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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.

This question is a generalization of this question 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. 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?

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Have you read the Salvatore-Longoni paper? front.math.ucdavis.edu/0401.5075 It's very close in spirit with your line of inquiry. –  Ryan Budney Mar 27 '12 at 20:15
Look at the intersection pairing on $H_2$ of the total space. From this you can read off the Euler number and, hence, the Chern class. –  Misha Mar 27 '12 at 21:15
I think some context is needed for the Salvatore-Longoni paper. For compact manifolds, the homotopy type of configuration spaces are homotopy invariants if the manifolds are highly connected and the number of points is small (open: just simply connected + many points), but S&L give a counterexample when the compact manifolds have fundamental group. –  Ben Wieland Mar 27 '12 at 22:53
@Ben, right, but simply-connected 4-manifolds are still below the (known) threshold for the homotopy-type of configuration spaces to be a homotopy-invariant of the input manifold. –  Ryan Budney Mar 28 '12 at 0:52
@Ben: I believe the case of simply-connected and two points is also still open. –  Mark Grant Mar 28 '12 at 15:39

2 Answers 2

up vote 16 down vote accepted

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 $H_3(C_2(E_k)) = \mathbb{Z}_{k/2} \oplus \mathbb{Z}_2$ .

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@Paolo: many thanks for this nice and detailed answer. –  alvarezpaiva Mar 29 '12 at 11:30
@ Juan-Carlos: thanks for asking this interesting question! –  Paolo Salvatore Mar 29 '12 at 19:56

This is a long comment, not an answer.

Let $E_k$ be the total space of the orientable bundle over $S^2$ with fiber $\mathbb R^2$ and Euler class $k$. $E_k = \mathbb R^2 \rtimes_k S^2$. Let $\pi : E_k \to S^2$ be the bundle projection.

$C_2 E_k = \{ (x,y) \in E_k^2 : x \neq y\}$ is the configuration space, with $p : C_2 E_k \to E_k$ the map $p(x,y)=x$.

Consider the composite of $\pi \circ p : C_2 E_k \to S^2$. It's a fibration and the fibers are homotopy-equivalent to $S^3 \vee S^2$, although that's not the most honest way of perceiving the fibers. The idea is to think of $\pi \circ p(x,y) = \pi(x)$ as a point in the $0$-section of $\pi$. If $\pi(y) \neq \pi(x)$ you homotope $y$ to $\pi(y)$, i.e. the 0-vector over $\pi(y)$. If $\pi(y) = \pi(x)$ you can't do this in $E_k$, so you can homotope $\pi(y)$ to be a unit vector in $\pi^{-1}(\pi(x))$. In other words, the fiber of $\pi \circ p$ over $\pi(x)$ looks like the sphere bundle of $\pi$ with all each circle fiber over points in a neighbourhood of $\pi(x)$ collapsed. So what's really going on is the fibers have as a deformation-retract a subspace that's $S^3$ union a $2$-cell, but the attachment map for the $2$-cell is along a great circle. The nice thing about this deformation-retract, is it's equivariant with respect to the monodromy. Precisely,

$$C_2 E_k \simeq (S^3 \cup e^2) \rtimes S^2$$

where the monodromy $SO_2 \to Aut(S^3 \cup e^2)$ is rotation about this great circle. Specifically, if you think of $S^3$ as the unit sphere in $\mathbb C^2$, and let the great circle be $S^1 \times \{0\} \subset S^3$, then its the action of $S^1$ on $S^3$ given by $(z, (z_1,z_2)) = (z_1,zz_2)$. The action is trivial on the $2$-cell attachment.

So as a space, it's $S^3 \rtimes S^2$ union a $D^2 \times S^2$ attached along the $S^1 \times S^2 \subset S^3 \rtimes S^2$ corresponding to where the monodromy is trivial.

I think the attaching map is null-homologous in $H_* (S^3 \rtimes_k S^2)$. So this means

$H_*(C_2 E_k)$ is free abelian, with ranks $1, 0, 2, 1, 1, 1, 0, 0, 0$ in dimensions 0 through 8 respectively. The $H_4(C_2 E_k)$ class is interesting, have you computed its self-intersection number?

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Thanks Ryan. I'll think about this for a while and see what I come up with. –  alvarezpaiva Mar 28 '12 at 15:31

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