I’m reading Foundations of Hyperbolic Manifolds by John G. Ratcliffe (third edition). In his proof of Theorem 1.4.1, he demonstrates that any two points on a Euclidean geodesic through the origin are given by linearly dependent vectors $ \beta(s) $ and $ \beta(l) $ where $ s \in [0, l] $. He shows linear dependence after making the claim for a geodesic arc $\beta$ where $\beta(0)=0$ that $|\beta(s)|=s$.
My confusion comes from this last statement. If $|\beta(s)|=s$, then it appears he is assuming geodesics in Euclidean space are straight lines. (This comes from his definition of geodesic lines as distance preserving maps $\beta : [0, l] \rightarrow R$ so that the length of the geodesic curve between $0$ and $\beta(s)$ is $s$. Thus the statement $|\beta(s)|=s$ says the length of the curve, $s$, is the same as the length of the vector $\beta(s)$, so the geodesic curve is a straight line). One might argue this is a well known fact, but it appears that this is what he is actually trying to prove. That is to say, he concludes (mid way through the proof) that $ \beta(s) $ and $ \beta(l) $ are linearly dependent, and thus all points on a Euclidean geodesic lie on a straight line. But in order to prove this, he appears to be making that assumption in the first place when stating $ |\beta(s)|=s $.
Can anyone see where I’m getting this wrong?
Included below is the full stated theorem and proof as stated in the book:
The proof references theorem 1.3.1 (Cauchy’s Inequality)
Best Answer
The section of the proof you are reading is the proof of the implication (1)$\implies$(2), and the proof starts, naturally enough, by assuming that $\alpha : [a,b] \to \mathbb E^n$ is a geodesic. By definition of a geodesic, it follows that the distance from $\alpha(a)$ to $\alpha(a+s)$, namely the quantity $| \alpha(a) - \alpha(a+s) |$, is equal to $s$ (for all $s \in [0,l] = [0,b-a]$).
Now they define a new function $\beta(s) = \alpha(a+s)-x$. It follows that \begin{align*} |\beta(s)| &= |\beta(s)-0| \\ &= |\beta(s)-\beta(0)| \\ &= |(\alpha(a+s)-x) - (\alpha(a+0)-x)| \\&= |\alpha(a+s)-\alpha(a)| \\& = s \end{align*}