2
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Richard Muniz Jun 3 '16 at 14:41
  • $\begingroup$ 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. $\endgroup$ – Willie Wong Jun 3 '16 at 15:40
  • $\begingroup$ That's amazing! Thank you very much! $\endgroup$ – Richard Muniz Jun 3 '16 at 17:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.