Vector Bundle Connection and Agreement a on Curve

Question

Problem 10.3 (Tu - Differential Geometry)

Let $\nabla$ be a connection on vector bundle $E \to M$, $p \in M$, and $X_p \in T_p M$. Show that if sections $s$ and $t$ agree on a curve through $p$ with initial vector $X_p$ then $\nabla_{X_p} s = \nabla_{X_p} t$.

Relevant excerpt from book

Questions

1. Since we know that a connection at a point is well-defined, isn't $\nabla_{X_p} s = (\nabla_X s)_p$ a definition? The book doesn't really explain what the LHS means otherwise.

2. I'm really unsure how to solve the problem. For example, I see no way to use the Leibniz rule, because any $f$ smooth on $M$ with $f(p) = 1$ will be nonzero in some neighborhood around p, so we cannot guarantee that $f (s-t)$ is zero in any neighborhood of $p$.

Update

Peek-a-boo answered Question 1.

Please review my proposed answer below, which addresses Question 2.

Answer 1

Let $s', t'$ be the sections that agree on a curve $c : I \to M$ and define $s := s' - t'$. By bilinearity, it suffices to show that $\nabla_{X_p} s = 0$.

Let $U$ be a framed open set about $p$ with local frame ${ e_1, \ldots, e_n}$. Note that $\nabla_{X_p}^U = \nabla_{X_p}$, so we will use the latter as a shorthand.

Then $s = \sum s^i e_i $ with $s^i$ smooth functions on $U$. Because $s(c(t)) = 0$ for each $t \in I$, clearly $(s^i \circ c)'(t) = 0$ for each $i$. Moreover, $s^i(p) = 0$ because $s(c(0)) = s(p) = 0$.

Now write \begin{align} \nabla_{X_p} s &= \nabla_{X_p} \sum s^i e_i \\ &= \sum (X_p s^i) e_i + s^i(p) \nabla_{X_p}e_i \\ &= \sum (s^i \circ c)'(0) e_i \\ &= 0 \end{align} $\square$

Answer 2

For 1, it is an implicit definition. It says that given $\nabla$ (acting on vector fields on $M$ and sections of $E$) there exists a unique operator $\tilde{\nabla}:TM\times\Gamma(E)\to E$ (which they have still denoted as $\nabla$) such that for each $p\in M$,

• $\tilde{\nabla}$ restricts to an $\Bbb{R}$-bilinear map $T_pM\times\Gamma(E)\to E_p$
• for each $h\in T_pM$ and $s\in \Gamma(E)$, and vector field $X\in \mathfrak{X}(M)$, \begin{align} X(p)=h\implies\tilde{\nabla}_hs=(\nabla_Xs)_p. \end{align}

So, the definition of $\tilde{\nabla}$ is as a unique map satisfying such and such property. The point is that you have to ensure well-definition of $\tilde{\nabla}$ because there are infintely many vector fields $X$ whose value at a particular point $p$ is equal to a given tangent vector $h\in T_pM$. The reason for the existence and uniqueness of such a $\tilde{\nabla}$ is simply because $\nabla$ is $C^{\infty}(M)$-linear in the vector field slot where $X$ goes.

To check that these depend on the section $s$ only through its values along a given curve with tangent $h$, simply write down the local coordinate formula involving the Christoffel symbols; I think that’s the fastest way to go (see e.g the second half of my answer here… I only talk about connections in the tangent bundle, but the same idea goes through for every vector bundle).