[Differential Geometry] On the Smooth Equivalence of Parametrised Differentiable Curves.

We say that two regular parametric curves ϕ\phi and ψ\psi are smoothly equivalent, if there is a strictly monotonically increasing differentiable function ss such that ψ(s(t))=ϕ(t)\psi(s(t)) = \phi(t). Reparameterisation means that you have parametrised ψ\psi by some new parameter s(t)s(t) that is a function of tt, instead of the original parameter, tt. For example, you can reparameterise a curve with its arc length, given as a function of tt.

Why does ss have to be monotonically increasing rather than decreasing? ss must be monotonically increasing, because we want to preserve the orientation (i.e. direction through which it travels) of the curve.

Are the follwing two curves ϕ(θ)=(cos(θ),sin(θ)),\phi(\theta) = (cos(\theta),sin(\theta)),

ψ(θ)=(cos(2θ),sin(2θ)),\psi(\theta) = (cos(2\theta),sin(2\theta)),

smoothly equivalent if the domain of ϕ\phi is [0,2π][0,2\pi], the domain of ψ\psi is [0,π][0,\pi]?

These two curves have the same trace, but are not considered to be equivalent curves. The curves have distinct velocities, which you can easily see when you calculate tangent vectors at each of the points on the curves, ψ(θ)=2ϕ(θ),\psi'(\theta)=2\cdot\phi'(\theta),

so ψ(θ)\psi(\theta) travels twice as fast as ϕ(θ)\phi(\theta). Note that this means that you cannot distinguish parametrised differentiable curve simply by looking at their image sets. Smoothly equivalent parametrised curves can be reparametrised from one to the other, and thus we obtain an equivalence class of regular curves.

Below is a list of parametrised differentiable curves that all trace out a unit circle in R2\mathbb{R}^2.

ϕ1(t)=cost+isint, t[0t2π], \phi_1(t) = \cos t + i\sin t, ~ t \in \left[0 \leq t \leq 2\pi \right],

ϕ2(t)=cos(2πt2)+isin(2πt2), t[0t1], \phi_2(t) = \cos(2\pi t^{2}) + i\sin(2\pi t^{2}),~ t \in \left[ 0 \leq t \leq 1 \right],

ϕ3(t)=cost+isint, t[0t4π], \phi_3(t) = \cos t + i\sin t, ~t \in \left[ 0 \leq t \leq 4\pi \right],

ϕ4(t)=cos(t)+isin(t), t[0t2π]. \phi_4(t) = \cos (-t) + i\sin(-t), ~t \in \left[ 0 \leq t \leq 2\pi \right].

Note that ϕ1(t) \phi_1(t) and ϕ2(t)\phi_2(t) are mutually smoothly equivalent, because they are reparametrisations of each other traced in the same direction. However, ϕ3(t)\phi_3(t) traces the unit circle twice while ϕ1(t)\phi_1(t) traces it once, so they are not smoothly equivalent. Also, ϕ4(t)\phi_4(t) traces the unit circle clockwise, while ϕ1(t)\phi_1(t) traces it anticlockwise, so they are not smoothly equivalent.

Note by Tasha Kim
1 year, 5 months ago

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