Let $\gamma(t)=[cos(at),sin(at),cos(bt),sin(bt)]$ be a curve in $\mathbb{R}^4$. We can think of the image of this curve to be a curve in $\mathbb{R}^3$, as a helix swirling about the z-axis, or rather, on the surface or a torus. I want to find the Frenet frame of $\gamma(t)$.
Since the curve is defined to be in $\mathbb{R}^4$, the standard method of finding a frenet frame in terms of $\{T,N,B\}$ will not work. We will have to used the generalized method of finding a frenet frame, which involves the gramm-schmidt process. I have tried to make this calculation and things got extremly awful, which made me think that perhaps there is a way of calculating the frenet frame using the fact that we know it's on the surface of a torus, thus it will be contained in $\mathbb{R}^3$.
I'm a little afraid to try to make this calculation by brute force again, if anyone has any insights to help me out here it would be much appreciated!!
Clarification: Well, in $\mathbb{R}^4$ we can think of this curve as parameterizing two circles in $\mathbb{R}^2 \times \mathbb{R}^2$. But we can think of $S^1 \times S^1$ as a Torus, which can be imbedded into $\mathbb{R}^3$
By definition, the Frenet frame of a curve in $\Bbb R^4$ (and this will generalize to higher dimensions) satisfying certain generic conditions is defined as follows. Start with an arclength parametrization $\alpha(s)$. Define an orthonormal frame $e_1,e_2,e_3,e_4$, depending on $s$, by \begin{alignat*}{4} \alpha' &= e_1 &&& \\ e_1' &= &\kappa_1 e_2 \\ e_2' &= -\kappa_1 e_1 & & &+\kappa_2 e_3 \\ e_3' &= & -\kappa_2 e_2 & && +\kappa_3 e_4 \\ e_4' &= &&&-\kappa_3 e_3 \end{alignat*} Note that for the Frenet frame of a curve in $\Bbb R^3$ to be well-defined, we must have $\kappa$ everywhere nonzero. For a curve in $\Bbb R^4$, we need $\kappa_1$ and $\kappa_2$ everywhere nonzero.