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]

315 questions

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### 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?

Jr.

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### 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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### 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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### 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

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### 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

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### 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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### What vector bundles are tangent bundles of smooth manifolds?

Given a smooth manifold $M$, we can naturally associate to it a vector bundle ${\rm T}M\rightarrow M$ called the tangent bundle of $M$. This operation induces a functor $\rm T:\rm Diff\rightarrow\rm Vect(Diff)$ from the category of smooth manifolds…

Dry Bones

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### 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…

B.Hueber

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### Change of two normal coordinates based on two nearby points?

Let $M$ be a manifold and $L(M)$ be the tangent frame bundle on $M$. Let $\Gamma$ be a linear connection on $L(M)$ which induces a covariant derivative $\nabla$ on $TM$.
Let $p, q$ be two distinguished and sufficiently close points on $M$, connected…

Dreamer

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### How to express the covariant derivative on $TM$ by the covariant derivative on $M\times M$?

Let $(M,g,\Gamma)$ be a Riemannian manifold with the Levi-Civita connection. $M\times M$ and $TM$ have natural metric structrures inherited from $(M,g,\Gamma)$. Let $\phi:TM \rightarrow M \times M$ be defined as follows:
$$ \phi(z,u) =…

Adam Latosiński

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### Differential using sheaves

I am trying to understand smooth real manifolds using sheaves and finding some trouble with differential of smooth maps.
Some notation:
Let $M$ be manifold, I denote by $\mathcal{O}_M$ it's sheaf of smooth $\mathbb{R}-$functions and by $\mathcal{T}M…

espacodual

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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

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### Global Trivializing Sections of Tangent bundle

I quote from Complex Geometry- An Introduction by Daniel Huybrechts.
Let $U\subseteq \mathbb{C}^n$ be an open subset. Thus, $U$ can in particular be considered as a $2n$-dimensional real manifold. For $x\in U$ we have real tangent space $T_xU$ at…

user312648

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### Show that $S^n \times \mathbb R$ is parallelizable

A manifold $M$ is said to be parallelizable if it admits $k$ linearly independent vector fields. I know that this is equivalent to the tangent space $TM$ being trivial. I am trying to show that $S^n\times \mathbb{R}$ is parallelizable, but have…

user106299

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### Triviality of complexified tangent bundle

Let be $M$ be a smooth manifold (possibly with boundary) and assume that the complexified tangent bundle $TM \otimes \mathbb{C}$ is trivial. Does this imply that $TM$ is stably trivial? This seems to be true for all two dimensional manifolds. Note…

user39598

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