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

• A
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

lavinia
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.

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!

WWGD
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.

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?

WWGD
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.

Infrared
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
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.

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.

WWGD
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.

Metmann
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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lavinia
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.

Metmann
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.

WWGD
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.

lavinia
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
lavinia
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
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 ##?

lavinia
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

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

lavinia
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.

lavinia
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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Metmann

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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lavinia
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
WWGD
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?

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

What embedding?

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