0
$\begingroup$

Let be $(M,g)$ a riemannian manifold with a singular riemannian $\mathcal{F}$ in $M$, see [1] the definition of singular riemannian foliation.

The riemannian metric on $M$ induces a distance on $M$, and consequentely a pseudo-distance on $M/\mathcal{F}$ by

$$d_{M/\mathcal{F}}(L_1,L_2):= \inf\{d(p_1,q_1)+d(p_2,q_2)+\dotsb+d(p_{n},q_{n})\} $$ where the infimum is taken over all finite sequences $(p_1, p_2,\dots, p_n)$ and $(q_1, q_2, \dots, q_n)$ with $p_1=L_1, q_n\in L_2$ and $q_i,p_{i+1}$ belong to the same leaf.

Is the pseudo-distance $d_{M/\mathcal{F}}$ in fact a distance?

$\endgroup$
  • $\begingroup$ If you only ask for $p_0 \in L_1$ and $p_k \in L_2$, do you let all other points $p_1, \dots, p_{k-1}$ move around freely on $M$? $\endgroup$ – Ben McKay Jan 11 '17 at 19:55
  • $\begingroup$ What is the definition of holonomy group of a metric singular foliation? $\endgroup$ – Ben McKay Jan 11 '17 at 20:03
  • $\begingroup$ Yes, the points move freely on $M$, we could think it as a "discrete path". I made some confusion about holonomy, I am going to edit the question. $\endgroup$ – melomm Jan 11 '17 at 20:19
  • $\begingroup$ Why isn't the infimum of the sums always acheived at $k=1$, by the triangle inequality? $\endgroup$ – Ben McKay Jan 11 '17 at 20:21
  • $\begingroup$ @BenMcKay is right in his comments. I did some corrections in the original question. $\endgroup$ – melomm Jan 11 '17 at 21:03
3
$\begingroup$

The pseudo-distance is not a distance for $M=\mathbb{R}^2/\mathbb{Z}^2$ with the foliation whose leaves are the images of the lines $L_{x_0}=\left\{(x_0+tx_1,ty_1) \, : \, t \in \mathbb{R}\right\}$ if $x_1$ and $y_1$ are not rational multiples, as the lines are dense in the torus, so get arbitrarily close to one another.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ All leaves are closed if the pseudo-metric is a metric on the leaf space. The map to the leaf space is always continuous, and so the leaves are closed just when the points of the leaf-space are closed, which is clear if the pseudo-metric is a metric. $\endgroup$ – Ben McKay Jan 11 '17 at 22:05

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.