Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
4  
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
1  
@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
add comment

1 Answer

up vote 4 down vote accepted

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.

share|improve this answer
    
that's right, I realize that I made a big confusion with notations, sorry. –  Jr. May 3 '12 at 1:48
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.