 Solution

No. For sufficiently smooth curves,is unchanged by proper Euclidean motions and transforms correctly under translations of the arc-length parameter, so it is a pointwise Euclidean invariant of a curve. It is not determined by the two numbers $\kappa (s)$ and $\tau (s)$.

For a concrete comparison near $s=0$, use the plane curve reconstructed from curvature construction withThe resulting unit-speed planar curves both have $\kappa (0)=1$ and $\tau (0)=0$, but their curvature derivatives at zero are respectively $0$ and $1$. Thus no single function $f$ can satisfy $Q(\gamma ,s)=f(\kappa (s),\tau (s))$ for all curves.

Indeed, the covariance conditions alone even permit nonlocal examples such as $Q(\gamma ,s)={\kappa }_{\gamma }(s+1)$. This is the distinction recorded by pointwise Euclidean invariants need not depend only on current curvature and torsion.