Given two complex line bundles over the complex projective line ${\mathbb CP}^1$, prove or disprove that their total spaces are *homeomorphic* if and only if their Chern numbers are equal up to sign.




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This question is a generalization of [this question](https://mathoverflow.net/questions/92337/) by zygund. In my answer I recalled an idea of Wu Wen-Tsün which can be found in the introduction to his remarkable book *A theory of embedding, immersion, and isotopy of polytopes in a Euclidean space*: in order to distinguish, up to homeomorphism, two topological spaces that are homotopy equivalent (like any two lines bundles over the projective line), it is fruitful to consider the homotopy type of their [configuration spaces](http://en.wikipedia.org/wiki/Configuration_space). Wu Wen-Tsün's book shows that this is particularly useful in considering embedding problems in Euclidean spaces. 


So my **real** question is: 


Given two complex line bundles over ${\mathbb CP}^1$ such that their two-point configuration spaces are homotopy equivalent, are their Chern numbers equal up to sign?



Configuration spaces and non homeomorphic vector bundles
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 $H\_3(C\_2(E\_k)) = \mathbb{Z}\_{k/2} \oplus \mathbb{Z}\_2$ .


