Tangent Bundle and Level set

[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.

 

[fresh_42]

$TS^1$ 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"?

 

[jbergman]

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?

 

[lavinia]

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.

 

[Korybut]

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

I think about this problem for a while again.

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 $\Phi : TS^1 \rightarrow T$.

 

[Korybut]

What is the book's name?

Book is Rudolph and Schmidt "Differential Geometry and Mathematical Physics"

 

[jbergman]

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

I think about this problem for a while again.

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.

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]

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

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.

 

[Korybut]

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?