Lenght of the curve in Riemannian metric.

Question

Let $M^{k}$ a submanifold, $h:U\to M^{k}$ a chart, and $\gamma:[a,b]\to h(U)\subset M^{k}$ a curve in $M^{k}$. Represent the curve in coordinates $(h,U)$ as $h^{-1}\circ\gamma(t)=(y^{1}(t),...,y^{k}(t))$. Show that the lenght of the curve can be calculate in function of coefficients of Riemannian metric $g_{ij}$ in the chart $(h,U)$ by the formula $$l(\gamma)=\int_{a}^{b}{\sum_{i,j=1}^{k}{g_{ij}(\gamma(t))\dfrac{dy^{i}(t)}{dt}\dfrac{dy^{j}(t)}{dt}dt}}$$

Hi, I have this exercises and yet I cannot conclude anything. The Lenght of the curve is the expresion $$l(\gamma)=\int_{a}^{b}{g_{\gamma(t)}(\dot{\gamma}(t)),\dot{\gamma}(t))^{\frac{1}{2}}dt}$$ Any hint or idea, thanks!

Answer 1

Define a metric on $U$ where $h: U\rightarrow M$ : $$ g_{ij}:= h^\ast g (e_i,e_j) $$

If $\gamma : [a,b] \rightarrow M$ is a curve, then $$h^{-1}\circ \gamma (t)= (y^1(t),\cdots, y^n(t)) $$ is a curve in $U$

So its tangent is $$ \frac{d}{dt} h^{-1}\circ \gamma =(\frac{d}{dt} y^1(t),\cdots, \frac{d}{dt}y^n(t)) $$ Hence $$ h^\ast g ( \frac{d}{dt} h^{-1}\circ \gamma , \frac{d}{dt} h^{-1}\circ \gamma ) = \frac{d}{dt}y^i\frac{d}{dt} y^j h^\ast g (e_i,e_j)=\frac{d}{dt}y^i\frac{d}{dt} y^j g_{ij} $$

So $$ L=\int_a^b \sqrt{\frac{d}{dt}y^i\frac{d}{dt} y^j g_{ij} } $$