There are various (equivalent?) descriptions of a universal finite-type knot invariant, e.g. https://arxiv.org/abs/q-alg/9603010. They take the form of formal power series valued in Feynman diagrams (certain oriented decorated trivalent graphs). Each such Feynman diagram corresponds to a configuration space integral, and in particular is a function on the space of knots. In https://arxiv.org/abs/q-alg/9603010 these functions are denoted $I(\Gamma)$.

To evaluate the universal finite-type invariant on a knot we need to understand the values of $I(\Gamma)(K)$ on some particular knot $K$ for a Feynman diagram $\Gamma$. Is there a combinatorial way to do this? I am thinking of this process as ``expanding'' $K$ into a formal sum of Feynman diagrams: the coefficient $I(\Gamma)(K)$ tells me how many times $\Gamma$ appears. (Actually, we have to divide by an automorphism factor.)

This seems like it should be possible, because I have a different way of ``expanding'' $K$. (Well, I'm not quite sure this works, but I suppose that's part of the question.) By using the Vassiliev skein relation, we can expand $K$ as a sum of singular knot diagrams (with double points). These correspond to chord diagrams, which in turn correspond to Feynman diagrams via the STU relation. If we do this for long enough, we should be able to write $K$ as a sum of Feynman diagrams, at least up to some fixed order $n$. The coefficient of $\Gamma$ in this expansion should be something like the value of $I(\Gamma)(K)$.

Can you turn the approach of the third paragraph into a way to compute $I(\Gamma)(K)$? If not, is there a way to compute $I(\Gamma)(K)$ that avoids actually working with configuration space integrals?

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    $\begingroup$ I think Drinfeld has an approach to the Kontsevich integral which is purely algebraic. See the discussion in section 4 of these notes: arxiv.org/abs/q-alg/9702009 $\endgroup$
    – Ian Agol
    May 6, 2021 at 17:10
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    $\begingroup$ Your question could be viewed as a little ill-defined. For example, I could take the perspective that "the" universal finite type invariant is the map on path-components for the Taylor tower from Embedding Calculus. From this perspective it is very computable, and fairly canonical. When you start constructing spanning sets for the invariants of a given type and start talking about relations, then it becomes more complicated. But you don't need to construct a basis for $\mathbb R$ over $\mathbb Q$ to "compute" a real-valued function. $\endgroup$ May 6, 2021 at 20:27

1 Answer 1


First of all, however your universal finite type invariant $Z$ is given the question of computing the coefficient of a given diagram is somewhat ill-defined, since those diagrams are not linearly independent. However it might be that $Z$ is given as a sum over all diagrams weighted by some coefficients given by a specific formula, and then you might want to compute that specific formula combinatorially. In the case you're talking about (or maybe a slight variant thereof), this is done I believe in Dylan Thurston, Integral Expressions for the Vassiliev Knot Invariants (https://arxiv.org/abs/math/9901110).

As pointed out in the comments, there is another construction of such an invariant, known as the Kontsevich integral, which can be computed combinatorially using a Drinfeld associator. This a certain formal power series in two non commuting variables. In this approach it doesn't quite makes sense to compute the coefficient of a single diagram; Rather, there is an algorithm to compute a Drinfeld associator up to a given degree $n$, and then you can compute the invariant associated with any knot up to chord diagrams having at most $n$ chords. Again what you get is a linear combination of diagrams which are not linearly independent so there is no canonically unique way to write down the result.

As for your third paragraph, unfortunately this is just now how this works: there is no unique way to write a know as a combination of singular not, and there is definitely no tautological map from singular knots to chord diagrams. Rather, there is a canonical map from diagrams with $n$ chords, to knots with $n$ singularities modulo the space of knots with $n+1$ singularities. This maps happens to be an isomorphism and a universal invariant is basically an inverse of that, but this is definitely something complicated however you choose to construct it.

  • $\begingroup$ Thanks for your answer, and especially the clarification about singular knots and chord diagrams. $\endgroup$ May 7, 2021 at 14:09
  • $\begingroup$ Section 4.5 of Dylan Thurston's undergrad thesis (linked) mentions a "combinatorial formula" which will be "expanded in a future paper". Do you know if that paper was ever written? $\endgroup$ May 7, 2021 at 14:10
  • $\begingroup$ I'm not sure tbh, I would suespect there are other references giving this kind of combinatorial construction btu I can't remember any (I'm more familiar with the Kontsevich integral side of the story). Also, I should have said it's still not quite known whether these two invariants (the one coming from configuration space integrals, and the Kontsevich integral) are equal, see eg arxiv.org/abs/math/0004094 $\endgroup$
    – Adrien
    May 7, 2021 at 15:46

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