Nonisomorphic vector bundles with diffeomorphic total spaces

What's an example of two non-isomorphic vector bundles $E,F$ over the same base such that the total spaces $Total(E), Total(F)$ are homeomorphic? Assume that rank of these bundles is the same as dimension of the base manifold.

(EDIT: Mike Miller brought up the excellent point that I need to look at bundles which are not isomorphic covered by any self-homeomorphism of the base, otherwise one can look at two codimension $2$ submanifolds on a manifold such that one can be taken to another by an ambient homeomorphism, with different fundamental classes, and then look at the corresponding complex line bundles. This cannot happen for bundles with the rank of the base, so easiest example comes in dimension $3$, say, eg the torus and the line bundles coming from two of the generator circles in $H_1$)

Notice that the Euler class of the bundle is an obstruction from this happening - if the Euler classes were different, the total spaces would have different (compactly supported) cohomology ring structure (the self-intersection number of the zero sections would be different)

The simplest example I can come up with which passes the Euler class test is $TS^5$ and $S^5 \times \Bbb R^5$. But $TS^5$ seems to have a complicated total space, so it's unclear how to see if they are or are not diffeomorphic (if not, how would one prove that?).

It seems to me that there is no easy general technique to determine the diffeomorphism type of the total space from the vector bundle once the Euler class is ruled out. So any general comment along the lines of this would be much appreciated, even as answers.

Solution 1:

Let $E, E'$ be vector bundles over a compact CW base with proper homotopy equivalent total spaces; say $f: E \to E'$ is a proper homotopy equivalence. Suppose also they are both equipped with a Riemannian metric for convenience. Then the unit sphere bundles are homotopy equivalent.

Because the zero section is compact, $f^{-1}(0_{E'})$ is compact, and hence contained in one of the disc bundles $E_{\leq r} = \{v \in E \mid \|v\| \leq r\}$. So the map from the $2r$-sphere $E_{2r} = \{v \in E \mid \|v\| = 2r\}$ to $E'$ misses the zero section, and by dividing out by the norms appropriately we obtain a map $f_{2r}: E_1 \to E'_1$. I claim this map is a homotopy equivalence; because a sphere bundle over a CW base is again a CW complex, it suffices to show it induces an isomorphism on homotopy groups. Note that because $f$ is a proper homotopy equivalence, it induces a bijection on mapping spaces $[X,E] \to [X,E']$ of proper homotopy classes of proper maps. Apply this to $X = S^k \times [1,\infty)$. Given a map $g: S^k \to E'_1$, extend it to a map $g(x,t) = tg(x)$, $S^k \times [1,\infty) \to E'_{\geq 1}$; there is thus an associated proper map $g': S^k \times [1,\infty) \to E_{\geq 1}$. For sufficiently large $t$, $f(g'(x,t))$ is never zero. In particular, this implies that the map $f_c$ above is surjective on homotopy groups. A similar argument shows injectivity on homotopy groups.

There is a partial converse to this: if there is a homotopy equivalence between the unit sphere bundles $E_1 \to E'_1$ that preserves, up to homotopy, the projection to the base, then $E$ and $E'$ are proper homotopy equivalent. This follows simply by showing that the mapping cones of $E_1 \to B$ and $E'_1 \to B'$ are homotopy equivalent relative to the map $E_1 \to E'_1$, thus getting a homotopy equivalence between $E_{\leq 1} \to E'_{\leq 1}$ that preserves the unit spheres. Then extend this to the whole vector bundle by scalar multiplication. But there's no reason that the proper homotopy equivalence above should preserve the projection to the base, so these are not obviously equivalent.

What you're asking is for examples of non-homeomorphic vector bundles whose compactly supported cohomology rings are isomorphic. (You actually asked for the weaker condition that the self-intersection of the zero section, aka the cup square of the fundamental class of the zero section, is zero.) We can apply the condition above. The simplest example I can give are the two $S^3$-bundles over $S^2$, determined by the two homotopy classes in $\pi_1(SO(3))$; one is trivial, and the other is not. To prove that they're not homotopy equivalent, I'm going to do some work with their Stiefel-Whitney classes. If $E_1$ is a fiber bundle, let $T_vE$ be the vertical tangent bundle (the tangents to the fibers); then there is a non-canonical isomorphism $TE_1 \cong T_vE_1 \oplus \pi^*(TB)$. Thus we have an equality of total Stiefel-Whitney classes $w(TE_1) = w(T_vE_1) \pi^*(w(TB))$. In the case that $E_1$ is a linear sphere bundle (aka the unit bundle of some vector bundle $E$), like these, we can calculate $T_vE_1$ quite easily. $T_vE$ is canonically isomorphic to $\pi^*E$; then $T_vE_1 \oplus N_vE_1 = T_vE$, and the normal bundle of $E_1$ is canonically trivial. So $w(T_vE_1) = w(T_vE) = pi^*(w(E))$. Thus $w(TE_1) = \pi^*(w(E)w(TB))$.

Now let $E$ be a rank $k$ vector bundle over some sphere $S^n$ with nonzero Stiefel-Whitney class $w_n$. Then $E_1$ is not homotopy equivalent to $S^n \times S^k$; the calculation above shows that $w(E_1)$ is nonzero, but it is for $S^n \times S^k$, which is stably parallelizable, and Stiefel-Whitney classes are a homotopy invariant. Thus the total spaces of the two $n$-plane bundles over $S^2$ (which are determined by their $w_2$) are not proper homotopy equivalent, and similarly one obtains examples over many other spheres.

(The simplest example, then, are the two 3-sphere bundles over $S^2$.)

I have absolutely no idea whether or not $TS^5$ and $S^5 \times \Bbb R^5$ are proper homotopy equivalent. I tried to carry out the calculation but ran into trouble; it could go either way. I would be unsurprised if there was some $n \neq 1, 3, 7$ such that $TS^n$ is proper homotopy equivalent to $S^n \times \Bbb R^n$. Maybe $n=11$ would be a good place to fiddle.

Solution 2:

Your question says diffeomorphic but let's do homeomorphic instead. Fiber bundles with fiber $\Bbb R^n$ are classified by maps into $B\text{Homeo}(\Bbb R^n)$. It would suffice to find an example where the map $BO(n) \to B\text{Homeo}(\Bbb R^n)$ was not injective on maps out of $X$, and we may as well remove the letter $B$. To do this let's start by doing a little calculation.

There is a fibration $\text{Homeo}(\Bbb R^n) \to \mathrm{TOP}(n)$, the space of germs of homeomorphisms of $\Bbb R^n$ fixing a point, given by sending a homeomorphism to its germ at $\infty$ (take one point compactification, stereographically project). The fiber is the space of homeomorphisms of $S^n$ fixing a neighborhood of $\infty$. Blowing up the point at infinity into a sphere, this is the space of homeomorphisms of $D^n$ that are the identity near the boundary, and that's contractible by Alexander's trick. So the map $\text{Homeo}(\Bbb R^n) \to \mathrm{TOP}(n)$ is a homotopy equivalence.

Milnor's paper introducing microbundles here gives examples of spaces for which the induced map after stabilizing - so $[X,O] \to [X, \mathrm{TOP}]$ - is not injective. (This is true for every space of the form $X_{n,q} = S^{4n-1} \cup_q D^{4n}$, where my attaching map is degree $q>1$.) Because $X$ is finite CW, this map must fail to be injective at some finite stage, and so there is a pair of rank $k \gg 0$ vector bundles over $X_{n,q}$ with homeomorphic total spaces. See lemma 9.1.