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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeJul 5th 2019
    • (edited Jul 6th 2019)

    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 KK, hence the isotopy-class of an embedding of differentiable manifolds of the circle into the 3-sphere

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

    With any choice of closed tubular neighbourhood embedding

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

    (where 𝔻 2\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:

    S 1×S 1 AfA S 3Int(K) g (po) S 1×𝔻 2 AKA S 3 \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 ff and gg denote the boundary inclusions and the right vertical map is the defining inclusion of the complement in S 3S^3

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

    of the interior of K\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

    S 1×S 1 AfA C pr 1 (hpo) S 1 AKA S 3 \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 KK 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 CC,

    2. a map f:S 1×S 1Cf \;\colon\; S^1 \times S^1 \longrightarrow C

    such that

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

    Now the above construction should define a function:

    ϕ:KnotsHomotopyKnots. \phi \;\colon\; Knots \longrightarrow HomotopyKnots \,.

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

    • CommentRowNumber2.
    • CommentAuthorDavidRoberts
    • CommentTimeJul 5th 2019

    Better: a homotopy 1-type C=BGC=\mathbf{B}G. And we probably want some finite presentation requirements on GG, 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.

    • CommentRowNumber3.
    • CommentAuthorRichard Williamson
    • CommentTimeJul 5th 2019
    • (edited Jul 5th 2019)

    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 1S 3S^{1} \rightarrow S^{3} is not just a homotopy but a homotopy such that h(,t)h(-, t) is an embedding for all tt.

    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!

    • CommentRowNumber4.
    • CommentAuthorCharles Rezk
    • CommentTimeJul 5th 2019

    There’s a little bit more going on here: the pair (C,S 1×S 1)(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.

    • CommentRowNumber5.
    • CommentAuthorUrs
    • CommentTimeJul 5th 2019

    @David, yeah, one thing to think about is to add more conditions on ff. On the other hand, there is already the strong condition that it does form a homotopy pushout to give S 3S^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 ϕ\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?

  1. 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.

    • CommentRowNumber7.
    • CommentAuthorRichard Williamson
    • CommentTimeJul 5th 2019
    • (edited Jul 5th 2019)

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

    • CommentRowNumber8.
    • CommentAuthorEgbertRijke
    • CommentTimeJul 5th 2019

    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×S 1K(G,1)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×S 1K(G,1)S^1\times S^1 \to K(G,1)?

    I would also be curious to know whether or not the conditions in HomotopyKnot\mathbf{HomotopyKnot} already imply that the complement C is a 11-type.

    • CommentRowNumber9.
    • CommentAuthorDavidRoberts
    • CommentTimeJul 6th 2019

    I agree with Egbert: the data is not just CC, but includes the map S 1×S 1CS^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 3S^3.

    • CommentRowNumber10.
    • CommentAuthorCharles Rezk
    • CommentTimeJul 6th 2019

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

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

    • CommentRowNumber11.
    • CommentAuthorEgbertRijke
    • CommentTimeJul 6th 2019
    • (edited Jul 6th 2019)

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

    S 1 1 S 1×S 1 D S 1 1\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!

    • CommentRowNumber12.
    • CommentAuthorUrs
    • CommentTimeJul 6th 2019

    Okay, great. Thanks, Charles.

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

    • CommentRowNumber13.
    • CommentAuthorUrs
    • CommentTimeJul 6th 2019
    • (edited Jul 6th 2019)

    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 ϕ\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?

  2. 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.

    • CommentRowNumber15.
    • CommentAuthorRichard Williamson
    • CommentTimeJul 6th 2019
    • (edited Jul 6th 2019)

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

    • CommentRowNumber16.
    • CommentAuthorUrs
    • CommentTimeJul 6th 2019

    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?)