I seek to clarify some confusion I have with regards to the notion of an isometry with the use of an example. This is related to a question I had asked previously. Consider a helicoid with an axis of length $l$ and pitch $p$ parametrised as $$x(s,v) = (v\cos{s}, v\sin{s}, ps) ,$$ where $ s\in [0,l]$ and $v \in [-a,a]$. In general I have the relation $ \frac{2\pi}{p}= \frac{l}{n}$, where $n$ is the number of turns of the helicoid around the axis.
The first fundamental form coefficients of this helicoid will be $$E = 1, F =0 , G = p^2+v^2.$$
Now I take another helicoid with same parametrization as before but this time, I take $ s \in [0, p]$ and $v \in [-a,a]$. Essentially the pitch is same as before, so the first fundamental coefficients dont change but the number of turns is reduced to 1. This seems to be a local isometry to me but the length of the axis is not preserved, so this violates the isometry condition that lengths of curves on the surfaces are preserved.
Alternatively one can say I am attempting to isometrically unscrew a helicoid of n turns to a helicoid with just one turn. But although it seems to be an isometry due to the matching first fundamental coefficients, it doesnt seem to preserve distances. The attached figure is meant to sort of convey what I am saying. Can anyone please clarify what is happening here??
The parametrization of the second is $$\big(v\cos(s/n), v\sin(s/n),ps\big).$$ Now compute the first fundamental form. Certainly no local isometry.