We know that for $n\geq 2$ the de Sitter space $\mathbb{S}^n_1(r)$ and the hyperbolic space $\mathbb{H}^n(r)$ have constant curvature $1/r^2$ and $-1/r^2$, respectively.
Looking at references such as here (p. 23), here (p. 4) and here (p. 6), the definitions given for the signed curvature produce $\kappa=1$ for $\alpha(s)=(\sinh s,\cosh s)$ and $\kappa=-1$ for $\beta(s)=(\cosh s,\sinh s)$, while I'd expect reversed results, bearing in mind my first comment. I'm having trouble accepting that $\mathbb{H}^1$ has positive curvature.
If $\alpha(s)=(x(s),y(s))$, defining ${\bf N}(s)=-\epsilon (y'(s),x'(s))$, where $\epsilon=1$ if $\alpha$ is spacelike and $-1$ if timelike, produces the expected curvatures, but the ordered basis $({\bf T},{\bf N})$ is always negative.
Is there a way around this situation? Am I forced to choose between having a positive basis or keeping my intuition?
