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Sep 8, 2015 at 22:37 answer added Will Sawin timeline score: 8
Sep 8, 2015 at 22:23 comment added Will Sawin And they are indeed topologically distinct $S^2$-bundles over $S^n$. The intersection form of $\mathbb P^1 \times \mathbb P^1$ is even, while the intersection form of $\mathbb P^2$ blown up at a point is odd.
Sep 5, 2015 at 18:05 comment added Vitali Kapovitch $\pi_4(S^3)\cong \lim\pi_{k+1}(S^k)$ is the stable homotopy group of spheres. By Pontrjagin construction $\lim\pi_{k+1}(S^k)$ is the same as the group of framed $1$-cobordisms $\Omega_1^{fr}$. a framing on a circle is just an element of $\pi_1(SO)=\pi_1(SO(3))=Z_2$. This gives a natural iso of $\pi_4(S^3)$ and $\pi_1(SO(3))$ which as Allen explained classifies $S^2$ bundles over $S^2$. Can’t speak to the rest of the question.
Sep 5, 2015 at 11:13 comment added Allen Knutson The relevant homotopy calculation for $S^2$-bundles over $M$ is $Map(M,BDiff(S^2)) = Map(M,BO(3))$, so for $M$ also $S^2$ it's $Map(S^2,BO(3)) = \pi_2(BO(3)) = \pi_1(O(3)) = \pi_1(SO(3)) = \pi_1(SU(2)/Z_2)$. I don't know how to get from there to $\pi_4(SU(2))$.
Sep 4, 2015 at 12:20 comment added Daniel Loughran Both del Pezzo surfaces of degree 8 are $\mathbb{P}^1$-bundles over $\mathbb{P}^1$, i.e. $S^2$-bundles over $S^2$ (these are special Hirzebruch surfaces). This seems relevant, but I don't know enough topology to know if it actually is.
Sep 4, 2015 at 11:36 history asked user25309 CC BY-SA 3.0