# A question about covariant derivative

Let $V,W$ smooth vector fields along a smooth curve $c:I \rightarrow M$ , where $M$ is a Riemannian manifold, if $\frac{d<V,W>}{dt}=0$ why we must have $<V,W>=$ constant?

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Correct me if I'm wrong, but I think this is just ordinary calculus. We are differentiating the function $t \mapsto \langle V(c(t)) , W(c(t))\rangle$, which goes from $I$ to $\mathbb R$. Hence if its derivative is zero, it is constant. –  user15464 May 2 '12 at 2:49
$<V,W>$ is a vector field, we must be carefull with this, if your claim is right every parallel vector field must be constant... –  Jr. May 2 '12 at 3:28
@Jr. Why is $<V,Ｗ>$ a vector field? What do you mean by $<\cdot,\cdot>$? A metric? –  Yuchen Liu May 2 '12 at 5:13
@user15464 you are right! –  Jr. May 3 '12 at 1:38

Assuming our notations all agree**, user15464 is right.

Let's recall some definitions:

• A vector field is a map $V\colon M \to TM$ with $V(p) \in T_pM$ for each $p \in M$.
• A vector field along a curve $c\colon I \to M$ is a map $V\colon I \to TM$ with $V(t) \in T_{c(t)}M$ for each $t \in I$.

So:

• If $V, W\colon M \to TM$ are vector fields, then $\langle V, W\rangle \colon M \to \mathbb{R}$ is a function via $\langle V, W\rangle(p) = \langle V_p, W_p \rangle.$

• If $V, W \colon I \to TM$ are vector fields along a curve, then $\langle V, W\rangle\colon I \to \mathbb{R}$ is a function via $\langle V, W\rangle(t) = \langle V(t), W(t) \rangle.$

In our case, $V$ and $W$ are vector fields along a curve $c$, and so $\langle V, W\rangle$ is a map $I \to \mathbb{R}$. This means that the only kind of derivative that makes sense in this context is the ordinary derivative from single-variable calculus $d/dt$.

That is, $\langle V, W\rangle$ is not a vector field. In particular, it does not make sense to talk about (say) the covariant derivative of $\langle V, W\rangle$, nor can one say that $\langle V, W \rangle$ is "parallel."

** I am assuming that $\langle \cdot, \cdot \rangle$ denotes the Riemannian metric on $M$, and that $d/dt$ denotes the ordinary derivative from single-variable calculus.

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that's right, I realize that I made a big confusion with notations, sorry. –  Jr. May 3 '12 at 1:48