Talk:Connected sum
 | This article is rated B-class on Wikipedia's content assessment scale. It is of interest to the following WikiProjects: | ||||||||||
| |||||||||||
Connected sum of knots requires orientation of the knots to be well defined. edit
For the connected sum of knots to be well defined, one has to consider oriented knots in 3-space. To define the connected sum for two oriented knots:
- Consider a planar projection of each knot and suppose these projections are disjoint.
- Find a rectangle in the plane where one pair of sides are arcs along each knot but is otherwise disjoint from the knots and so that the arcs of the knots on the sides of the rectangle are oriented around the boundary of the rectangle in the same direction.
- Now join the two knots together by deleting these arcs from the knots and adding the arcs that form the other pair of sides of the rectangle.
The resulting connected sum knot inherits an orientation consistent with the orientations of the two original knots, and the oriented ambient isotopy class of the result is well-defined, depending only on the oriented ambient isotopy classes of the original two knots. In this manner, oriented ambient isotopy classes of oriented knots form a commutative unique factorization monoid.
If one does not take into account the orientations of the knots, the connected sum operation is not well defined on isotopy classes of (nonoriented) knots. To see this, consider two noninvertible knots K, L which are not equivalent (as unoriented knots); for example take the two pretzel knots K = P(3,5,7) and L = P(3,5,9). Let K+ and K- be K with its two inequivalent orientations, and let L+ and L- be L with its two inequivalent orientations. There are four oriented connected sums we may form:
- A = K+ # L+
- B = K- # L-
- C = K+ # L-
- D = K- # L+
The oriented ambient istotopy classes of these four oriented knots are all distinct. And, when one considers ambient isotopy of the knots without regard to orientation, there are two distinct equivalence classes: { A ~ B } and { C ~ D }. To see that A and B are unoriented equivalent, simply note that they both may be constructed from the same pair of disjoint knot projections as above, the only difference being the orientations of the knots. Similarly, one sees that C and D may be constructed from the same pair of disjoint knot projections.
-- Chuck 14:48, 18 May 2007 (UTC)
- Hi, Chuck. The article would be much improved if you were to add this stuff to it. Joshua R. Davis 18:11, 18 May 2007 (UTC)