Let $ (M,g) $ be a closed parallelizable Riemannian manifold. Whenever one comes across the standard trick of choosing a local basis $(e_1, \ldots, e_n)$ of $TM$ for some computation in local coordinates, when are we allowed to freely extend $e_i$ to a global nowhere zero vector field $X_i$ such that $(X_1, \ldots, X_n)$ is a global basis of $TM$? It seems to me this should give no problem when we're dealing with a coordinate independent computation, for example with tensor fields. Is there some subtlety one must be careful with when wanting to extend a local frame to a global frame in general?

Perhaps to give a specific example, if $(e_1, \ldots, e_n)$ is our local frame on a coordinate chart $U$, and the coefficients of the metric tensor $g$ are $g_{ij} = g(e_i, e_j)$, can we really find a global frame $(X_1, \ldots, X_n)$ such that $g(X_i, X_j)$ coincides with $g_{ij}$ over $U$? That is, the components of the metric tensor can be globally defined smooth functions?


1 Answer 1


It really depends on what the computation is. But obviously we cannot just extend a local chart to a global one, even if the manifold has trivial tangent bundle. The major issue is given a chart, we have the Lie bracket $[\frac{\partial }{\partial x_i}, \frac{\partial } {\partial x_j}] = 0$ (which plays a key role in many computations, explicitly or implicitly). But if $X_1, \cdots, X_n$ form a trivilization of $TM$, all we know is they are linearly independent everywhere, not necessarily $[X_i, X_j]=0$.

In fact, $[X_i, X_j]\equiv 0$ for all $i,j $ is also a sufficient condition for a local chart to exists everywhere such that $X_i=\frac{\partial }{\partial x_i}$. So this is probably the condition you are looking for.

Especially, when we wrote $g_{ij}$ for Riemannian manfiold, we are assuming they come from a chart, i.e. $X_i=\frac{\partial }{\partial X_i}$. Otherwise many properties may not hold. For example, the Levi-Civita connection satisfies $\nabla_X Y-\nabla_Y X = [X, Y]$. In the case of $[X, Y]=0$, we have the special property $\nabla_X Y=\nabla_Y X$.

  • $\begingroup$ I'm not necessarily hoping that all $[X_i, X_j] = 0$, though I see now how that would be helpful. The toy example I'm working with is trying to see if I can express the coefficients of an affine connection on a parallelizable manifold as global functions, and then doing the same for the torsion tensor. $\endgroup$ Commented Apr 24 at 2:18
  • $\begingroup$ One also has that $[X_i, X_j] = 0$ for all $i,j$ if and only if $M$ is a torus, which I just found here: mathoverflow.net/questions/239116/…. Interesting result $\endgroup$ Commented Apr 24 at 2:24
  • $\begingroup$ You definitely can define the coefficients of an affine connection on a parallelizable manifold as global functions, but those coefficients are not necessarily Christoffel symbols, so not always convenient to work with. $\endgroup$ Commented Apr 24 at 2:36
  • $\begingroup$ Just to be clear, by "not necessarily Christoffel symbols", you just mean that the given affine connection need not be torsion free, right? I'm just working with general affine connections so I don't expect to need Christoffel symbols specifically, if this is the case. $\endgroup$ Commented Apr 24 at 2:40
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    $\begingroup$ Yes, we can choose an orthonormal global frame by Gram-Schmidt, but it's still problematic to call them "$g_{ij}$". For example, you cannot use them to calculate the curvature (tensor) as usual. Being parallelizable is a topological condition, and there could be very different Riemannian structures on the manifold with different curvatures. Therefore having a orthonormal global frame doesn't tell you much about the metric. $\endgroup$ Commented Apr 24 at 3:29

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