# tangent bundle

Let $M$ be a differentiable manifold. Let the tangent bundle $TM$ of $M$ be(as a set) the disjoint union^{} ${\coprod}_{m\in M}{T}_{m}M$ of all the tangent spaces^{} to $M$, i.e., the set of pairs

$$\{(m,x)|m\in M,x\in {T}_{m}M\}.$$ |

This naturally has a manifold structure^{}, given as follows. For $M={\mathbb{R}}^{n}$, $T{\mathbb{R}}^{n}$ is obviously isomorphic^{} to ${\mathbb{R}}^{2n}$, and is thus obviously a manifold. By the definition of a differentiable manifold, for any $m\in M$, there is a neighborhood $U$ of $m$ and a diffeomorphism $\phi :{\mathbb{R}}^{n}\to U$. Since this map is a diffeomorphism, its derivative^{} is an isomorphism^{} at all points. Thus $T\phi :T{\mathbb{R}}^{n}={\mathbb{R}}^{2n}\to TU$ is bijective^{}, which endows $TU$ with a natural structure of a differentiable manifold. Since the transition maps for $M$ are differentiable^{}, they are for $TM$ as well, and $TM$ is a differentiable manifold. In fact, the projection $\pi :TM\to M$ forgetting the tangent vector and remembering the point, is a vector bundle. A vector field on $M$ is simply a section^{} of this bundle.

The tangent bundle is functorial in the obvious sense: If $f:M\to N$ is differentiable, we get a map $Tf:TM\to TN$, defined by $f$ on the base, and its derivative on the fibers.

Title | tangent bundle |
---|---|

Canonical name | TangentBundle |

Date of creation | 2013-03-22 13:58:59 |

Last modified on | 2013-03-22 13:58:59 |

Owner | bwebste (988) |

Last modified by | bwebste (988) |

Numerical id | 5 |

Author | bwebste (988) |

Entry type | Definition |

Classification | msc 58A32 |

Related topic | VectorField |

Related topic | LieAlgebroids |