# Riemannian foliations and their leaf space

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?

• 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 $M$? – Ben McKay Jan 11 '17 at 19:55
• What is the definition of holonomy group of a metric singular foliation? – Ben McKay Jan 11 '17 at 20:03
• 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. – melomm Jan 11 '17 at 20:19
• Why isn't the infimum of the sums always acheived at $k=1$, by the triangle inequality? – Ben McKay Jan 11 '17 at 20:21
• @BenMcKay is right in his comments. I did some corrections in the original question. – melomm Jan 11 '17 at 21:03

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.