Ten tangle problems, composed by David A. Krebes:
1. Can a two-component unlink intersect each of a pair of disjoint
balls in a square knot tangle?
2. Does the granny knot tangle G embed in a separable 2-component
link with both components appearing in G? Same question for
(T_2)* + (T_3)* instead of G.
3. Does the genus-one tangle appearing at the end of my original
paper embed in an unknot?
4. Show that the only invertible tangles are the integral tangles. A
tangle T is invertible if there is a tangle T' such that T + T' =
T_0. Necessary condition: q = \pm 1.
5. Classify, to n crossings, the tangles that are not
moving-endpoint-isotopic to a tangle that has a diagram with
fewer crossings.
6. Characterize the tangles which are mirror images of themselves.
This will be a very large class, since knotting a strand with an
amphichiral knot such as the figure eight does not destroy this
property. Necessary condition: p = 0 or q = 0.
7. Show that there are no tangles T such that T is equivalent to T*.
Necessary condition: p/q = -q/p; p^2 = -q^2, ie. p=q=0. Solved.
8. If a tangle has a separable link, ie. if there is a ball inside
the larger ball whose interior intersects the tangle but whose
boundary does not, then p=q=0. Is the converse true? ie. Does
p=q=0 imply that the tangle contains a separable link?
9. Prove that any two tangles (B,T) and (B,T') with a method of
comparison, for instance a homeomorphism from T to T', which are equivalent
with respect to the embedding problem are
moving-endpoint-isotopic.
10. Call a tangle "non-alternating" if it is not
moving-endpoint-isotopic to a tangle with an alternating
diagram. Can a non-alternating tangle embed in an
alternating link?
NOTES
-I do not say "unsolved" problems because Danny Ruberman feels that he has some
sort of handle on 1,2,4 and 8. I have solved 7. I have found an
example of a 6-tangle for 10, and I suspect there are 4-tangles as well.
-1: I am not requiring that the union of these two balls be all of
S^3. They are disjoint; their boundaries are disjoint. They are
separable in the sense of separable links.
-p and q are defined in my paper "An Obstruction to Embedding
4-Tangles in Links" (32 pages, 17 figures)
(http://xxx.lanl.gov/abs/math.GT/9902119) (Journal of Knot Theory and
its Ramifications Vol. 8 No. 3 May 99, 321-352). They are, up to sign,
the determinants of certain links derived from the tangle.
-4 would be a nice result. Seems very intuitive. The context is that
of my original "Obstruction" paper. An integral tangle is a sum of a
non-negative number of single-crossing tangles T_1 or T_{-1}, with T_0
as the identity. The integral tangles are parametrized by the
integers; we use the notation T_n for the appropiate integer n. The
balls are cubes, the endpoints are in standard position and a "tangle"
is a fixed-endpoint isotopy (as opposed to "moving-endpoint isotopy";
see below) class of a one-manifold with the prescribed endpoints.
"Tangle sum" is concatenation followed by a horizontal compression.
"\pm 1" denotes +1 or -1.
-3 is one Danny Ruberman and I have been working on.
-I omitted the question of whether the 5-crossing oriented tangle
based on T_3^* + T_2^* embeds in the unknot because Danny believes he
has a proof that it does not.
-"moving-endpoint-isotopic" means images of each other under an
isotopy in which the boundary of the intermediate tangles is allowed
to move (but stay on the boundary sphere of course.)
-9: The "method of comparison" is a way of matching up components and
orientations of those components. "Equivalence with respect to the
embedding problem" means that for every (unoriented) link L in S^3 and
orientation preserving embedding j of B in S^3 with j(B)\cap L = j(T),
there is an orientation-preserving embedding j' of B in S^3 such that
j'(B)\cap L = j'(T') = j(T) and j'(h(x))=j(x) for all points x in T,
and vice versa (exchange the roles of T and T' and replace h with
h^{-1}). The converse to the statement is easily seen to be true.
Note that this problem really sits in a more general setting: The
genus of B, the number of boundary components of the tangles, the
dimension of the tangles, and the dimension of B can all be tweaked.
This result would say that structure and behaviour are in some sense
equivalent.