# What is the construction of charts for the tangent bundle on the unit circle?

• I
• Korybut
In summary, the problem is to show that the map from the tangent space ##T_m M## to ##\mathbb{R}^n## is a diffeomorphism, using the charts constructed on the manifold ##T##. This is challenging because the map is abstract and not explicitly defined, unlike the charts on the manifold which can be easily constructed. However, the fact that ##T## is a manifold allows for the construction of a map from ##T## to ##S^1## and a map from the inverse image of an open set on ##S^1## to ##U_\alpha \times \mathbb{R}##, which may imply the existence of the desired diffeomorphism.
Korybut
TL;DR Summary
How one show isomorphism?
Hello there!
Reading the textbook on differential geometry I didn't get the commentary. In Chapter about vector bundles authors provide the following example

Let ##M=S^1## be realized as the unit circle in ##\mathbb{R}^2##. For every ##x\in S^1##, the tangent space ##T_x S^1## can be identified with subspace of vectors orthogonal to ##x##. This yields a bijection ##\Phi## from ##TS^1## onto subset

##T=\{ (x,X)\in S^1 \times \mathbb{R}^2\, :\, x \perp X \}## of ##\mathbb{R}^4##.

This is the level set of the smooth mapping

##F:\mathbb{R}^4 \rightarrow \mathbb{R}^2\, ,\;\; F(x,X):=( \| x \|^2,\; x\cdot X)##

at the regular value ##c=(1,0)##. Hence it carries a smooth structure. One can check that ##\Phi## is a diffeomorphism with respect to this structure. (To see this, let ##pr_k:\mathbb{R}^2 \rightarrow \mathbb{R}## denote the natural projection to the ##k##-th component and choose the charts on ##S^1## and ##T## to be restrictions of ##pr_k## and ##pr_k \times pr_k##, respectively, ##k=1,2##.) Thus, ##TS^1## can naturally be identified with ##T##.

Diffeomorphism in this particular example seems completely tautological however I don't get this constructions of charts with ##pr_k##. I understand how one can build the charts on ##S^1## using projections ##pr_k## nonetheless how this should help with diffeomorphism proof is not clear.

##TS^1## is defined as a vector bundle, and ##T## as a subspace in ##\mathbb{R}^4.##

These are two different kinds of objects. However, we want them to diffeomorphic. Both have their own topology. The topology of the vector bundle is defined by the product topology of total space and base space, and the homeomorphic and diffeomorphic trivializations (charts). The topology and differential structure of ##T## is the subspace topology inherited by the Euclidean topology of ##\mathbb{R}^4.##

These are formally two different structures. Thus it has to be shown that both are the same, although defined differently. And the projections are all we have.

The difficulty with those examples is their lack of difficulty. The statement seems so obvious that it is hard to tell what has to be shown at all. The only chance to do so is to proceed step by step along with the definitions: which structures do we have, which do we need in order for the properties to make sense (homeomorphic, diffeomorphic, continuous, etc.), and what exactly are those mappings that are called "naturally identical"?

Korybut
Korybut said:
Summary:: How one show isomorphism?

Hello there!
Reading the textbook on differential geometry I didn't get the commentary. In Chapter about vector bundles authors provide the following example

Let ##M=S^1## be realized as the unit circle in ##\mathbb{R}^2##. For every ##x\in S^1##, the tangent space ##T_x S^1## can be identified with subspace of vectors orthogonal to ##x##. This yields a bijection ##\Phi## from ##TS^1## onto subset

##T=\{ (x,X)\in S^1 \times \mathbb{R}^2\, :\, x \perp X \}## of ##\mathbb{R}^4##.

This is the level set of the smooth mapping

##F:\mathbb{R}^4 \rightarrow \mathbb{R}^2\, ,\;\; F(x,X):=( \| x \|^2,\; x\cdot X)##

at the regular value ##c=(1,0)##. Hence it carries a smooth structure. One can check that ##\Phi## is a diffeomorphism with respect to this structure. (To see this, let ##pr_k:\mathbb{R}^2 \rightarrow \mathbb{R}## denote the natural projection to the ##k##-th component and choose the charts on ##S^1## and ##T## to be restrictions of ##pr_k## and ##pr_k \times pr_k##, respectively, ##k=1,2##.) Thus, ##TS^1## can naturally be identified with ##T##.

Diffeomorphism in this particular example seems completely tautological however I don't get this constructions of charts with ##pr_k##. I understand how one can build the charts on ##S^1## using projections ##pr_k## nonetheless how this should help with diffeomorphism proof is not clear.
I also don't quite follow what the chart maps are here?

What is the book's name?

Korybut
It took a while to try to understand the question.

I think the problem is to use the chart definition of differentiable to check that this map is a diffeomorphism. In principle ##T## and the tangent. bundle to the circle might have different differentiable structures. The statement of the problem appeals to a theorem, basically the Implicit Function Theorem, to conclude ##T## is a smooth manifold but it does not provide a differentiable structure as an atlas of charts.

The coordinate projection maps define four coordinate charts on the circle in ##R^2##.
Their product defines coordinate charts on the level manifold ##T##.

If the unit circle in the plane is parameterized as (cosα,sinα) then the projection onto the first coordinate is cosα and this is a homeomorphism of either the upper or lower half circle onto an open interval.

If one parameterizes the manifold ##T## as (cosα,sinα,rsinα,-rcosα) then the product of say the first and the third projection maps is (cosα,rsinα) and this is a homeomorphism from an open subset of ##T## onto an open subset of ##R^2##. (Proof?)Note: For manifolds that are embedded in Euclidean space one can define the tangent bundle as the subset of ##R^{n}×R^{n}## of points ##(x,v)## where ##v## is tangent to the manifold at the point ##x##. I think this can make one ask whether the problem is tautological.

Last edited:
Korybut and jbergman
First of all I would like to thank for all the answers.

On manifold there is homeomorphic map ##\kappa## from open subset ##U## to ##\mathbb{R}^n## and this is something touchable (don't know how to put it correctly. But it is quite clear how to manipulate with these maps).

On the other hand there is a map from tangent space ##T_m M## (equivalence classes of curves which pass through point ##m## and have the same derivative) to ##\mathbb{R}^n##. This is super abstract! This map exists but I don't know how to explicitly write it. This is my main obstruction to show diffeomorphism.

Nonetheless set ##T## is a manifold. I can build the charts. And there is no any abstract things. I can easily construct map (using this projections ##pr## in each chart)
##\pi : T \rightarrow S^1##
and map (again using projections ##pr##)
## \chi_\alpha : \pi^{-1}(U_\alpha) \rightarrow U_\alpha \times \mathbb{R}##
Where ##U_\alpha## are open sets of ##S^1##.

Does this imply that I implicitly build this diffeomorphism ##\Phi : TS^1 \rightarrow T##.

jbergman said:
What is the book's name?
Book is Rudolph and Schmidt "Differential Geometry and Mathematical Physics"

Korybut said:
First of all I would like to thank for all the answers.

On manifold there is homeomorphic map ##\kappa## from open subset ##U## to ##\mathbb{R}^n## and this is something touchable (don't know how to put it correctly. But it is quite clear how to manipulate with these maps).

On the other hand there is a map from tangent space ##T_m M## (equivalence classes of curves which pass through point ##m## and have the same derivative) to ##\mathbb{R}^n##. This is super abstract! This map exists but I don't know how to explicitly write it. This is my main obstruction to show diffeomorphism.
Just to address this first question, the Tangent Bundle is a vector bundle, often denoted ##TM## this is a smooth manifold which is just the disjoint union of all the tangent spaces often described by an ordered pair ##(p, v)##.
$$TM = \coprod_{p \in M} T_p M$$

with a projection map ##\pi(p,v)=p##.

The charts for this manifold that give it a smooth structure are just built off the charts of M. Given any open set and smooth chart ##(U, \varphi)## on M, we define a corresponding chart ##\varphi' : \pi^{-1}(U) \rightarrow \mathbb{R}^{2n}##, with ##\varphi'(v_p) = (\varphi(p), v)## where ##v## here actually has n components of the vector.

So that is the abstract way to define a tangent bundle. Now, in your original post since the circle is already embedded in R^2, one can just describe the tangent vectors in a more concrete fashion as just the tangent vectors to the circle at a point of different lengths using the more basic notion of vectors.

I think the problem was meant to work from that POV. When I have more time, I will look back at the original problem.

Korybut said:
Nonetheless set ##T## is a manifold. I can build the charts. And there is no any abstract things. I can easily construct map (using this projections ##pr## in each chart)
##\pi : T \rightarrow S^1##
and map (again using projections ##pr##)
## \chi_\alpha : \pi^{-1}(U_\alpha) \rightarrow U_\alpha \times \mathbb{R}##
Where ##U_\alpha## are open sets of ##S^1##.

Does this imply that I implicitly build this diffeomorphism ##\Phi : TS^1 \rightarrow T##.

Korybut
jbergman said:
I think the problem was meant to work from that POV. When I have more time, I will look back at the original problem.
Yes. Using "physical intuition" everything is pretty clear but I would like to elaborate this simple example in a rigorous way.

After even more thinking I come to the conclusion that this statement
Korybut said:
Nonetheless set T is a manifold. I can build the charts. And there is no any abstract things. I can easily construct map (using this projections pr in each chart)
π:T→S1
and map (again using projections pr)
χα:π−1(Uα)→Uα×R
Where Uα are open sets of S1.

Does this imply that I implicitly build this diffeomorphism Φ:TS1→T.
is not true. Providing such maps (##\pi## and ##\chi_\alpha##)implies that set ##T## is vector bundle with base ##S^1## but not necessarily a tangent bundle.

I believe problem is solved but I would like to receive some comments from the experts if I miss something.

Set ##T## is given embedding in ##\mathbb{R}^4## and projection ##\pi## is very simple
##\pi(x_1,x_2,X_1,X_2)=(x_1,x_2)##
For obvious reasons
##\pi(T)=S^1##

Now I want to build the charts on ##S^1##. I will call ##U_x## set where angle that parametrizes this cirlce belongs to ##(0,\pi)##.
##\kappa^T_x : \pi^{-1}(U_x) \rightarrow \mathbb{R}\times \mathbb{R}##
##\kappa^T_x \Big(x,\sqrt{1-x^2},X,-\frac{xX}{\sqrt{1-x^2}}\Big)=(x,X)##

Another set ##U_y## where angle belongs to ##(\frac{\pi}{2},\frac{3\pi}{2})##. Corresponding map
##\kappa^T_y :\pi^{-1}(U_y)\rightarrow \mathbb{R}\times \mathbb{R}##
##\kappa^T_y\Big(-\sqrt{1-y^2},y,\frac{yY}{\sqrt{1-y^2}},Y\Big)=(y,Y)##

Transition for intersection ##U_x## and ##U_y##
##(x,X)\rightarrow \Big(x,\sqrt{1-x^2},X,-\frac{xX}{\sqrt{1-x^2}}\Big) \rightarrow \Big(-\sqrt{1-y^2},y,\frac{yY}{\sqrt{1-y^2}},Y\Big)=(y(x),Y(x,X))##
where
##y(x)=\sqrt{1-x^2}##
##Y(x,X)=-\frac{x}{\sqrt{1-x^2}} X=(\sqrt{1-x^2})^\prime X##

So this is indeed the tangent bundle. Not just an arbitrary vector bundle.

I wonder if there maybe some subtleties which I might miss or a simpler way to reach the goal?

## 1. What is a tangent bundle?

A tangent bundle is a mathematical concept that represents the collection of all tangent spaces at each point of a manifold. It is a vector bundle, meaning that each tangent space is associated with a point on the manifold and is itself a vector space.

## 2. How is the tangent bundle related to the level set?

The tangent bundle and the level set are both important concepts in differential geometry. The tangent bundle is used to study the behavior of a manifold at each point, while the level set is used to study the behavior of a function on the manifold. The tangent bundle can be thought of as the "space" in which the level set exists.

## 3. What is the purpose of the tangent bundle?

The tangent bundle is an important tool in differential geometry and is used to study the behavior of functions and curves on a manifold. It allows for the calculation of derivatives and tangent vectors at each point, which are essential for understanding the local behavior of a manifold.

## 4. How is the tangent bundle constructed?

The tangent bundle is constructed by taking the union of all the tangent spaces at each point on a manifold. This results in a new manifold that is twice the dimension of the original manifold. The tangent bundle can also be constructed using a more abstract approach, known as the jet bundle construction.

## 5. What are some applications of the tangent bundle and level set?

The tangent bundle and level set have many applications in mathematics and physics. In mathematics, they are used in the study of differential geometry, topology, and dynamical systems. In physics, they are used in the study of classical mechanics, quantum mechanics, and general relativity. They are also used in computer graphics and computer vision for image processing and shape analysis.

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