Processing math: 100%
Not signed in (Sign In)

Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

  • Sign in using OpenID

Site Tag Cloud

2-category 2-category-theory abelian-categories adjoint algebra algebraic algebraic-geometry algebraic-topology analysis analytic-geometry arithmetic arithmetic-geometry book bundles calculus categorical categories category category-theory chern-weil-theory cohesion cohesive-homotopy-type-theory cohomology colimits combinatorics complex complex-geometry computable-mathematics computer-science constructive cosmology definitions deformation-theory descent diagrams differential differential-cohomology differential-equations differential-geometry digraphs duality elliptic-cohomology enriched fibration foundation foundations functional-analysis functor gauge-theory gebra geometric-quantization geometry graph graphs gravity grothendieck group group-theory harmonic-analysis higher higher-algebra higher-category-theory higher-differential-geometry higher-geometry higher-lie-theory higher-topos-theory homological homological-algebra homotopy homotopy-theory homotopy-type-theory index-theory integration integration-theory k-theory lie-theory limits linear linear-algebra locale localization logic mathematics measure-theory modal modal-logic model model-category-theory monad monads monoidal monoidal-category-theory morphism motives motivic-cohomology nlab noncommutative noncommutative-geometry number-theory object of operads operator operator-algebra order-theory pages pasting philosophy physics pro-object probability probability-theory quantization quantum quantum-field quantum-field-theory quantum-mechanics quantum-physics quantum-theory question representation representation-theory riemannian-geometry scheme schemes set set-theory sheaf simplicial space spin-geometry stable-homotopy-theory stack string string-theory superalgebra supergeometry svg symplectic-geometry synthetic-differential-geometry terminology theory topology topos topos-theory tqft type type-theory universal variational-calculus

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

Welcome to nForum
If you want to take part in these discussions either sign in now (if you have an account), apply for one now (if you don't).
    • 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 K, hence the isotopy-class of an embedding of differentiable manifolds of the circle into the 3-sphere

    K:S1AAAS3.

    With any choice of closed tubular neighbourhood embedding

    K:S1×𝔻2AAAAS3

    (where 𝔻2 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:

    S1×S1AfAS3Int(K)g(po)S1×𝔻2AKAS3

    Here f and g denote the boundary inclusions and the right vertical map is the defining inclusion of the complement in S3

    CS3Int(K)

    of the interior of 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

    S1×S1AfACpr1(hpo)S1AKAS3

    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:S1×S1C

    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:

    ϕ:KnotsHomotopyKnots.

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

    • CommentRowNumber2.
    • CommentAuthorDavidRoberts
    • CommentTimeJul 5th 2019

    Better: a homotopy 1-type C=BG. 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.

    • 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 S1S3 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!

    • CommentRowNumber4.
    • CommentAuthorCharles Rezk
    • CommentTimeJul 5th 2019

    There’s a little bit more going on here: the pair (C,S1×S1) 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 f. On the other hand, there is already the strong condition that it does form a homotopy pushout to give S3. 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 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 C 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 S1×S1K(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 S1×S1K(G,1)?

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

    • CommentRowNumber9.
    • CommentAuthorDavidRoberts
    • CommentTimeJul 6th 2019

    I agree with Egbert: the data is not just C, but includes the map S1×S1C, 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 S3.

    • CommentRowNumber10.
    • CommentAuthorCharles Rezk
    • CommentTimeJul 6th 2019

    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:S1×S1D so that the homotopy pushout of f along pr1:S1×S1S1 is contractible. Then then take C:=S3D with the map S1×S1fDC.

    I claim such a map f exists with D=S2S1. Decompose the torus as a union T1T2 of two tubes along boundary circles (so that pr1 projects each tube to a semicircle inside S1). Define f so that it sends the boundary circles to the basepoint of D, sends T1 surjectively to S2D by an obvious quotient map, and sends T2 to S1D by pr2. The fundamental group and homology groups of the pushout of f along pr1 are calculated to be trivial.

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

    Another way to see that the pushout with D is contractible, is to see that D is the mapping cone of the map S1S1×S1 which sends x to (x,*). The homotopy pushout S1S1×S1D is then contractible by the pasting lemma of pushouts:

    S11S1×S1DS11

    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 C 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 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?)