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Let $\nabla$ be an affine connection on a pseudo-Riemannian manifold $(M,g)$. Let $c:[0,1] \rightarrow M$ be a differentiable curve and consider vector fields $Y,Z$ along $c$. Is it true that the expression

$$\frac{d}{dt}|_{t=0} \; g(Y,Z)-g \left( \frac{\nabla}{dt}|_{t=0} \; Y,Z \right)-g \left( Y, \frac{\nabla}{dt}|_{t=0} \; Z \right)$$

does depend only on the tangent vectors $Y(0)$ and $Z(0)$?

My motivation for this claim comes from the fact that, for all vector fields $X,Y,Z \in \Gamma(TM)$ and all points $p \in M$, the covariant derivative of the metric tensor $$(\nabla g)(X,Y,Z)(p):= X_{p}g(Y,Z)-g \left( \nabla_{X_{p}}Y,Z_{p} \right)-g \left( Y_{p}, \nabla_{X_{p}}Z \right)$$ depends only on the tangent vectors $X_{p},Y_{p},Z_{p}$. If one chooses a curve $\tilde{c}:[0,1] \rightarrow M$ such that $\tilde{c}(0)=p$, $\tilde{c}'(0)=X_{p}$ one is led to the first expression above, so it seems that my question can be answered in the affirmative, if the vector fields $Y,Z$ along $\tilde{c}$ are of the form $\tilde{Y} \circ \tilde{c}$, $\tilde{Z} \circ {c}$ with $\tilde{Y}, \tilde{Z} \in \Gamma(TM)$.

But what about the general case?

I might add that the original exercise I wanted to solve was to prove that the metric tensor is parallel (i.e. $\nabla g =0$), if and only if for every curve $c$ every parallel transport along $c$ is an isometry. However, I came across my question above while attempting to prove this and found the question interesting in its own right.

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I think you've skipped some steps somewhere. In order to compute $X_p g(Y, Z)$, for example, one must know $Y$ and $Z$ in a neighbourhood of $p$. – Zhen Lin May 22 '12 at 18:33
That's right. But $\nabla g$ does not depend on the values $X,Y$ and $Z$ take in a neighborhood of $p$. This is the case, because $\nabla g$, unlike $Xg(Y,Z)$, gives rise to a $(3,0)$ tensor field on $M$. The point why this happens is that $(\nabla g)(f \cdot X,Y,Z)=f \cdot (\nabla g)(X,Y,Z)$ for all smooth functions $f: M \rightarrow \mathbb{R}$ and similarly in the other two arguments. – Nils Matthes May 23 '12 at 8:46
Then, the answer to your question is just yes – isn't it? Because your first expression is equal to $(\nabla g)(\dot{c}(0), Y_p, Z_p)$. – Zhen Lin May 23 '12 at 10:01
@Zhen Lin: I think what you said should be right. I just have to check what I said in the previous comment in the situation where $Y,Z$ are vector fields along a curve $c$ with $c'(0)=X_{p}$. Thanks anyways, I feel that some wrong ideas and intuition don't haunt me any longer. I'll answer my own question soon, when I feel I understood everything in question. – Nils Matthes May 23 '12 at 16:37
up vote 0 down vote accepted

After having thought a bit about the above problem, and thanks to the remarks of Zhen Lin, I think I am in a position to answer my question in the affirmative.

It is true that, whenever you have an $r$-multilinear map

$$A: \underbrace{\Gamma(TM) \times ... \times \Gamma(TM)}_{r-times} \longrightarrow \Gamma(TM)$$ such that for all $f \in C^{\infty}(M)$ we also have $$A(X_{1},...,X_{i-1},fX_{i},X_{i+1},...,X_{r})= fA(X_{1},...,X_{r})$$

in every component of $A$ then there exists an $(r,1)$-tensor field $B$ such that $A(X_{1},...,X_{r})(p)=B_{p}(X_{1}(p),...,X_{r}(p))$. In particular the value of $A(X_{1},...,X_{r})(p)$ does depend only on the tangent vectors $X_{1}(p),...,X_{r}(p)$.

The same, however is true, when we consider a (smooth) curve $c:[0,1] \rightarrow M$ and replace $\Gamma(TM)$ by $\Gamma_{c}(TM)$ ($=$vector fields along $c$) in the above reasoning. One then simply has to check that

$$\frac{d}{dt}|_{t=0} \; g(fY,Z)-g \left( \frac{\nabla}{dt}|_{t=0} \; fY,Z \right)-g \left( fY, \frac{\nabla}{dt}|_{t=0} \; Z \right)= f \left( \frac{d}{dt}|_{t=0} \; g(Y,Z)-g \left( \frac{\nabla}{dt}|_{t=0} \; Y,Z \right)-g \left( Y, \frac{\nabla}{dt}|_{t=0} \; Z \right) \right)$$


$$ \frac{d}{dt}|_{t=0} \; g(Y,fZ)-g \left( \frac{\nabla}{dt}|_{t=0} \; Y,fZ \right)-g \left( Y, \frac{\nabla}{dt}|_{t=0} \; fZ \right)= f \left( \frac{d}{dt}|_{t=0} \; g(Y,Z)-g \left( \frac{\nabla}{dt}|_{t=0} \; Y,Z \right)-g \left( Y, \frac{\nabla}{dt}|_{t=0} \; Z \right) \right) $$

for all $f \in C^{\infty}([0,1])$, but this is easy.

Hence indeed $$\frac{d}{dt}|_{t=0} \; g(Y,Z)-g \left( \frac{\nabla}{dt}|_{t=0} \; Y,Z \right)-g \left( Y, \frac{\nabla}{dt}|_{t=0} \; Z \right)$$ does depend only on $Y(0)$ and $Z(0)$.

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