Proof that Connection is local operator.

Question

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.

Answer 1

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.