I can't seem to understand what my professor wants to say with his notation. So I hope that some of you can help me.
We're looking at some variational curve $\gamma_t : I \to \mathbb{R}^2, \ t \in (-\varepsilon, \varepsilon)$ that maps curves
$$ (-\varepsilon, \varepsilon) \times I \to \mathbb{R}^2, \ (t,s) \mapsto \gamma_t(s). $$
Then, we write $\gamma_t'(s) := \frac{d}{ds} \gamma_t(s)$ and $\dot{\gamma}_t(s) = \frac{d}{dt} \gamma_t(s)$.
Main Culprit: My professor writes the variational vector field $\dot{\gamma}_t = \frac{d}{dt} \gamma_t$ in some local form $$ \dot{\gamma}_t = (\alpha + i \beta) \frac{\gamma'}{\|\gamma' \|}.$$
What is this mess? What exactly should $\alpha$ and $\beta$ be? Just what does he want to say with this equation?
(The goal is to find extrema of the energy $E(\gamma) = \int_a^b \kappa(s)^2 \ ds$.)
Best Answer
Here's the thing: If $N$ is a unit normal vector field along a surface $M$ in $\Bbb R^3$, you can define $J_p(v) = N(p)\times v$ for each $p \in M$. This satisfies $J_p^2 = -{\rm Id}$ (an almost-complex structure), so that if $v \in T_pM$ is not the zero vector, then $\{v,J_pv\}$ is a basis for $T_pM$. We can regard $J$ as "multiplication by $\rm i$", so your professor means $$\dot{\gamma}_t = \alpha \frac{\gamma'}{\|\gamma'\|} + \beta \frac{J\gamma'}{\|\gamma'\|}.$$He is just writing $\dot{\gamma}_t$ as a combination of convenient basis elements.