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In dimension $k=1$, i.e. for embeddings $\sqcup_k S^1\hookrightarrow S^3$, and using a totally unlinked embedding as your basepoint $e$, this is what's called the ring group or the loop group. It is closely related to the braid group and has been studied a ton, but two places with references you could follow are Brendle-Hatcher "Configuration spaces of rings and wicketsConfiguration spaces of rings and wickets" and Brownstein-Lee "Cohomology of the group of motions of n strings in 3-space".

The fundamental group $\pi_1(\text{Emb}(\sqcup_k S^1,S^3),e)$ can been identified with McCool's "symmetric automorphism group". This is all the automorphisms of a free group $\langle x_1,\ldots,x_k\rangle$ which take each generator $x_i$ to a conjugate of some generator $x_j$. (A loop around one component of the link has to go to a loop around some component of the link.)

This is the image of $\pi_0(\text{Diff}(S^3\setminus\sqcup_k S^1))$ in $\text{Aut}(\pi_1(S^3\setminus\sqcup_k S^1))$, but since $S^3\setminus\sqcup_k S^1$ is not aspherical, this doesn't give us that $\pi_1(\text{Emb}(\sqcup_k S^1,S^3),e)=\pi_0(\text{Diff}(S^3\setminus\sqcup_k S^1))$ yet. I would be glad to see an argument that a diffeomorphism acting trivially on $\text{Aut}(\pi_1(S^3\setminus\sqcup_k S^1))$ must be isotopic (or even homotopic) to the identity.

In dimension $k=1$, i.e. for embeddings $\sqcup_k S^1\hookrightarrow S^3$, and using a totally unlinked embedding as your basepoint $e$, this is what's called the ring group or the loop group. It is closely related to the braid group and has been studied a ton, but two places with references you could follow are Brendle-Hatcher "Configuration spaces of rings and wickets" and Brownstein-Lee "Cohomology of the group of motions of n strings in 3-space".

The fundamental group $\pi_1(\text{Emb}(\sqcup_k S^1,S^3),e)$ can been identified with McCool's "symmetric automorphism group". This is all the automorphisms of a free group $\langle x_1,\ldots,x_k\rangle$ which take each generator $x_i$ to a conjugate of some generator $x_j$. (A loop around one component of the link has to go to a loop around some component of the link.)

This is the image of $\pi_0(\text{Diff}(S^3\setminus\sqcup_k S^1))$ in $\text{Aut}(\pi_1(S^3\setminus\sqcup_k S^1))$, but since $S^3\setminus\sqcup_k S^1$ is not aspherical, this doesn't give us that $\pi_1(\text{Emb}(\sqcup_k S^1,S^3),e)=\pi_0(\text{Diff}(S^3\setminus\sqcup_k S^1))$ yet. I would be glad to see an argument that a diffeomorphism acting trivially on $\text{Aut}(\pi_1(S^3\setminus\sqcup_k S^1))$ must be isotopic (or even homotopic) to the identity.

In dimension $k=1$, i.e. for embeddings $\sqcup_k S^1\hookrightarrow S^3$, and using a totally unlinked embedding as your basepoint $e$, this is what's called the ring group or the loop group. It is closely related to the braid group and has been studied a ton, but two places with references you could follow are Brendle-Hatcher "Configuration spaces of rings and wickets" and Brownstein-Lee "Cohomology of the group of motions of n strings in 3-space".

The fundamental group $\pi_1(\text{Emb}(\sqcup_k S^1,S^3),e)$ can been identified with McCool's "symmetric automorphism group". This is all the automorphisms of a free group $\langle x_1,\ldots,x_k\rangle$ which take each generator $x_i$ to a conjugate of some generator $x_j$. (A loop around one component of the link has to go to a loop around some component of the link.)

This is the image of $\pi_0(\text{Diff}(S^3\setminus\sqcup_k S^1))$ in $\text{Aut}(\pi_1(S^3\setminus\sqcup_k S^1))$, but since $S^3\setminus\sqcup_k S^1$ is not aspherical, this doesn't give us that $\pi_1(\text{Emb}(\sqcup_k S^1,S^3),e)=\pi_0(\text{Diff}(S^3\setminus\sqcup_k S^1))$ yet. I would be glad to see an argument that a diffeomorphism acting trivially on $\text{Aut}(\pi_1(S^3\setminus\sqcup_k S^1))$ must be isotopic (or even homotopic) to the identity.

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Tom Church
Tom Church
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In dimension $k=1$, i.e. for embeddings $\sqcup_k S^1\hookrightarrow S^3$, and using a totally unlinked embedding as your basepoint $e$, this is what's called the ring group or the loop group. It is closely related to the braid group and has been studied a ton, but two places with references you could follow are Brendle-Hatcher "Configuration spaces of rings and wickets" and Brownstein-Lee "Cohomology of the group of motions of n strings in 3-space".

The fundamental group $\pi_1(\text{Emb}(\sqcup_k S^1,S^3),e)$ can been identified with McCool's "symmetric automorphism group". This is all the automorphisms of a free group $\langle x_1,\ldots,x_k\rangle$ which take each generator $x_i$ to a conjugate of some generator $x_j$. (A loop around one component of the link has to go to a loop around some component of the link.)

This is the image of $\pi_0(\text{Diff}(S^3\setminus\sqcup_k S^1))$ in $\text{Aut}(\pi_1(S^3\setminus\sqcup_k S^1))$, but since $S^3\setminus\sqcup_k S^1$ is not aspherical, this doesn't give us that $\pi_1(\text{Emb}(\sqcup_k S^1,S^3),e)=\pi_0(\text{Diff}(S^3\setminus\sqcup_k S^1))$ yet. I would be glad to see an argument that a diffeomorphism acting trivially on $\text{Aut}(\pi_1(S^3\setminus\sqcup_k S^1))$ must be isotopic (or even homotopic) to the identity.