gravity, etc. etc.

Granted. But inaccuracies can be misleading. I just wanted to dispel the possible implication that bundles are defined only for manifolds.

• Metmann
Gold Member
Does the bundle depend ( up to bundle morphism I guess) on the embedding, or is the bundle dependent only on homotopy/isotopy (and then what are the morphisms in the category of bundles)? How about the dependence on the ambient space in which the object is embedded?

Gold Member
Does the bundle depend ( up to bundle morphism I guess) on the embedding...

What embedding?

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Gold Member
I mean say the tangent bundle of a knotted circle. The circle may be embedded in different ways, including some embeddings where it has non-trivial knots, e.g. the trefoil is an embedding of the circle. Is the tangent bundle of the trefoil , or any other embedding of the circle the same as that of the standard ##S^1:= \{x: ||x||=1\} ##? EDIT: A more dramatic case: Is the bundle of Alexander's Horned Sphere the same as that of the standard sphere as defined? Is the bundle preserved by homotopy, i.e., do homotopic embeddings of the same space have isomorphic tangent bundles? Or do we use pullback bundles to study TM when we have different, possibly non-homotopic or non-isotopic embeddings?

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Homework Helper
I am not an expert, but I would say the tangent bundle of a smooth manifold is defined intrinsically, independent of an embedding. Then when you have a smooth embedding, that bundle also embeds as a sub bundle of the ambient embedding space. So every embedding of the circle into any euclidean space will have as (embedded) tangent bundle just the embedded copy of the intrinsic tangent bundle of the circle as its tangent bundle, hence all tangent bundles of all embedded circles are isomorphic, namely they are all isomorphic to the intrinsic tangent bundle.

As for the horned sphere, which I don't really know at all well, it looks from the pictures as if the embedding is only topological and not smooth, hence the tangent bundle would probably not embed along with it, and there would not be an obvious (to me) definition of an embedded tangent bundle. I.e. I suspect that the horned sphere is not a smooth submanifold of euclidean space so i don't know how to define its tangent bundle. there is such a thing as a topological tangent bundle, but i don't know what its definition is.

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• WWGD
Homework Helper
as far as homotopy goes, there is a theory of classifying spaces of bundles. I.e. as i recall distantly, given an embedding of a manifold X into euclidean space E of high dimension, for any n plane bundle on X, not just the tangent bundle, one gets an induced map from the manifold X to the grassmannian of all n dimensional subspaces of E. Then I believe that different embeddings of X, equipped with the same bundle, into E define homotopic maps from X to that grassmannian. conversely any map from X into the grassmannian pulls back a bundle to X indued from the universal n plane bundle on the grassmannian, and for any two homotopic maps the pull back bundles are isomorphic. something like that, it has been a long time, and not my area. the standard source in the 60's were xeroxed notes by milnor. you can google classifying spaces for vector bundles, or just see the 6.28 here:
https://www.ma.utexas.edu/users/dafr/M392C-2012/Notes/lecture6.pdf

• WWGD
Gold Member
A smooth embedding is by definition diffeomophism. If the embedding is not smooth then the image may not have a tangent bundle. For instance a square is a non-smoothly embedded circle and does not have well defined tangent spaces at the corners. The Alexander horned sphere is not a smooth embedding as mathwonk said.

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Gold Member
A smooth embedding is by definition diffeomophism. If the embedding is not smooth then the image may not have a tangent bundle. For instance a square is a non-smoothly embedded circle and does not have well defined tangent spaces at the corners. The Alexander horned sphere is not a smooth embedding as mathwonk said.
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.

Gold Member
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.

• WWGD
Staff Emeritus
Homework Helper
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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.

• lavinia
Homework Helper
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.

Gold Member
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.

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