Length of a curve on a Riemannian manifold

Question

I have a question of definition: take $(M,g)$ a riemannian manifold, $\gamma:I\rightarrow M$ a smooth path in M. We define the length of $\gamma$ as follows: $$l(\gamma) = \int_I \vert\vert\gamma'(t)\vert\vert dt = \int_I \sqrt {g_{\gamma(t)}\left(\frac{d\gamma}{dt},\frac{d\gamma}{dt}\right)}dt$$ My question is how is $\frac{d\gamma}{dt}$ interpreted as a tangent vector? I have encountered this definition problem quite a few times so I decided to ask about it.

I know that tangent vectors can be seen as equivalence classes of curves. i.e. $[\gamma]$ can act on the set of smooth functions as follows: $$ [\gamma](f) = \frac{d}{dt}(f\circ\gamma)\vert_{t=0}$$

Is this what $\frac{d\gamma}{dt}$ means? I can easily show linearity but it does not seem evident to me how to show that $\frac{d\gamma}{dt}$ respects Leibniz's identity.

This is just the product rule for the derivative of functions of one variable. — Moishe Kohan, Mar 16 '17 at 13:41

score 1 · Answer 1 · answered Mar 16 '17 at 13:41

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Any derivative respects the Leibniz rule by elementary calculus. $\frac{d\gamma}{dt}$ can be viewed either as the naive tangent vector of the curve $\gamma$ or as a derivation as you described; these two definitions are equivalent.

answered Mar 16 '17 at 13:41

Mikhail Katz

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So $\frac{d\gamma}{dt} \in T_M$ sends $f \in C^\infty(M)$ to $R$ as follows: $\frac{d\gamma}{dt} (f) = \frac{d}{dt}(f\circ\gamma)\vert_{t = 0}$.
How do you show that this respects Leibniz?
– user405156 Mar 16 '17 at 14:02
If you let $F=f\circ\gamma$ and $ G=g\circ\gamma$ you get two ordinary functions of a real variable $t$ for which the Leibniz rule is proved in calculus. A good reference is Keisler's Elementary Calculus textbook. – Mikhail Katz Mar 16 '17 at 14:13

Length of a curve on a Riemannian manifold

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