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I ran into the following sanity check. Is the following statement true?

Every smooth fiber bundle (with compact fiber) over $S^2$ can be extended to a smooth fiber bundle over $\mathbb{C}P^\infty$ (or just $\mathbb{C}P^n$).

For the case of $\mathbb{C}P^2$, if the bundle is a principal bundle, then it can be extended, since $\pi_3(BG)=\pi_2(G)=0$. So I thought possible counterexamples might come from a fiber $F$, with $\pi_2(\text{Diff}(F))\ne 0$, for example $\text{Diff}(S^1\times S^2)\simeq O(2)\times O(3)\times \Omega SO(3)$, but seems like the fiber bundle can still be extended, since $S^3\rightarrow S^2\rightarrow B(\text{Diff}(S^1\times S^2))$ is still nullhomotopic.

Update: The counterexample is provided in Oscar's post, when the fiber is $(D^7,\partial D^7)$. Now my question is, if the fiber is a closed manifold, is the following statment true?

Every smooth fiber bundle(with closed fiber) over $S^2$ can be extended to a smooth fiber bundle over $\mathbb{C}P^\infty$ (or just $\mathbb{C}P^n$).

In fact, this should be my original question, but somehow I wrote "compact" but with "closed" in mind. I am really sorry about that.

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  • $\begingroup$ If you distinguish between a bundle and a principal bundle, it's better to define a bundle first. $\endgroup$ – Alex Degtyarev Oct 21 '16 at 8:25
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No it isn't, but I had to dig quite deep to get a counterexample. Let us look at smooth $(D^7, \partial D^7)$-bundles over $S^2$, i.e $D^7 \to E \overset{\pi} \to S^2$ with an identification $\partial E \cong S^2 \times \partial D^7$. These are classified by a map $$f : S^2 \to BDiff_\partial(D^7)$$ to the classifying space of the group of diffeomorphisms of $D^7$ relative to the boundary. I will show that there exists an $f$ which does not extend to $\mathbb{CP}^2$.

Smoothing theory gives an identification $$BDiff_\partial(D^7) \simeq \Omega^7\left(\frac{TOP(7)}{O(7)}\right)$$ and by page 246 of Kirby--Siebenmann the map $$\Omega^7\left(\frac{TOP(7)}{O(7)}\right) \to \Omega^7\left(\frac{TOP}{O}\right)$$ is 2-connected, so there is a surjection $$\pi_2(BDiff_\partial(D^7)) \to \pi_9(\tfrac{TOP}{O}) \quad \quad(*).$$ To find an example of such a bundle which does not extend over $\mathbb{CP}^2$, it is therefore enough to show that the map $$- \circ \eta : \pi_9(\tfrac{TOP}{O}) \to \pi_{10}(\tfrac{TOP}{O}) \quad \quad(**)$$ given by precomposition with the (suitably suspended) Hopf map is not trivial.

By surgery theory $\pi_n(\tfrac{TOP}{O}) \cong \Theta_n$, the group of homotopy $n$-spheres, and a homotopy $n$-sphere has a unique Spin structure, determining a $\pi_n(\tfrac{TOP}{O}) \cong \Theta_n \to \Omega^{Spin}_n$, which we can further compose with the Atiyah--Bott--Shapiro map $$\alpha : \Omega^{Spin}_n \to KO_n.$$ Applying this construction to both sides, the map $(**)$ above becomes the map $$- \circ \eta : KO_9 = \mathbb{Z}/2 \to KO_{10} = \mathbb{Z}/2,$$ which is an isomorphism. Finally, it is well-known that $$\pi_n(\tfrac{TOP}{O}) \cong \Theta_n \to \Omega^{Spin}_n \to KO_n$$ is surjective for $n \equiv 1,2 \mod 8$.

So choose a $f' \in \pi_9(\tfrac{TOP}{O}) \cong \Theta_9$ which is non-trivial under the Atiyah--Bott--Shapiro map, and use that $(*)$ is surjective to lift it to a $f \in \pi_2(BDiff_\partial(D^7))$. This has the desired property.

Addendum

The following gambit can be used to get a closed fibre example. The product of the natural maps $SO(n+1) \to Diff^+(S^n)$ and $Diff_\partial(D^n) \to Diff^+(S^n)$ gives a map $$SO(n+1) \times Diff_\partial(D^n) \to Diff^+(S^n)$$ which is a weak homotopy equivalence (though not a homomorphism). Thus the map $$BDiff_\partial(D^7) \to BDiff^+(S^7)$$ is (split) injective on all homotopy groups, so the example above also gives an oriented $S^7$-bundle over $S^2$ which cannot extend to $\mathbb{CP}^2$.

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    $\begingroup$ Wow, thank you for the counterexample. Is it possible to find a counterexample with closed fiber? Actually that should be my original question, but somehow I wrote "compact", sorry about that. $\endgroup$ – ZZY Oct 21 '16 at 16:31

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