Let $A$ and $B$ be skew lines in $\mathbb{R}^3$.  Choose four points $a_1, a_2, a_3, a_4$ on $A$ and four points $b_1, b_2, b_3, b_4$.  For all $i,j \in [4]$ draw a line segment from $a_i$ to $b_j$.  Since $A$ and $B$ are skew, none of these line segments intersect each other.  Thus, this is a straight line drawing of $K_{4,4}$ in $\mathbb{R}^3$ without crossings.

> Is this drawing of $K_{4,4}$ **knotted**?

Recall that a drawing of a graph $G$ is *knotted* if some cycle of $G$ is drawn as a non-trivial knot.  I suspect that the answer is no, but could not prove it.  The motivation for this problem comes from a [paper][1] of David Wood and myself, where we determine the maximum number of copies of a fixed tree in various sparse graph classes.  A negative answer to the above question would essentially solve the problem for the class of *knotless* graphs.  These are the graphs that have a knotless embedding in $\mathbb{R}^3$.  

One approach would be to use the classification of knots with small [stick number][2], but I am hoping there is a more elegant solution. The answer might depend on the positions of the points on $A$ and $B$.  I would be happy if at least one choice of points yields a knotless drawing.   

  


  [1]: https://arxiv.org/abs/2009.12989
  [2]: https://en.wikipedia.org/wiki/Stick_number