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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! :)

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What conditions do you have on $L$? It seems you are assuming that no three lines are coplanar. –  j.c. Oct 11 '12 at 18:24
    
So $L$ is just any configuration of lines; the graph $G'$ I'm trying to make is not meant to be complete or anything; I'm starting with some lines in $\mathbb{R}^{3}$, if two intersect, draw the plane determined by them and my intention is two map this situation into something like $\mathbb{R}^{2}$, where in that case the two vertices correspoding to the two lines are united by a Jordan arc. If you have two pairs of lines which intersect, then draw the two corresponding Jordan arcs - I want them to intersect in this new $G'$ because the planes determined by each pair in $R^{3}$ intersect... –  Cosmin Pohoata Oct 11 '12 at 18:32
    
What do you mean by `two planes intersect exactly once'? They should either be parallel or intersect by a line... –  Ilya Bogdanov Oct 11 '12 at 18:33
    
Yes; work in the projective space, sorry. So any two planes intersect in precisely one line (which may be at infinity) - I want to translate that into $G'$ where the arcs meet at exactly one point –  Cosmin Pohoata Oct 11 '12 at 18:49
    
Please recognize also that your definition of thrackle differs from a classical one. In a classical definition, it is assumed that the intersection of any two edges (while it is not in a common vertex) is transversal... –  Ilya Bogdanov Oct 11 '12 at 19:10
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1 Answer

up vote 2 down vote accepted

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 showed that a thrackle on $n$ vertices has at most $2n-3$ edges.

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Thank you :). //Sorry for the late acceptance of your answer. I tried to convince myself that you might be wrong for a while :D. The thing I intended was to connected the joints problem/conjecture with the partial results known for the Conway thrackle conjecture (to get a new proof of the joints problem). –  Cosmin Pohoata Oct 19 '12 at 22:48
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