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 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, 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.

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 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 exactly 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, 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.