Can we homotope away a (non-trivial) knot.?

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In summary, It is possible to un-knot any knot by using an ambient homotopy, which is a continuous transformation of the knot in three-dimensional space. This can be achieved by starting with a complicated embedding of the knot and a standard embedding, and then using a straight-line homotopy to connect the two. By extending this homotopy to an open neighborhood of the knot and patching it to the whole space, we can prove the fact that any knot can be unknotted.
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Bacle
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The answer is pretty clearly yes, but , embarrassingly, I cannot find a good

argument to this-effect. Any hints, please.? (I know the answer is no for

isotopies, i.e., we cannot isotope-away a knot.). I have thought of embedding

the knot in S^3 , and avoiding the point-at-infinity, but this is not working too well.

Thanks.
 
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What you are after here is a mathematical proof of the obvious fact that any knot can be unknotted if we allow self intersections... that is so say, by an ambient homotopy.

So, we start with an complicated embedding p:S^1--> R^3 and the standard embedding j:S^1-->R^3. Call K_1:=j(S^1) and K_2:=p(S^1)

Then we can define a straight-line homotopy H:K_1 x I -->R^3, [itex]H(x,t) = x + t(p(j^{-1}(x))-x)[/itex] between K_1 and K_2. But since K_1 and K_2 are compact and embedded, there exists an open epsilon-tubular neigborhood T_1 of K_1 and T_2 of K_2. This allows us to extend the straight-line homotopy H to T_1 in an obvious way. (Namely, every point of T_1 can be uniquely written in the form x+a*n(x) where n(x) is the unit normal to K_1 at x and |a|<epsilon and analogously for K_2. Define H':T_1 x I -->R^3 by H(x+a*n(x),t)=[x+a*n(x)] + t([y+a*n(y)]-[x+a*n(x)]), where [itex]y = p(j^{-1}(x))[/itex].)

Finally, patch H to the whole of R^3 by a bump function: Let [itex]\lambda:\mathbb{R}^3\rightarrow [0,1][/itex] be continuous with value 0 on R^3 - T_1 and value 1 on K_1 (see Urysohn's Lemma) and define H'':R^3 x I --> R^3 by parts by:

[itex]H''(x,t)=H'(x,\lambda(x)t)[/itex] if x is in T_1
[itex]H''(x,t)=x[/itex] if x is in R^3 - T_1
 

Related to Can we homotope away a (non-trivial) knot.?

What is a non-trivial knot?

A non-trivial knot is a mathematical concept that refers to a knot that cannot be untied without cutting or passing the rope through itself.

What is a homotopy?

Homotopy is a mathematical concept that describes the continuous deformation of one mathematical object into another. In the case of knots, it refers to the continuous deformation of one knot into another without cutting or gluing parts of the knot.

Can any knot be homotoped away?

No, not all knots can be homotoped away. Non-trivial knots, by definition, cannot be homotoped away because they cannot be continuously deformed into a simple circle without breaking or passing the rope through itself.

What is the significance of homotoping a knot away?

Homotoping away a knot has significant implications in knot theory and topology. It allows us to prove that certain knots are equivalent or that they are not equivalent, based on their homotopy class.

Can homotopy be used in real-world applications?

Yes, homotopy has applications in various fields, including physics, biology, and computer science. In physics, it is used to study the properties of DNA molecules, while in computer science, it is used for data compression and image recognition. Homotopy also has applications in studying fluid dynamics and the behavior of materials.

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