0
$\begingroup$

To derive the local expression for the affine connection. Take a vector fields $X$ and $Y$, $X$ has local expression $X= \sum_j X^j\frac{\partial}{x_j}$, $Y$ has local expression $Y= \sum_j Y^j\frac{\partial}{x_j}$, then $\nabla_XY = \nabla_{\sum_j X^j\frac{\partial}{x_j}}\sum_j Y^j\frac{\partial}{x_j}$, why does this equality hold?

My confusion is that $X= \sum_j X^j\frac{\partial}{x_j}$ only locally, and $\sum_j X^j\frac{\partial}{x_j}$ is only a vector field on an open subset of the manifold, but affine connections are for vector fields defined on the whole manifold, I do not really see why the equality would hold, it does not make any sense to me.

To be more explicit. Take charts $(U,\phi)$ of $M$, then $\forall p \in U$, $X_p = \sum_j X^j\frac{\partial}{x_j}|_p$, which is obvious, then we would consider the vector field $\sum_j X^j\frac{\partial}{x_j}$ which is defined on $U$. So we examine $\nabla_{\sum_j X^j\frac{\partial}{x_j}}\sum_j Y^j\frac{\partial}{x_j}$, the confusion arises because $\nabla$ takes vector fields defined on $M$ rather than vector fields defined on $U$ as its input.

1 Answers 1

1

The claim here is that in that local coordinate system, the field $\nabla_X Y$ is equal to the thing on the right. That's completely obvious, in the sense that $X$ is equal to the thing in the subscript (within this patch) and $Y$ is equal to the thing on the right (within this patch), so the left and right hand sides are equal just by substitution.

You might well ask "Well, what if we have two overlapping patches? Then we'd have two DIFFERENT expressions for $\nabla_X Y$. How do we know that they're equal?" And the answer is "That depends on how your source defines that covariant derivative." It might be that they're equal because the covariant derivative has been shown to be well-defined everywhere through some coordinate-invariant means; it might be because through explicit computation, the two different coordinate expressions have been shown equal. There are probably other proofs as well, but I can't say which one applies without knowing which definition you've used.

  • 0
    I sill do not quite follow, can you see my edit? – 2017-01-16
  • 0
    As I said, I'd need to know what definition of $\nabla_X Y$ you're using. But in general, for any definition I've ever seen, if $X$ and $X'$ are vector fields that agree on some open set $V$, and $Y$ and $Y'$ agree on $V$ as well, then $\nabla_X Y = \nabla_{X'} Y'$ on $U$. So by extending $\sum_j X^j\frac{\partial}{x_j}|_p$ to all of $M$ via a smooth partition of unity, you get a field $X'$ that agrees with $X$ on some open set $V$ within $U$, and that's enough to show that the the thing you've computed *is* the covariant derivative you wanted. Tell me your def'n of covar. derivative, OK? – 2017-01-16
  • 0
    I am so sorry. The definition that I am using is from do carmo's Riemannian Geometry. https://en.wikipedia.org/wiki/Affine_connection#Formal_definition_on_the_frame_bundle it is the same as in the wikipedia formal definition as differential operator – 2017-01-16
  • 0
    That definition only depends on local properties of the vector field, so any two fields that agree locally will lead to the same covariant derivative, locally. – 2017-01-16
  • 0
    I see. Thank you so much! – 2017-01-16