Background
Recall that a (oriented) knot is a smoothly embedded circle $S^1$ in $\mathbb R^3$, up to some natural equivalence relation (which is not quite trivial to write down). The collection of oriented knots has a binary operation called connected sum: if $K_1,K_2$ are knots, then $K_1 \\# K_2$ is formed by spatially separating the knots, then connecting them by a very thin rectangle, which is glued on so that all the orientations are correct. Connect sum is commutative and associative, making the space of knots into a commutative monoid. In fact, by a theorem of Schubert, this is the free commutative monoid on countably many generators. A ($\mathbb C$-valued) knot invariant is a $\mathbb C$-valued function on this monoid; under "pointwise" multiplication, the space of knot invariants is a commutative algebra $I$, and $\\#$ makes $I$ into a cocommutative bialgebra. I.e. $I$ is a commutative monoid object in $(\text{CAlg})^{\rm{op}}$, where $\text{CAlg}$ is the category of commutative algebras.
Warm-up question: Any knot $K$ defines an algebra morphism $I \to \mathbb C$, i.e. a global point of $I \in (\text{CAlg})^{\rm{op}}$. Are there any other global points?
Finite type invariants
Recall that a singular knot is a smooth map $S^1 \to \mathbb R^3$ with finitely many transverse self-intersections (and otherwise it is an embedding), again up to a natural equivalence. Any knot invariant extends to an invariant of singular knots, as follows: in a singular knot $K_0$, there are two ways to blow up any singularity, and the orientation determines one as the "right-handed" blow-up $K_+$ and the other as the "left-handed" blow-up $K_0$. Evaluate your knot invariant $i$ on each blow-up, and then define $i(K_0) = i(K_+) - i(K_-)$. Note that although the connect-sum of singular knots is not well-defined as a singular knot, if $i\in I$ is a knot-invariant, then it cannot distinguish different connect-sums of singular knots. Note also that the product of knot invariants (i.e. the product in the algebra $I$) is not the point-wise product on singular knots.
A Vassiliev (or finite type) invariant of type $\leq n$ is any knot invariant that vanishes on singular knots with $> n$ self-intersections. The space of all Vassiliev invariants is a filtered bialgebra $V$ (filtered by type). The corresponding associated-graded bialgebra $W$ (of "weight systems") has been well-studied (some names: Kontsevich, Bar-Natan, Vaintrob, and I'm sure there are others I haven't read yet) and in fact is more-or-less completely understood (e.g. Hinich and Vaintrob, 2002, "Cyclic operads and algebra of chord diagrams", MR1913297, where its graded dual $A$ of "chord diagrams" is described as a sort of universal enveloping algebra). In fact, this algebra $W$ is Hopf. I learned from this question that this implies that the bialgebra $V$ of Vassiliev invariants is also Hopf. Thus it is a Hopf sub-bialgebra of the algebra $I$ of knot invariants.
I believe that it is an open question whether Vassiliev invariants separate knots (i.e. whether two knots all of whose Vassiliev invariants agree are necessarily the same). But perhaps this has been answered — I feel reasonably caught-up with the state of knowledge in the mid- to late-90s, but I don't know the literature from the 00s.
Geometrically, then, $V \in (\text{CAlg})^{\rm{op}}$ is a commutative group object, and is a quotient (or something) of the monoid-object $I \in (\text{CAlg})^{\rm{op}}$ of knot invariants. The global points of $V$ (i.e. the algebra maps $V \to \mathbb C$ in $\text{CAlg}$) are a group.
Main Questions
Supposing that Vassiliev invariants separate knots, there must be global points of $V$ that do not correspond to knots, as by Mazur's swindle there are no "negative knots" among the monoid $I$. Thus my question.
Main question. What do the global points of $V$ look like?
If Vassiliev invariants do separate knots, are there still more global points of $V$ than just the free abelian group on countably many generators (i.e. the group generated by the free monoid of knots)? Yes: the singular knots. What else is there?
What can be said without knowing whether Vassiliev invariants separate knots?
