# Can we apply Reidemeister moves to self crossings of one component only of a “trivial” link and keep the others unchanged?

Suppose $$L=L_1 \cup L_2 \cup L_3$$ be a classical link of three components. Suppose $$L$$ is an unlink, that is $$L$$ can be splitted into three simple closed curves. Assume that $$L$$ has a diagram in 2-plane such that

• There are some crossings between $$L_1$$ and $$L_2$$ And between $$L_1$$ and $$L_3$$.
• There are some crossings between $$L_2$$ and $$L_3$$.
• There are some self intersections of each $$L_i, i=1,2,3$$.

I have two questions:

Q1: Can we apply a finite sequence of Reidemeister moves to the self crossings of $$L_1$$ only so that all self crossings of $$L_1$$ are eliminated without affecting the other crossings of $$L$$. If the answer is yes, then this can be generalized to $$L_2$$ and $$L_3$$. This means we can first eliminate self crossings of $$L$$ and then we may deal with the crossings between the components. Is this possible always?

Q2: Can we first apply a finite sequence of Reidemeister moves to split $$L_1$$ from the link $$L$$ without affecting the crossings which are not between $$L_1$$ and $$L_i, i=1,2,3$$. If so, then we may split the components of the link first and then deal with the self crossings of each component.

I know that the answer of both of my questions is no if the link $$L$$ is not trivial link. The whitehead link with five crossings is a counterexample. How about the case of an unlink?

So, I believe the answer is no for each question. Here is not so much as a proof, but a couple candidate links for you to consider. I think it is easy to see these are indeed unlinks. But I got away with two components in each. Making similar links with 3 components should not be hard, but I leave that to you. The first link has twists in each component that cannot be removed without some type II moves or type III moves first, between the shared crossings.

The second question is harder. I think this link does it though. You cannot unlink the red, keeping the blue fixed without a type II first. If you fix the red, then you will need a few moves between red and blue to get to a type I move with the blue before you can go any further.

Now, this is in now way a proof. I am not sure how you might have to go about showing these are in fact, counter-examples. Perhaps looking at the Reidemeister graph and somehow showing that every path to the trivial diagram must go though a "bad" point, or something. Cooking up counter-examples to conjectures like this is a good way to build intuition into how crappy diagrams really are in knot theory. But its what we got. Good luck.

Edit

So upon further reflection, I am now sure that my link for Question 2 does not hold up. You can slide the blue link down the red and off of it without any blue only Reidemeister move.

One definition of the unlink is a link where each component simultaneously bounds an embedded disk and these disks are all disjoint. This allows us to pick any component $$L_1$$ and isotope it along this disk to a small unknot component which has no crossings with any other components. This lets us keep the other components fixed the whole time we are running this isotopy. But I don't see how to guarantee that the crossings of $$L_1$$ are not subjected to Reidemeister moves or not in this.

Quick summary: We can fix any components of an unlink we like and isotope the rest away.

• I think the answer is yes for the second question. Since each of the three components bounds an embedded disk in 3-space, we may move the closed disk of the component $L_1$ freely in 3-space so that all crossings between $L_1$ and the other components are eliminated, that is $L_1$ is splitted from $L$. In general, we may split the components first and then eliminate the self crossings of each component. Am I right? Aug 17, 2020 at 23:53
• @user113715 This is where it gets sticky. Maybe it doesn't matter for your situation, but the isotopy we have from the disk might produce Reidemeister moves between the crossing of just $L_1$. You can definitely split the components like you say, but I can't see a guarantee that the diagram of $L_1$ alone will be unmodified after you split it. But if you only care about getting them apart, leaving the other components unchanged, you are good. Aug 18, 2020 at 13:03