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!

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    $\begingroup$ 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 $\endgroup$ – Adrien Feb 14 '19 at 17:06
  • $\begingroup$ 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)$\in\mathcal{K}_{n}$ ? $\endgroup$ – Meths Feb 14 '19 at 18:12
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    $\begingroup$ 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. $\endgroup$ – Adrien Feb 14 '19 at 18:15
  • $\begingroup$ 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). $\endgroup$ – Meths Feb 14 '19 at 18:28

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