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I'm trying to understand why ambient isotopies are defined the way they are for knot theory.

Definitions: my text defines an isotopy between embeddings $f_1, f_2$ from $X$ to $Y$ as a continious map $F: X \times [0,1] \to Y \times [0, 1]: (x, t) \mapsto (f(x, t), t)$ such that $f(x, 0) = f_0(x)$ and $f(x, 1) = f_1(x)$. The intuition behind this is that $F$ continiously deforms the two embeddings into each other. This makes sense to me.

An ambient isotopy between $f_0$ and $f_1$ is defined as a continious map $H: Y \times [0, 1] \to Y \times [0, 1]: (y, t) \mapsto (h(y, t), t)$ such that $h(y, 0) = y$ and $h(f(x, 0), 1) = f(x, 1)$. The intuition is that $H$ defines an isotopy (by $F: (x, t) \mapsto (h(f(x, 0), t), t)$) that also preserves the ambient space around the embeddings. My precise questions are:

  1. Why are two knots equal if they are ambient isotopic?
  2. Why do we define $h$ the way we do?
  3. Why does the $F$ defined by $H$ respect the ambient space?
user388557
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    What's your reference ? – Kelvin Lois Jun 09 '18 at 10:32
  • @Sou Sorry, what does 'your reference' mean? My level of knowledge is one course in topology (20 hours a week, for 10 weeks, where we covered the basic topological properties such as homeomorphisms, constructions of topologies and some topological invariants) and the basic definitions of knot theory. – user388557 Jun 09 '18 at 10:35
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    I mean what textbook did you use ? – Kelvin Lois Jun 09 '18 at 10:38
  • @Sou Burde and Zieschang for knot theory, but I'm only a couple of pages in, and lecture notes for topology – user388557 Jun 09 '18 at 11:03
  • Perhaps this will help you understand a bit why this definition is used: https://math.stackexchange.com/questions/1336275/which-two-knots-are-isotopic-but-not-ambient-isotopic/1336396#1336396 – Cronus Sep 25 '21 at 07:21

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Imagine that you have a sheet of thin rubber lying on your desk, so it's basically $I \times I$. It's taped down around its edge. You draw on it, in magic marker, a circle.

If you now put a finger in the center of the rubber sheet and tug it a little to one side, your circle may distort a bit, perhaps becoming egg-shaped. To a topologist, this is still a "circle in a rectangle". One reason is that it's easy to undo the operation -- you ease up on the sheet and it deforms back to what it looked like before. Such an operation is called an "isotopy" of the ambient space. In this example, $Y$ is the rubber sheet, and the motion induced by your finger is a sequence of mappings of $Y$ to itself: each piece of the rubber sheet moves (over a brief period of time) to a new location that happens to be a location where some rubber sheet was before.

What about the drawing? Well, that's an embedding of the circle into $Y$, and as $Y$ 'slides around within itself', the drawing gets carried along. You could define the knot to simply be the point set where the ink happens to be, but I assume that your text has made clear that things get simpler if we regard the map (i.e., the parameterization of the point-set by a circle) rather than the point-set itself. So when I say the drawing corresponds to the knot, what I really mean is that a parameterization of the drawing corresponds to the map $f$ in your definition.

With a homotopy of $f$, you could change the image of $f$ from a circle to, say, a figure $8$, or a single point. With an isotopy of $Y$, you can only change it to a distorted circle. That seems to better capture the notion that two knots are "the same" (in the sense that the knot used to tie your shoelaces is the same as the one used to tie my shoelaces, assuming that in each case we glue the tips together to get a circular shoelace (i.e., an embedding of $S^1$ rather than of $I$), even though your shoes and mine are not in exactly the same location.

John Hughes
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  • What if you require the homotopy to remain an injective function from $S^1$ to $\Bbb{R}^3$ at all times, would you still be able to get a figure 8 knot from a regular circle? Can you perhaps direct me to a proof/visualization of this fact? – Cronus Sep 24 '21 at 03:21
  • I didn't say anything about a figure-8 knot; I was talking about the appearance of the numeral 8 in most standard fonts, i.e., a 2D shape that crosses itself at one point. Since this is not the image of an injective (continuous) function from $S^1$ to $\Bbb R^3$, it clearly can't be arrived-at by the kind of homotopy you describe. – John Hughes Sep 25 '21 at 01:51