Mathematics

Questions tagged [tangent-bundle]

The tangent $TX$ of a smooth (real or complex) manifold is defined as disjoint union of all the tangent space at the points of $X$. This the first and natural example of vector bundle.

425 questions
41
votes
4 answers

Why is the tangent bundle orientable?

Let $M$ be a smooth manifold. How do I show that the tangent bundle $TM$ of $M$ is orientable?
Lonely Penguin
  • 4,236
23
votes
1 answer

Why is the Sasaki metric natural?

Let $(M,g)$ be a Riemannian manifold with $\text{dim}(M)=n$. Then, there is a "natural" metric $\tilde{g}$ on the tangent bundle $TM$, so that $(TM,\tilde{g})$ is a Riemannian manifold, called the Sasaki metric, where a line element is written (with…
user3658307
  • 10,894
18
votes
2 answers

Why is the abstract functorial definition of the tangent bundle not widely accepted?

The following quote from page 595 of Spivak's Calculus exemplifies my viewpoint on definitions: It is an important part of a mathematical education to follow a construction of the real numbers in detail, but it is not necessary to refer ever again…
Baylee V
  • 987
13
votes
1 answer

What do the Fibers tell us about torsion? (Geometric Intuition)

Edit: I provided a full answer to the question based on my current understanding of the topic. Any form of proof-reading would be highly appreciated. Alternative answers are also welcome. Inspired by this wonderful post: I am trying to gain a…
Pellenthor
  • 989
12
votes
1 answer

Are $T\mathbb{S}^2$ and $\mathbb{S}^2 \times \mathbb{R}^2$ different?

I have seen the claim that $T\mathbb{S}^2$ and $\mathbb{S}^2 \times \mathbb{R}^2$ are not diffeomorphic, but I have only ever seen the proof that they are not isomorphic as vector bundles (which is a cute application of the hairy ball theorem). How…
D. Thomine
  • 11,383
10
votes
2 answers

Example of a parallelizable smooth manifold which is not a Lie Group

All the examples I know of manifolds which are parallelizable are Lie Groups. Can anyone point out an easy example of a parallelizable smooth manifold which is not a Lie Group? Are there conditions on a parallelizable smooth manifold which forces it…
Partha Ghosh
  • 1,469
  • 8
  • 23
9
votes
0 answers

For Hirzebruch Surfaces, does the tangent exact sequence split?

Consider the projection $\pi:\mathbb{F}_n \rightarrow \mathbb{P}^1$ . Is it true that the following exact sequence $$0 \rightarrow T_{\pi} \rightarrow T_{\mathbb{F}_n} \rightarrow \pi^*T_{\mathbb{P}^1} \rightarrow 0$$ splits? Since $n=0$ case is…
Changho Han
  • 111
9
votes
1 answer

Basis of the space of vector fields on smooth manifolds

Let $M$ be a smooth ( $C^\infty$ ) manifold. Let $\mathfrak{X}(M)$ be a set of all vector fields on $M$ and let $\mathfrak{F}(M)$ be a set of all real smooth functions on $M$ . $\mathfrak{X}(M)$ is a real vector space and it is also a module over…
blue
  • 846
9
votes
2 answers

Second derivatives, Hamilton and tangent bundle of tangent bundle TTM

I'm learning the Hamilton formalism of classical mechanics, where a second order differential equation is formalized as two first order differential equations on the cotangent bundle of the configuration manifold. I find the concept of tangent…
akreuzkamp
  • 123
9
votes
2 answers

Why is the tangent bundle defined using a disjoint union?

In textbooks about differential geometry, one finds often the disjoint union in the definition of the tangent bundle (e.g. in "Lee: Introduction to smooth manifolds", or "Amann, Escher: Analysis III"):…
user674359
9
votes
1 answer

Coordinate-free proof of non-degeneracy of symplectic form on cotangent bundle

It's relatively straightforward to provide a coordinate-free definition of the symplectic form on a cotangent bundle; the usual way to do this is to construct the tautological 1-form $$\lambda(\xi) = \langle D\pi(\xi), \pi'(\xi)\rangle,$$ where…
Thurmond
  • 1,145
8
votes
0 answers

Explicit proof that tangent bundle of the 2-sphere, TS^2, is parallelizable

Let $T S^2$ denote the tangent bundle of the 2-sphere $S^2 = \{x \in \mathbb{R}^3 : \|x\|=1\}$ . In this paper , Fodor proves that $T S^2$ is parallelizable, using machinery that I do not understand --- see Theorems 2.5 and 3.2 therein. Question : A…
ccriscitiello
  • 847
8
votes
0 answers

Are $\mathbb{R}P^3$ and $T^1S^2$ isometric?

It is well-known that 3-dimensional real projective space $\mathbb{R}P^3$ is diffeomorphic to $T^1S^2$ , the unit tangent bundle of the 2-sphere. However, I could not find any reference to whether these spaces are also isometric as Riemannian…
TilBe
  • 171
8
votes
0 answers

Proof details of the fact that the unit tangent bundle is compact in $TM$ if $M$ is a compact manifold

Let $M$ to be a manifold $m$-dimensional with a smooth hermitian metric $g$. The tangent bundle of $M$ is given by $TM= \bigcup_{p\in M} T_{p}M$, and the unit tangent bundle is given by $S=\{x \in TM: \|x\|_{g(p)} =1$ where $x \in T_{p}M\} $. (1) In…
Pedro do Norte
  • 273
8
votes
1 answer

Relation between Riemannian metric and Hermitian structure.

Let $M$ be a complex manifold with almost complex structure $J$, and $TM$ it's real tangent space, $TM^{\mathbb{C}}=TM \otimes \mathbb{C}$ its complexified tangent space. Now I'm getting confused with the metrics on this space, and so I will try and…
user500074
1
2 3
…
28 29