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Pictures below are from the do Carmo's Riemannian Geometry. I can't understand the 2.6 Remark. In my view, the affine connection is a local notion since $X(y_k)(p)$ depends on a neighborhood of $p$.

According to the 2.6 Remark, operator $R_p$ depends only on $p$. I agree with it, since it is obvious from the equation (1) and (2). But in my view, curvature of $p$ should depends on a neighborhood of $p$, since looking only for one point $p$, how to know whether it is bent? It is inconsistent with intuition.

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5201314
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Enhao Lan
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1 Answers1

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Well, you are mixing two ideas. One is that the curvature tensor itself depends on a neighbourhood of $p$, which is true. Here we mean the association $p \mapsto R_p$, or the functions $p \mapsto (R_{i j}^k)_p$.

And the other refers to the map $(X_p,Y_p,Z_p)\mapsto R_p(X_p,Y_p,Z_p)$. What remark 2.6. is saying is that this map only depends on the definition of $X, Y$ and $Z$ on $p$, not how do you extend this vectors to vector fields in a neighbourhood of $p$.

A. J. Pan-Collantes
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  • Thanks， whether $X(f)(p)$ is also a local notion , where $f$ is a functio on $M$ and $X\in T_pM$ is a vector. – Enhao Lan Aug 21 '21 at 07:00
  • @lanse7pty Remember a tangent vector (at $p$) is a derivation (at $p$). Since you are only having a single tangent vector instead of $X$ being a tangent vector field, it doesn't make sense to have the $(p)$ at the end. – user10354138 Aug 21 '21 at 09:04
  • @user10354138 Thanks, I think I understand you. Only when $X$ is vector field, the $(p)$ of $X(f)(p)$ is suitable, is it ? So, if $X$ is a vector field and $f$ is a function, whether $X(f)$ is a local notion ? – Enhao Lan Aug 21 '21 at 10:14
  • It is local for $f$. More precisely, $X(f)$ is tensorial in $X$ but not $f$ (it depends on the $1$-jet of $f$). – user10354138 Aug 21 '21 at 10:33