Constructing a chart with coord. basis equal to given basis at one pt.

[center o bass]

Suppose we have a manifold $M$ and at $p \in M$ we have a basis for the tangent space of vectors $X_i$. Since $M$ is a manifold, there exists a local chart $(U,\phi)$ about $p$. Now the question is, given such a chart , how can we construct a new chart in a such that $X_i = \left. \frac{\partial}{\partial x^\mu} \right|_{p}$.

I know there is a theorem that says given commuting vector fields we can find a chart such that these vector fields locally is the coordinate basis of that chart.

However, I want to prove in a transparent manner the less general statement above; that we can construct a coordinate system such that the vector $X_i$ _in the tangent space at p_ is the coordinate basis $\left. \tfrac{\partial}{\partial x^\mu} \right|_{p}$ just at p.

 

[Ben Niehoff]

Use the standard construction for Riemann normal coordinates, but forget the fact that the basis is orthonormal.

 

[pasmith]

Suppose we have a manifold $M$ and at $p \in M$ we have a basis for the tangent space of vectors $X_i$. Since $M$ is a manifold, there exists a local chart $(U,\phi)$ about $p$. Now the question is, given such a chart , how can we construct a new chart in a such that $X_i = \left. \frac{\partial}{\partial x^\mu} \right|_{p}$.

I know there is a theorem that says given commuting vector fields we can find a chart such that these vector fields locally is the coordinate basis of that chart.

However, I want to prove in a transparent manner the less general statement above; that we can construct a coordinate system such that the vector $X_i$ _in the tangent space at p_ is the coordinate basis $\left. \tfrac{\partial}{\partial x^\mu} \right|_{p}$ just at p.

Let $(U,\phi)$ be a chart on $M$ whose domain contains $p$, and consider the chart $(U, A \circ \phi)$ where $A : \mathbb{R}^n \to \mathbb{R}^n$ is linear and invertible. This chart is smoothly compatible with the original chart $(U,\phi)$.

The transition function $A$ then pushes forward to a linear map $A_{*} : T_pM \to T_pM$, giving $$\left.\frac{\partial}{\partial x^i}\right|_p = \frac{\partial \tilde x^j}{\partial x^i} <br /> \left.\frac{\partial}{\partial \tilde x^j}\right|_p = <br /> (A_{*})_i{}^j \left.\frac{\partial}{\partial \tilde x^j}\right|_p$$ where $(x^i)$ are the coordinate functions of the chart $(U,\phi)$ and $(\tilde x^j)$ are those of the chart $(U, A \circ \phi)$. Since $\tilde x^j = A^j{}_i x^i$ we have that $$\frac{\partial \tilde x^j}{\partial x^i} = (A_{*})_i{}^j = A^j{}_i.$$ If ${X_j}$ is a basis for $T_pM$ we may then set $$\left.\frac{\partial}{\partial \tilde x^j}\right|_p = X_j$$ in the above to obtain $$\left.\frac{\partial}{\partial x^i}\right|_p = A^j{}_i X_j.$$