# Holonomic basis

Is the following definition correct?

Given a differentiable manifold $$M$$ and an ordered basis $$\{e_j^m\}$$ of the tangent space $$T_m M$$ with $$m\in M$$ (they are vectors and not vector fields). An holonomic basis is given iff the following holds:

$$[e_i^m,e_j^m]=0\quad \forall i,j\,.$$

• If they are vectors (and not fields), how do you calculate the commutator? Also, if we are restricting our attention to a single point, then any basis at that point is holonomic. – Bence Racskó Sep 4 at 21:32
• The definitions of holonomic and non-holonomic bases refer to vector fields, as does the commutator. As far as I know, none of those things are defined for vectors at a point. The linked Wikipedia article is about vector fields. So I wouldn't say that every basis (at a point) is trivially holonomic. I would say that "holonomic" is simply undefined for vectors at a point. I can easily construct an example in which $A$ is a holonomic basis of vector fields and $B$ is a non-holonomic basis of vector fields, and the basis vectors in $A$ and $B$ are the same at a particular point. – Chiral Anomaly Sep 7 at 20:41