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Even though they are linear ODE, for most connections given explicitly by some functions $\Gamma^i_{jk}$ on a domain, one cannot perform their integration. …

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 …

Many manifolds have curvature-free (i.e., flat) connections on their tangent bundles. … For example, any orientable $3$-manifold $M$ is parallelizable, i.e., its tangent bundle is trivial, so it carries a flat connection (in fact, many flat connections). …

NB: I'm combining my previous comments into an answer, because I believe that this is better than leaving them scattered. As another commenter has pointed out, the skew-symmetric part of the Ricci t …

The standard way to discuss the geometry of connections and geodesic flow 'invariantly' (by which, I assume you mean 'without reference to coordinates') is to exploit the underlying geometric features …

(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 …

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 …

Part of the confusion is that your positional notational convention is not the standard one. In most books, what you are writing as $R_{ab}{^c}_d$ would be written as $R{^c}_{dab}=-R{^c}_{dba}$ (thoug …

Élie Cartan proposes such interpretations in his fundamental paper Sur les variétés à connexion affine et la theorie de la relativité généralisée (Ann. Ec. Norm. 40 (1923), 325–412 and 41 (1924), 1–25 …

The answer depends on the dimension. When $n=2$, Ricci-flatness of a connection implies that it is flat, so, in that case, yes, you get holonomy reduction locally. However, when $n>2$, Ricci-flatnes …

This gives a couple of $1$-parameter families of connections with the same constant sectional curvature $s$ for each $s$ in $\mathbb{R}$, and a traceless example for each $s\ge0$. … Now, acting by the orthogonal group $\mathrm{O}(4)$, you see that each of these curvatures (which are invariant under $\mathrm{O}(4)$) comes from a space of connections (respectively, traceless connections …

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 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 …

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 $\nabla$ …