Could you please provide me with the precise definition of isotropic curves of a conformal structure on a manifold $M$?
If there is such a definition, then can I say the following: if $c$ is an isotropic curve corresponding to a Riemannian/Lorentzian metric $g$ on $M$, then $c$ is also an isotropic curve for any Riemannian/Lorentzian metric of the form $e^u\times g$, where $u\in C^{\infty}(M,\mathbb{R})$ is a smooth function on $M$?
3.One or two examples please?
Here is the semi-detail of the context I've come across the term 'isotropic':
Consider the $3$-dimensional anti-de-Sitter space $AdS_3^{*}=\{\vec{x}\in R^4: x_1^2+x_2^2-x_3^2-x_4^2=-1\}$, with its Lorentzian metric of signature (2,1), and its universal Lorentzian cover $AdS_3=\mathbb{D}\times \mathbb{R}$ with the Lorentz metric, which is a warped product of the hyperbolic metric $h$ on $\mathbb{D}$ and the Euclidean metric $dt^2$on $\mathbb{R}$, and is given by: $g=h-(\frac{1+r^2}{1-r^2})^2 dt^2$. Now, note that, if you consider $m=\frac{g}{(\frac{1+r^2}{1-r^2})^2}$, then the Lorentz metric $m$, which is conformal to $g$, extends to $\partial{\mathbb{D}}\times \mathbb{R}$.
The paper talks about the isotropic curves of this conformal Lorentz structure on $\partial{\mathbb{D}}\times \mathbb{R}$, and states that 'isotropic curves' of this structure foliates $\partial{\mathbb{D}}\times \mathbb{R}$.
I am unable to follow what they mean by these 'isotropic curves' in this context. A detailed explanation of the last paragraph will be greatly appreciated!
In Lorentzian geometry, isotropic/null/lightlike vectors $v$ are those that satisfy $g(v,v)=0$, and isotropic/null/lightlike curves $\gamma$ are those whose tangent vectors are everywhere isotropic; i.e. $g(\dot \gamma,\dot \gamma)=0$.
In your example it looks like $(\mathbb{D},h)$ is the Poincaré disc model $$\left(\{(x,y) \in \mathbb{R}^2 : x^2 + y^2 < 1\}, \frac{4}{(1-r^2)^2}(dx^2 + dy^2)\right)$$ where $r=\sqrt{x^2 + y^2}$ is the radial coordinate. Note that this does not extend smoothly to $\partial \mathbb{D}$ because $1-r^2=0$ there. However, if we modify it with the factor you mentioned, we get a $(1+r^2)^2$ instead, which can be extended to all of $\mathbb{R}^2$.
Likewise the product metric $$ m = \frac{4}{(1+r^2)^2}(dx^2 + dy^2) - dt^2$$ extends to all of $\mathbb{R}^2 \times \mathbb{R}$, and in particular to the surface $\partial \mathbb D \times \mathbb{R}$ (by taking the usual induced metric on a submanifold). Using polar coordinates $(r,\theta)$ on $\mathbb R ^2 $ so that the submanifold is $\{r=1\}$, this metric turns out to be simply $d\theta^2 - dt^2$ on the cylinder $S^1 \times \mathbb{R}$, which has two foliations by null curves: in coordinates $(\theta, t)$,
$$ \{\gamma_{\theta_0}^\pm(t) = (\theta_0 \pm t, t) : \theta_0 \in S^1\}.$$
Intuitively these are just helices on the infinite cylinder with angle $\pi/4$ to the horizontal/vertical axes.
Since the condition $g(\dot\gamma,\dot\gamma)=0$ is invariant under conformal changes (i.e. multiplying $g$ by a positive factor), the null curves are the same for any conformally related metric; so they are determined from the conformal structure only.