# Coordinates on Riemannian manifolds

Let $M$ be an n-dimensional manifold endowed with a Riemannian metric. Suppose we have a coordinate chart, say $(U,\varphi)$ where $U\subset M$ and $\varphi\colon U\rightarrow {\mathbb R}^n$, and let $\partial_1,\ldots\partial_n$ denote the corresponding coordinate vector fields. Suppose $\partial_n$ is bounded. Is it possible to construct a new coordinate chart $(U,\varphi')$ (same $U$) in such a way that the following two conditions:

1. At each point the subspace generated by $\{\partial_1',\ldots,\partial_{n-1}'\}$ is the same as that generated by $\{\partial_1,\ldots,\partial_{n-1}\}$.
2. $||\partial_n'||=1$ at all points,

are satisfied?

I have no idea. Condition 2 seems to be a little strong.

Not always.

Let $M = \mathbb{R}^2 \setminus \mathbb{R}^+$ with the induced metric (plane with a slit). Let $U = M$ and $\varphi$ be the polar coordinate map $\varphi = (r,\theta)$. Let $r = x^1$ and $\theta = x^2$.

Your condition 1 requires that the level sets of $x^2$ in the primed coordinate system be the same as the level sets of $\theta$. But this means regardless of what you choose as the coordinate function $x^1$, you must have $\|\partial_2'\| \to \infty$ as "$r\to \infty$", due to the infinite separation of leaves of $\theta$ near infinity.

Observing that defining the distribution of $\{ \partial_1, \ldots, \partial_{n-1}\}$ is in fact equivalent to defining the level sets of the coordinate $x^n$ (modulo topological concerns), you can rephrase your question as the following:

Give a Riemannian manifold $U$ with metric $g$, and a non-degenerate function $v:U\to\mathbb{R}$ (in the sense that $\mathrm{d}v \neq 0$), can you find a vector field $\eta$ on $U$ such that $g(\eta,\eta) = 1$ and $\eta(v) = 1$.

The objection above the fold comes from the minmax characterisation:

$$\|\mathrm{d}v\| = \max_{\eta: \|\eta\| = 1} \eta(v)$$

so that if $\|\mathrm{d}v\| < 1$ the problem cannot be solved; now, the $v$ should be a function with the same level sets as $x^n$, so if $\|\mathrm{d}x^n\|$ is not bounded below (from zero) initially, there is no hope of rescaling $v$ to satisfy the characterization.

Now, what if $\mathrm{d}v$ is bounded from below? by rescaling we can assume that it is bounded from below by 1. Next, since in the original question the level sets of $v$ are coordinated, we can assume that there exists also a non-vanishing vector field $\tau$ such that $\tau(v) = 0$ on our manifold $U$. Since $\tau$ is non-vanishing we can further assume that $g(\tau,\tau) = 1$.

Then to construct the vector field $\eta$ we can take the ansatz

$$\eta = \frac{1}{\|\mathrm{d}v\|^2}\mathrm{d}v^\sharp + \alpha \tau$$

But construction $\eta(v) = 1$. And we just need to solve for $\alpha$ using the algebraic equation

$$\|\eta\| = 1 \iff \frac{1}{\|\mathrm{d}v\|^2} + \alpha^2 = 1 \iff \alpha = \pm \sqrt{ 1 - \frac{1}{\|\mathrm{d}v\|^2}}$$

• Thank you, that is really helpful! . What if $||\partial_n||$ is bounded? , I'm interested in take an atlas on a compact manifold changing a given atlas to one with these properties in each chart. – Richard Muniz Jun 3 '16 at 14:41
• It is more or less doable. It is more convenient to phrase the condition in terms of $\|\mathrm{d} x^n\|$ being bounded from below instead of $\|\partial_n\|$ bounded from above. See my edit above. You can take $\tau$ to be $\partial_1$ in the original chart. – Willie Wong Jun 3 '16 at 15:40
• That's amazing! Thank you very much! – Richard Muniz Jun 3 '16 at 17:05