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Suppose I have a manifold that is a $n$-disk bundle over $S^n$, i.e a $2n$-dimensional manifold with a single critical point of index $0$ and $n$. I am trying to understand which of these manifolds have almost complex structures and the space of almost complex structures on such manifolds.

The smooth type of these manifolds is determined by the framing of the $n$-handle. This is a torsor over $\pi_{n-1}(O(n))$, i.e., if we fix an initial framing, the other framings come about by multiplying by $\pi_{n-1}(O(n))$. There is a unique almost complex structure on the ball $B^{2n}$. In our fixed initial framing, this almost complex structure represents an element $\alpha \in \pi_{n-1}(O(2n)/U(n))$. The disk bundle obtained using this initial framing has an almost complex structure if $\alpha\in \pi_{n-1}(O(2n)/U(n))$ vanishes. When we change the framing by $\beta \in \pi_{n-1}(O(n))$, the resulting almost complex structure represents $\beta \alpha \beta^{-1} \in \pi_{n-1}(O(2n)/U(n))$; here we consider the image of $\beta$ in $\pi_{n-1}(O(2n))$ by the inclusion $O(n) \hookrightarrow O(2n)$ and act on $\alpha$ by conjugation; this is just the change of basis formula. The new disk bundle represented by $\beta$ will have an almost complex structure if $\beta \alpha \beta^{-1} \in \pi_{n-1}(O(2n)/U(n))$ vanishes. Has the conjugation action on homotopy groups been calculated? (More generally, what is the group structure on $\pi_k(O(n))$ coming from multiplication of matrices?) More precisely, for what $\beta$ do $\beta \alpha \beta^{-1} \in \pi_{n-1}(O(2n)/U(n))$ vanish?

Hence the existence of an almost complex structure on the disk bundle seems to depend on the smooth type of the bundle. Is this correct? However if an almost complex structure exists, then all almost complex structures are related by $\pi_n(O(2n)/U(n))$, which is independent of the smooth type.

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  • $\begingroup$ As pointed out in JHF's answer, the existence is independent of the smooth structure. Note that $B^{2n}$ has a canonical almost complex structure and to extend it to n-handle, the structure needs to be contractible along the n-1 attaching sphere $S^{n-1} \subset S^{2n-1} = \partial B^{2n}$, which is clearly the case since $S^{n-1}$ bounds a ball $B^n$. The set of almost complex structures is determined by $\pi_n(O(2n)/U(n))$ which is also independent of the smooth structure. $\endgroup$ – user39598 Aug 7 '17 at 20:08
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I think an Eckmann-Hilton argument would show that the group structure on $\pi_k(O(n))$ coming from the fact that $O(n)$ is a group agrees with that coming from the fact that $S^k$ is a cogroup, at least when $k \geq 1$. Moreover, this operation will be abelian. So the conjugation action is trivial.

So, we see that the choice of framing is not important. This should help answer the rest of your questions.

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  • $\begingroup$ So it doesn't depend on the smooth type of the disk bundle? Is this because they are all homotopy equivalent to the zero section $S^n$? $\endgroup$ – user39598 Aug 4 '17 at 17:12

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