On scannable properties of the original knot from a knot shadow

The knots that lie above all shadows

“…Moreover, we only consider two particular types of tangles, illustrated in Figure 6. A tangle is of Type I if it consists of a string λ with endpoints (−1, −1, 0) and (−1, 1, 3), and a string ρ with endpoints (1, −1, 0) and (1,1,3), and it is of Type II if it consists of a string λ with endpoints (−1, −1, 0) and (−1, 1, 3), and a string ρ with endpoints (1, 1, 0) and (1, −1, 3).…”

“…As in the proof of Lemma 2.1 we assume that lk(T ) = m/2, as the arguments in the case lk(T ) = −m/2 are totally analogous. It is easy to see that there exist strings α and β in 3-space, disjoint from the interior of the cylinder ∆×[0, 3], such that (i) the endpoints of α (respectively, β) are (−1, −1, 0) and (−1, 1, 3) (respectively, (1, 1, 0) and (1, −1, 3)); (ii) the projection of α ∪ β onto the xy-plane is S \ U ; and (iii) the vertical projections of α and β are the strands a and b, respectively, shown in Figure 8(b).…”

“…It is easy to see that there exist strings α and β in 3-space, disjoint from the interior of the cylinder ∆×[0, 3], such that (i) the endpoints of α (respectively, β) are (−1, −1, 0) and (−1, 1, 3) (respectively, (1, −1, 0) and (1, 1, 3)); (ii) the projections of α and β onto the xy-plane are A and B, respectively; and (iii) the vertical projections of α and β are the strands a and b, respectively, shown in Figure 9 The strategy to prove the proposition is to show that S is the shadow of a link that satisfies certain properties. We start by letting M be the set of all links M that have S as their shadow, and such that: (i) the part of M that projects to U is a tangle T of Type I; and (ii) the part of M that projects to A ∪ B is α ∪ β.…”

“…In [3], Hanaki investigates the following related question: given a shadow S, what can be said about invariants of a knot that projects to S? Hanaki's work illustrates very well the difficulty of Question 1.1, with a running example of a shadow with 9 crossings for which it is not easy to determine whether or not it resolves into the torus knot T 2,7 .…”