I was working on something and stumbled upon the following situation. I have in front of me a configuration $L$ of lines in $\mathbb{R}^{3}$ and say I consider the graph $G$ having as vertex set $L$ with an "edge" between two lines $l\_{i}$ and $l\_{j}$ if they interesect (again, in $\mathbb{R}^{3}$). Consequently, if we have such an edge, we can associate with it the plane determined by the lines $l\_{i}$ and $l\_{j}$. Now, note that these "edges", as planes, have the special property that every two of them intersect exactly once. Thus, what I'm trying to do is embed this graph in $\mathbb{R}^{2}$ or on a surface to get a new graph $G'$, isomorphic with $G$, which is a so-called "thrackle", i.e. it has points as vertices, and they are joined by Jordan arcs or maybe some algebraic curves, so that every two such edges/arcs intersect exactly once. Can I do this?


Any insight is more than welcomed! :)



About a graph embedding from R^3 to...
No. Say $L$ consists of $n$ lines, all concurrent but no three coplanar. Then $G$ is the complete graph on $L$, isomorphic to $K\_n$. But for $n>3$, $K\_n$ has no thrackle. For example, [Lovász, Pach, and Szegedy](http://www.ams.org/mathscinet-getitem?mr=1476318) showed that a thrackle on $n$ vertices has at most $2n-3$ edges.


