# Subalgebras and ideals of algebra of Vassiliev invariants

Let $$\mathcal{K}$$ be the commutative monoid whose elements are (isotopy) equivalence classes of knots with composition under the connected knot sum, and $$\mathbb{Z}\mathcal{K}$$ be the corresponding monoid algebra. Consider the 'tautological knot invariant' which sends a knot to itself, and whose extension under the Vassiliev skein relation acts as a linear function $$\mathcal{K}\rightarrow\mathbb{Z}\mathcal{K}$$.

Chmutov (Introduction to Vassiliev knot invariants) claims that,

(i) The images of of knots with $$n$$ double points span a subalgebra $$\mathcal{K}_{n}$$ of $$\mathbb{Z}\mathcal{K}$$

(ii) $$\mathcal{K}_{n}$$ is an ideal of $$\mathbb{Z}\mathcal{K}$$

Could anybody shed some light on how this is so? I can't immediately see how the required properties are preserved under the connected knot sum. Thanks!

• This is definitely not a subalgebra (and I don't think he claims it is), but this is fairly easy to see it's an ideal, basically because the connected sum of a knot with $n$ singularities, with a non-singular ordinary knot, is again a knot with $n$ singularities – Adrien Feb 14 '19 at 17:06
• He claims the subalgebra structure on p.75 immediately after the definition of a Goussarov filtration. Also I understand your argument, but surely it has to hold at the level of: (non-singular ordinary knot) # (the image of a knot with $n$ singularities)$\in\mathcal{K}_{n}$ ? – Meths Feb 14 '19 at 18:12
• I think he means subalgebra as in "closed under multiplication", ie non unital, which an ideal always is. And sure, you're right, that was just a short version, basically you can make a connected sum of an ordinary knot and a singular one, which will depends on the point you choose on the singular knot, but its image won't, and this operation does indeed gives a singular knot with the same number of singularities. – Adrien Feb 14 '19 at 18:15
• Okay, I think I've got it: let $k^{(n)}$ be a knot with $n$ singularities and $v$ be the tautological invariant extended via the Vassiliev skein relation. Then $v(k^{(n)}\#k^{(0)})=v(\tilde{k}^{(n)})$, but also $v(k^{(n)}\#k^{(0)})=v(k^{(n)})\#k^{(0)}$ (the last bit following by thinking about how v acts on that connected sum a little bit). – Meths Feb 14 '19 at 18:28