# Riemannian manifolds, definition of the covariant derivative.

I am reading the book by Lee, Introduction to Riemannian manifolds which I think is very nice. However among the things I don't understand is the following :

Theorem 4.24 (Covariant Derivative Along a Curve).

Let $$M$$ be a smooth mani- fold with or without boundary and let $$\nabla$$ be a connection in $$TM$$ . For each smooth curve $$\gamma : I \to M$$ , the connection determines a unique operator $$D_t:\mathfrak{X}(\gamma)\to\mathfrak{X}(\gamma)$$,

called the covariant derivative along $$\gamma$$, satisfying the following properties:

(i) LINEARITY OVER $$\mathbb{R}$$: $$D_t(aV+bW)=aD_tV+bD_tW$$ for $$a,b\in\mathbb{R}$$:

(ii) PRODUCT RULE :

$$D_t(f V)=f'+ D_t V$$ for $$f\in C^\infty(I)$$

(iii) If $$V \in \mathfrak{X}(M)$$ is extendible, then for every extension $$\tilde{V}$$ of $$V$$

$$D_t V(t)= \nabla_{\gamma'(t)}\tilde{V}$$.

There is an analogous operator on the space of smooth tensor fields of any type along $$\gamma$$ .

But in property (iii), If it was "$$V\in \mathfrak{X}(M)$$ and $$\gamma'$$ are both extendible, then for every extension $$\tilde{V}$$ and $$\tilde{\gamma'}, D_tV(t)=(\nabla_{\tilde{\gamma'}}\tilde{V})(\gamma(t))$$", it would make more sense to me, because here $$\gamma'(t)$$ is just a tangent vector and not a vector field. There is something I am not getting right.

• The covariant derivative $\nabla_XY$ depends only on the value of $X$ at a point, not in a neighborhood. (It is "tensorial in $X$").
– Neal
Oct 21 '20 at 14:50

I think you want use Proposition 4.5: $$\nabla_X Y |_p$$ depends only on the value of $$X$$ at $$p$$ (as well as the values of $$Y$$ in a neighborhood of $$p$$). Then, for a tangent vector $$v \in T_p M$$ we interpret the expression $$\nabla_v Y$$ by arbitrarily extending $$v$$ to a vector field in a neighborhood of $$p$$; the proposition implies that the result is independent of extension.
This means that $$\nabla_W \,\cdot\restriction_p$$ only depends on $$W(p)$$, i.e., no matter how you extend $$W(p)$$ to a vector field, the connection will be the same at $$p$$.
Recall that in the other entry this is no longer true, but still is local, which means that if $$V,\tilde V$$ are vector fields, that coincide in a vicinity of $$p$$, then $$\nabla_W V(p) = \nabla_W \tilde V (p)$$.