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Prove that connection $\nabla $ on a Riemannian manifold $M$ is compatible with metric iff

$$Xg(Y,Z)=g(\nabla_XY,Z)+g(Y,\nabla_XZ),$$

for every smooth vector fields $X,Y,Z$.

I am confused about how to prove compatibility from this equation. Any help is appreciated it. By my text it should be obvious (Do Carmo).

  • 1
    How is 'compatible with metric' defined?2012-09-30
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    If for any smooth curve $c$ and parallel vector fields $V,W$ along $c$ the function $g(V(t),W(t))$ is constant along $c$ we say that connection $\nabla$ is compatible with metric $g$.2012-10-01

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This is somewhat delicate to do with complete detail and involves exactly the kind of issues on which Do Carmo doesn't bother to expand.

Let $\gamma : [0,1] \rightarrow M$ be a smooth curve and let $P,P'$ be two smooth parallel vector fields along $\gamma$. Define $f(t) = g(P(\gamma(t)), P'(\gamma(t)))$. The function $f : [0,1] \rightarrow M$ is smooth. We will show that the derivative $f'(t)$ is identically zero.

Let $t_0 \in [0,1]$. Choose some vector field $Z \in \Gamma(M)$ such that $Z(\gamma(t_0)) = \dot{\gamma}(t_0)$. Consider three cases:

(1) $\dot{\gamma}(t_0) \neq 0$. Then, for small enough $\epsilon > 0$, the image $\gamma([t_0 - \epsilon, t_0 + \epsilon])$ is an embedded compact submanifold of $M$ with boundary. This implies that $P,P'$ (restricted to the image) are defined on a compact submanifold of $M$ and so can be extended to vector fields $X,X' \in \Gamma(M)$, defined $\textbf{globally}$ on $M$, such that $X(\gamma(t)) = P(\gamma(t)), X'(\gamma(t)) = P'(\gamma(t))$ for $t \in [t_0 - \epsilon, t_0 + \epsilon]$. Thus, $$ f'(t_0) = \dot{\gamma}(t_0)g(X,X') = \left. Zg(X,X')\right|_{p=\gamma(t_0)} = \left. g(\nabla_Z X, X')\right|_{p=\gamma(t_0)} + \left. g(X, \nabla_Z X')\right|_{p=\gamma(t_0)} = g(\left. \frac{DP}{dt} \right|_{t=t_0}, X'(\gamma(t_0))) + g(X(\gamma(t_0)), \left. \frac{DP'}{dt} \right|_{t=t_0}) = 0. $$

(2) $\dot{\gamma}(t_0) = 0$, but there is a sequence $t_n \rightarrow t_0$ such that $\dot{\gamma}(t_n) \neq 0$. For example, you can think about a curve $\gamma$ that traces a line segment from $t = 0$ to $t_0 = \frac{1}{2}$, slows down smoothly as $t$ approaches $\frac{1}{2}$, and then turns back on its trace. In this case, you can't necessarily extend the vector fields $P, P'$ to $\textbf{global}$ vector fields, but you get from continuity of $f'$ and from the previous analysis that $f'(t_0) = 0$.

(3) $\dot{\gamma}(t) \equiv 0$ in a neighborhood of $t_0$. This implies that $\gamma(t)$ is constant around $t_0$. Since parallel transport along a constant curve is constant, the function $f(t)$ itself is constant around $t_0$ and in particular $f'(t_0) = 0$.

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    How to show the other direction is true?2017-01-16
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    Read Proposition 3.3 of Do Carmo's text.2017-01-16
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    Thank you very much!! Do Carmo never addresses this kind of issue as you did, can you recommend any reference that has more rigorous treatment to Riemannian Geometry? Where do you learn this sort of technique from?2017-01-16
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    Actually, I didn't notice it back then but Do Carmo does handle this issue rigorously using Proposition 3.2. I think that the person who asked the question also missed it. Anyway, I strongly recommend Lee's Riemannian Manifold - Introduction to Curvature (and also his Smooth Manifolds book which fills many of the differential topology gaps in Do Carmo's book).2017-01-16
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    Do you mean that using proposition 3.2 can also prove the direction that you proved in this answer?2017-01-16
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    @Keith: Yep, he shows first an important intermediate step which makes everything easier.2017-01-16
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    I am so sorry, but can I ask one more question? If I am gonna use De Carmo's proposition 3.2, I would proceed as first picking a vector field such that $V(\gamma(t_0)) =\dot \gamma(t_0)$, then take any arbitrary vector fileds$V,V'$ along $\gamma$, compute $\frac{d \langle V,V' \rangle}{dt}|_{t_0}=Z(t_0)\langle V,V' \rangle$, for this equality to hold, I need to view $V,V'$ as vector fields on $M$, can we always extend vector fields along a curve to vector fields on $M$?2017-01-17
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    I'm not sure what exactly you want to prove and under which assumptions, but you cannot always extend a vector field along a curve to a vector field on $M$ (think of the case where the curve is constant). However, you can always write locally a vector field $V(t)$ along $\gamma$ as $V(t) = f^i(t) X_i(\gamma(t))$ where the vector fields $X_i$ are vector fields on some open neighborhood of $M$ containing (part of) the image of $\gamma$. Then you can use the product rule and the relation between $\frac{DX_i}{dt}$ and $\nabla_{\dot{\gamma}} X_i$.2017-01-17