# Connection on vector bundle : Prove that $(\nabla_X Y)(p)=0$ if $X(p)=0$.

Recently i tried to learn more about connection using Jeffrey Lee's book Manifolds and Differential Geometry (i already read John Lee's Riemannian Manifolds before). After several pages in Jeffrey Lee's book (proposition 12.5), i realized maybe some step missing from a proof about connection in Lemma 4.2 in John Lee's book Riemannian Manifold. And i think its quite confusing for a beginner like me. Here is the lemma 4.2 in John Lee's Riemannian Manifold :

$\textbf{Lemma}.$Let $E \rightarrow M$ be a vector bundle over $M$ and $\nabla : \mathfrak{X}(M) \times \Gamma(E) \rightarrow \Gamma(E)$ be a connection on $E$. For any $X \in \mathfrak{X}(M)$ and $Y \in \Gamma(E)$ and $p \in M$, the value $(\nabla_XY)(p)$ only depends on the value of $X$ at $p$ and the value of $Y$ in neighbourhood of $p$.

The second claim is fine. What i'm worried about is the first claim. That is $(\nabla_X Y)(p)$ depends on $X$ at $p$. This is the proof from the book :

$\textbf{Proof}$ By linearity it is sufficient to show that $(\nabla_XY)(p)=0$ whenever $X_p=0$. Choose a coordinate neighbourhood of $p$ and write $X=X^i \partial_i$ in coordinates on $U$, with $X^i(p)=0$, then $$(\nabla_XY)(p) = (\nabla_{X^i\partial_i} Y)(p) = X^i(p) (\nabla_{\partial_i}Y)(p) = 0 . (\nabla_{\partial_i}Y)(p) = 0$$ The first equality, we used lemma $4.1$, which allows us to evaluate $(\nabla_XY)(p)$ by computing locally in $U$; the second , we used linearity of $\nabla$ over $C^{\infty}(M)$in $X$. $\qquad$ $\square$

What bug me is that the argument involving lemma 4.1 which is we can evaluate $(\nabla_XY)(p)$ locally in $U$. Lemma 4.1 says that $(\nabla_XY)(p)$ just depends on the value of $X$ in an arbitrarily small neighbourhood of $p$. I don't see why this is implies that we can evaluate $(\nabla_XY)(p)$ locally in $U$. For any ways, the first argument for $\nabla$ must be a global section $X \in \mathfrak{X}(M)$, so we can't replace $X$ in $(\nabla_XY)(p)$ with $X=X^i \partial_i$, because $X=X^i \partial_i$ is just local expression of $X$ in $U$. This is different from Jeffrey Lee's approach which is by definition of $\textbf{natural covariant derivatives}$ first before proving above statement, which i found more a bit more clear.

Anyone can help me with this ? any help will be appreciated. Thanks.

If your problem is that you want a global section, but you agree that it only depends on the values locally, then you can take your vector field $X=X^i \partial_i$ defined on a specific neighbourhood $U$, take a compact set with non-empty interior $K \subset U$, and then extend $X$ to a vector field $\widetilde{X}$ on $M$ which coincides with $X$ on $K$ (by multiplying with a bump function).

Clarifying:

Whatever definition of $\nabla$ or vector fields you are using, you must be able to show that given $V \subset M$ open set, there exists a connection $\nabla^V$ in $V$ satisfying $$(\nabla^V_X Y)(q)=(\nabla_{\widetilde{X}} \widetilde{Y})(q),$$ where $q \in V$ and $\widetilde{X},\widetilde{Y}$ are extensions of $X,Y$ respectively. This can be seen in the section on affine connections on Helgason, for instance, and is essentially what is being said by Lee when he says that you can evaluate a connection locally.

Therefore, coupling this with my previous statement, we must have $$(\nabla^{\mathrm{int}(K)}_{X|_{\mathrm{int}(K)}}Y|_{\mathrm{int}(K)})(q)=(\nabla_XY)(q),$$ and you can perform the computation done in your question on the left side.

• I already tried to do that before, but i can't get the components functions out from the $\nabla_{\tilde{X}}\tilde{Y}$ because to do that i think it needed a global chart for $\tilde{X}$. Oct 21, 2017 at 13:46
• I just realized that how about extend each $X^i\partial_i$ (for $i=1,\dots,n$) separately so that we get $n$ global sections and therefore we can get components out of $\nabla$ ? Oct 21, 2017 at 13:53
• @Sou I added clarification. Oct 21, 2017 at 15:07
• Thank you for the complete answer and some reference. I've seen the same results on Jeffrey Lee's book. So, with this way we can use the argument of linearity in $X$ because now the connection defined in chart on $Int(K)$ ? Oct 21, 2017 at 15:14
• @Sou Yes.${}{}{}$ Oct 21, 2017 at 15:54

I think i can provide an alternative by avoiding the construction of new local connection in the nbhd of a point (as @Aloizio explain above). Its achieved by extending $$X^i$$ and $$\partial_i$$ separately to $$M$$ so that the restrictions of them on $$U$$ is equal to $$X$$.

$$\textbf{Construction :}$$

Let $$p \in M$$ and $$(U,x)$$ is a local chart s.t $$p \in U$$. Let $$\{\partial_i\}$$ be the local frame on $$U$$. The representation of $$X \in \mathfrak{X}(M)$$ in $$U$$ is $$X=X^i \partial_i, \quad \text{with} \quad X^i : U \rightarrow \mathbb{R} \quad \text{and} \quad \partial_i : U \rightarrow TM$$

Choose a neighbourhood $$V$$ of $$p$$ s.t $$p \in V \subset \bar{V} \subset U$$

By restrict each $$X^i$$ and $$\partial_i$$ to $$\bar{V}$$, we have a local section $$X|_{\bar{V}} = X^i|_{\bar{V}} \partial_i|_{\bar{V}}$$ on $$\bar{V}$$. With

$$X^i|_{\bar{V}} : \bar{V} \rightarrow \mathbb{R} \quad \text{and} \quad \partial_i|_{\bar{V}} : \bar{V}\rightarrow TM$$

And by partition of unity argument, we can find their extension on $$M$$ s.t they're agree with the original one on $$\bar{V}$$ and their support lie inside $$U$$. Denote these extensions of $$X^i|_{\bar{V}}$$ and $$\partial_i|_{\bar{V}}$$ by $$\widetilde{X}^i$$ and $$E_i$$ respectively. That is we have $$\widetilde{X}^i : M \rightarrow \mathbb{R}, \quad s.t \quad \widetilde{X}^i|_{\bar{V}} = X^i|_{\bar{V}}, \qquad \text{supp} (\widetilde{X}^i) \subset U$$ $$E_i : M \rightarrow TM, \quad s.t \quad E_i|_{\bar{V}}= \partial_i|_{\bar{V}}, \qquad \text{supp}(E_i) \subset U$$ For a fix $$i$$, now we have new global section $$\widetilde{X}^i E_i : M \rightarrow TM \quad \text{(no summation)}.$$

After we do this for all $$i=1,\dots, n$$, we add them and we have $$\widetilde{X} := \widetilde{X}^1 E_1 + \cdots + \widetilde{X}^n E_n,$$ which is obviously the extension of $$X=X^i\partial_i$$ on $$V$$. By Lemma 4.1 and linearity over $$X$$, $$(\nabla_X Y)(p) = (\nabla_{\widetilde{X}} Y)(p) = (\nabla_{\widetilde{X}^1 E_1 + \cdots + \widetilde{X}^n E_n} Y)(p) = \sum_{i=1}^n \widetilde{X}^i(p) (\nabla_{E_i} Y)(p)$$ and by construction, $$\widetilde{X}^i(p) = X^i(p) = 0$$. Therefore $$(\nabla_X Y)(p)=0$$.