^^ [[Schutz - 2 - Differentiable Manifolds and Tensors|2. Differentiable Manifolds and Tensors]]
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# Basis Vectors and Basis Bector Fields
Any collection of $n$ linearly independent vectors in the [[Schutz - 2.7 - Vectors and Vector Fields#^4e794a|tangent space]] $T_P$ is a basis for $T_P$. By choosing a basis in each $T_P \, \forall \, P \in M$, we arrive at a basis for vector fields.
> **Coordinate Basis**
> If we have a coordinate system $\{x_i\}$ in a neighborhood $U$ of $P$, then the coordinates define the *coordinate basis* $\{\partial/\partial x_i\}$ at all points in $U$.
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One could use another basis, say $\{\pmb{e}_i\}$, where $i$ is a label that distinguishes the basis vectors from one another and does not denote a component. At a point $P$, an arbitrary vector $\pmb{V}$ can be written as:
$
\pmb{V} = \sum V^i \frac{\partial}{\partial x^i} = \sum V'^i \pmb{e}_i
$
Where $\{V^i\}$ are the components of $\pmb{V}$ in the basis $\{\partial/\partial x^i\}$ and $\{V'^i\}$ are the components in the basis $\{\pmb{e}_i\}$. The components are related to one another by vector transformation laws. If $\pmb{V}$ and the bases $\{\partial/\partial x^i\}$ and $\{\pmb{e}_i\}$ are regarded as vector fields instead, then the components $\{V^i\}$ and $\{V'^i\}$ are of the field $V$ are *functions* on $M$.
> **Differentiable Vector Field**
> A [[Schutz - 2.7 - Vectors and Vector Fields#^a9b237|vector field]] is said to be *differentiable* if its component functions $\{V^i\}$ are differentiable.
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Note that above we assumed that the vectors $\{\partial/ \partial x^i\}$ of an arbitrary coordinate system are all linearly independent. This is just a condition for the coordinates to be "good" coordinates, i.e. they provide a bijection from some neighborhood $U$ of $P$ onto a region $V$ of $\mathbb{R}^n$. The proof that the $\{x^i\}$ are good coordinate is simple and given in the book on Page 35.