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One question if some knot theorists (or at least topologists) are reading along: I can understand why knot theorists are so interested in finding invariants. B

by wolfgke 10y ago

One question if some knot theorists (or at least topologists) are reading along: I can understand why knot theorists are so interested in finding invariants. But now let's define a "dinvariant" ("dual invariant" or "different invariant"): A dinvariant assigns to each knot also some object such that if the knots are different (or topological space are different in their class where they come from (say: are different simplical complexes or different CW complexes), the dinvariant will assign different values. On the other hand, if the knots are equivalent, the assigned values might not be equal. What I want to know is: Why doesn't there seem to exist a theory of dinvariants for knots (or topological spaces)?

nhatcher 10y ago
That is extremely difficult but that is, indeed, the goal. The main business of algebraic topology is doing exactly that.
- wolfgke 10y ago
  As far as I understand it the far goal is to find invariants that are also dinvariants. But why don't we build a systematic theory of dinvariants (similar to the theory of invariants) to get a much better understanding of them?
  - nhatcher 10y ago
    Because we can't find them :S. In the case of knots, we would rather have polynomials associated with knots that satisfy your requirement. But we just don't know how to do that. It is the same with topological spaces and homotopy theory or cohomology. Whenever somebody finds a new algebraic invariant that is capable of differentiating between simmilar structures that's a big deal. Like the Jones polynomials did (Or the Donaldson polynomials in the 80's for four manifolds)
    - wolfgke 10y ago
      > Whenever somebody finds a new algebraic invariant that is capable of differentiating between simmilar structures that's a big deal. Not every dinvariant needs to be an invariant. :-)
      - nhatcher 10y ago
        Oh! shit, I misunderstood your proposal completely. That is definitely not the goal of Algebraic Topology... Sorry about that. If you say two knots are equivalent, in mathematics, that means they are equal, one and the same. You can't have a tool that assigns different structures to the same thing. But I see where you are coming from. In mathematics we have the concept of a presentation (of a group or a vector space, for example). You can imagine a presentation of a knot like in a two dimensional representation (typical one) and assign an invariant to that. That wouldn't be an invariant of a knot, that would be an invariant of a knot, as you mention, but it would be an invariant of a knot plus some extra structure. Those objects are also studied in many instances in which the original is two difficult. As simple example imagine a knot plus an orientation. Invariants of that structure might be dinvariants of the knot.
        wolfgke 10y ago
        > If you say two knots are equivalent, in mathematics, that means they are equal, one and the same. As far as I know a knot is a continuous embedding of S^1 into R^3. Equal means exactly same subset of R^3. Equivalence of knots means that there is a suitable ambient isotopy between the nots (I just looked this up in wikipedia: https://en.wikipedia.org/wiki/Knot_theory#Knot_equivalence https://en.wikipedia.org/wiki/Knot_theory#Knot_equivalence). So equality is a much stronger criterion than equivalence. So 'equal => equivalence', but the other direction 'equivalence => equal' trivially does not hold (just apply some transition in R^3; these knots are equivalent but not equal).
        nhatcher 10y ago
        No, there is only one unknot and only one 3_1 knot, for example. Am embedding is not a knot, the knot is the embedding + the equivalence relation. So, just the embedding would be a presentation of the knot.
Varinius 10y ago
Invariants are useful because they allow you to distinguish equivalent knots, which is otherwise really hard to do. If two knots have different invariants, then they are surely different. Dually, you would want the following: if two knots have identical dinvariants, then they are surely the same. Since "if the knots are equivalent, the assigned values might not be equal", dinvariants cannot accomplish this function, and that makes them mostly useless. (full disclosure: lawyer who did his undergraduate degree in math, not a topologist)
- wolfgke 10y ago
  > Invariants are useful because they allow you to distinguish equivalent knots, which is otherwise really hard to do. If two knots have different invariants, then they are surely different. > if two knots have identical dinvariants, then they are surely the same. Since "if the knots are equivalent, the assigned values might not be equal", dinvariants cannot accomplish this function, and that makes them mostly useless. That doesn't make them useless. Dinvariants just serve a different purpose: - invariants serve the purpose of distinguishing knots that are different - dinvariants serve the purpose of detecting that knots that look very different are actually the same
tgb 10y ago
I think it comes down to the difficulty of finding such things. The goal is usually to find "enough" invariants that all objects can be distinguished, but we settle for just whatever invariants we can happen to find. Finding an invariant is pretty easy: it's straight-forward to show that something is invariant under a certain transformation since you just apply a transformation (in this case, usually a homotopy or similar) and see what it can do to the invariant. For knot theory, you can often check just what it does under the Reidemeister moves, giving you just a couple things to check. How would you go about showing that something is a dinvariant? You would need to take two objects that are distinct and show that their dinvariants are distinct. I.e. use the fact that a homotopy between the two knots does not exist. This is much more difficult, though it does get done. In particular, it's done whenever we have a complete classification of all objects of a certain type (say orientable surfaces, classified by the single invariant (and also dinvariant) their genus). Though generally here we would be actually approaching this from an invariant perspective but just showing that once you have enough invariants, they collectively become a dinvariant.

by wolfgke 10y ago

nhatcher 10y ago

That is extremely difficult but that is, indeed, the goal. The main business of algebraic topology is doing exactly that.

wolfgke 10y ago
As far as I understand it the far goal is to find invariants that are also dinvariants. But why don't we build a systematic theory of dinvariants (similar to the theory of invariants) to get a much better understanding of them?
- nhatcher 10y ago
  Because we can't find them :S. In the case of knots, we would rather have polynomials associated with knots that satisfy your requirement. But we just don't know how to do that. It is the same with topological spaces and homotopy theory or cohomology. Whenever somebody finds a new algebraic invariant that is capable of differentiating between simmilar structures that's a big deal. Like the Jones polynomials did (Or the Donaldson polynomials in the 80's for four manifolds)
  - wolfgke 10y ago
    > Whenever somebody finds a new algebraic invariant that is capable of differentiating between simmilar structures that's a big deal. Not every dinvariant needs to be an invariant. :-)
    - nhatcher 10y ago
      Oh! shit, I misunderstood your proposal completely. That is definitely not the goal of Algebraic Topology... Sorry about that. If you say two knots are equivalent, in mathematics, that means they are equal, one and the same. You can't have a tool that assigns different structures to the same thing. But I see where you are coming from. In mathematics we have the concept of a presentation (of a group or a vector space, for example). You can imagine a presentation of a knot like in a two dimensional representation (typical one) and assign an invariant to that. That wouldn't be an invariant of a knot, that would be an invariant of a knot, as you mention, but it would be an invariant of a knot plus some extra structure. Those objects are also studied in many instances in which the original is two difficult. As simple example imagine a knot plus an orientation. Invariants of that structure might be dinvariants of the knot.
      - wolfgke 10y ago
        > If you say two knots are equivalent, in mathematics, that means they are equal, one and the same. As far as I know a knot is a continuous embedding of S^1 into R^3. Equal means exactly same subset of R^3. Equivalence of knots means that there is a suitable ambient isotopy between the nots (I just looked this up in wikipedia: https://en.wikipedia.org/wiki/Knot_theory#Knot_equivalence https://en.wikipedia.org/wiki/Knot_theory#Knot_equivalence). So equality is a much stronger criterion than equivalence. So 'equal => equivalence', but the other direction 'equivalence => equal' trivially does not hold (just apply some transition in R^3; these knots are equivalent but not equal).
        nhatcher 10y ago
        No, there is only one unknot and only one 3_1 knot, for example. Am embedding is not a knot, the knot is the embedding + the equivalence relation. So, just the embedding would be a presentation of the knot.

Varinius 10y ago

Invariants are useful because they allow you to distinguish equivalent knots, which is otherwise really hard to do. If two knots have different invariants, then they are surely different. Dually, you would want the following: if two knots have identical dinvariants, then they are surely the same. Since "if the knots are equivalent, the assigned values might not be equal", dinvariants cannot accomplish this function, and that makes them mostly useless. (full disclosure: lawyer who did his undergraduate degree in math, not a topologist)

wolfgke 10y ago
> Invariants are useful because they allow you to distinguish equivalent knots, which is otherwise really hard to do. If two knots have different invariants, then they are surely different. > if two knots have identical dinvariants, then they are surely the same. Since "if the knots are equivalent, the assigned values might not be equal", dinvariants cannot accomplish this function, and that makes them mostly useless. That doesn't make them useless. Dinvariants just serve a different purpose: - invariants serve the purpose of distinguishing knots that are different - dinvariants serve the purpose of detecting that knots that look very different are actually the same

tgb 10y ago

I think it comes down to the difficulty of finding such things. The goal is usually to find "enough" invariants that all objects can be distinguished, but we settle for just whatever invariants we can happen to find. Finding an invariant is pretty easy: it's straight-forward to show that something is invariant under a certain transformation since you just apply a transformation (in this case, usually a homotopy or similar) and see what it can do to the invariant. For knot theory, you can often check just what it does under the Reidemeister moves, giving you just a couple things to check. How would you go about showing that something is a dinvariant? You would need to take two objects that are distinct and show that their dinvariants are distinct. I.e. use the fact that a homotopy between the two knots does not exist. This is much more difficult, though it does get done. In particular, it's done whenever we have a complete classification of all objects of a certain type (say orientable surfaces, classified by the single invariant (and also dinvariant) their genus). Though generally here we would be actually approaching this from an invariant perspective but just showing that once you have enough invariants, they collectively become a dinvariant.