Projection of an invariant almost complex structure to a non-integrable one

Question

My apologies in advance if my question is obvious or elementary.

We identify elements of $S^3$ with their quaternion representation $x_1 + x_2i + x_3j + x_4k$. We consider two independent vector fields $S_1(a) = ja$ and $S_2(a) = ka$ on $S^3$. On the other hand $P: S^3\to S^2$ is a $S^1$-principal bundle with the obvious action of $S^1$ on $S^3$. Then the span of $S_1, S_2$ is the standard horizontal space associated to the standard connection of the principal bundle $S^3 \to S^2$. Then each horizontal space has an almost complex structure $J$. This is the standard structure associated to $S_1, S_2$ coordinates which is defined by $J(S_1) = S_2, J(S_2) = -S_1$.

Is this structure invariant under the action of $S^1$? If yes, we can define a unique almost complex structure on $S^2$ which is $P$ related to the structure on the total space. Now is this structure on $S^2$ integrable?

As a similar question, is there an example of a principal bundle $P\to X$, with $P$ a real manifold, $X$ a complex manifold, and a connection which admits an invariant almost complex structure projecting to a non-integrable structure?

Answer 1

First of all, every almost complex structure on a two-dimensional manifold is integrable, see here.

Let $\pi : P \to X$ be a smooth principal $G$-bundle equipped with a connection. The connection determines a horizontal subbundle $H$ of $TP$; moreover, $H \cong \pi^*TX$. If $X$ admits an almost complex structure $J$, then so does $H$ via the above isomorphism; moreover, the almost complex structure $J'$ on $H$ constructed in this way is invariant under the action of $G$. In particular, $J'$ projects to $J$ as you put it.

So to come up with an example, just choose a smooth principal $G$-bundle $\pi : P \to X$ where is $X$ is a complex manifold. Now let $J$ be a non-integrable almost complex structure on $X$ (here we need $\dim_{\mathbb{R}} X > 2$), then $J'$ will project to $J$ which is non-integrable.