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This can be answered by obstruction theory for the fibration $$ F=SO(4)/U(2) \to BU(2) \to BSO(4) $$ where the fibre is actually a 2-sphere: $F=S^2$. Start with the tangent bundle of an oriented 4-man …

There is a simple way to understand the implication "geometric implies algebraic homotopy" if you remember that $\Omega^*(M \times I)$ is the (projective) tensor product of $\Omega^*(M)$ and $\Omega^* …

The natural starting point of this story are E-orientations on, say closed, manifolds M. That's just a fundamental class $[M^n] \in E_n(M)$ such that cap product induces a (Poincare duality) isomorphi …

There is a beautiful explanation of the Chern character that my student Fei Han proved in his thesis: The Chern character is given by the map that "crosses with the circle". The hard part is to explai …

John, if you look at chapter 8 of Freedman-Quinn's book on topological 4-manifolds, you'll find the following computation of the homotopy groups of Top(4)/O(4): $\pi_3 = Z/2$ and $\pi_i = 0$ for $i=0 …