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Recently i tried to studying Riemannian Geometry using John Lee's book Riemannian Manifolds : An Introduction to Curvature. There is an exercise about completing the proof in Lemma 4.1. Its about showing that a connection is a local operator. I've already done that, but i was not really sure that my proof is valid.

Let $\mathfrak{X}(M)$ space of smooth vector fields and $\Gamma(E)$ be the space of smooth sections on $\pi :E \rightarrow M$. And $$\nabla : \mathfrak{X}(M) \times \Gamma(E) \rightarrow \Gamma(E) $$ be a connection in a bundle $E$. Show that $\nabla$ is a local operator. I.e

1).Show that $\nabla_X Y (p) = 0 $ if $Y \in \Gamma(E)$ is vanishes on a nbhd $U$ of $p$.

$\textbf{Proof from the book} :$

Let $p \in U \subset M$, $\mathfrak{X}(M)$ , $Y \in \Gamma(E)$ s.t $Y(p)\equiv 0$ for all $p \in U$. Choose a smooth bump function

$$\varphi : M \rightarrow \mathbb{R}, \qquad \text{supp }\varphi \subset U,\qquad \varphi(p)=1$$

So $\varphi Y \equiv 0$ for all $M$. Therefore by linearity

$$\nabla_X(\varphi Y) (p) = \nabla_X(0 \cdot \varphi Y)(p) = 0 \cdot \nabla_X(\varphi Y)(p) = 0$$

and product rule gives $$0 = \nabla_X(\varphi Y)(p) = (X\varphi )(p)Y_p + \varphi(p) \nabla_X Y(p) = 0+ 1.\nabla_X Y(p) = \nabla_X Y(p) \qquad \qquad \square$$

When i'm trying to prove it, i end up choose bump function differently.

$\textbf{My proof} :$

The subset $M \smallsetminus \{p\}$ is open and $M \smallsetminus U$ is closed. Choose a smooth bump function $\varphi$ for $M \smallsetminus U$ supported in $M \smallsetminus \{p\}$. Because $\varphi \equiv 1$ on $M \smallsetminus U$ and supp $ \varphi \subset M \smallsetminus \{p\}$, $\varphi Y \equiv Y$ in all $M$. So $$\nabla_X Y(p) = \nabla_X (\varphi Y)(p) = (X \varphi)(p) Y_p + \varphi(p)\nabla_X Y(p) = 0 \qquad \qquad \qquad \square$$

Is this this valid ? Thank you.

$\textbf{EDIT}$ : I just want to tell that i accidentally i encounter the similar technique to choose bump function as i did above in Helgason's book Differential Geometry, Lie Groups, and Symmetric Spaces.

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For all smooth vector fields $X$, $\nabla_X$ is a linear operator over the space of smooth sections. Hence if $Y$ is the zero section, $\nabla_X Y=0$.

What you need to prove then is that $\nabla$ is indeed a linear operator. That proof will depend on which definition of $\nabla$ was given to you.

In some books (especially in introductory Riemannian geometry books) $\nabla$ is given axiomatically (e.g. it verifies linearity axioms and Leibniz rule). In more advanced geometry books, it is a special case of exterior covariant derivative with respect to a connection form on a principal bundle (where linearity of $\nabla$ follows naturally).

What I just said is true globally and locally. Consider an open set $U$ as a submanifold (i.e. a manifold) and you are all set if you want to get local properties.

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  • $\begingroup$ Athough not only about Riemannian geometry, I recommend Kobayashi & Nomizu's "Foundations of Differential Geometry". That's where I learned about connections. Then if you have an interest for physics, gauge theory (YM, YMH, CS, etc.) is a good way to play concretely with connections. I'm talking about connections here because the OP's questions was about covariant derivative. I don't have a specifically Riemannian geometry book in mind. This is a wide topic and it depends what aspects of Riem. geom. you are interested in (geodesic flow ? Einstein's equations ? and so on...). $\endgroup$
    – Noé AC
    Sep 12, 2017 at 13:37
  • $\begingroup$ I interested in Lorentzian geometry. I trying to read Beem, Ehrlich, and Easley book Global Lorentzian Geometry, but that is too much for me right now. So i'm thinking to get a sense for Riemann geometry first. I tried to read Kobayashi & Nomizu's book before but i realized that its not for beginner. But i'll try again later $\endgroup$ Sep 12, 2017 at 14:06
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    $\begingroup$ Here's a quote from Paul Halmos : "Don't just read it; fight it! Ask your own questions, look for your own examples, discover your own proofs. Is the hypothesis necessary? Is the converse true? What happens in the classical special case? What about the degenerate cases? Where does the proof use the hypothesis?". $\endgroup$
    – Noé AC
    Sep 12, 2017 at 14:18
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    $\begingroup$ Thats true. It need patient and take alot of time. I'm from physics before. So when i came to mathematics, its feels really different about how i suppose to read text. $\endgroup$ Sep 12, 2017 at 14:29

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