Q:

Rotation by $90°$ in differential geometry

Let $f: \Omega \subset \mathbb{R}^2  \rightarrow \mathbb{R}^3$ be a parametrized surface and $\nabla_{c'}c'$ be the covariant derivative of a curve $c:I \rightarrow \Omega$ that is parametrized by arc-length. Now, we define the geodesic curvature by $\kappa_g:=||\nabla_{c'}c'||$. Then we have for the first fundamental form $g(c', \nabla_{c'}c')=0$, as the second derivative is orthogonal to the first one. Now I want to define a matrix $J$ such that $\kappa_g  Jc' = \nabla_{c'}c'$. In $\mathbb{R}^3$ this would be just a rotation matrix of $90^°$ to the left or right (QUESTION 1: it is not immediately clear, in which direction we have to rotate, is this correct?), but QUESTION 2: how does this matrix look for vectors in $\Omega$?
Assume we define a function $\alpha: I \rightarrow \mathbb{R}$ where $v$ is a parallel vector field along $c$ such that $\cos(\alpha(t)) = g(v(t),\gamma'(t))$, then I get:
$$\frac{d}{dt} ( \cos(\alpha(t)) = - \sin(\alpha(t)) \alpha'(t) = D_{c'}g(v,c') = g(\nabla_{c'}v,c')+ g(v,\nabla_{c'}c') = \kappa_g g(v,J c') = \kappa_g \cos(\alpha(t)-\frac{\pi}{2}) = \kappa_g \sin(\alpha) \Rightarrow \kappa_g = -\alpha'(t),$$
QUESTION 3: I somehow suspect that I have a sign wrong in this last equation, does anybody see through this?
if anything is unclear, please let me know.

A:

As a preamble to this response, I want to comment that we should be careful about saying "the matrix" $J$. It's true that for any $t$ you can define a linear automorphism $J_t$ of $T_{c(t)} f(\Omega)$ by your formula, but there's no obvious "natural" basis for $T_{c(t)} f(\Omega)$, and you'll need to choose one to actually write down an honest-to-goodness matrix $J_t$.
Answer 1. $\nabla_{c'} c'$ points in the direction the curve $c$ is curving, which could obviously be to either the "left" or "right" (which only make sense once you fix some orientation on $\Omega$ anyway). In particular there's no way, for example, to write down an automorphism $J$ that only depends on $c(t), c'(t)$, and $T_{c(t)} f(\Omega)$, although this data does (as you say) reduce $J$ to one of two choices, both rotations by $\pi/2$. 
Answer 2. You have an isomorphism $df : T_x \mathbb{R}^2 \to T_{f(x)} f(\Omega)$ which allows you to move your automorphism $J$ over to $T_x \mathbb{R}^2$. Specifically, let $J_t$ be the transformation of $T_{c(t)} f(\Omega)$ you've defined. Then if $p \in \mathbb{R}^2$ is given by $p = f^{-1}(c(t))$ and $J'_t$ is the transformation
$$
  J'_t = df^{-1}_{p} \circ J \circ df_p : T_p \mathbb{R}^2 \to T_p \mathbb{R}^2,
$$ 
then $J'_t$ satisfies the obvious analogue of your formula. If you want to actually write down a matrix, again it will depend on the specific curve $c(t)$ in principle (because of the left/right problem). However, you can at least translate some "rotate by $\pi/2$ matrix" back to $\mathbb{R}^2$ using $df$. Here's a summary of how one might do this:
Write $f = (f_1, f_2, f_3)$ and let $D$ be the matrix
$$
D = \left( \begin{array}{cc}
\frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\
\frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \\
\frac{\partial f_3}{\partial x} & \frac{\partial f_3}{\partial y} \end{array} \right)
$$
Since $f$ is giving a parameterization of a surface, this matrix $D$ has a left inverse $C$ ($C$ is a $2 \times 3$ matrix with $CD = 1$).
A normal vector to $f(\Omega)$ at a point $f(x)$ is easy to find; one can take the cross product of the columns in the above matrix (since these columns span the tangent space to the surface). Then you can write down the ($3 \times 3$) matrix for a $\pi/2$ rotation about this normal in $\mathbb{R}^3$; call this matrix $R$. The corresponding "rotation" in $T_x \Omega = \mathbb{R}^2$ is then evidently given by $CRD$. (I suspect that writing down an actual formula for this matrix in terms of the partial derivatives of $f$ is both unpleasant and not particularly useful. But if you have an actual parameterization and point you're interested in, you can compute the derivatives and then it's just relatively easy linear algebra.)
Answer 3. Because you know $J$ is a rotation by $\pi/2$ you can say
$$
g(v, Jc') = \cos(\alpha(t) \pm \pi/2)
$$
but there's no way of distinguishing whether the sign should be $+$ or $-$, so you only get to conclude $\kappa_g = |\alpha'(t)|$.

