Let me introduce a definition first. If $M$ is a smooth $n$-manifold, then a connection in the tangent bundle $TM$ is a map $\nabla:\mathfrak{X}(M)\times\mathfrak{X}(M)\to\mathfrak{X}(M)$ that takes in smooth vector fields on $M$ and satisfies:

  1. $\nabla_{f_1 X_1+f_2 X_2}\ Y=f_1 \nabla_{X_1}\ Y+f_2 \nabla_{X_2}\ Y$
  2. $\nabla_X\ (a_1 Y_1+a_2 Y_2)=a_1\nabla_X Y_1+a_2\nabla_X Y_2$
  3. $\nabla_X\ (fY)=f\nabla_X\ Y+(Xf)Y$

with obvious meaning of each symbol. For your information, this definition of a connection in $TM$ can be found in Lee's RM. After ushering in the definition, Lee defines the connection coefficients $\Gamma_{ij}^k$ of $\nabla$ w.r.t. a smooth local frame $\{E_i\}_{i=1}^n$ for $TM$ over an open subset $U$ of $M$ by setting $$\nabla_{E_i}\ E_j=\Gamma_{ij}^k E_k.$$ Now that's where my question comes in. What is meant by $\nabla_{E_i}\ E_j$? After all, these $E_i$'s are ONLY defined locally on $M$ (they may not live in $\mathfrak{X}(M)$). I'm stuck in this question for a long time and try to make it out. Please help me out if you know something about it. Thank you.

  • $\begingroup$ To me, the best hope is to extend local fields $E_i$ to global ones $\tilde{E}_i$ and show that the choice of extension is irrelevant. That is, if $E_i$ can be extended to $\tilde{E}_i$ and $\bar{E}_i$, can we assure that $\nabla_{\tilde{E}_i}\ \tilde{E}_j=\nabla_{\bar{E}_i}\ \bar{E}_j$? $\endgroup$
    – Boar
    Aug 19, 2021 at 5:15
  • 1
    $\begingroup$ doesn't Lee have some propositions immediately after defining connections that $(\nabla_XY)(p)$ depends only on $X(p)$ and the values of $Y$ along a curve $\gamma$ whose tangent at $p$ is $X(p)$. Even if he doesn't prove the fully strengthened version of this dependence immediately, I surely recall him proving that $\nabla_XY$ is local wrt $X$ and $Y$ (i.e depends only on values of $X$ and $Y$ in an open neighborhood of $p$) $\endgroup$
    – peek-a-boo
    Aug 19, 2021 at 5:45
  • $\begingroup$ Yeah, he did. Lee has stated in a proposition that $(\nabla_X\ Y)_p$ depends only on the values of $Y$ in a nbd of $p$ and the value of $X$ at $p$. The thing is, can we extend each $E_i$ to a global vector field? There is indeed a lemma in Lee's ISM that helps me extend a vector field along a closed set $A\subseteq M$. But these $E_i$'s are all defined on an open set. $\endgroup$
    – Boar
    Aug 19, 2021 at 6:31
  • $\begingroup$ In doing $(\nabla_E\ F)_p$ with $E,F$ defined on an open set $U$, we can take care of $E$ by considering the closed set $\{p\}$ and using the extension lemma. How about the disturbing field $F$? $\endgroup$
    – Boar
    Aug 19, 2021 at 6:49
  • $\begingroup$ We can certainly find vector fields $X_i$ defined on $M$, such that there is a neighborhood $U$ of $p$ such that $X_i|_U=E_i|_U$. $\endgroup$
    – peek-a-boo
    Aug 19, 2021 at 7:31

1 Answer 1


Look at the second paragraph on page 91 of Introduction to Riemannian Manifolds (2nd edition). It says

Thanks to Propositions 4.3 and 4.5, we can make sense of the expression $\nabla_v Y$ when $v$ is some element of $T_p M$ and $Y$ is a smooth local section of $E$ defined only on some neighborhood of $p$. To evaluate it, let $X$ be a vector field on a neighborhood of $p$ whose value at $p$ is $v$, and set $\nabla_v Y = \nabla_X Y|_p$. Proposition 4.5 shows that the result does not depend on the extension chosen. Henceforth, we will interpret covariant derivatives of local sections of bundles in this way without further comment.

Proposition 4.3 shows that a connection on a vector bundle $E\to M$ automatically determines a unique connection on the restriction of $E$ to any open subset of $M$; this is what justifies applying the connection to a vector field defined only on an open subset. Proposition 4.5 then says that the value of $\nabla_X Y$ at a point $p$ depends only on the value of $X$ at $p$; this is what justifies the notation $\nabla_v Y$ for a vector $v\in T_p M$.

To actually compute the value of $\nabla_v Y$ at a point $p\in M$, just choose a local frame defined on a neighborhood of $p$, compute the connection coefficients in that local frame, and use the formula you displayed in your post. There's no need to extend anything to a global vector field.

  • $\begingroup$ Hi professor, could I ask your assistance here, please? Unfortunately I did not completely understand integration over the boundary of a manifolds with corners and so I have thought you can help me. Excuse me for the bother. $\endgroup$ Sep 7, 2021 at 17:26

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