# Why do metrics act on tangent vectors?

Consider a manifold $M$ and a curve $\gamma : \mathbb{R} \supseteq I \rightarrow M$. The length of this curve is defined as

$$L = \int \sqrt{g(X,X)}_{\gamma(t)} dt$$

where $g$ is the metric tensor, $X$ is the tangent vector to the curve and $t$ is the parameter of the curve.

I do not understand why we require a tangent space in order to calculate lengths of curves. I understand a metric tensor takes in two tangent vectors and spits out a number which allows us to calculate lengths and angles on manifolds, but why exactly do we require a metric defined in this way? Why is there no notion of length that is independent of tangent spaces and is in terms of the manifold alone?

• There is a way of defining the length of curves in a general metric space. See this question: math.stackexchange.com/questions/153892/… When the curve is smooth in a manifold, both definitions coincide. Commented Sep 20, 2018 at 18:04
• As Eduardo says you can always define length of a curve in a metric space. Now when the curve is differentiable (in $\mathbb{R}^n$) the length coincide with the integral of the norm of the derivative. This is why we have this formula in general in Riemannian manifolds. Why the tangent space ? Because you want to differentiate the curve. Now, the intuition of why you want to differentiate comes from physics: the length is the integral of the norm of the speed. Commented Sep 20, 2018 at 19:34