I know a differentiable manifold $M$ of dimension $n$ is parallelizable if there exist (smooth of course) vector fields $\{X_i\}_{j=1}^n$ which are linearly independent in $T_pM$ at each point $p \in M$.

  1. What are few examples of parallelizable manifolds of dimension at least ? What is an example which is compact? What are some nonparalleizable ones?
  2. How do I show that a manifold $M$ as above is parallelizable if and only if the tangent bundle $TM$ is trivial?

I am a total beginner with these smooth manifold things, so any help or references to a book that can help answer my quetsions would be greatly appreciated!

Some thoughts so far. I give some examples in $\mathbb{R}^3$. A cylinder is certainly paralleizable by taking vector fields $X_1$ constant, parallel to the direction of the axis and $X_2$ constant, pointing around the circumference of the cylinder. Another example is a torus; take vector fields $X_1$ constant, pointing along the circumference of the "big circle" and $X_2$ constant, looping in and out of the "donut hole." The torus is compact, and the cylinder is also compact if it is finite. As for nonparallelizable manifolds $S^2$ is an example; any attempt to parallelize $S^2$ leads to points on the sphere at which the vector field cannot be nonzero, which isn't good if we want to get the tangent spce at those points by combining tangent vectors from the vector fields. This is true by the Hairy Ball Theorem.

  • 1
    $\begingroup$ The standard examples that are in each textbook include spheres: they are parallelizable in dimensions $0,1,3,7$ and not in other dimensions. The fact that these spheres are paralelizable can be shown by complex, quaternionic resp. octonionic multiplication of some fixed bases of $T_e S^k$ (realized in $\mathbb{R}^{k+1}$). In even-dimensional spheres, there is not even one nowhere zero vector field on the sphere ("Hairy ball theorem"). $\endgroup$ Dec 16, 2014 at 22:44
  • $\begingroup$ note that the examples you give (torus, cylinder) are lie groups, which are always parallelizable $\endgroup$
    – yoyo
    Dec 16, 2014 at 22:49

3 Answers 3


Some good $($introductory$)$ sources, in general, for all things smooth manifolds:

  • Topology from the Differentiable Viewpoint, by Milnor
  • Differential Topology, by Guillemin-Pollack
  • Differential Forms and Applications, by Do Carmo
  • A Comprehensive Introduction to Differential Geometry, Vol. 1, by Spivak
  • Introduction to Smooth Manifolds, by Lee
  • Foundations of Differentiable Manifolds and Lie Groups, by Warner
  • Brian Conrad's Differential Geometry Notes


Easy examples of parallelizable manifolds are $\mathbb{R}^n$, the tori $\mathbb{R}^n/\mathbb{Z}^n$ $($the points are cosets of the additive subgroup $\mathbb{Z}^n$ of $\mathbb{R}^n$, and the charts are open subsets of $\mathbb{R}^m$ that contain at most one point in each equivalence class, with each point mapping to its equivalence class$)$, and the argument used here to show $SO(n,\mathbb{R})$ is parallelizable $($$SO(n)$ is parallelizable$)$ actually shows that any Lie group is parallelizable. $($If $X_i$ are linearly independent tangent vectors at the identity, then for any paths $\gamma_i(t)$ with $\gamma_i'(0) = X_i$ we note that the paths $g\gamma(t)$ are smooth, and we will say that $V_i(g)$ is the equivalence class of $g\gamma_i'(0)$. Then each $V_i$ is a smooth vector field with no zeros, and the vector fields $V_i$ parallelize our Lie group.$)$

There are a lot of obstructions to parallelizability. One is the Euler characteristic. It turns out that any surface with nonzero Euler characteristic cannot have even a single smooth non-vanishing vector field. With algebraic topology one can show that a closed manifold with nonzero Euler characteristic cannot have a non-vanishing vector field. This includes genus $g$ surfaces for any $g \neq 1$. Another obstruction to paralleizability is non-orientability: every parallelizable manifold is orientable. Perhaps this is easier to see: given a parallelizable manifold with vector fields $X_1, \dots, X_n$ forming a basis for the tangent space at each point, we can always reverse the orientation of a connected chart so that they form a positively oriented basis at some point, and by continuity, at all points in each chart $($since determinants and the vector fields $X_i$ are continuous, and our charts are connected$)$. Then the transition maps will always have positive determinant, since the basis $X_1, \dots, X_n$ at each point is positively oriented at all points in all charts, and gets taken to itself via transition maps. So for example, the Klein bottle, Möbius band, $\mathbb{P}^n(\mathbb{R})$ for even $n > 1$ cannot be parallelized.


Suppose the tangent bundle of $M$ is trivial; then our diffeomorphism $\rho^{-1}: M \times \mathbb{R}^n \to TM$ gives us vector fields $\rho^{-1}(M \times \{e_i\})$, $1 \le i \le n$ where $e_i$ are the standard basis coordinates. By hypothesis, the vectors $\rho^{-1}(p, e_i)$ are smooth functions of $p$ in local coordinates and are linearly independent at each $p \in M$. Conversely, suppose we have $n$ vector fields that are linearly independent at each point; then the map that takes each point $(p, \xi)$, $\xi = (\xi_1, \dots, \xi_n) \in \mathbb{R}^n$, to the tangent vector $\sum \xi_i X_i(p)$ is easily checked to be bijective, smooth, and have derivative of rank $2n$ at all points, and is thus a diffeomorphism $M\mathbb{R}^n \to TM$.


I'll explain the relationship between trivialization of the tangent bundle and having $n$ linearly independent vector fields:

A vector field is a smooth map $X: M \rightarrow TM$ such that $X(p) \in TM_p$. That is, it's a smooth assignment of a tangent vector at $p$ to each point $p \in M$.

Suppose we have $n = \dim(M)$ vector fields $\{X_i\}_{i=1}^n$ which are linearly independent at every point, i.e. such that for all $p \in M$, the $n$ vectors $\{X_i(p)\}_{i=1}^n$ in $TM_p$ are linearly independent. Then we can define a map $f: M \times \mathbb{R}^n \rightarrow TM$ given by $f(p,\sum_{i=1}^n a_i e_i) = \sum_{i=1}^n a_i X_i(p) \in TM_p$.

This map is smooth because the $X_i$ are, and is a linear isomorphism when restricted to $\{p\}\times \mathbb{R}^n$ for any $p$. In fact, this map $f$ is a diffeomorphism: it's bijective, and we can compute its derivative well enough to know its derivative must be an isomorphism at every point, and thus by the inverse function theorem it's a diffeomorphism.

$$df_{p,v} = \left(\begin{array}{c|c} I_n & 0 \\ \hline * & A(p) \end{array}\right)$$

(In coordinates $\phi: U \rightarrow M$, $U \subset \mathbb{R}^n$ near $p$, giving a local trivialization $U \times \mathbb{R}^n \rightarrow TM$ via $(p,v) \mapsto d\phi_p(v) \in TM_p$. Here $A(p)$ is invertible and has columns given by the $X_i(p)$ expressed in the basis coming from $d\phi: TU_p (\cong \mathbb{R}^n) \rightarrow TM_p$.)

Conversely suppose we have a trivialization, i.e. a diffeomorphism $\Phi: M \times \mathbb{R}^n \rightarrow TM$ such that $\Phi(p,-): \{p\} \times \mathbb{R}^n \rightarrow TM_p$ is a linear isomorphism. Then we get vector fields by taking $X_i = \Phi(-,e_i): M \rightarrow TM$. The $n$ of these are linearly independent and are vector fields as described above.


Every parallelizable manifold is obviously orientable, hence you get an easy to check obstruction : non-orientable manifolds are not parallelizable.
This immediately shows that, for example, all even-dimensional projective spaces $\mathbb P^{2n}(\mathbb R)$ are not parallelizable.
Moreover there is a perfect (albeit a bit more advanced) criterion for orientability: a manifold $M$ is orientable if and only if its first Stiefel-Whitney class is zero: $w_1(M)=0\in H^1(M,\mathbb Z/2)$ .


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