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If $E$ has rank $0$ or $1$, 'yes', otherwise, 'no'. Just do a dimension count. You'll find that you have (many) more unknowns than equations and, for general $h$ and $h''$, there will be no solution …

An obvious necessary local condition is that $d\omega=0$. On the open set $U\subset X$ on which $\omega^2\not=0$, one has the further condition that the only ${\frak su}(2)$-valued connection that co …

Unfortunately, you need to be careful, as this is false as stated, unless you allow the chart to be only Lipschitz. Consider, for example, my answer to this question, where a smooth example in dimens …

A slightly different point of view for answering this question is the following one: First, if $M$ is simply connected, then $E\to M$ admits a flat connection if and only if $E$ is trivial, so in this …

This is more of a comment than an answer, but it's too long for a comment, so I'm putting it here. It sounds as though you are asking what sorts of groups of diffeomorphisms there are acting transiti …

The $\nabla$-compatible metrics on $E$ are the positive-definite $\nabla'$-parallel sections of $S^2(E^*)$, where $\nabla'$ is the connection on $S^2(E^*)$ induced by $\nabla$. When $M$ is connected …

The answer is 'basically, no'. The tensor bundles that you list, such as $T^{(n,m)}M$ and its quotients (such as $\Lambda^p(TM)$, etc.), are first order prolongations of $\mathrm{Diff}(M)$, whereas $ …

You could try É. Cartan's papers on infinite pseudogroups (mostly appearing 1904-05). In particular, see Paragraph 57 of Sur la structure des groupes infinis de transformation (suite). There, for exa …

Remark: I assume that you want $A$ to be a non-trivial bundle. Otherwise, of course, any parallelizable compact manifold would be an example. In particular, any compact Lie group would be an exampl …

The difference is that, for a vector bundle, there is usually no natural Lie group action on the total space that acts transitively on the fibers. The fact that all of the fibers are, individually Li …

These questions are all answered in terms of the exterior algebra over the ring $R$ of the free module $M=R^n$. Namely, the skew-symmetric matrices naturally live in $\Lambda^2(M) \simeq R^N$ where $ …

The answer is already 'no' in dimension $4$. The generic almost complex structure compatible with a symplectic structure in dimension $4$ does not admit any pseudoholomorphic functions (in your sense …

The answer is 'no'. For example, just take $M$ to be $\mathbb{R}^n$ (for $n>1$), and $E = M\times \mathbb{R}^r$ for some $r>1$. Let $\omega$ be any $1$-form on $M$, and define a connection $\nabla$ …

Yes, there is such a twistor fibration over each $S^{2n}$, and the resulting manifold is a complex manifold endowed with a holomorphic $n$-plane field transverse to the fibers of the mapping. Namely, …

You're really asking an algebra question about how the various representations of $\mathrm{Spin}(8)$ interact. There are lots of places where you can read about this, but here is a set of notes that …