# Which local computations can be done globally for parallelizable manifolds?

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?

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 $$\nabla_X Y=\nabla_Y X$$.
• 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. Commented Apr 24 at 2:18
• 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 Commented Apr 24 at 2:24
• 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. Commented Apr 24 at 3:29