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Let $K$ be a knot in 3-space. We know that if $K$ is not prime (composite) then $K$ can be represented as a connected sum of two knots, both are non-trivial. My question is : does this apply to any arbitrary projection of $K$ in a plane? In other words, if I take any projection of $K$, then I can decompose it into two non-trivial knots by the decomposition operation or in some projections, we have to apply some Reidemeister moves first before applying the decomposition operation.

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I will make an assumption about what you mean by "decompose it into two non-trivial knots by the decomposition algorithm." Let's call a diagram composite if there is a simple closed curve in the plane intersecting the diagram exactly twice away from the crossings such that both the interior and exterior of that simple closed curve contain crossings of the diagram. I will interpret your question as follows:

Rephrased question: Are all diagrams of composite knots composite?

If I've interpreted your question correctly, then the answer is no. Here is a counterexample of the connected sum of two trefoils.

enter image description here

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  • $\begingroup$ I don't understand why the counterexample is a counterexample? I see that we can apply the decomposition algorithm twice to obtain the prime factors (the two trefoils). Do we have cases that we cannot factorize or decompose by the decomposition algorithms which you explained correctly. $\endgroup$
    – user113715
    Jan 19 '21 at 1:57
  • $\begingroup$ I'm not sure what you mean by "decomposition algorithm." Every composite knot will have diagrams that are composite (as defined above) and diagrams that are not composite. There will always be sequences of Reidemeister moves transforming the not composite diagrams into the composite diagrams. One can find not composite diagrams where this sequence of Reidemeister moves is as long as one desires. $\endgroup$ Jan 19 '21 at 2:04
  • $\begingroup$ Good. So not any diagram of composite knot is composite. Your counterexample is composite because we can find two simple closed curves, each cut the diagram in two arcs . The interior of the curves have no crossing points but the exterior have. We apply the operation to each of these curves to decompose the composite knot. The operation is cut the two segments and join the end points of the first segment to the endpoints of the other without introducing any crossing. $\endgroup$
    – user113715
    Jan 19 '21 at 4:15

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