For any vector fields $V,W$ along a curve $a$, $\frac{d}{dt} \langle V,W \rangle = \langle D_tV,W \rangle+ \langle V,D_tW \rangle$

Question

For any vector fields $V,W$ along a curve $a$, $\frac{d}{dt} \langle V,W \rangle = \langle D_tV,W \rangle+ \langle V,D_tW \rangle$, if the connection is compatible with the given metric, i.e. for any vector fields $X,Y,Z$, $X\langle Y,Z \rangle= \langle \nabla_XY,Z \rangle+\langle X,\nabla_XZ \rangle$.

I propose a proof as follow:

$\frac{d}{dt}\langle V,W \rangle = \dot a(t)\langle V,W \rangle = \langle \nabla_{\dot a(t)}V,W \rangle + \langle V,\nabla_{\dot a(t)}W \rangle = \langle D_tV,W \rangle + \langle V,D_tW \rangle$.

But, I do not think this is right, because $\nabla_{\dot a(t)}V = D_tV$ only if $V$ is extendible, here $V$ is only assumed to be a vector field along $a$. How should I fix this problem?

Actually $\frac{d}{dt}\langle V,W \rangle = \dot a(t)\langle V,W \rangle$ also requires that they are extendible vector fields.

Answer 1

We have that $V$ and $W$ are defined along a curve $a(t)$, say $$ V(t) = Y(a(t)), \; \; \; \; W(t) = Z(a(t)), $$ then we have that $$ \frac{DV}{dt} = \nabla_{\frac{da}{dt}}Y, $$ so let $X(p) = \frac{dc}{dt}|_{t=t_0}$ and assuming $X\langle Y , Z \rangle = \langle \nabla_X Y , Z \rangle + \langle Y , \nabla_{X} Z \rangle$ then we have $$ \begin{align} \frac{d}{dt} \left\langle U , V \right\rangle\bigg|_{t=t_0} &= \frac{d}{dt} \left\langle Y , Z \right\rangle\bigg|_{t=t_0} \\ &= X(p)\left\langle Y , Z\right\rangle_p \\ &= \left\langle \nabla_X(p) Y , Z \right\rangle_p + \left\langle Y , \nabla_X(p) Z\right\rangle_p \\ &= \left\langle \nabla_{\frac{da}{dt}} Y , Z \right\rangle_p + \left\langle Y , \nabla_{\frac{da}{dt}} Z \right\rangle_p \\ &= \left\langle \frac{DV}{dt} , W \right\rangle_p + \left\langle V , \frac{DW}{dt}\right\rangle_p. \end{align} $$

Since the result is just being applied in a small enough neighbourhood around $t_0$ we just need to be able to chose vector fields that agree locally, to do that consider a compact interval $J\subset I$ and a coordinate system $(x,U)$ on $M$ such that $a(J) \subset U$ then we have

Lemma Let $g(t)$ be a differentiable function on an open interval containing $J$. Then there exists an $\epsilon > 0 $ and a function $G \in C^{\infty}(M)$ such that $$ G(a(t)) = g(t), \; \; \forall t, \; \; |t-t_0| < \epsilon $$

Sketch of proof: There is some coordinate function $x^i : U \rightarrow \mathbb{R}^n$ such that $t \mapsto x^{i}(a(t))$ has nonzero derivative at $t=t_0$, therefore we can invert this to get a function $\eta^i$ such that $t = \eta^i (x^i(a(t))$ in some small interval around $t_0$, so the function $q \rightarrow g(\eta^i(x^i(q))$ is defined and differentiable for $q \in U$ near to $a(t_0)$, so we can then select a function $\tilde{G} \in C^{\infty}(U)$ such that $\tilde{G}(q) = q(\eta^i(x^i(q)))$ for all $q$ in some neighbourhood of $\gamma(t_0)$, also we clearly have $$ \tilde{G}(a(t)) = g(t), $$ This can then be extended to some function $G \in C^{\infty}(M)$. Infact using parition of unity arguments it can be extended to a function $G$ agreeing with $g(t)$ along all of $t \in J$. $\square$

So if the vector field $V(t)$ along $a(t)$ is written as $V = \sum_i V^i(t) \frac{\partial}{\partial x_i}$ then we can use the Lemma above to write the component functions $V^i : J \rightarrow \mathbb{R}$ as $C^{\infty}$ functions $Y^{i}(p)$ agreeing along $a(t)$ giving a vector field $Y \in \mathfrak{X}(M)$ with local coordinate expression $\sum_i Y^i \frac{\partial}{\partial x_i}$ such that around $a(t_0)$ we have $V(t) = Y(a(t))$.

Answer 2

Here is an explanation for submanifolds in the Euclidian space (eg surfaces in $E_3$), which explains the geometric meaning of the covariant derivative.

If $M\subset E$ is a submanifold and $a(t)$ a curve on $M$, the operator ${D\over dt}$ on vector fields on $M$ is defined as follows: if $X$ is such a vector field, extend it in a vector field on a neigbourhood of $M$, still denoted by $X$.

Then ${D\over dt} X (a(t_0))$ is the orthogonal projection of the usual derivative ${d\over dt} X (a(t_0))$ onto the tangent space $T_{a(t_0)}M$.

Therefore, if $Y$ is another vector field tangent to $M$,

$<{d\over dt} X (a(t_0)),Y(a(t_0)>=<{D\over dt} X (a(t_0)),Y(a(t_0)>$,

whence the result, as standard calculus in $E$ tells you that : $d/dt<X(a(t)),Y(a(t))>=<d/dtX(a(t)),Y(a(t))>+<X(a(t)),d/dt Y(a(t))>$

Answer 3

The proof you have written is not correct. And it is not correct because the reason you have mentioned. Typically the proof use an orthonormal basis at a point on $a$ and then extend the orthonormal basis (ONB) to an ONB on the whole of $a$ using parallel transport (here you have to use compatibility). Then writing everything with respect to this ONB and using the product rule at both side we prove this theorem. See Proposition 3.2 of Chapter 2 of Do Carmo's Riemannian geometry book.