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Suppose one has a knot $K$ embedded in $\mathbb{R}^3$; but view $\mathbb{R}^3$ as a 3-flat in $\mathbb{R}^4$. Of course $K$ is not a knot in $\mathbb{R}^4$. I am wondering if there has been any study of how many "moves" are needed to unravel $K$ using the 4th dimension?

One might make this a sharper question in several ways. Here is one attempt. Say $K$ is represented as a 3D polygon of $n$ vertices in $\mathbb{R}^3$. A move consists of rotating a subchain of $C \subset K$ with endpoints $(a,b)$ into 4D and then back again at some new orientation into the 3-flat containing $K$. The endpoints $a$ and $b$ remain fixed, while $C$ is replaced by $C'$. Call the result knot $K'$.

Q1. Can every knot $K \subset \mathbb{R}^3$ be unraveled by these moves?

Answer: Yes, by Ian Agol's convincing argument.

Q2. How many such moves are needed to untangle $K$, as a function of some measure of $K$'s complexity, say, its crossing number?

These are entirely naïve questions, and I am not sure I have formulated them coherently. If not, apologies for the distraction!

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One has to exercise some care: if "there isn't room" on the other side of the linkage for the subchain, then there might be problems. For sufficiently loose knots, any crossing should be invertible with the aid of the 4th dimension by tweaking just one strand. If you need two strands to switch places, send one to the positive half space and the other to the negative half space and reposition there. I suspect less than 2n moves are needed where n is the number of crossings. Gerhard "Ask Me About System Design" Paseman, 2012.01.24 –  Gerhard Paseman Jan 24 '12 at 22:31
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Regarding the question in your preamble, I think the unknotting number of $K$ could be thought of as a measure of "how many "moves" are needed to unravel $K$ using the 4th dimension". –  Oliver Jan 25 '12 at 2:31

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I think your moves suffice. One may prove that your moves may rotate the polygonal knot into a convex planar polygon by induction.

As a warmup, suppose we have a polygonal knot in the plane. Consider its convex hull, one gets a convex polygon. If the knot is convex, then it lies on the boundary of this polygon. Otherwise, there is an edge $e$ in the polygon which is not an edge of the knot, and whose endpoints are vertices of the knot. Take an arc (chain of edges) of the knot bounded by these vertices, and rotate it to the other side of the line containing the edge $e$. Repeat this process until you obtain a convex polygon. So this shows that any planar polygon may be chain-rotated to be convex.

Now take a 3D polygonal knot $K$ with $n$ segments. If $n=3$, then we are done, so assume $n>3$. Take the convex hull of the knot, and assume it is nonplanar. The boundary is a polyhedron, and there is some edge $e$ of the polyhedron which is not an edge of the knot. Take the two endpoints of this edge, these must be vertices of the polygonal knot. The interior of $e$ might meet other vertices of the polygonal knot, so take two vertices $v_1,v_2$ of the knot in the edge $e$ which have no other vertices between them. Take a supporting plane for the polyhedron along the edge $e$, and choose an arc $a$ of the knot with the vertices $v_1,v_2$ as endpoints. Rotate this arc in 4D to lie on the other side of the support plane. Now the knot has been reduced to a geometric connect sum: there is a plane meeting the knot in two vertices of the polygon, with a rotated copy of $a$ lying on one side, and $K-a$ lying on the other. Take the two knots which this is a connect sum of, by taking $a\cup e$ and $K-a\cup e$. Each of these knots may be rotated into a convex polygon in the plane by induction, so perform the rotations individually, making sure that they don't interfere with the other side by genericity. Once we have rotated both knots to be convex planar, then we may rotate their connect sum to be planar, and therefore convex. So by induction, any polygonal knot may be chain-rotated to be convex planar.

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Just a remark: your 2D "warmup" is called the Erdős–Nagy theorem: en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Nagy_theorem . –  Joseph O'Rourke Jan 25 '12 at 1:41
    
"making sure that they don't interfere with the other side": How does genericity ensure this, if I may ask? –  Joseph O'Rourke Jan 25 '12 at 1:43
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Joseph: The track of the isotopy corresponding to the move is 2-dimensional. The other part of the knot that might interfere is 1-dimensional. 1+2 < 4, so generically these two submanifolds 4-space will not intersect. –  Kevin Walker Jan 25 '12 at 2:03
    
Thanks, Kevin! $\mbox{}$ –  Joseph O'Rourke Jan 25 '12 at 2:09
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Kevin is correct, but I should probably say a bit more since the rotation is supposed to be rigid. Consider the supporting plane $P$ containing the edge $e$. The unit normal bundle to the segment $e$ is a 2-sphere. Any unit vector together with $P$ spans a 3-dimensional subspace and the unit vector defines a half-space in this subspace; for example when this is an orthogonal vector to $P$ in $R^3$, one obtains $R^3$. So take the unit vector pointing to the side of $P$ containing $K$, and rotate it to the unit vector in the side not containing $K$. –  Ian Agol Jan 25 '12 at 2:18

There is a canonical way of unknotting knots in $\mathbb R^3$ once you consider them to be in the larger ambient space, $\mathbb R^4$. It has a fairly slick formulation. I think once you adapt it to your situation it gives a very small number of moves -- I suppose the number should be proportional to the bridge number of the knot diagram? I suppose it really depends on what precisely your moves are. Here is the construction:

Let $K_3$ be the space of embeddings (PL or smooth) $\mathbb R \to \mathbb R^3$ which agree with the standard inclusion $t \longmapsto (t,0,0)$ outside of $[-1,1]$. Let $K_4$ be the corresponding space when the target is turned to $\mathbb R^4$. Then the inclusion map $K_3 \to K_4$ is null-homotopic. In fact, there's two (seemingly distinct) null-homotopies. Here is a rough sketch of how they're constructed.

Let $b : \mathbb R \to \mathbb R$ be a bump function whose peak occurs around $b(-1)$, and which decreases from $b(-1)$ to zero along $[-1,1]$. Then if $f \in K_3$, let $if \in K_4$ be its inclusion. Take the straight line homotopy from $if$ to $if + be_4$ where $e_4 = (0,0,0,1)$. Let $x : \mathbb R \to \mathbb R^4$ be the standard inclusion. The next part of the contraction is to take the straight-line homotopy from $if+ be_4$ to $x + be_4$. The next part is to take the straight line homotopy from $x+ be_4$ to $x$.

You can do the same type of null-homotopy using a bump function whose peak is at $b(1)$, and which increases from zero to $b(1)$ along $[-1,1]$.

It's a curiosity that if you put these two constructions together you get a map $K_3 \to \Omega K_4$ (loop space on $K_4$). I haven't found a way to determine if this map is non-trivial.

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@Ryan I think your bump function can easily be parametrized in terms of the cosine ;) –  Scott Carter Jan 25 '12 at 0:24

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