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The answer is yes. The third homology group distinguishes these spaces, namely $H_3(C_2(E_k))=\mathbb{Z}_k$ (up to extension).

This shows that configuration spaces of homotopic open manifolds are easier to tell apart than configuration spaces of homotopic closed manifolds. The latter have the same additive homology, and one needs finer invariants, like Massey products, to tell them apart.

Here is the argument: The space $C_2(E_k)$ has the homotopy type of a pushout $$A_1 \leftarrow A_{12} \rightarrow A_2$$ The space $A_1$ corresponds to pairs $(x,y) \in C_2(E_k)$ such that $\pi(x)$ and $\pi(y)$ are close to each other in $S^2$ (say $<\pi(x),\pi(y)> \geq 0$), and fibers are distinct (use parallel transport). Up to homotopy $A_1$ is the fiberwise configuration space, a bundle $A_1 \to S^2$ with fiber $F_2(\mathbb{R}^2) \simeq S^1$, that is the sphere bundle of $E_k$. The space $A_{12}$ corresponds to the case $<\pi(x),\pi(y)>=0$ and is the pullback of $A_1$ along the projection $SO(3) \to S^2$. Finally $A_2$ corresponds to the case when $\pi(x)$ and $\pi(y)$ are close to antipodal (negative scalar product), and is homotopically a bundle $A_2 \to S^2$ with fiber $(\mathbb{R}^2)^2$, so $A_2 \simeq S^2$. The bundle $A_1$ has clutching function $S^1 \times S^1 \to S^1$ $(z,u) \mapsto z^ku$ and $A_{12}$ has clutching function $S^1 \times (S^1 \times S^1) \to S^1 \times S^1 $ $(z,u,v) \mapsto (z^k u, z^2 v)$. Apply the Mayer Vietoris sequence to compute the homology of $A_1, A_{12}$, and the induced map. Then $H_1(A_1)=\mathbb{Z}/k, \, H_2(A_1)=0,\, H_3(A_1)=\mathbb{Z}$; in the odd case $H_1(A_{12})=\mathbb{Z}, \, H_2(A_{12})=0, \, H_3(A_{12})=\mathbb{Z}, \, H_4(A_{12})=\mathbb{Z} .$ In the even case $H_1(A_{12})$ and $H_2(A_{12})$ have an extra $\mathbb{Z}/2$ summand. The map $A_{12} \to A_1$ induces multiplication by $k$ (odd) or $k/2$ (even) on $H_3$. Apply again Mayer Vietoris to compute the homology of $C_2(E_k)$.

This gives the answer. In the even case either $H_3(C_2(E_k))=\mathbb{Z}_k$ or $H_3(C_2(E_k)) = \mathbb{Z}_{k/2} \oplus \mathbb{Z}_2$ .

The answer is yes. The third homology group distinguishes these spaces, namely $H_3(C_2(E_k))=\mathbb{Z}_k$ (up to extension).

This shows that configuration spaces of homotopic open manifolds are easier to tell apart than configuration spaces of homotopic closed manifolds. The latter have the same additive homology, and one needs finer invariants, like Massey products, to tell them apart.

Here is the argument: The space $C_2(E_k)$ has the homotopy type of a pushout $$A_1 \leftarrow A_{12} \rightarrow A_2$$ The space $A_1$ corresponds to pairs $(x,y) \in C_2(E_k)$ such that $\pi(x)$ and $\pi(y)$ are close to each other in $S^2$ (say $<\pi(x),\pi(y)> \geq 0$), and fibers are distinct (use parallel transport). Up to homotopy $A_1$ is the fiberwise configuration space, a bundle $A_1 \to S^2$ with fiber $C_2(\mathbb{R}^2) \simeq S^1$, that is the sphere bundle of $E_k$. The space $A_{12}$ corresponds to the case $<\pi(x),\pi(y)>=0$ and is the pullback of $A_1$ along the projection $SO(3) \to S^2$. Finally $A_2$ corresponds to the case when $\pi(x)$ and $\pi(y)$ are close to antipodal (negative scalar product), and is homotopically a bundle $A_2 \to S^2$ with fiber $(\mathbb{R}^2)^2$, so $A_2 \simeq S^2$. The bundle $A_1$ has clutching function $S^1 \times S^1 \to S^1$ $(z,u) \mapsto z^ku$ and $A_{12}$ has clutching function $S^1 \times (S^1 \times S^1) \to S^1 \times S^1 $ $(z,u,v) \mapsto (z^k u, z^2 v)$. Apply the Mayer Vietoris sequence to compute the homology of $A_1, A_{12}$, and the induced map. Then $H_1(A_1)=\mathbb{Z}/k, \, H_2(A_1)=0,\, H_3(A_1)=\mathbb{Z}$; in the odd case $H_1(A_{12})=\mathbb{Z}, \, H_2(A_{12})=0, \, H_3(A_{12})=\mathbb{Z}, \, H_4(A_{12})=\mathbb{Z} .$ In the even case $H_1(A_{12})$ and $H_2(A_{12})$ have an extra $\mathbb{Z}/2$ summand. The map $A_{12} \to A_1$ induces multiplication by $k$ (odd) or $k/2$ (even) on $H_3$. Apply again Mayer Vietoris to compute the homology of $C_2(E_k)$.

This gives the answer. In the even case either $H_3(C_2(E_k))=\mathbb{Z}_k$ or $H_3(C_2(E_k)) = \mathbb{Z}_{k/2} \oplus \mathbb{Z}_2$ .

The answer is yes. The third homology group distinguishes these spaces, namely $H_3(C_2(E_k))=\mathbb{Z}_k$ (up to extension).

This shows that configuration spaces of homotopic open manifolds are easier to tell apart than configuration spaces of homotopic closed manifolds. The latter have the same additive homology, and one needs finer invariants, like Massey products, to tell them apart.

Here is the argument: The space $C_2(E_k)$ has the homotopy type of a pushout $$A_1 \leftarrow A_{12} \rightarrow A_2$$ The space $A_1$ corresponds to pairs $(x,y) \in C_2(E_k)$ such that $\pi(x)$ and $\pi(y)$ are close to each other in $S^2$ (say $<\pi(x),\pi(y)> \geq 0$), and fibers are distinct (use parallel transport). Up to homotopy $A_1$ is the fiberwise configuration space, a bundle $A_1 \to S^2$ with fiber $F_2(\mathbb{R}^2) \simeq S^1$, that is the sphere bundle of $E_k$. The space $A_{12}$ corresponds to the case $<\pi(x),\pi(y)>=0$ and is the pullback of $A_1$ along the projection $SO(3) \to S^2$. Finally $A_2$ corresponds to the case when $\pi(x)$ and $\pi(y)$ are close to antipodal (negative scalar product), and is homotopically a bundle $A_2 \to S^2$ with fiber $(\mathbb{R}^2)^2$, so $A_2 \simeq S^2$. The bundle $A_1$ has clutching function $S^1 \times S^1 \to S^1$ $(z,u) \mapsto z^ku$ and $A_{12}$ has clutching function $S^1 \times (S^1 \times S^1) \to S^1 \times S^1 $ $(z,u,v) \mapsto (z^k u, z^2 v)$. Apply the Mayer Vietoris sequence to compute the homology of $A_1, A_{12}$, and the induced map. Then $H_1(A_1)=\mathbb{Z}/k, \, H_2(A_1)=0,\, H_3(A_1)=\mathbb{Z}$; in the odd case $H_1(A_{12})=\mathbb{Z}, \, H_2(A_{12})=0, \, H_3(A_{12})=\mathbb{Z}, \, H_4(A_{12})=\mathbb{Z} .$ In the even case $H_1(A_{12})$ and $H_2(A_{12})$ have an extra $\mathbb{Z}/2$ summand. The map $A_{12} \to A_1$ induces multiplication by $k$ (odd) or $k/2$ (even) on $H_3$. Apply again Mayer Vietoris to compute the homology of $E_k$.

This gives the answer. In the even case either $H_3(C_2(E_k))=\mathbb{Z}_k$ or $H_3(C_2(E_k)) = \mathbb{Z}_{k/2} \oplus \mathbb{Z}_2$ .

The answer is yes. The third homology group distinguishes these spaces, namely $H_3(C_2(E_k))=\mathbb{Z}_k$ (up to extension).

This shows that configuration spaces of homotopic open manifolds are easier to tell apart than configuration spaces of homotopic closed manifolds. The latter have the same additive homology, and one needs finer invariants, like Massey products, to tell them apart.

Here is the argument: The space $C_2(E_k)$ has the homotopy type of a pushout $$A_1 \leftarrow A_{12} \rightarrow A_2$$ The space $A_1$ corresponds to pairs $(x,y) \in C_2(E_k)$ such that $\pi(x)$ and $\pi(y)$ are close to each other in $S^2$ (say $<\pi(x),\pi(y)> \geq 0$), and fibers are distinct (use parallel transport). Up to homotopy $A_1$ is the fiberwise configuration space, a bundle $A_1 \to S^2$ with fiber $F_2(\mathbb{R}^2) \simeq S^1$, that is the sphere bundle of $E_k$. The space $A_{12}$ corresponds to the case $<\pi(x),\pi(y)>=0$ and is the pullback of $A_1$ along the projection $SO(3) \to S^2$. Finally $A_2$ corresponds to the case when $\pi(x)$ and $\pi(y)$ are close to antipodal (negative scalar product), and is homotopically a bundle $A_2 \to S^2$ with fiber $(\mathbb{R}^2)^2$, so $A_2 \simeq S^2$. The bundle $A_1$ has clutching function $S^1 \times S^1 \to S^1$ $(z,u) \mapsto z^ku$ and $A_{12}$ has clutching function $S^1 \times (S^1 \times S^1) \to S^1 \times S^1 $ $(z,u,v) \mapsto (z^k u, z^2 v)$. Apply the Mayer Vietoris sequence to compute the homology of $A_1, A_{12}$, and the induced map. Then $H_1(A_1)=\mathbb{Z}/k, \, H_2(A_1)=0,\, H_3(A_1)=\mathbb{Z}$; in the odd case $H_1(A_{12})=\mathbb{Z}, \, H_2(A_{12})=0, \, H_3(A_{12})=\mathbb{Z}, \, H_4(A_{12})=\mathbb{Z} .$ In the even case $H_1(A_{12})$ and $H_2(A_{12})$ have an extra $\mathbb{Z}/2$ summand. The map $A_{12} \to A_1$ induces multiplication by $k$ (odd) or $k/2$ (even) on $H_3$. Apply again Mayer Vietoris to compute the homology of $C_2(E_k)$.

This gives the answer. In the even case either $H_3(C_2(E_k))=\mathbb{Z}_k$ or $H_3(C_2(E_k)) = \mathbb{Z}_{k/2} \oplus \mathbb{Z}_2$ .

The answer is yes. The third homology group distinguishes these spaces, namely $H_3(C_2(E_k))=\mathbb{Z}_k$ (up to extension).

This shows that configuration spaces of homotopic open manifolds are easier to tell apart than configuration spaces of homotopic closed manifolds. The latter have the same additive homology, and one needs finer invariants, like Massey products, to tell them apart.

Here is the argument: The space $C_2(E_k)$ has the homotopy type of a pushout $$A_1 \leftarrow A_{12} \rightarrow A_2$$ The space $A_1$ corresponds to pairs $(x,y) \in C_2(E_k)$ such that $\pi(x)$ and $\pi(y)$ are close to each other in $S^2$ (say $<\pi(x),\pi(y)> \geq 0$), and fibers are distinct (use parallel transport). Up to homotopy $A_1$ is the fiberwise configuration space, a bundle $A_1 \to S^2$ with fiber $F_2(\mathbb{R}^2) \simeq S^1$, that is the sphere bundle of $E_k$. The space $A_{12}$ corresponds to the case $<\pi(x),\pi(y)>=0$ and is the pullback of $A_1$ along the projection $SO(3) \to S^2$. Finally $A_2$ corresponds to the case when $\pi(x)$ and $\pi(y)$ are close to antipodal (negative scalar product), and is homotopically a bundle $A_2 \to S^2$ with fiber $(\mathbb{R}^2)^2$, so $A_2 \simeq S^2$. The bundle $A_1$ has clutching function $S^1 \times S^1 \to S^1$ $(z,u) \mapsto zu^k$ and $A_{12}$ has clutching function $S^1 \times (S^1 \times S^1) \to S^1 \times S^1 $ $(z,u,v) \mapsto (z^k u, z^2 v)$. Apply the Mayer Vietoris sequence to compute the homology of $A_1, A_{12}$, and the induced map. Then $H_1(A_1)=\mathbb{Z}/k, \, H_2(A_1)=0,\, H_3(A_1)=\mathbb{Z}$; in the odd case $H_1(A_{12})=\mathbb{Z}, \, H_2(A_{12})=0, \, H_3(A_{12})=\mathbb{Z}, \, H_4(A_{12})=\mathbb{Z} .$ In the even case $H_1(A_{12})$ and $H_2(A_{12})$ have an extra $\mathbb{Z}/2$ summand. The map $A_{12} \to A_1$ induces multiplication by $k$ (odd) or $k/2$ (even) on $H_3$. Apply again Mayer Vietoris to compute the homology of $E_k$.

This gives the answer. In the even case either $H_3(C_2(E_k))=\mathbb{Z}_k$ or $H_3(C_2(E_k)) = \mathbb{Z}_{k/2} \oplus \mathbb{Z}_2$ .

The answer is yes. The third homology group distinguishes these spaces, namely $H_3(C_2(E_k))=\mathbb{Z}_k$ (up to extension).

This shows that configuration spaces of homotopic open manifolds are easier to tell apart than configuration spaces of homotopic closed manifolds. The latter have the same additive homology, and one needs finer invariants, like Massey products, to tell them apart.

Here is the argument: The space $C_2(E_k)$ has the homotopy type of a pushout $$A_1 \leftarrow A_{12} \rightarrow A_2$$ The space $A_1$ corresponds to pairs $(x,y) \in C_2(E_k)$ such that $\pi(x)$ and $\pi(y)$ are close to each other in $S^2$ (say $<\pi(x),\pi(y)> \geq 0$), and fibers are distinct (use parallel transport). Up to homotopy $A_1$ is the fiberwise configuration space, a bundle $A_1 \to S^2$ with fiber $F_2(\mathbb{R}^2) \simeq S^1$, that is the sphere bundle of $E_k$. The space $A_{12}$ corresponds to the case $<\pi(x),\pi(y)>=0$ and is the pullback of $A_1$ along the projection $SO(3) \to S^2$. Finally $A_2$ corresponds to the case when $\pi(x)$ and $\pi(y)$ are close to antipodal (negative scalar product), and is homotopically a bundle $A_2 \to S^2$ with fiber $(\mathbb{R}^2)^2$, so $A_2 \simeq S^2$. The bundle $A_1$ has clutching function $S^1 \times S^1 \to S^1$ $(z,u) \mapsto z^ku$ and $A_{12}$ has clutching function $S^1 \times (S^1 \times S^1) \to S^1 \times S^1 $ $(z,u,v) \mapsto (z^k u, z^2 v)$. Apply the Mayer Vietoris sequence to compute the homology of $A_1, A_{12}$, and the induced map. Then $H_1(A_1)=\mathbb{Z}/k, \, H_2(A_1)=0,\, H_3(A_1)=\mathbb{Z}$; in the odd case $H_1(A_{12})=\mathbb{Z}, \, H_2(A_{12})=0, \, H_3(A_{12})=\mathbb{Z}, \, H_4(A_{12})=\mathbb{Z} .$ In the even case $H_1(A_{12})$ and $H_2(A_{12})$ have an extra $\mathbb{Z}/2$ summand. The map $A_{12} \to A_1$ induces multiplication by $k$ (odd) or $k/2$ (even) on $H_3$. Apply again Mayer Vietoris to compute the homology of $E_k$.

This gives the answer. In the even case either $H_3(C_2(E_k))=\mathbb{Z}_k$ or $H_3(C_2(E_k)) = \mathbb{Z}_{k/2} \oplus \mathbb{Z}_2$ .