Till now, the symmetry group of all chiral noninvertible knots was trivial, but chiral noninvertible knots of the form 6^{*}r_{1}.r_{2} 0:r_{3} 0.r_{4} with r_{1} = r_{3} or r_{2} = r_{4} have the symmetry group Z_{2}. Knots of the form 6^{*}r_{1}.r_{2}:r_{3} 0.r_{4} 0 derived from 6^{*}2.2:2 0.2 0 are revertible iff r_{1} = r_{3} and tangle type of (r_{2},r_{4}) is not ([0],[¥]) or ([¥],[0]), r_{2} = r_{4} and tangle type of (r_{1},r_{3}) is not ([0],[¥]) or ([¥],[0]), noninvertible achiral iff r_{1} = r_{2} and r_{3} = r_{4}, and chiral otherwise. Among the chiral knots of this form, knots with r_{1} = r_{3} or r_{2} = r_{4} have the symmetry group Z_{2}, and a trivial symmetry group otherwise. Knots of the form 6^{*}r_{1}.r_{2}.r_{3}:r_{4} derived from 6^{*}2.2.2:2 are chiral noninvertible iff r_{1} ¹ r_{3}, and revertible otherwise. There is an interesting connection between the knots derived from 6^{*}2.2:2 0 and 6^{*}2.2 0:2 0. For every n they generate the same number of knots, and the derived knots from those two classes mutually correspond with regard to all kinds of symmetries and invertibility. The same holds for the following pairs: 6^{*}2.2 0::2 0 and 6^{*}2.2.2, 6^{*}2.2.2.2 and 6^{*}2.2.2 0:2, 6^{*}2.2 0.2.2 0 and 6^{*}2.2.2:2 0, 6^{*}2.2.2 0.2 and 6^{*}2.2.2:2 The similar connection exists between 6^{*}2.2 0:2 0.2 and 6^{*}2.2:2 0.2 0 if we consider only the number of knots derived, the number of chiral noninvertible knots among them, and their distribution according to the symmetry groups (Z_{2} or trivial). In the same way it is possible to find noninvertibility criteria for all knots derived from different basic polyhedra replacing vertices by Rtangles. Next class of noninvertible polyhedral knots will be obtained by replacing vertices of a basic polyhedron by pretzel (stellar) tangles. For example, all knots of the form 6^{*}p_{1}.r_{1}, where p_{1} is the pretzel tangle of the pretzel type [¥], and r_{1} is Rtangle of the type [0], are chiral noninvertible. We already mentioned that a pretzel tangle p_{1} will be of the type [¥] if it consists of an arbitrary number of Rtangles of the types [1], [¥], and from exactly one Rtangle of the type [0]. First examples are 11crossing chiral noninvertible knots 6^{*}(3,2).2, 6^{*}(2,3).2, 6^{*}(2 1,2).2, and 6^{*}(2,2 1).2. We conclude that all knots derived from the basic polyhedron 6^{*} with one pretzel tangle p_{1} and one Rtangle r_{1} will be chiral noninvertible. The first class of polyhedral knots with pretzel tangles that requires symmetry discussion are knots of the form 6^{*}p_{1}.p_{2}, where p_{1}, p_{2} are pretzel tangles of the types ([¥],[¥]), or ([0],[0]), respectively. If p_{1}, p_{2} are pretzel tangles of different types, [0] and [¥], the corresponding knots are chiral noninvertible. Knots of the form 6^{*}p_{1}.p_{2} where p_{1}, p_{2} are pretzel tangles of the types ([¥], [¥]) are achiral noninvertible iff p_{1}, p_{2} are mutually palindromic pretzel tangles, and chiral noninvertible otherwise. For example, the knot 6^{*}(3,2).(2,3) is achiral noninvertible, and 6^{*}(3,2).(3,2) is chiral noninvertible. The same criterion holds for the knots of the same form 6^{*}p_{1}.p_{2}, where p_{1}, p_{2} are pretzel tangles of the types ([0],[0]).
In the same way we can conclude that all knots of the form
6^{*}p_{1}.r_{1} 0, where p_{1} is a pretzel tangle, and r_{1} is
Rtangle, are chiral noninvertible.
Knots of the form 6^{*}p_{1}.p_{2} 0, where p_{1}, p_{2} are pretzel tangles of different types [0], [¥], are chiral noninvertible. If the pretzel tangles p_{1}, p_{2} are of the same type ([0],[0]), or ([¥],[¥]), knots of the form 6^{*}p_{1}.p_{2} 0 are chiral noninvertible iff p_{1} ¹ p_{2}. For example, knot 6^{*}(3,2).(3,2) 0 is invertible, and 6^{*}(5,2).(3,2) 0 is chiral noninvertible. In general, for every basic polyhedron infinite classes of noninvertible knots derived from it can be recognized.
Replacement of the tangle of particular reduced type ([0_{k}],
[1_{k}], or [¥_{k}]) by a tangle of the same type preserves
the number of components. Hence we have the following main
conjecture:
Conjecture Given a noninvertible knot K, every replacement of a tangle of the specific type by a tangle of the same type, respecting symmetry conditions, gives a noninvertible knot and preserves chirality.
In other words: noninvertibility is typedependent and
symmetrydependent property, and not only the property of
particular knots.
