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This question is moved from math stackexchange which I posted several days ago without an answer.

Background(ignore this paragraph if you know finite type invariants well): Recall that a finite type invariant of degree $n$ is an invariant $V$ such that $V^{(n+1)}=0$ where $V^{(n+1)}$ is the extension of $V$ to $(n+1)$-singular knots. Fix a base field $\mathbb F$ of characteristc $0$. Let $\mathcal V_n$ be the space of type $n$ invariants and $\mathcal K_n$ be the linear space spanned by the complete resolution of $n$-singular knots(these are commone zeros of type $n-1$ invariants), and let $\mathcal V$ and $\mathcal K$ be the corresponding filtered space of finite type invariants and knots. The relations not distinguished by the whole finite type invariants $\mathcal V=\bigcup\mathcal V_n$ is $\bigcap\mathcal K_n=\text{ker} (\mathcal K\to\hat{\mathcal K})$, and Vassiliev conjecture is that $\mathcal K\to\hat{\mathcal K}$ is injective. Note that every finite type invariant factors through the Kontsevich integral $Z:\mathcal K\to\hat{\mathcal A}$, which takes value in the completion of space of chord diagrams mod 1T/4T relations, by a weight system(a function on $\hat{\mathcal A}$ that vanishes on $\bigoplus_{k\geq n}\mathcal A_k$ for some $n$), so the whole finite type invariants are equally powerful as $Z$, and we can show that $\hat Z:\hat{\mathcal K}\cong\hat{\mathcal A}$(because you can show that $Z:\mathcal K/\mathcal K_{n+1}\cong\bigoplus_{k\leq n}\mathcal A_k$), and Vassiliev conjecture is that $Z$ injective.

My question: What is the relation of Vassiliev's conjecture with the following conditions(at least when $\mathbb F=\mathbb R$ so that the limit is defined):

(1) Every invariant factors through a universal finite type invariant, say $\mathcal K\to\hat{\mathcal K}$. As a result, we can write $V$ as $f(\sum_{n}V_n)$, with $f$ a linear functional on $\hat\bigoplus\mathcal V_n$ and $\sum_{n}V_n\in \hat\bigoplus\mathcal V_n$ a formal sum.

(2)[Taylor expansion] Every invariant $V$ has an expansion $V=\sum_n V_n$ of finite type invariants for $V_n$ type $n$.

(3)Every invariant $V$ can be wrtten as $V=\lim_{n\to \infty} V_n$ for $V_n$ type $n$.

(4)[Stone-Weierstrass]Every invariant $V$ can be approximated by finite type invariants pointwise, namely $V=\lim_{n\to \infty} V_n$ for finite type $V_n$(not necassarily type $n$)

(Rmk) I do think Vassiliev conjecture is equivalent to (1), and I don't think it implies (2),(3),(4) even if an invariant $V$ factors through $Z:\mathcal K\to \hat{\mathcal K}=\hat {\mathcal A}=\prod_n \mathcal{A}_n$, here is an annoying problem of convergence. I have no idea of wether (2),(3),(4) implies Vassiliev's conjecture or not.

I doubt (2),(3),(4), might be false: are there known examples of an invariant that cannot expand as power series of finite type invariants or as a limit of finite type invariants? (It's known that coefficients of Jones polynomial satisfies(3),hence (4))

Update This paper shows that (4) is equivalent to the conjecture that finite type invariants separate knots.

This question is moved from math stackexchange which I posted several days ago without an answer.

Background(ignore this paragraph if you know finite type invariants well): Recall that a finite type invariant of degree $n$ is an invariant $V$ such that $V^{(n+1)}=0$ where $V^{(n+1)}$ is the extension of $V$ to $(n+1)$-singular knots. Fix a base field $\mathbb F$ of characteristc $0$. Let $\mathcal V_n$ be the space of type $n$ invariants and $\mathcal K_n$ be the linear space spanned by the complete resolution of $n$-singular knots(these are commone zeros of type $n-1$ invariants), and let $\mathcal V$ and $\mathcal K$ be the corresponding filtered space of finite type invariants and knots. The relations not distinguished by the whole finite type invariants $\mathcal V=\bigcup\mathcal V_n$ is $\bigcap\mathcal K_n=\text{ker} (\mathcal K\to\hat{\mathcal K})$, and Vassiliev conjecture is that $\mathcal K\to\hat{\mathcal K}$ is injective. Note that every finite type invariant factors through the Kontsevich integral $Z:\mathcal K\to\hat{\mathcal A}$, which takes value in the completion of space of chord diagrams mod 1T/4T relations, by a weight system(a function on $\hat{\mathcal A}$ that vanishes on $\bigoplus_{k\geq n}\mathcal A_k$ for some $n$), so the whole finite type invariants are equally powerful as $Z$, and we can show that $\hat Z:\hat{\mathcal K}\cong\hat{\mathcal A}$(because you can show that $Z:\mathcal K/\mathcal K_{n+1}\cong\bigoplus_{k\leq n}\mathcal A_k$), and Vassiliev conjecture is that $Z$ injective.

My question: What is the relation of Vassiliev's conjecture with the following conditions(at least when $\mathbb F=\mathbb R$ so that the limit is defined):

(1) Every invariant factors through a universal finite type invariant, say $\mathcal K\to\hat{\mathcal K}$. As a result, we can write $V$ as $f(\sum_{n}V_n)$, with $f$ a linear functional on $\hat\bigoplus\mathcal V_n$ and $\sum_{n}V_n\in \hat\bigoplus\mathcal V_n$ a formal sum.

(2)[Taylor expansion] Every invariant $V$ has an expansion $V=\sum_n V_n$ of finite type invariants for $V_n$ type $n$.

(3)Every invariant $V$ can be wrtten as $V=\lim_{n\to \infty} V_n$ for $V_n$ type $n$.

(4)[Stone-Weierstrass]Every invariant $V$ can be approximated by finite type invariants pointwise, namely $V=\lim_{n\to \infty} V_n$ for finite type $V_n$(not necassarily type $n$)

(Rmk) I do think Vassiliev conjecture is equivalent to (1), and I don't think it implies (2),(3),(4) even if an invariant $V$ factors through $Z:\mathcal K\to \hat{\mathcal K}=\hat {\mathcal A}=\prod_n \mathcal{A}_n$, here is an annoying problem of convergence. I have no idea of wether (2),(3),(4) implies Vassiliev's conjecture or not.

I doubt (2),(3),might be false: are there known examples of an invariant that cannot expand as power series of finite type invariants or as a limit of finite type invariants? (It's known that coefficients of Jones polynomial satisfies(3),hence (4))

Update This paper shows that (4) is equivalent to the conjecture that finite type invariants separate knots.

This question is moved from math stackexchange which I posted several days ago without an answer.

Background(ignore this paragraph if you know finite type invariants well): Recall that a finite type invariant of degree $n$ is an invariant $V$ such that $V^{(n+1)}=0$ where $V^{(n+1)}$ is the extension of $V$ to $(n+1)$-singular knots. Fix a base field $\mathbb F$ of characteristc $0$. Let $\mathcal V_n$ be the space of type $n$ invariants and $\mathcal K_n$ be the linear space spanned by the complete resolution of $n$-singular knots(these are commone zeros of type $n-1$ invariants), and let $\mathcal V$ and $\mathcal K$ be the corresponding filtered space of finite type invariants and knots. The relations not distinguished by the whole finite type invariants $\mathcal V=\bigcup\mathcal V_n$ is $\bigcap\mathcal K_n=\text{ker} (\mathcal K\to\hat{\mathcal K})$, and Vassiliev conjecture is that $\mathcal K\to\hat{\mathcal K}$ is injective. Note that every finite type invariant factors through the Kontsevich integral $Z:\mathcal K\to\hat{\mathcal A}$, which takes value in the completion of space of chord diagrams mod 1T/4T relations, by a weight system(a function on $\hat{\mathcal A}$ that vanishes on $\bigoplus_{k\geq n}\mathcal A_k$ for some $n$), so the whole finite type invariants are equally powerful as $Z$, and we can show that $\hat Z:\hat{\mathcal K}\cong\hat{\mathcal A}$(because you can show that $Z:\mathcal K/\mathcal K_{n+1}\cong\bigoplus_{k\leq n}\mathcal A_k$), and Vassiliev conjecture is that $Z$ injective.

My question: What is the relation of Vassiliev's conjecture with the following conditions(at least when $\mathbb F=\mathbb R$ so that the limit is defined):

(1) Every invariant factors through a universal finite type invariant, say $\mathcal K\to\hat{\mathcal K}$. As a result, we can write $V$ as $f(\sum_{n}V_n)$, with $f$ a linear functional on $\hat\bigoplus\mathcal V_n$ and $\sum_{n}V_n\in \hat\bigoplus\mathcal V_n$ a formal sum.

(2)[Taylor expansion] Every invariant $V$ has an expansion $V=\sum_n V_n$ of finite type invariants for $V_n$ type $n$. This equivalent to every invariant $V$ can be wrtten as $V=\lim_{n\to \infty} V_n$ for $V_n$ type $n$.

(3)[Stone-Weierstrass]Every invariant $V$ can be approximated by finite type invariants pointwise, namely $V=\lim_{n\to \infty} V_n$ for finite type $V_n$(not necassarily type $n$)

(Rmk) I do think Vassiliev conjecture is equivalent to (1), and I don't think it implies (2),(3), even if an invariant $V$ factors through $Z:\mathcal K\to \hat{\mathcal K}=\hat {\mathcal A}=\prod_n \mathcal{A}_n$, here is an annoying problem of convergence. I have no idea of wether (2),(3) implies Vassiliev's conjecture or not.

I doubt (2),(3), might be false: are there known examples of an invariant that cannot expand as power series of finite type invariants or as a limit of finite type invariants? (It's known that coefficients of Jones polynomial satisfies(2),hence (3))

Update This paper shows that (3) is equivalent to the conjecture that finite type invariants separate knots.

This question is moved from math stackexchange which I posted several days ago without an answer.

Background(ignore this paragraph if you know finite type invariants well): Recall that a finite type invariant of degree $n$ is an invariant $V$ such that $V^{(n+1)}=0$ where $V^{(n+1)}$ is the extension of $V$ to $(n+1)$-singular knots. Fix a base field $\mathbb F$ of characteristc $0$. Let $\mathcal V_n$ be the space of type $n$ invariants and $\mathcal K_n$ be the linear space spanned by the complete resolution of $n$-singular knots(these are commone zeros of type $n-1$ invariants), and let $\mathcal V$ and $\mathcal K$ be the corresponding filtered space of finite type invariants and knots. The relations not distinguished by the whole finite type invariants $\mathcal V=\bigcup\mathcal V_n$ is $\bigcap\mathcal K_n=\text{ker} (\mathcal K\to\hat{\mathcal K})$, and Vassiliev conjecture is that $\mathcal K\to\hat{\mathcal K}$ is injective. Note that every finite type invariant factors through the Kontsevich integral $Z:\mathcal K\to\hat{\mathcal A}$, which takes value in the completion of space of chord diagrams mod 1T/4T relations, by a weight system(a function on $\hat{\mathcal A}$ that vanishes on $\bigoplus_{k\geq n}\mathcal A_k$ for some $n$), so the whole finite type invariants are equally powerful as $Z$, and we can show that $\hat Z:\hat{\mathcal K}\cong\hat{\mathcal A}$(because you can show that $Z:\mathcal K/\mathcal K_{n+1}\cong\bigoplus_{k\leq n}\mathcal A_k$), and Vassiliev conjecture is that $Z$ injective.

My question: What is the relation of Vassiliev's conjecture with the following conditions(at least when $\mathbb F=\mathbb R$ so that the limit is defined):

(1) Every invariant factors through a universal finite type invariant, say $\mathcal K\to\hat{\mathcal K}$. As a result, we can write $V$ as $f(\sum_{n}V_n)$, with $f$ a linear functional on $\hat\bigoplus\mathcal V_n$ and $\sum_{n}V_n\in \hat\bigoplus\mathcal V_n$ a formal sum.

(2)[Taylor expansion] Every invariant $V$ has an expansion $V=\sum_n V_n$ of finite type invariants for $V_n$ type $n$.

(3)Every invariant $V$ can be wrtten as $V=\lim_{n\to \infty} V_n$ for $V_n$ type $n$.

(4)[Stone-Weierstrass]Every invariant $V$ can be approximated by finite type invariants pointwise, namely $V=\lim_{n\to \infty} V_n$ for finite type $V_n$(not necassarily type $n$)

(Rmk) I do think Vassiliev conjecture is equivalent to (1), and I don't think it implies (2),(3),(4) even if an invariant $V$ factors through $Z:\mathcal K\to \hat{\mathcal K}=\hat {\mathcal A}=\prod_n \mathcal{A}_n$, here is an annoying problem of convergence. I have no idea of wether (2),(3),(4) implies Vassiliev's conjecture or not.

I doubt (2),(3),(4), might be false: are there known examples of an invariant that cannot expand as power series of finite type invariants or as a limit of finite type invariants? (It's known that coefficients of Jones polynomial satisfies(3),hence (4))

Update This paper shows that (4) is equivalent to the conjecture that finite type invariants separate knots.

This question is moved from math stackexchange which I posted several days ago without an answer.

Background(ignore this paragraph if you know finite type invariants well): Recall that a finite type invariant of degree $n$ is an invariant $V$ such that $V^{(n+1)}=0$ where $V^{(n+1)}$ is the extension of $V$ to $(n+1)$-singular knots. Fix a base field $\mathbb F$ of characteristc $0$. Let $\mathcal V_n$ be the space of type $n$ invariants and $\mathcal K_n$ be the linear space spanned by the complete resolution of $n$-singular knots(these are commone zeros of type $n-1$ invariants), and let $\mathcal V$ and $\mathcal K$ be the corresponding filtered space of finite type invariants and knots. The relations not distinguished by the whole finite type invariants $\mathcal V=\bigcup\mathcal V_n$ is $\bigcap\mathcal K_n=\text{ker} (\mathcal K\to\hat{\mathcal K})$, and Vassiliev conjecture is that $\mathcal K\to\hat{\mathcal K}$ is injective. Note that every finite type invariant factors through the Kontsevich integral $Z:\mathcal K\to\hat{\mathcal A}$, which takes value in the completion of space of chord diagrams mod 1T/4T relations, by a weight system(a function on $\hat{\mathcal A}$ that vanishes on $\bigoplus_{k\geq n}\mathcal A_k$ for some $n$), so the whole finite type invariants are equally powerful as $Z$, and we can show that $\hat Z:\hat{\mathcal K}\cong\hat{\mathcal A}$(because you can show that $Z:\mathcal K/\mathcal K_{n+1}\cong\bigoplus_{k\leq n}\mathcal A_k$), and Vassiliev conjecture is that $Z$ injective.

My question: What is the relation of Vassiliev's conjecture with the following conditions(at least when $\mathbb F=\mathbb R$ so that the limit is defined):

(1) Every invariant factors through a universal finite type invariant, say $\mathcal K\to\hat{\mathcal K}$. As a result, we can write $V$ as $f(\sum_{n}V_n)$, with $f$ a linear functional on $\hat\bigoplus\mathcal V_n$ and $\sum_{n}V_n\in \hat\bigoplus\mathcal V_n$ a formal sum.

(2)[Taylor expansion] Every invariant $V$ has an expansion $V=\sum_n V_n$ of finite type invariants for $V_n$ type $n$. This equivalent to every invariant $V$ can be wrtten as $V=\lim_{n\to \infty} V_n$ for $V_n$ type $n$.

(3)[Stone-Weierstrass]Every invariant $V$ can be approximated by finite type invariants pointwise, namely $V=\lim_{n\to \infty} V_n$ for finite type $V_n$(not necassarily type $n$)

(Rmk) I do think Vassiliev conjecture is equivalent to (1), and I don't think it implies (2),(3), even if an invariant $V$ factors through $Z:\mathcal K\to \hat{\mathcal K}=\hat {\mathcal A}=\prod_n \mathcal{A}_n$, here is an annoying problem of convergence. I have no idea of wether (2),(3) implies Vassiliev's conjecture or not.

I doubt (2),(3), might be false: are there known examples of an invariant that cannot expand as power series of finite type invariants or as a limit of finite type invariants? (It's known that coefficients of Jones polynomial satisfies(2),hence (3))

This question is moved from math stackexchange which I posted several days ago without an answer.

Background(ignore this paragraph if you know finite type invariants well): Recall that a finite type invariant of degree $n$ is an invariant $V$ such that $V^{(n+1)}=0$ where $V^{(n+1)}$ is the extension of $V$ to $(n+1)$-singular knots. Fix a base field $\mathbb F$ of characteristc $0$. Let $\mathcal V_n$ be the space of type $n$ invariants and $\mathcal K_n$ be the linear space spanned by the complete resolution of $n$-singular knots(these are commone zeros of type $n-1$ invariants), and let $\mathcal V$ and $\mathcal K$ be the corresponding filtered space of finite type invariants and knots. The relations not distinguished by the whole finite type invariants $\mathcal V=\bigcup\mathcal V_n$ is $\bigcap\mathcal K_n=\text{ker} (\mathcal K\to\hat{\mathcal K})$, and Vassiliev conjecture is that $\mathcal K\to\hat{\mathcal K}$ is injective. Note that every finite type invariant factors through the Kontsevich integral $Z:\mathcal K\to\hat{\mathcal A}$, which takes value in the completion of space of chord diagrams mod 1T/4T relations, by a weight system(a function on $\hat{\mathcal A}$ that vanishes on $\bigoplus_{k\geq n}\mathcal A_k$ for some $n$), so the whole finite type invariants are equally powerful as $Z$, and we can show that $\hat Z:\hat{\mathcal K}\cong\hat{\mathcal A}$(because you can show that $Z:\mathcal K/\mathcal K_{n+1}\cong\bigoplus_{k\leq n}\mathcal A_k$), and Vassiliev conjecture is that $Z$ injective.

My question: What is the relation of Vassiliev's conjecture with the following conditions(at least when $\mathbb F=\mathbb R$ so that the limit is defined):

(1) Every invariant factors through a universal finite type invariant, say $\mathcal K\to\hat{\mathcal K}$. As a result, we can write $V$ as $f(\sum_{n}V_n)$, with $f$ a linear functional on $\hat\bigoplus\mathcal V_n$ and $\sum_{n}V_n\in \hat\bigoplus\mathcal V_n$ a formal sum.

(2)[Taylor expansion] Every invariant $V$ has an expansion $V=\sum_n V_n$ of finite type invariants for $V_n$ type $n$. This equivalent to every invariant $V$ can be wrtten as $V=\lim_{n\to \infty} V_n$ for $V_n$ type $n$.

(3)[Stone-Weierstrass]Every invariant $V$ can be approximated by finite type invariants pointwise, namely $V=\lim_{n\to \infty} V_n$ for finite type $V_n$(not necassarily type $n$)

(Rmk) I do think Vassiliev conjecture is equivalent to (1), and I don't think it implies (2),(3), even if an invariant $V$ factors through $Z:\mathcal K\to \hat{\mathcal K}=\hat {\mathcal A}=\prod_n \mathcal{A}_n$, here is an annoying problem of convergence. I have no idea of wether (2),(3) implies Vassiliev's conjecture or not.

I doubt (2),(3), might be false: are there known examples of an invariant that cannot expand as power series of finite type invariants or as a limit of finite type invariants? (It's known that coefficients of Jones polynomial satisfies(2),hence (3))

Update This paper shows that (3) is equivalent to the conjecture that finite type invariants separate knots.