Say that we have a Riemannian manifold $(M,g)$ equipped with a linear connection $\nabla$. We know that $g$ is a positive definite, symmetric 2-tensor field on $M$. We say that $\nabla$ is compatible with $g$ if $ \nabla_X \langle Y,Z \rangle = \langle \nabla_X Y, Z \rangle + \langle Y, \nabla_X Z \rangle$. Where $\langle Y,Z \rangle$, for any vector fields $Y,Z$, can be understand as smooth function derived from $g$, that is $g(Y,Z): M \rightarrow \mathbb{R} $, defined by $g(Y,Z)(p) = g_p (Y_p,Z_p)$ which is can be shown to be smooth if $g$ is. Now i asked to show that the above criterion (or definition) for $\nabla$ to be compatible with $g$ is equivalent to \begin{equation} \frac{d}{dt}\langle V,W \rangle = \langle D_t V, W \rangle + \langle V, D_t W \rangle \qquad \forall V,W \, \text{are vector fields along a curve} \, \gamma : I \rightarrow M \end{equation} What i dont really understand is that how we interpret the function $\langle V,W \rangle : I \rightarrow \mathbb{R}$ above. How this is related or derived from $g$ ? My guess is that we interpret this as the smooth function $g_{\gamma}(V,W)$ where $g_{\gamma}$ understand as the induced metric on $\gamma(I)$. But to have induced metric on $\gamma(I)$ we have to consider it as a submanifold of $M$ which is put another restriction to the curve. I dont sure this is true because at the beginning the lemma (John Lee Riemannian Manifold), it doesn't say anything about $\gamma$ except that it is a smooth map. Can anyone help me ? Thank you.

(If anyone want to follow the notation, its from John Lee's books, Introduction to Smooth Manifold and Riemannian Manifold.)

  • $\begingroup$ What do you mean when you write $D_t$? $\endgroup$ Aug 21, 2017 at 10:43
  • $\begingroup$ Its a covariant derivatives along a curve. Its a operator on space of smooth vector field along $\gamma$. $\endgroup$ Aug 21, 2017 at 10:47
  • $\begingroup$ Do you only want to know how to interpret $\langle V,W\rangle:I\to \mathbb{R}$ or do you want to have a proof? $\endgroup$ Aug 21, 2017 at 11:01
  • $\begingroup$ I know how to do that. But i just feel uneasy about the function. I thought the proof more convincing if i know the precise interpretation of the function. $\endgroup$ Aug 21, 2017 at 11:03

1 Answer 1


You have got $g(V,W)_p$ that is, varying $p$, a smooth function of $M$. Now, for each $t\in\mathbb{R}$, you get a value through the curve: $t\mapsto g(V,W)_{\gamma(t)}$. This is only the composition of the previous function by the curve $\gamma$. Notice that we cannot use the induced metric: the fields $V,W$ are fields on $T_{M}$, if we were to use the induced metric we would need to use only the components of these fields tangent to the curve, but that's not the case.

However, the operator $D_t$ is a pullback of the connection through the curve, and we could say it is an "induced" connection.

  • 1
    $\begingroup$ Thanks a lot @Alessio Di Lorenzo. Thats really help. $\endgroup$ Aug 21, 2017 at 12:17

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