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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeJul 5th 2019
   • (edited Jul 6th 2019)
   • PermaLink
   Author: Urs
   Format: MarkdownItex[[Egbert Rijke]] has the following idea, and it makes me wonder. I'll spell it out in some detail for possible bystanders, but in the end it's quite an immediate observation that possibly has occurred to others before. Accordingly, the answer to the question at the end might be clear to those who have thought about this before, or may in any case have been discussed elsewhere. Pointers are welcome. $\,$ Consider a [[knot]] $K$, hence the [[isotopy]]-class of an [[embedding of differentiable manifolds]] of the [[circle]] into the [[3-sphere]] $$ K \;\colon\; S^1 \overset{ \phantom{AAA} }{\hookrightarrow} S^3 \,. $$ With any choice of closed [[tubular neighbourhood]] [[embedding of differentiable manifolds|embedding]] $$ \mathbf{K} \;\colon\; S^1 \times \mathbb{D}^2 \overset{\phantom{AAAA}}{\hookrightarrow} S^3 $$ (where $\mathbb{D}^2$ denotes the closed 2-disk) we get a presentation of the [[3-sphere]] as the union of the closed solid torus with the complement of that tubular neighbourhood, hence an [[attaching space]] [pushout](https://ncatlab.org/nlab/show/Top#UniversalConstructions) of topological spaces of this form: $$ \array{ S^1 \times S^1 &\overset{ \phantom{A} f \phantom{A} }{\longrightarrow}& S^3 \setminus Int(\mathbf{K}) \\ {}^{\mathllap{ g }}\Big\downarrow &{}^{(po)}& \Big\downarrow \\ S^1 \times \mathbb{D}^2 &\underset{ \phantom{A} \mathbf{K} \phantom{A}}{\longrightarrow}& S^3 } $$ Here $f$ and $g$ denote the [[boundary]] inclusions and the right vertical map is the defining inclusion of the [[complement]] in $S^3$ $$ C \;\coloneqq\; S^3 \setminus Int(\mathbf{K}) $$ of the [[interior]] of $\mathbf{K}$. Since the left vertical map is manifestly a [[cofibration]] in the [[standard model structure on topological spaces]], this exhbits a [[homotopy pushout]] of the corresponding [[homotopy types]] $$ \array{ S^1 \times S^1 &\overset{ \phantom{A} f \phantom{A} }{\longrightarrow}& C \\ {}^{\mathllap{ pr_1 }}\Big\downarrow &{}^{(hpo)}& \Big\downarrow \\ S^1 &\underset{ \phantom{A} K \phantom{A} }{\longrightarrow}& S^3 } $$ where now the left map is equivalently the projection onto the first circle factor. For **example** if $K$ is the [[unknot]] then this is the familiar presentation of the 3-sphere as the [[join of topological spaces|join]] of two circles. But we may consider the homotopical data of the second diagram in its own right: Say that a **homotopy knot** is the evident equivalence class of the following data: 1. a [[homotopy 1-type]] $C$, 1. a map $f \;\colon\; S^1 \times S^1 \longrightarrow C$ such that * the homotopy pushout of $f$ along projection to the first factor is the 3-sphere, as in the previous diagram. Now the above construction should define a function: $$ \phi \;\colon\; Knots \longrightarrow HomotopyKnots \,. $$ **Question:** Is $\phi$ a bijection? If not, how does it fail to be one?

   Egbert Rijke has the following idea, and it makes me wonder.

   I’ll spell it out in some detail for possible bystanders, but in the end it’s quite an immediate observation that possibly has occurred to others before. Accordingly, the answer to the question at the end might be clear to those who have thought about this before, or may in any case have been discussed elsewhere. Pointers are welcome.

   $\,$

   Consider a knot $K$, hence the isotopy-class of an embedding of differentiable manifolds of the circle into the 3-sphere

   $$K \;\colon\; S^1 \overset{ \phantom{AAA} }{\hookrightarrow} S^3 \,.$$

   With any choice of closed tubular neighbourhood embedding

   $$\mathbf{K} \;\colon\; S^1 \times \mathbb{D}^2 \overset{\phantom{AAAA}}{\hookrightarrow} S^3$$

   (where $\mathbb{D}^2$ denotes the closed 2-disk)

   we get a presentation of the 3-sphere as the union of the closed solid torus with the complement of that tubular neighbourhood, hence an attaching space pushout of topological spaces of this form:

   $$\array{ S^1 \times S^1 &\overset{ \phantom{A} f \phantom{A} }{\longrightarrow}& S^3 \setminus Int(\mathbf{K}) \\ {}^{\mathllap{ g }}\Big\downarrow &{}^{(po)}& \Big\downarrow \\ S^1 \times \mathbb{D}^2 &\underset{ \phantom{A} \mathbf{K} \phantom{A}}{\longrightarrow}& S^3 }$$

   Here $f$ and $g$ denote the boundary inclusions and the right vertical map is the defining inclusion of the complement in $S^3$

   $$C \;\coloneqq\; S^3 \setminus Int(\mathbf{K})$$

   of the interior of $\mathbf{K}$.

   Since the left vertical map is manifestly a cofibration in the standard model structure on topological spaces, this exhbits a homotopy pushout of the corresponding homotopy types

   $$\array{ S^1 \times S^1 &\overset{ \phantom{A} f \phantom{A} }{\longrightarrow}& C \\ {}^{\mathllap{ pr_1 }}\Big\downarrow &{}^{(hpo)}& \Big\downarrow \\ S^1 &\underset{ \phantom{A} K \phantom{A} }{\longrightarrow}& S^3 }$$

   where now the left map is equivalently the projection onto the first circle factor.

   For example if $K$ is the unknot then this is the familiar presentation of the 3-sphere as the join of two circles.

   But we may consider the homotopical data of the second diagram in its own right:

   Say that a homotopy knot is the evident equivalence class of the following data:

   1. a homotopy 1-type $C$,

   2. a map $f \;\colon\; S^1 \times S^1 \longrightarrow C$

   such that

   • the homotopy pushout of $f$ along projection to the first factor is the 3-sphere, as in the previous diagram.

   Now the above construction should define a function:

   $$\phi \;\colon\; Knots \longrightarrow HomotopyKnots \,.$$

   Question: Is $\phi$ a bijection? If not, how does it fail to be one?

2. • CommentRowNumber2.
   • CommentAuthorDavidRoberts
   • CommentTimeJul 5th 2019
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexBetter: a homotopy **1-type** $C=\mathbf{B}G$. And we probably want some finite presentation requirements on $G$, since we know that fundamental groups of knot complements have the Wirtinger presentation, which is finite, coming from a knot diagram. One might even get more specific with the allowable finite presentations, but this is a start.

   Better: a homotopy 1-type $C=\mathbf{B}G$. And we probably want some finite presentation requirements on $G$, since we know that fundamental groups of knot complements have the Wirtinger presentation, which is finite, coming from a knot diagram. One might even get more specific with the allowable finite presentations, but this is a start.

3. • CommentRowNumber3.
   • CommentAuthorRichard Williamson
   • CommentTimeJul 5th 2019
   • (edited Jul 5th 2019)
   • PermaLink
   Author: Richard Williamson
   Format: MarkdownItexSome initial thoughts on #1, without having thought about it much: 1) The story misses the concept of embeddings and of knot isotopy, so I think has little chance of being right unfortunately. That is, an equivalence of a pair of knots $S^{1} \rightarrow S^{3}$ is not just a homotopy $h : S^{1} \times I \rightarrow S^{3]$ but a homotopy such that $h(-, t)$ is an embedding for all $t$. 2) Specifically, knot complements can be homotopy equivalent without the knots being equivalent. Even if one requires the complements to be _homeomorphic_, it is a deep theorem of Gordon-Luecke that the knots must be equivalent. My gut feeling is that this kind of thing is not possible: knot theory is simply not a homotopy invariant subject!

   Some initial thoughts on #1, without having thought about it much:

   1) The story misses the concept of embeddings and of knot isotopy, so I think has little chance of being right unfortunately. That is, an equivalence of a pair of knots $S^{1} \rightarrow S^{3}$ is not just a homotopy but a homotopy such that $h(-, t)$ is an embedding for all $t$.

   2) Specifically, knot complements can be homotopy equivalent without the knots being equivalent. Even if one requires the complements to be homeomorphic, it is a deep theorem of Gordon-Luecke that the knots must be equivalent.

   My gut feeling is that this kind of thing is not possible: knot theory is simply not a homotopy invariant subject!

4. • CommentRowNumber4.
   • CommentAuthorCharles Rezk
   • CommentTimeJul 5th 2019
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   Author: Charles Rezk
   Format: MarkdownItexThere's a little bit more going on here: the pair $(C, S^1\times S^1)$ is a 3-manifold with boundary. The right homotopical approximation to a manifold is a "Poincare complex", i.e., a homotopy type which "has Poincare duality". Likewise the right homotopical approximation to a manifold with boundary is a "relative Poincare complex". So you maybe you can add that to the mix. Even with this, I would agree with Richard that there is not much likelyhood of getting an equivalence between knots and homotopy knots. Especially since surgery doesn't work in dimension 3 at all, so the usual ways of turning homotopy theoretic information into results about manifolds won't work here.

   There’s a little bit more going on here: the pair $(C, S^1\times S^1)$ is a 3-manifold with boundary. The right homotopical approximation to a manifold is a “Poincare complex”, i.e., a homotopy type which “has Poincare duality”. Likewise the right homotopical approximation to a manifold with boundary is a “relative Poincare complex”. So you maybe you can add that to the mix.

   Even with this, I would agree with Richard that there is not much likelyhood of getting an equivalence between knots and homotopy knots. Especially since surgery doesn’t work in dimension 3 at all, so the usual ways of turning homotopy theoretic information into results about manifolds won’t work here.

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeJul 5th 2019
   • PermaLink
   Author: Urs
   Format: MarkdownItex@David, yeah, one thing to think about is to add more conditions on $f$. On the other hand, there is already the strong condition that it does form a homotopy pushout to give $S^3$. This implies a lot. The question is: How much, if anything, does one have to require in addition? @Richard: Yes, the first thought that comes to mind is that homotopy of knot embedding maps is clearly much coarser than isotopy. But then, the definition of "homotopy knots" above does not in fact involve homotopy of knot embeddings, instead it involves conditions on the knot complements. Maybe this is still too coarse, but it does not seem to be as obvious. In either case, what I'd hope to see is some concrete arguments about how $\phi$ fails to be a bijection. If it's not injective, can we produce a concrete example of a pair of knots that get send to the same "homotopy knot"-data?

   @David, yeah, one thing to think about is to add more conditions on $f$. On the other hand, there is already the strong condition that it does form a homotopy pushout to give $S^3$. This implies a lot. The question is: How much, if anything, does one have to require in addition?

   @Richard: Yes, the first thought that comes to mind is that homotopy of knot embedding maps is clearly much coarser than isotopy. But then, the definition of “homotopy knots” above does not in fact involve homotopy of knot embeddings, instead it involves conditions on the knot complements. Maybe this is still too coarse, but it does not seem to be as obvious.

   In either case, what I’d hope to see is some concrete arguments about how $\phi$ fails to be a bijection. If it’s not injective, can we produce a concrete example of a pair of knots that get send to the same “homotopy knot”-data?

6. • CommentRowNumber6.
   • CommentAuthorRichard Williamson
   • CommentTimeJul 5th 2019
   • PermaLink
   Author: Richard Williamson
   Format: MarkdownItexHi again Urs, this was what I intended to answer in 2). Just take the square knot and granny knot for example: homotopy equivalent complements, not equivalent as knots.

   Hi again Urs, this was what I intended to answer in 2). Just take the square knot and granny knot for example: homotopy equivalent complements, not equivalent as knots.

7. • CommentRowNumber7.
   • CommentAuthorRichard Williamson
   • CommentTimeJul 5th 2019
   • (edited Jul 5th 2019)
   • PermaLink
   Author: Richard Williamson
   Format: MarkdownItexFor prime knots and $C$ a 1-type it is injective I think, though this is a serious theorem.

   For prime knots and $C$ a 1-type it is injective I think, though this is a serious theorem.

8. • CommentRowNumber8.
   • CommentAuthorEgbertRijke
   • CommentTimeJul 5th 2019
   • PermaLink
   Author: EgbertRijke
   Format: MarkdownItexI agree that some knots have equivalent complements, and that the square knot and granny knot is the simplest such example. However, the type of maps $S^1\times S^1 \to K(G,1)$ might have many connected components. That is to say, there might be many non-homotopic maps from the torus to the complement of the square/granny knot. Richard, are you claiming that not only their complements are the same, but also their boundary inclusions $S^1\times S^1 \to K(G,1)$? I would also be curious to know whether or not the conditions in $\mathbf{HomotopyKnot}$ already imply that the complement C is a $1$-type.

   I agree that some knots have equivalent complements, and that the square knot and granny knot is the simplest such example. However, the type of maps $S^1\times S^1 \to K(G,1)$ might have many connected components. That is to say, there might be many non-homotopic maps from the torus to the complement of the square/granny knot. Richard, are you claiming that not only their complements are the same, but also their boundary inclusions $S^1\times S^1 \to K(G,1)$?

   I would also be curious to know whether or not the conditions in $\mathbf{HomotopyKnot}$ already imply that the complement C is a $1$-type.

9. • CommentRowNumber9.
   • CommentAuthorDavidRoberts
   • CommentTimeJul 6th 2019
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   Author: DavidRoberts
   Format: MarkdownItexI agree with Egbert: the data is not just $C$, but includes the map $S^1\times S^1 \to C$, and equivalence of two things as given in #1 should include compatibility of these, such that the induced map on pushouts is (homotopic to) the identity of $S^3$.

   I agree with Egbert: the data is not just $C$, but includes the map $S^1\times S^1 \to C$, and equivalence of two things as given in #1 should include compatibility of these, such that the induced map on pushouts is (homotopic to) the identity of $S^3$.

10. • CommentRowNumber10.
    • CommentAuthorCharles Rezk
    • CommentTimeJul 6th 2019
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    Author: Charles Rezk
    Format: MarkdownItexEgbert, I don't think $C$ will be forced to be a 1-type. To see this, it's enough to have a map $f\colon S^1\times S^1\to D$ so that the homotopy pushout of $f$ along $\mathrm{pr}_1\colon S^1\times S^1\to S^1$ is contractible. Then then take $C:= S^3\vee D$ with the map $S^1\times S^1\xrightarrow{f} D\to C$. I claim such a map $f$ exists with $D=S^2\vee S^1$. Decompose the torus as a union $T_1\cup T_2$ of two tubes along boundary circles (so that $\mathrm{pr}_1$ projects each tube to a semicircle inside $S^1$). Define $f$ so that it sends the boundary circles to the basepoint of $D$, sends $T_1$ surjectively to $S^2\subset D$ by an obvious quotient map, and sends $T_2$ to $S^1\subset D$ by $\mathrm{pr}_2$. The fundamental group and homology groups of the pushout of $f$ along $\mathrm{pr}_1$ are calculated to be trivial.

    Egbert, I don’t think $C$ will be forced to be a 1-type. To see this, it’s enough to have a map $f\colon S^1\times S^1\to D$ so that the homotopy pushout of $f$ along $\mathrm{pr}_1\colon S^1\times S^1\to S^1$ is contractible. Then then take $C:= S^3\vee D$ with the map $S^1\times S^1\xrightarrow{f} D\to C$.

    I claim such a map $f$ exists with $D=S^2\vee S^1$. Decompose the torus as a union $T_1\cup T_2$ of two tubes along boundary circles (so that $\mathrm{pr}_1$ projects each tube to a semicircle inside $S^1$). Define $f$ so that it sends the boundary circles to the basepoint of $D$, sends $T_1$ surjectively to $S^2\subset D$ by an obvious quotient map, and sends $T_2$ to $S^1\subset D$ by $\mathrm{pr}_2$. The fundamental group and homology groups of the pushout of $f$ along $\mathrm{pr}_1$ are calculated to be trivial.

11. • CommentRowNumber11.
    • CommentAuthorEgbertRijke
    • CommentTimeJul 6th 2019
    • (edited Jul 6th 2019)
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    Author: EgbertRijke
    Format: MarkdownItexAnother way to see that the pushout with $D$ is contractible, is to see that $D$ is the mapping cone of the map $S^1\to S^1\times S^1$ which sends $x$ to $(x,\ast)$. The homotopy pushout $S^1\sqcup^{S^1\times S^1} D$ is then contractible by the pasting lemma of pushouts: $$\array{& S^1 & \rightarrow & 1 & \\ & \downarrow & & \downarrow & \\ & S^1\times S^1 & \rightarrow & D & \\ & \downarrow & & \downarrow & \\ & S^1 & \rightarrow & 1 }$$ because the vertical composite on the left is just the identity. Thank you for this nice example!

    Another way to see that the pushout with $D$ is contractible, is to see that $D$ is the mapping cone of the map $S^1\to S^1\times S^1$ which sends $x$ to $(x,\ast)$. The homotopy pushout $S^1\sqcup^{S^1\times S^1} D$ is then contractible by the pasting lemma of pushouts:

    $$\array{& S^1 & \rightarrow & 1 & \\ & \downarrow & & \downarrow & \\ & S^1\times S^1 & \rightarrow & D & \\ & \downarrow & & \downarrow & \\ & S^1 & \rightarrow & 1 }$$

    because the vertical composite on the left is just the identity. Thank you for this nice example!

12. • CommentRowNumber12.
    • CommentAuthorUrs
    • CommentTimeJul 6th 2019
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    Author: Urs
    Format: MarkdownItexOkay, great. Thanks, Charles. So let's demand $C$ to be a 1-type (have edited accordingly in #1) and ask the same question again.

    Okay, great. Thanks, Charles.

    So let’s demand $C$ to be a 1-type (have edited accordingly in #1) and ask the same question again.

13. • CommentRowNumber13.
    • CommentAuthorUrs
    • CommentTimeJul 6th 2019
    • (edited Jul 6th 2019)
    • PermaLink
    Author: Urs
    Format: MarkdownItexOh, I now see that I had missed Charles's #4 above, probably our messages overlapped. So these extra conditions on C and f are about making $\phi$ closer to being surjective. Maybe more urgent is the question of it being injective. Can we answer the question in #8 regarding that specific example?

    Oh, I now see that I had missed Charles’s #4 above, probably our messages overlapped.

    So these extra conditions on C and f are about making $\phi$ closer to being surjective. Maybe more urgent is the question of it being injective. Can we answer the question in #8 regarding that specific example?

14. • CommentRowNumber14.
    • CommentAuthorRichard Williamson
    • CommentTimeJul 6th 2019
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    Author: Richard Williamson
    Format: MarkdownItexIt's a good question. My guess is that, yes, it is a counterexample, but I do not off the top of my head have a proof. For links with more than one component there will definitely be counterexamples, because these are not necessarily determined by their complements even up to homeomorphism.

    It’s a good question. My guess is that, yes, it is a counterexample, but I do not off the top of my head have a proof. For links with more than one component there will definitely be counterexamples, because these are not necessarily determined by their complements even up to homeomorphism.

15. • CommentRowNumber15.
    • CommentAuthorRichard Williamson
    • CommentTimeJul 6th 2019
    • (edited Jul 6th 2019)
    • PermaLink
    Author: Richard Williamson
    Format: MarkdownItexIf one allows wild knots, there are counterexamples as well I think. (Edited: I initially suggested one, but it may have been faulty).

    If one allows wild knots, there are counterexamples as well I think. (Edited: I initially suggested one, but it may have been faulty).

16. • CommentRowNumber16.
    • CommentAuthorUrs
    • CommentTimeJul 6th 2019
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    Author: Urs
    Format: MarkdownItexThanks for the input! One lesson I take from the discussion so far is that the answer is apparently neither evident nor well-known. I would re-iterate to Egbert the suggestion to forward this to MathOverflow. (I can do it, if you prefer?)

    Thanks for the input!

    One lesson I take from the discussion so far is that the answer is apparently neither evident nor well-known. I would re-iterate to Egbert the suggestion to forward this to MathOverflow. (I can do it, if you prefer?)