# How can one define ordinary derivative of a vector field along a curve?

Definition 1: A curve $C$ on a manifold $M$ is a smooth function $C:(a,b) \rightarrow M$.

Definition 2: A vector field $v^a=v^a(t)$ along a curve $C$ is an assignment of vectors on the tangent space at each point $C(t)$ on the curve.

Definition 3: Let $v^b$ be a vector field on $M$. The derivative operator $\partial_a v^b$ is defined by taking partial derivative at each component of $v^b$, given that a fixed coordinate system is chosen.

Definition 4: $v^a$ is said to be parallelly transported along the curve $C$ if $t^a \nabla_a v^b=0$.

In Wald's General Relativity, the derivative operator $\nabla_a$ is defined on vector field on the manifold, but still the above is valid by defining similar thing for vector fields along a curve.

Theorem: Suppose a coordinate system is fixed. The following are equivalent.

(1) A vector $v^a$ is parallelly transported along a curve $C$.

(2) $t^a\partial_a v^b+t^a {\Gamma^c}_{ab} v^c=0$ [Tensor form]

(3) $\displaystyle\frac{dv^\nu}{dt}+\sum_{\mu,\lambda=1}^n t^\mu {\Gamma^nu}_{\lambda\nu}v^\lambda=0$ [Component form]

This confuses me a lot. In particular, we are unable to define the derivative operator on vector field along a curve in the manner in definition 3. To make sense of taking partial derivatives at each component, formally speaking, it is defined as partial derivatives of the composition of the components and the parametrization of the manifold (which makes sense because the components themselves are formally functions from $M$ to $\mathbb{R}$).

However, the curve may have self-intersection and two different tangent vectors at the same point might be defined for a vector field along a curve, which means the components now cannot be a function from $M$ to $\mathbb{R}$. So we cannot make composition with parametrizations now in order to allow us to define partial derivatives. Therefore, I am unable to understand the meaning of statement (2) in the above theorem.

Can anyone give me some clue? Thank you.

You should think of eq. (2) (and (3) for that matter) as an equation of functions defined on the interval $(a,b)$. In particular you check its validity at each parameter $t\in(a,b)$ and it does not matter whether $C$ has self intersections or not.

Your main trouble seems to be the interpretation of $t^a \partial_a v_b$. The point is that $\partial_a v_b$ is not well defined, but only the whole term $t^a\partial_a v_b$ is.

Here is an analogy to make clear what is happening: Suppose $f$ is a (scalar) function defined on $\{y=0\}\subset \mathbb{R}^2$. Surely $\partial_xf$ is well defined, but coming from multivariable calculus you might think it's the same thing as $$\partial_x f = e_x\cdot \nabla f, \tag{1}$$ where $e_x$ is the unit vector pointing in $x$-direction. However, $\nabla f = (\partial_x f , \partial_y f)$ does not make sense by itself. In which sense is equation $(1)$ meaningful then? Suppose you extend $f$ to a function $\tilde f$ on all of $\mathbb{R}^2$, then $\nabla \tilde f$ is defined on all of $\mathbb{R}^2$ and you will find that on $\{y=0\}$ you have $$\partial_x f = e_x\cdot \nabla \tilde f \tag{2},$$ in particular the result is independent of the extension $\tilde f$ that you've chosen.

So how to compute $t^a \partial_a v_b$? First note, that for fixed index $b$ this is a scalar function of one variable $s\in (a,b)$.

(i) Fix $s_0$ and choose a neighbourhood $I=(s_0-\epsilon,s_0 + \epsilon)$ such that $C\restriction_I$ does not have any self intersections.

(ii) Find a vector field $\tilde v^c$ which is defined on a neighbourhood $U$ of $C(I)$ in $M$ and extends $v^c$.

(iii) Compute $\partial_a\tilde v_b$ (makes sense on all of $U$) and then contract it with $t^a$ to obtain $t^a \partial_a\tilde v_b$ (only defined on C(I)).

(iv) Put $t^a \partial_a v_b(s) := t^a \partial_a\tilde v_b(s)$ for $s\in I$.

Much in the same spirit as in the example you will find that the computation does not depend on the actual extension $\tilde v^c$ that you've chosen. If $C$ has self intersections (say $C(s_0) = C(s_1)$), then the values of $t^a \partial_a v_b(s_0)$ and $t^a \partial_a v_b(s_1)$ may differ. But that is OK, after all it is a function on $(a,b)$ and not on $C(a,b)$.

• You are right, it's only necessary to precompose the Christoffel symbols with $C$. Then both $v^a$ and $\Gamma_{ab}^c\circ C$ are functions of one parameter $t\in(a,b)$. The rest of my answer should still make sense. – Jan Bohr Aug 27 '18 at 10:01
• Why can we precompose the vector field with $C$? C $C$ is a function from $(a,b)$ to $M$, while the vector field is no way a function from $C(a,b)$. – Jerry Aug 27 '18 at 10:03
• A vector field along $C$ is a map $V: (a,b) \rightarrow TM$ with $V(t)\in T_{C(t)}M$ for each $t\in(a,b)$. Once you've chosen coordinates around some point $C(t_0) \in M$ you locally write it as $V(t) = v^a(t) \partial_a$, where $v^a:(t_0-\epsilon,t_0+\epsilon) \rightarrow \mathbb{R}$. – Jan Bohr Aug 27 '18 at 10:06
• But how do you take partial derivative of the component if the components are just function of $t$ but not a function from the trace of the curve $C(a,b)$? – Jerry Aug 27 '18 at 10:06
• Well the point is that $\nabla_X V$ (the covariant derivative of $V$ along the vector field $X$) only makes sense if $X$ is tangent to $C$. (Think of a vector valued function defined on the $x$-axis of $\mathbb{R}^2$ - taking the derivative with respect to $y$ is meaningless). Now if $X$ is tangent to $C$, then $\nabla_X V$ is well defined, and eq. (3) tells you how to compute it. – Jan Bohr Aug 27 '18 at 10:11