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If the second covariant derivative of every vector field $Z$ is symmetric in the sense that $\nabla(\nabla Z)$ (which is a section of $TM\otimes T^*M\otimes T^*M$) is a section of the sub bundle $TM\o …

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 something of an exercise is unwinding the definitions, but there's an interesting twist to it as well, so here's an outline of an answer: Let $G$ be a connected $3$-dimensional Lie group (not …

Using the structure equations, it is not difficult to show that, if $f:(M,g)\to(N,h)$ is a diffeomorphism of (not necessarily complete) connected surfaces that is affine in the OP's sense, i.e., $\nab …

Remark on connections with the same geodesics: I realize that I didn't respond to the OP's confusion about connections with the same geodesics vs. compatible with a metric $g$ but with torsion. … \quad \square $$ Finally, we examine when two $g$-compatible connections have the same geodesics: Lemma 3: If $g$ is a nondegenerate (pseudo-)Riemannian metric, and $\nabla$ and $\nabla^*$ are linear connections …

Your first question is a bit ambiguous. Are you asking whether, for each $p\in U$, the matrices $S_k(p)$ span the Lie algebra of $\mathrm{Hol}^0_p(\nabla)$ or are you asking whether, after taking the …

I will assume, though you didn't say, that the ground field is $\mathbb{R}$. (For all I know, the argument below might fail when the ground field is finite, etc.) Yes, when $n>2$, it works for $m > …

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 $\nabla$ …

The affine group of $(M,\nabla)$ is a Lie group $G$ by Kobayashi's theorem that shows that the automorphism group of any affine connection is a Lie group (see Kobayashi and Nomizu's Foundations of Dif …

Yes. Just take the Levi-Civita connection of any left-invariant Riemannian metric on the Lie group. The metric is complete, so any two points can be joined by a geodesic (Hopf-Rinow). Thus, the geo …

It's easy to check (in local coordinates) that this mapping from connections to sections of the given bundle has all of the desired properties. … Then, in a natural way, the space of affine connections is identified with the sections of $J^1(F^*(M))/\mathrm{GL}_n(\mathbb{R})$, while the space of torsion-free affine connections is identified with …

(The connections $\nabla_i$ are equal outside the union of the interiors of $D_1$ and $D_2$.) Then $\nabla$ is not the Levi-Civita connection of any metric on $\mathbb{R}^2$. … Note that the above example shows that the condition of being locally Riemannian is not a closed condition on germs of torsion-free connections in the plane (since it can hold on the complement of a point …

NB: In what follows, to save typing, I will be working on a manifold $M$, but I will write $T$, $T^*$, etc. to denote the bundles $TM$, $T^*M$, etc. and let $M$ be understood. It seems that the OP w …

Yes. Take the Levi-Civita connection of any conformal metric $g = e^{2u}(dx^2+dy^2)$ of positive curvature, say. Then, by (local) Gauss-Bonnet, the holonomy around any smooth closed loop $\gamma$ is …

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 …