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Suitable curvature functions determine a regular Frenet curve up to a Euclidean rigid motion. In the plane one signed curvature suffices; in three dimensions positive curvature and torsion determine a unit-speed curve, as in the fundamental theorem of regular space curves. The result is local on the parameter interval and assumes the regularity needed by the frame equations.

For a regular parameter $t$, the curvature and torsion of a space curve areThe torsion of a space curve requires $\kappa \ne 0$. Reparameterize by arc length $s$ and set $T={\alpha }_s$, $N=T_s/\kappa$, $B=T\times N$. Differentiating the pairwise inner products of this orthonormal Frenet frame shows that its derivative matrix is skew symmetric. Since $T_s=\kappa N$, defining $\tau =N_s\cdot B$ gives the Frenet-Serret formulasThe fundamental theorem of regular space curves says that prescribed smooth $\kappa >0$ and smooth $\tau$ on an interval determine a unit-speed curve uniquely up to an orientation-preserving Euclidean rigid motion. Solve this frame's linear differential equation with an initial oriented orthonormal frame, then integrate ${\alpha }_s=T$; uniqueness proves the assertion.

Under the intended convention that the shared parameter $s$ is arc length, equality on an open interval gives equal position and equal Frenet frames at one point. Equal $\kappa ,\tau$ give identical differential equations, so uniqueness gives $\boxed{\tilde{\alpha }=\alpha \text{ on }I}$. The printed “regular smooth” hypothesis alone does not establish this convention. Without common arc length, an orientation-preserving smooth change of speed that is the identity on $J$ but not elsewhere reparameterizes a circle into a counterexample: both curves have $\kappa =1$, $\tau =0$ and agree on $J$, but differ elsewhere as parameterized curves.

For an oriented surface with unit normal $n$, define normal curvature ${\kappa }_n=T_s\cdot n$ and geodesic curvature ${\kappa }_g=T_s\cdot (n\times T)$. ThusA unit-speed curve is a geodesic exactly when ${\kappa }_g=0$. If also ${\kappa }_n\equiv 0$, then $T_s=0$, so it is a straight line. A vertical ruling of a circular cylinder is such a geodesic, although the cylinder contains no open piece of a plane. A nonconstant regular straight line cannot be a simple closed curve. Hence no planar patch is required, and such a geodesic cannot be simple and closed.

Finally let the principal curvatures be ${\kappa }_1,{\kappa }_2$. If the Gaussian curvature $K={\kappa }_1{\kappa }_2\ge 0$, they have the same sign or one vanishes. Euler formula for normal curvature makes ${\kappa }_n$ a weighted average of them. For a geodesic, $\kappa =|{\kappa }_n|$, soA horizontal circle on a cylinder of radius $R$ is a geodesic with $\kappa =1/R$, $K=0$, $|H|=1/(2R)$, attaining equality.

The fundamental theorem of regular space curves states that smooth functions $k>0$ and $\tau$ on an interval determine a smooth arc-length-parametrized curve in ${\mathbb{R}}^3$, unique up to a proper Euclidean motion. Consequently two such curves $\alpha$ and $\tilde{\alpha }$ are related parameterwise by a proper Euclidean motion if and only iffor every $s$.

The fundamental theorem of regular space curves states that smooth functions $\kappa (s)>0$ and $\tau (s)$ determine a unit-speed regular curve in ${\mathbb{R}}^3$ with that curvature and torsion, uniquely up to a proper Euclidean motion of Euclidean three-space. Existence follows by solving the Frenet-Serret formulas for an oriented orthonormal frame and then integrating its tangent vector.

LetThis is another unit-speed embedded curve. The assumed identities say that $\alpha$ and $\beta$ have the same positive curvature and the same torsion. By the uniqueness part of the fundamental theorem of regular space curves, there is a proper Euclidean motion $\varphi$ such thatSince $s\mapsto \pm s+{\sigma }_0$ is a bijection of $\mathbb{R}$, this motion preserves $\alpha (\mathbb{R})$.

It is nontrivial. For the plus sign, ${\sigma }_0\ne 0$ and $\varphi =\mathrm{id}$ would imply $\alpha (s)=\alpha (s+{\sigma }_0)$, contradicting embeddedness. For the minus sign, the identity would imply $\alpha (s)=\alpha (-s+{\sigma }_0)$ for every $s$, and injectivity would force $s=-s+{\sigma }_0$ for every $s$, which is impossible. Thus the required nontrivial proper Euclidean motion exists.