I would like to find the arc length of a manifold $a\left( \theta \right)$. The manifold can be regarded as a curve embedded in an N-dimensional complex space.
The manifold is given as $a\left( \theta \right) = {e^{-jr\pi \sin \theta }} = \left[ {\begin{array}{*{20}{c}} {{e^{-j\pi \sin \theta }}}\\ {{e^{-j2\pi \sin \theta }}}\\ \vdots \\ {{e^{-jN\pi \sin \theta }}} \end{array}} \right]$ , and $r = {\left[ {1,2, \cdots ,N} \right]^T}$.
The related reference defines the arc length s as $s = \int_0^\theta {\left\| {\frac{{da\left( \theta \right)}}{{d\theta }}} \right\|d\theta } $, and the result is given as $s\left( \theta \right) = \pi \left\| r \right\|\left( {1 - \cos \theta } \right)$.
The problem is that I know little about differential geometry, and I don't know how to derivate the above formulation. I need help with the derivation.
This is a differential geometry question, but you don't need differential geometry to solve it. Just apply the definition $$ \left\|{da(\theta) \over d\theta}\right\| = \sqrt{ {da(\theta) \over d\theta} \cdot \overline{da(\theta) \over d\theta}} = \sqrt{\sum_{i=1}^N {da_i(\theta) \over d\theta}\overline{da_i(\theta) \over d\theta}}$$ and you'll get a simple expression for the integrand that you can then integrate.