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Math Help - How is the derivative of a vector field Y on a curve (almost) the covariant of Y?

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    How is the derivative of a vector field Y on a curve (almost) the covariant of Y?

    This two-part problem is from O'Neill's Elementary Differential Geometry, section 2.5. It is in the second statement of the second part where I'm unclear, beginning with "Thus."

    Full problem:

    Let $W$ be a vector field defined on a region containing a regular curve $a$. Then $W\circ a$ (i.e. $W$ composed with $a$) is a vector field on $a$ called the restriction of $W$ to $a$.

    1. Prove that $\nabla_{a^{\prime}(t)}W = (W(a))^{\prime}(t)$.

    2. Deduce that the straight line in Definition 5.1 (below) may be replaced by any curve with initial velocity $v$. Thus the derivative $Y^{\prime}$ of a vector field $Y$ on a curve $a$ is (almost) $\nabla_{a^{\prime}}Y$.

    Definition 5.1. Let $W$ be a vector field on $\mathbb{R}^3$, and let $v$ be a tangent vector to $\mathbb{R}^3$ at the point $p$. The the covariant derivative of $W$ with respect to $v$ is the tangent vector $(W(p+tv))^\prime(0)$ at the point $p$.

    The following definition is useful for part (2) since it distinguishes between a vector field and a vector field on a curve.

    Definition 2.2. A vector field on a curve $a$ from $I$ to $\mathbb{R}^3$ is a function that assigns to each number $t$ in $I$ a tangent vector $Y(t)$ to $\mathbb{R}^3$ at the point $a(t)$.

    My attempt, and where I'm stuck:

    Part (1) was fairly straight-forward, using the definitions of covariant derivative and what it means to differentiate a composition.

    Part (2) has two parts. My approach to the first part is the following, and I believe it to be correct. The idea is to define a function on curves $a(t)$ and show that it agrees with the covariant derivative with respect to a vector $v$ at a point $p$ for all curves $a(t)$ such that $a(0)=p$ and $a^\prime(0)=v$. Part (1) can be used to show the function is well-defined and that it indeed equals the covariant.

    The second part, starting at "Thus" is where I'm having trouble. It's with the use of the word "almost." To me, if $Y$ is a vector field on a curve $a$, then using the definition of covariant, straight-line or otherwise, makes no sense because $Y\circ a$ is not defined (i.e., $Y$ is not defined on $\mathbb{R}^3$, only $\mathbb{R}$). So I thought this might be the almost part. However, what's 'almost' about it? I was thinking that perhaps given $Y$ on $a$ that a vector field $\overline{Y}$ could be defined such that $\overline{Y}\circ a = Y$, and that then $Y^\prime$ would equal the covariant of $\overline{Y}$ instead of $Y$. However, I don't think I can 'always' define such a $\overline{Y}$.
    Last edited by Dignified; August 11th 2010 at 09:30 PM.
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