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.
Questions tagged [tangent-bundle]
418 questions
40
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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
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23
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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 …
user3658307
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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
- 710
13
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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
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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
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10
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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
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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
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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
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9
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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
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9
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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…
user674359
9
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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
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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
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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\}…
Pedro do Norte
- 273
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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
7
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2 answers
What do the fibers of the double tangent bundle look like?
Consider the tangent bundle $\pi:TM\to M$ for some smooth manifold.
As outlined in the Wikipedia page, we can then consider the double tangent bundle via the projection $\pi_*:TTM \to TM$, with $\pi_*$ the pushforward of the canonical projection…
glS
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