@Bilateral: If you look at the final section of those notes, you'll see several applications of Theorem 4. I don't know any other place where Theorem 4 is stated explicitly in the form that I gave in those notes, but I confess that I don't know the EDS literature of the 1930-40s that well. Because the proof of Theorem 4 via Cartan-Kähler is so straightforward, I believe that some version of Theorem 4 was known to Élie Cartan (and I'm sure that he would have regarded it as 'obvious' after Kähler's extension of Cartan's existence theorem to differential ideals generated in arbitrary degree).

@AliTaghavi: A simple example, is to let $X = S^3$ and $B=S^2$ and let $\pi:S^3\to S^2$ be the Hopf map. The tangent bundle to $S^2$ is irreducible over $S^2$, but the pullback via $\pi$ is trivial over $S^3$.

@Bilateral: That's a good question. Basically, my reason for formulating it on the coframe bundle $P$ is that this can be done without making any arbitrary choices (such as, for example, local coordinates on $M$) and the solutions to your problem are exactly the same as sections $e:P\to M$ of this bundle that are integral manifolds of the EDS that I described. I don't see any formulation that is more natural than that. Moreover, the 'hidden' compatibility conditions in the equations you have written are uncovered naturally by closing the EDS under exterior derivative. I'll illustrate below.

The manifold on which this problem is posed as an exterior differential system is the coframe bundle $\pi:P\to M$, where an element $u\in P$ is an isomorphism $u:T_{\pi(u)}M\to\mathbb{R}^n$. Your equations are then interpreted as $2$-forms on $P$ and you seek a section $e:M\to P$ such that $e:M\to P$ is an integral manifold of the ideal generated by these $2$-forms. The $2$-form $\mathrm{d}(F_k e^k)$ is of the form $\tfrac12\,F_{jk}\, e^j\wedge e^k$, where the functions $F_{jk}=-F_{kj}$ are defined on $P$, and your desired section $e$ must take values in the zero locus of these functions, etc.

Oh, also, note that $\mathrm{Sp}(n)$ contains $\mathrm{Sp}(1)\times\cdots\times\mathrm{Sp}(1)$ ($n$ times) as a maximal rank subgroup (of rank $n$), and $\mathrm{Sp}(m)$ is simply-connected for all $m$.

Are you assuming that $n>1$? After all, $\mathrm{Spin}(7)$ contains $\mathrm{Spin}(6)$ as a maximal rank subgroup and $\mathrm{G}_2$ contains $\mathrm{SU}(3)$ as a maximal rank subgroup, and all of these groups are simply-connected. Also, $\mathrm{Spin}(5)$ contains $\mathrm{SU}(2)\times\mathrm{SU}(2)$.

That's correct. Moreover, the Riemannian isometry group, which has two components, is generated by the projective unitary group together with any anti-holomorphic isometry.

@IvinBabu: I don't have a copy of Chow's book handy, so I can't know exactly what you are referencing. However, there are many examples in dimension at least 4 of almost Hermitian structures $(J,\omega)$ with $\mathrm{d}\omega=0$ but with $J$ not integrable.