Defintion of length of a $C^1$-curve in Riemannian manifold

Question

Let $(M,g)$ be a Riemannian manifold. We know that if $\gamma:[a, b] \to M$ is a smooth (infinitely differentiable) curve, then $L(\gamma)$, length of $\gamma$ is defined by $$L(\gamma):=\int_{a}^b \sqrt{g_{\gamma(t)}(\gamma'(t),\gamma'(t))}\, dt.$$

Why does this definition will extend to $C^1$ curve ? What will be the justification? I cannot understand it completely as in the definition of a vector field, Riemannian metric we assume smoothness.

Answer 1

Here’s all you need:

• you need the object $g$ to be well-defined. So, a-priori, you need only assume $g$ is a rough section of the $(0,2)$ tensor bundle, satisfying the hypotheses for a pointwise inner product. A-priori, no smoothness of $g$ needed.
• To talk about the curve $\gamma’$ you need $\gamma$ to be once differentiable (not even continuous differentiability).

Just with these mere assumptions, you can define the integrand $\lambda(t)=g_{\gamma(t)}[\gamma’(t),\gamma’(t)]$. This is a non-negative function, so you can take its square root, and now you have a well-defined function $\sqrt{\lambda}:[a,b]\to\Bbb{R}$. Now, you integrate, so you have to assume integrability of $\sqrt{\lambda}:[a,b]\to\Bbb{R}$… so really its a very small amount of hypotheses.

Just to make everything simple, one could assume $g$ is a continuous metric tensor field, that $\gamma$ is $C^1$, so that $\gamma’$ is continuous, so that by composition, $\lambda$ and $\sqrt{\lambda}$ are continuous, and so integrable on $[a,b]$. We don’t even need $M$ to be $C^{\infty}$.

Just because textbooks often only define things for smooth manifolds, or smooth vector fields/forms does not mean definitions/theorems/concepts don’t extend more generally. Smoothness assumptions are merely there so you don’t have to deal with analysis generalities/technicalities on a first introduction to the geometry.