Let $M$ be differentiable n-manifold. Suppose that at $p \in M$ we are given a basis of tangent space $T_pM$ denoted as $(X^1,\ldots,X^n)$. Can we construct a coordinate representation $(x^1,\ldots,x^n)$, such that $\left. \frac{\partial}{\partial x^i} \right|_{p} = X^i$ for all $i=1,\ldots,n$? Should it be clear that such coordinate chart exists?
Coordinate representation for given tangent space
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4Take any coordinates around p and use a linear change of basis. – 2012-12-22
1 Answers
Yes. First, for some arbitrarily small $\epsilon > 0$ choose curves $\gamma^i :(-\epsilon,\epsilon) \to M$ with the property that $\frac{d}{dt}\gamma^{i}(t)|_{t=0} = X^i$. This can always be done for instance by choosing $\gamma^i$ to be the unique geodesic through $p$ and with initial value $\frac{d}{dt}\gamma^{i}(t)|_{t=0} = X^i$. Then consider the map $F:(-\epsilon,\epsilon)^n \to M$ given by $(x^1, \ldots,x^n) \mapsto (\gamma^1(x^1), \ldots, \gamma^n(x^n)).$ By the implicit function theorem this map is invertible near $p$ (since the $X^i$ are a basis) and thus there is a neighborhood $U \subset M$ of $p$ and a diffeomorphism $F^{-1} : U \to \mathbb R^n$. By the definition of $\frac{\partial}{\partial x^i}$ we see that $\frac{\partial}{\partial x^i} = X^i$ as desired.
The coordinates constructed in this way are often called normal coordinates, geodesic coordinates, exponential coordinates or some other variant.