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 $M={ℝ}^n$, $T{ℝ}^n$ is obviously isomorphic to ${ℝ}^{2n}$, 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 $\phi :{ℝ}^n\to U$. Since this map is a diffeomorphism, its derivative is an isomorphism at all points. Thus $T\phi :T{ℝ}^n={ℝ}^{2n}\to TU$ 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