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

    1. Prove that $\displaystyle $\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 $\displaystyle $v$$. Thus the derivative $\displaystyle $Y^{\prime}$$ of a vector field $\displaystyle $Y$$ on a curve $\displaystyle $a$$ is (almost) $\displaystyle $\nabla_{a^{\prime}}Y$$.

    Definition 5.1. Let $\displaystyle $W$$ be a vector field on $\displaystyle $\mathbb{R}^3$$, and let $\displaystyle $v$$ be a tangent vector to $\displaystyle $\mathbb{R}^3$$ at the point $\displaystyle $p$$. The the covariant derivative of $\displaystyle $W$$ with respect to $\displaystyle $v$$ is the tangent vector $\displaystyle $(W(p+tv))^\prime(0)$$ at the point $\displaystyle $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 $\displaystyle $a$$ from $\displaystyle $I$$ to $\displaystyle $\mathbb{R}^3$$ is a function that assigns to each number $\displaystyle $t$$ in $\displaystyle $I$$ a tangent vector $\displaystyle $Y(t)$$ to $\displaystyle $\mathbb{R}^3$$ at the point $\displaystyle $a(t)$$.

    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 $\displaystyle $a(t)$$ and show that it agrees with the covariant derivative with respect to a vector $\displaystyle $v$$ at a point $\displaystyle $p$$ for all curves $\displaystyle $a(t)$$ such that $\displaystyle $a(0)=p$$ and $\displaystyle $a^\prime(0)=v$$. 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 $\displaystyle $Y$$ is a vector field on a curve $\displaystyle $a$$, then using the definition of covariant, straight-line or otherwise, makes no sense because $\displaystyle $Y\circ a$$ is not defined (i.e., $\displaystyle $Y$$ is not defined on $\displaystyle $\mathbb{R}^3$$, only $\displaystyle $\mathbb{R}$$). So I thought this might be the almost part. However, what's 'almost' about it? I was thinking that perhaps given $\displaystyle $Y$$ on $\displaystyle $a$$ that a vector field $\displaystyle $\overline{Y}$$ could be defined such that $\displaystyle $\overline{Y}\circ a = Y$$, and that then $\displaystyle $Y^\prime$$ would equal the covariant of $\displaystyle $\overline{Y}$$ instead of $\displaystyle $Y$$. However, I don't think I can 'always' define such a $\displaystyle $\overline{Y}$$.
    Last edited by Dignified; Aug 11th 2010 at 08:30 PM.
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