3 ms·

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.

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.